I
OF T^E
University of California.
aiKT OF^
Class
Elementary
Pure Geometry
With Mensuration
A COMPLETE COURSE
OF GEOMETRY FOR SCHOOLS
BY
E. BUDDEN, M.A. Oxon., B.Sc. Lond.
MACCLESFIELD GRAMMAR SCHOOL ;
KORMERLY SCHOLAR OF WINCHESTER COLLEGE AND OF NEW COLLEGE, OXFORD
W. & R. CHAMBERS, Limited
LONDON AND EDINBURGH
1904
^M'hHlvAt^
Price, without Answers, it.; with Answers, it. ed.; Answers only, Sd.
Arithmetical Examples.
By W. Wallace Dunlop, M.A., Headmaster of Daniel Stewait's College,
Edinburgh ; and W, ■Woodbde.n, Author of Chambers's CommerciaZ Arithmetic.
208 pages ; with Answers, 256 pages.
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Arithmetic.
THEORETICAL AND PRACTICAL.
By John Sturgeon Mackay, M.A , LL.D., Fellow of the Royal Society ol
Bdinburgh ; Head Mathematical Master in the Edinbm-gh Academy.
472 pages.
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Chambers's Elementary Algebra.
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Elementary Pure Geometry
WITH MENSURATION.
By E. BuDDEN, M.A. Oxon., B.Sc. Lond., Macclesfield Grammar School;
formerly Scholar of Winchester College and of New College, Oxford.
288 pages. Parts I. and II. together, 197 pages, 2s.
PREFACE.
In Parts I. and II. of this book, which cover the whole of
Eucl. I.-YI. and the geometrical part of Trigonometry — i.e. to
solution of triangles — I have folloAved generally the Cambridge
Previous Syllabus and the recommendations of the Mathematical
Association ; but much additional matter has been introduced,
with modern methods suitable to this stage of the subject.
Part I. consists of experimental geometry, angle and parallels,
symmetrical and congruent figures, elementary areas — i.e. all
Eucl. I. — decimal measurement, similar and similarly situated
figures (linear properties, EucL VI. 2-18), and a short account of
simple loci.
Part II. contains circle properties (Eucl. III. and IV.), with
centre of similitude, radical axis, and tangent circles ; areas of parts
of divided lines and of similar figures (Eucl. II., VI. 1, 19...);
the methods of multiplication (similitude) and rotation, maxima
and minima, envelopes, and loci ; and it has a chapter on Trigono-
metry, covering the whole ground to solution of triangles.
Part III. contains an account of modern projective geometry
in the plane, elementary geometrical conies, and solid geometry,
including the mensuration of cylinder, pyramid, cone, and sphere.
I have sought to make improvements in the following directions :
1. Instruments are used from the very beginning, and parallels,
perpendiculars, circles, triangles, &c. drawn and some of their pro-
perties arrived at and stated without any formal proof whatever.
2. Experimental geometry leads up to the definitions of the
plane, straight line, angle, perpendicular, and direction. These,
with definitions of figures and the experimental treatment of the
area of the parallelogram, constitute the Introduction and Chapter I.
IV PREFACE.
3. The order has been so arranged that related properties are
brought together, and in most cases are dealt with on the same page.
Thus the four cases of congruent triangles are taken consecutively
in two pages and by one method — viz. direct superposition. The
parallelogram and similar triangles immediately follow these.
4. The general plan has been arranged upon the fundamental
principle that symmetry precedes congruence. The properties of
the isosceles triangle, and the complete cases of congruent triangles,
cannot be established until it is shown that au angle is reversible
— i.e. that the two faces of a plane angle are congruent. By
proving this we can establish the elementary properties of the
triangle and circle (Chapter II.) independently, and so gain a good
knowledge of the triangle before the more difficult case of two
triangles is approached. The failure to prove this invalidates the
proofs of most of the fundamental theorems in the ordinary text-
books.
It was the proof of this reversibility of the angle that led
me to my definition of the plane, and with it, as I had always
anticipated, to that of the straight line also. Laplace's defini-
tion of a plane, though sufficient as a test (and so used by the
great engineer Whitworth), does not give the straight line as the
intersection of two planes.
My double treatment of ratio in Chapters III., Y., and YII.
deserves a special notice. The mensuration of figures requires the
numerical treatment of ratio, which is, moreover, easier to under-
stand than the purely geometrical treatment ; and for this to be
formally rigorous our definition of ratio should be number or
measure, and should include irrational numbers, since without
these such expressions as sin 41°, considered as numbers, are
unintelligible. I have treated this part of the work completely
but simply by the method of decimal scales, a sufficient explanation
being given in two short notes in Chapters III. and V.
Pure geometry must be independent of the theory of number ;
and at the end of Chapter YII., in order to complete the account
of descriptive geometry, as distinct from that which is partly
numerical, I have given a purely geometrical definition and
treatment of ratio. My proofs of the propositions of Eucl. II.
are also purely geometrical — i.e. not derived from mensuration.
PREFACE. V
Thus, by substituting for the numerical definition of ratio in
Chapter III. the geometrical one of Chapter VII., the book
gives a complete course of strictly pure elementary geometry.
The whole of this treatment of ratio is, so far as I know,
original; though the numerical ratio is only a modification of
Dedekind's Schnitt, his complete system of fractions being
replaced by terminating decimals.
The use of the decimal scale, however, makes the process much
simpler, and as a matter of fact I arrived at my result by a quite
different route from that which Dedekind follows. Students
interested in Euclid's method should consult Professor M. J. ^I.
Hill's Euclid, V. and VI.
I have borrowed the description of multiplication and rotation
from Pedersen's Methodes et Tlieories ; and my knowledge of
modern geometry is derived chiefly from Mulcahy, Townsend,
Chasles, and Cremona. Students may consult Russell's Projective
Geometry. The derivation of pole and polar from Pascal's theorem,
the form of proof that a conic is a perspective of a circle, and of
the converse, and the always real construction for throwing any
five points on to a circle, are my own.
I have kept to the focus and directrix definition of a conic
because it is so much easier to derive the form and simpler
properties of the curve in this way. My construction of the
conic from what I have called the focircle, leads quite naturally
to the modern treatment of the curve. And further, the focal
properties of the conic can be studied simultaneously with the
beginning of Chapter VII., so that Chapters VII. and VIII. can
be taken concurrently. I hope that these chapters may help to
bring modem geometry within the range of the higher mathe-
matical classes of our schools.
I have not hesitated to modify Euclid's proofs, constructions,
and order in the interest of simplicity ; nor to introduce new
symbols and new names where found desirable. The latter I
have tried to make so that their meaning is obvious when once
learnt — e.g. right bisector, mean part of a line. I have used the
symbol ||| for 'is similar to,' to suggest the connection of similar
figures with parallelism ; and as it is also the symbol ^ for con-
gruence turned up, and very easy for boys to write (much easier,
VI PREFACE.
for instance, than - ), it seems tlie obvious one. Geometrical
symbols are used for verbs ( || 'is parallel to,' &c.), literal abbre-
viations (which can be coined as desired) for nouns or adjectives.
Numerous examples are given. Those at the end of each
chapter are carefully graded (with here and there a difficult
one), and should be taken concurrently with the text. General
examples for revision are given at the end of Chapter Y.
Answers are given to most of the practical questions in Part I.
to serve as checks on the drawing. In the other parts — e.g. in
tangency of circles — the drawings are easily checked without
numerical aid.
Much care has been taken by the publishers in setting the type
and reproducing the figures. The important parts of a figure given
in the statement, upon which a proof or construction depends, are
represented by thickened lines or points. Mere lines of construction
or aids to proof are, in general, dotted ; and the finally constructed
figure is shown by a thin continuous line.
Important statements (theorems, corollaries, constructions, and a
few worked examples) have been set throughout in heavier type ;
and in the definitions the thing defined is distinguished by heavier
type.
Short notes and exercises are set in smaller type ; notes gener-
ally are meant primarily for the teacher, who will exercise his
discretion in using or leaving them.
The general plan of headlines; the heavy, medium, and small
type ; and the thickened, dotted, and thin lines, make the book
very easy to use for purpose of reference — the proper function of a
text-book.
My thanks are due to my friends, Mr A. E. Holme of Dewsbury
(who communicated to me the construction for a perpendicular by
set-square), Mr K. W. Batho of St Paul's, Mr G. H. Hughes of
Marlborough, and Mr K. Baron of Macclesfield, for assistance with
proofs and examples ; and to the Committees of the British and
Mathematical Associations for making it possible to write a text-
book of geometry on a definite plan.
E. BUDDEjS".
Macclesfield, 1904.
SUGGESTIONS FOR THE TEACHER.
1. Complete lines are used for the more important parts of
a figure, broken lines or parts of lines — e.g. arcs — for the less
important. Pupils should be encouraged to adopt this distinction.
2. The instructions and experimental work in the Introduction
and in Chapter I. on the use of instruments, on the experimental
derivation of plane, straight line, bisectors, perpendiculars, and
areas, and the notes in Chapters III., V. on ratio should be taken
orally with beginners, at the teacher's discretion.
3. The constructions in the Introduction as far as a simple
plain scale of inches or centimetres, and the simpler constructions
at the end of Chapter II., should be mastered without any formal
proof, before the study of formal geometry is attempted.
4. Definitions, and all statements of theorems, should be learnt
by heart, as soon as reached in the ordinary course of the book.
The examples accompanying these, or to be found at the end of
the various chapters, should be taken pari passu. They have been
carefully graded, though here and there, for convenience of refer-
ence, an example from a later stage has been introduced before its
true place. Congruent triangles should not be used where proof
by symmetry is simpler.
5. The constructions in any chapter should in general be taken
in advance of the theorems — their formal proof may be taken as
soon as sufficient theory has been mastered. It is hoped that the
hints given in the Introduction and in Chapters III., lY., and V.
may be helpful in the solution of the more difficult problems.
6. An accuracy of 1 to 2 per cent, is all tha-t can be expected
without the conveniences and exact instruments of a drawing-office.
If" should be taken as correct for J^".
7. Four-figure tables should be used in Chapter YI. ; their use
should be briefly explained orally.
CONTENTS.
PART L
PAGE
INTRODUCTION.— INSTEUMENTS, Simple Constructions and
Measurement— Experimental Geometry— Scales 1
CHAPTER I. — Preliminary Definitions and Theorems-
Constructions OF Simple Figures — Experimental
Treatment of Area 23
CHAPTER II.— Intersecting Lines, Parallels, Triangle,
Circle, and Corresponding Constructions 42
CHAPTER III. — Congruent Triangles, Parallelogram,
Ratio, Proportional Division, Similar Figures,
Areas, and Corresponding Constructions — Solution
of Problems— Loci 59
PART II.
CHAPTER IV.— The Circle— Chord, Tangent, Angle, and
Rectangle Properties, and Corresponding Con-
structions — Multiplication — Experimental Solid
Geometry 93
CHAPTER v.— Rectangles by Algebraic Form— Mensura-
tion OF Areas, and Constructions — Maxima and
Minima— Inscribed Figures — Rotation and Multipli-
cation—Loci—Envelopes—General Examples— Ratio ...124
CHAPTER VI.— Elementary Trigonometry, Geometrical,
to Solution of Triangles 162
PART IIL
CHAPTER VII.— Modern Geometry— Inversion —Cross Ratio
AND Involution— Pole and Polar— Plane Projection
—Ratio by Geometry 190
CHAPTER VIII. — CoNics, treated partly by Modern
Geomp:try ..219
CHAPTER IX.— Elementary Solid Geometry of the Plane
AND THE Simpler Solid Figures, with Mensuration-
Notes ON Straight Line, Angle, Direction 253
ANSWERS TO EXAMPLES 278
USTTRODUOTIOK
INSTRUMENTS-EXPERIMENTAL GEOMETRY-SCALES.
The pupil should be provided with pencil, pencil compass for
drawing circles, divider for setting off lengths, straight-edge for
ruling straight lines, inch and centimetre scales divided into tenths,
inch divided into eighths, and preferably also into hundredths by
diagonal division. (The pencil used either for the compass or for
ruling straight lines should be hard — e.g. HHH — and its end
cut like that of a table-knife. This furnishes an excellent draw-
ing point.) A protractor for drawing angles, and set-squares * for
drawing parallels and perpendiculars, are also required.
^
2 units
Lengths are measured by means of a scale. This is a straight-
edge divided into suitable units (inch, centimetre, &c.), with at
least one unit subdivided into suitable parts, tenths, eighths, &c.,
as required.
' Measure or set off a length by a scale.'
If PQ is a length to be measured, adjust the divider points (not
the compass) to its ends, put the divider on the scale with one end
on a unit division, say 2, and the other in the divided unit, say
at 3. Then if this is divided into tenths,
PQ = 2.3 units.
Similarly, to set off a length 2-3 units, set one end of the
divider on the 2nd unit, and the other on the 3rd tenth, and prick
off the length PQ in any required position.
Ex. 1. Set off straight lines of 3-2",t 2-5 cm., 1^", 3^ cm.
Ex. 2. Draw circles, radii 1^", 3-3 cm., 2.2".
* Two are generally found suflScient. f " means inch or inches.
P.O. A
INSTRUMENTS.
[iNTROD.
MEASURE OF ANGLES.
The protractor is primarily a semicircle divided into 180 equal
parts ; the angle at the centre O formed by radii of one of these
parts is a degree (°), so that the angular space round a point O
(using a whole circle) contains 360 degrees. An angle of 90° is
called a right angle. In the figure the semicircle is shown divided
into arcs of 10° each, and one of them (40°-50°) divided into
degrees.
* Measure or construct an angle by protractor.'
If AOB is an angle to be measured, place the centre of the pro-
tractor at O, and the base along OA; note the division on the
circle which coincides with OB. In the figure the division is 45,
and angle AOB is 45°.
To construct an angle of 45° at a point O, set the protractor
with its centre at O and its base along one side OA of the angle,
mark with divider a point opposite the 45 division, and rule the
straight line OB through this point with a straight-edge. Then
angle AOB is 45°.
Ex. Draw a straight line AB, 2" long, at A make angles of 30°, 48°,
57°, 90°, 108°, 156° from AB.
The rectangular form of protractor is easily understood from the
semicircular form.
The use of the scale of chords for constructing angles will be
found after Construction 4, Chapter II.
INTROD.]
INSTRUMENTS.
SET-SQUARE-PARALLEL— PERPENDICULAR
Set-squares are right-angled triangles, used with a straight-edge
for drawing parallels, perpendiculars, (and one or two special
angles).
' Construct a parallel to a straight line from a given point.*
Draw a straight line AB, and mark a
point P.
Put one edge of the set-square along AB,
bring a straight-edge SS along another side
BC of the set-square, and hold the straight-
edge quite firm with the left hand.
Slide the set-square along the straight-edge
until the first edge AB traverses the point P,
in the position DE.
Hold the set-square firm, and rule the line
PE.
PE is parallel to AB.
Note. This can be proved as soon as Def. 15, Chapter I., is readied.
'Construct a perpendicular to a straight line from a given
point.'
Draw a straight line AB, and mark a point Q (above figure).
Put the hjrpotenuse (longest side) of the set-square along AB,
bring a straight-edge SS along another side BC of the set-square,
and hold the straight-edge quite firm.
Xow turn the set-square * so that its third side AC is along the
straight-edge, and slide it along until the hypotenuse traverses Q in
the position FG.
Hold the set-square firm, and rule the line QG.
QG is perpendicular to AB.
Note. This can be proved by Theorem 14, Chapter II.
Ex. Draw a straight line AB, and draw AC, making angle BAC 70°.
Draw BD parallel to AC and measure angle ABD. Also draw BE
perpendicular to AB.
The set-square must be turned round, not turned over.
4 instruments. [introd.
Division of Lines.
The set-square is also used to divide a straight line into any
number of equal jDarts by drawing parallels. ^
Thus, if AB in the figure is to be y^
divided into 7 equal parts : ^y^\ '
Draw another line AC, and with com- ^y^^' ' i '»
pass or divider set oif 7 equal parts along ^'^ ' < ' i I
it as far as C, say. ^ ^
Draw parallels to BC through the points of division; these
divide AB into 7 equal parts. (Theorem 34, Chapter III.)
EXAMPLES— I.
1. Draw a straight line 10 cm. long and measure it in inches. How
many inches in a centimetre ? Centimetres in an inch ?
2. Make a straight line 1^" long. Make angles of 70° and 50° at its
ends, forming a triangle. Measure the third angle.
3. Make a straight line AB, 5 cm. ; with centres A, B, radius 5 cm.,
draw two circles meeting in C ; join AC, BC. What do you know
about the straight lines AC, BC, AB ?
4. Measure the angle ACB in Ex. 3.
5. Make a straight line AB, 1^"; make AC perpendicular to it, 2";
measure BC.
6. If a straight line AB points east, draw straight lines AC, AD, AE
pointing N.E., S.W., and S. Are any two of these parts of one straight
line ?
7. In question 6, what is the number of degrees in the angles BAC,
BAD, BAE?
8. Draw a straight line AB, If"; at A make an angle BAC, 40°.
Take a point D in AB, 1|" from A ; draw DE parallel to AC.
9. Measure the angle BDE in Ex. 8. Measure also ADE.
10. Draw a straight line 2^" ; divide it into 5 equal parts.
11. Draw a straight line OA, 6". From O mark off successive lengths
of 1" ; divide the first inch into quarters.
12. Using the scale of Ex. 11, set off a length of 2|", and draw a circle
with this radius.
13. Bisect (by parallels) a line 5 cm. long, and draw a perpendicular
to it through the mid point.
14. Take a point A on the perpendicular of Ex. 1.3, 3 cm. from one end
of the first line, and measure its distance from the other end.
INTROD.] EXPERIMENTAL GEOMETRY. 6
EXPERIMENTAL GEOMETRY.
The Straight Line.
Eold a sheet of fairly stiff paper, and mark the fold as
accurately as you can.
Using the fold as guide, rule a pencil line !
on paper. Place your straight-edge along the i
line. Is the line straight 1 \
Set your fold along the straight-edge. Is '
the fold straight ?
If a fold is made in this manner, and the folded sheet pressed
out on a flat surface, the fold is practically straight. A very good
ruler can be extemporised by twice folding a sheet of foolscap.
Set the edges of two set-squares together ; can you see daylight
between them 1 If you can, the edges cannot both be straight.
Rule a line, using a straight-edge with the edge on the right of
the instrument ; turn the instrument over so that the same edge
is now on the left of the instrument, and see if it just fits along
the line. If it does, the straight-edge is accurate ; if not, not.
* A straight line can have one position only when two points
on it are fixed.'
The Angle.
r
The corner of an ordinary sheet of paper is formed by hoo
straight lines which meet at the corner.
Put your straight-edge along one edge of the paper ; the
other edge of the paper crosses the straight-edge, and is part
of a different straight line from the first edge.
The figure of two straight lines which end at a common point
is an angle (the Latin word for corner).
Look at the figure of a set-square. How many corners has it ?
How many angles? Its figure is called a triangle (i.e. three-
corner). The straight edges are its sides. How many sides has
a triangle ?
Look at a corner of your compass-box. How many angles are
there at the corner ?
6
EXPERIMENTAL GEOMETRY.
[iNTROD.
Size of an Angle.
Divide a piece of paper into two parts. Cut or fold one piece
across through a corner. Is your new angle at the corner greater
or smaller than the old 1
Since an angle can be greater or smaller, it is a magnitude —
i.e. it lias size.
Try the angles of your set-square against a corner of your paper.
Two of them should be smaller than the angle of
the paper; the third should fit exactly. This
angle is therefore equal to the angle of the
paper.
Measure each of these equal angles with your protractor. How
many degrees in each ?
Bisector.
Fold across a corner of your paper so that the fold passes
through the corner, and one side comes on the other.
Mark the fold, and unfold and flatten out the paper.
The fold divides the original angle into two
angles ; what do you know about these ? Why ?
Measure each of them, and also the corner angle,
with your protractor.
*A line which divides a figure into two equal parts is a
bisector of the figure.'
The fold above is the bisector of the angle of the corner.
The Right Angle — Perpendicular.
Fold over an edge AB of paper so that one part CB of the edge
comes exactly on to the other CA, and mark
the fold CD. Unfold again and flatten out.
The fold forms two angles, one with each
part of AB ; and these are equal; so that
the fold bisects the angle formed at C by^
the opposite parts CA, CB of the line. We therefore say
opposite parts of one straight line form an angle. This angle is
generally called two right angles, and sometimes a straight angle.
The line CD which bisects tlie straight angle, or two right
angles, at C is called a perpendicular to AB ; and the angles formed
by a straight line and a perpendicular to it are right angles.
introd.] experimental geometry. 7
Adjacent Angles — Opposite Angles.
Lay the protractor with its edge along AB (Jast figure) and its
centre at C. Measure the right angles at C. How many degrees
in a right angle? in two right angles 1 in a ^
straight angle ?
Draw a straight line AB, 4" or 5" long.
At a point C near its middle, draw another
line CD, r long. '^ c "
Lay the base of the protractor along AB, centre at C, and
measure the two angles formed. What is their sum ? How
many right angles 1
Draw two straight lines as before, but make
them cross. With protractor measure the angles
marked X, Y ; Y, Z ; Z, W. ^^
What is the sum of two adjacent (side by /
side) angles, as X, Y ?
How does X compare with the opposite angle Z ? Y with W ?
*If two straight lines meet, two adjacent angles always
make up two right angles; two opposite angles are always
equal.'
Parallel Straight Lines — Direction.
Draw by set-square and straight-edge two parallel lines 4" or 5"
long, and another line to cut these near
their middle.
Measure the angles marked X ; these are
towards the same parts, because they face
the same way.
Pick out and mark with Y, Z, &c., other
pairs of angles towards the same parts ;
measure each pair.
Draw a second line across the parallels, and repeat.
'When a straight line crosses two parallels, the angles
towards the same parts are always eaual.'
Are any angles equal which are not towards the same parts ?
Two lines which make equal angles towards the same parts with
any third line have the same direction ; so that * parallels are
lines in the same direction.'
It is shown in Chapter I. that parallel lines do not meet
8 experimental geometry. [introd.
Non-Parallels.
Draw two straight lines 4" or 5" long to form an angle, and a
third line to cut these.
Measure the angles marked +, X, towards
the same parts.
Are these equall Are other angles to-
wards the same parts equal? Draw a
second line crossing the two straight lines,
and repeat.
Two lines which make unequal angles towards the same parts
with any third line have different directions, and are non-parallel.
It is shown in Chapter II. that non-parallel lines in one plane
meet.
EXAMPLES-II.
1. Mark two points 4" apart on a sheet of paper ; mark by folding the
straight line joining them, and bisect this by folding again. Measure
the tAvo parts.
2. Draw a straight line AB, 2" long. Mark its mid point M, and
draw by set-square a perpendicular to AB at M. Take a point C on the
perpendicular 1" from M, and measure CA, CB. What do you notice?
Take a point D, 1^" from M, and repeat.
3. Draw a straight line AB, 1|" long ; mark points on it C, D, V'
from A and B. Draw perpendiculars at C, D. Are these parallel?
Why?
4. Make a straight line AB ; mark a point P on it, and draw a straight
line througli P at an angle of 73° to AB, and crossing it. Mark in
degrees the four angles at P.
5. Draw two parallels AB, CD by set-square ; mark a point P
on AB, and draw PQR to make an angle of 45° with AB, and cross
the parallels at P, Q. Write down in each angle at Q its number of
degrees.
6. Fold over a corner of your paper so as to bisect the angle, and
unfold again ; mark a point P on the fold 2" from the corner. Draw
through P a perpendicular to the fold to meet the sides in A, B.
Measure PA, PB.
7. Draw two lines crossing at an angle of 51°. Write down the
number of degrees in each of the other three angles.
8. Are two perpendiculars to a line always parallel ? Why ? Di-aw
two perpendiculars PC, QD to a line AB from points P, Q on it ; and
draw CE perpendicular to PC. Is CE parallel to AB ? Why ?
INTROD.]
/" ^ Of TM^-
EXPERIMEltoAL GEOMEfRY^'
Names of an Angle.
An angle is conveniently named by a letter at its point, as tlie
angle A ; but sometimes three or more lines meet at a point, as, for
instance, when an angle A is bisected by AD.
In that case any one of three angles at A might be ^^,^
meant by A ; so we put a letter on each of the sides as
well as at the point of the angles to distinguish them. The angles
are then named
BAG, i.e. angle at A with sides AB, AC ;
BAD, M A M AB, AD;
CAD, .1 A M AC, AD.
Note that the point of the angle is
indicated by the middle letter.
It is often convenient, however, to
X, Y, &c., in an angle to distinguish it.
angle X, the angle Y, &c.
Using A as the point at which the lines
BC and DE meet in the figure, write in three
letters the angles X, Y, Z. d -, /. " E
What is the value of BAE + BAD ?
What angle is equal to BAE? to BAD?
put a mark, such as
We then speak of the
Mark points
The Circle.
Draw a circle, radius 1". IMark the centre O.
P, Q, R on the circle, and measure PC, QO, RO.
'All points on a circle are equidistant (i.e. equally distant)
from the centre.'
Fold over a paper circle* so that the two parts exactly fit, and
mark the fold AB. Measure the length of the
fold, mark its mid i^oint O, and unfold again.
Mark points P, Q, R on the circumference, and
measure the distances PC, QO, RO. Are they
equal"? What point is O of the circle? What
fraction of AB is each of AC, PC, RO ?
If you know that a point is 1" from a given
point O, on what curve must the point be ? ISIark
a point O, and draw a curve every point of which is |" from O.
* Two sizes of circular filter-pajjers or jam-covers will be found useful.
10
EXPERIMENTAL GEOMETRY.
[iNTROD.
• R
Locus.
A curve or line containing all points of a given kind, as, for
example, all points 2" from a given point, is the locus of those
points. Sometimes a part only of a curve is the
locus, because only so7ne points, not all, on the
curve are of the given kind.
The locus of all points P which are f '' from O is
the circle, centre O, radius f ''.
Any point Q inside the circle is nearer to the
centre, and any point R outside the circle is farther
from the centre, than P.
This locus enables us to draw triangles whose sides are known.
Ex. Construct a triangle, sides 2 cm., 3 cm., 4 cm.
Set off one side AB, 4 cm.; tlien if C is the
third point, C is 2 cm. from A, and is on the
locus circle, centre A, radius 2 cm.; and it is
also on the locus circle, centre B, radius 3 cm.
Draw these circles meeting at C. Join AC,
CB.
ABC is the triangle.
EXAMPLES— III.
1. Draw a plan of the space on which a goat can feed when tethered
by a rope 10 ft. long. (Use 1" for 10 ft.)
2. Two posts are 10 ft. and 15 ft. from one end of a lane, and are 20 ft.
apart. Represent on a plan (1" to 10 ft.). If the lane is parallel to the
line of posts, draw its plan. (Draw the lane as a straight line. )
3. Draw a triangle, sides |", 1", 1|", and measure the greatest angle.
4. Draw an equilateral (equal-sided) triangle, each side 2-5 cm.
Measure an angle.
5. Draw an isosceles triangle (two sides equal), the equal sides 3-2 cm.,
the third side 1 -8 cm.
6. Draw a circle, radius 1", and mark a point P on it. Find two other
points Q, R on it |" and 1^" from P. Can you find a point on the circle
3" from P ?
7. Draw a straight line AB, and a point P about 1" from AB over its
middle point. Find a point in the straight line 1^" from P. How many
such points can you find ?
INTROD.]
EXPERIMENTAL GEOMETRY.
11
The Chord of a Circle.
Draw a circle, centre O, radius 2 cm. Mark two points P, Q on
it, and join PQ.
The straight line PQ is a chord of the circle ;
either part of the curve from P to Q is an arc of
the circle.
Draw a chord PR through the centre O.
A chord of a circle through the centre, as PR,
is a diameter.
If a circle is folded over a diameter, one side fits exactly on the
other, so that a diameter bisects the circle.
Fold over a paper circle so that one side fits exactly on the
other, and mark the fold AB.
AB is a diameter, and its mid point O the centre.
Fold over again at a point M in AB so that the
part MA of the first fold fits exactly along MB,
and mark the new (double) fold MP.
Unfold and flatten out as in the figure.
What are the angles atM? IsMP=MQ?
* A chord perpendicular to a diameter is bisected by it.'
The Isosceles Triangle.
Fold over a piece of paper at a point D in a straight-cut edge, so
that one part of the edge DB conies on the other DC.
Cut across AC so as to form a twofold
triangle, and unfold ; .*. side AB = ?
The figure formed is an isosceles"^ triangle ABC.
The fold AD is the bisector of angle A;
what are the angles at D ?
Also angle B exactly fits angle C ;
.'. angle B= 1
' The angles at the base of an isosceles triangle are equal.'
If BD is 2'\ calculate DC. If the angle BAC is 37°, calculate
angles BAD, DAC.
If angle B is 711°, ^hat is C?
If a triangle ABC is equilateral, what angles are equal ?
Ex. Make an isosceles triangle ABC, equal sides AB, AC 1^", third
side 2". Measure its angles. Which are equal ?
* That is, equal-sided.
12 EXPERIMENTAL GEOMETRY. [iNTROD.
Two Circles.
Take two paper circles, slide one over the other ; in how many
points do the two curves meet ?
' Two circles never meet in more than two points.'
Fold each so as to mark a diameter, place
them to overlap with these diameters in one
line, and fasten them together with gummed
paper.
Fold both over simultaneously about the
common diameter OQ ; one point B where they meet comes
exactly on the other point A.
Unfold ; the chord AB is bisected at M, and the angles at M
are right angles.
AB is the common chord of the two circles.
'The line of centres of two circles bisects at right angles
their common chord.'
The Angles of a Triangle.
Draw a triangle ABC ; and at B make DBE parallel to AC, and
prolong CB, AB to cross DE, making the angles X, Y, Z.
Then angle X = A (same parts), . ,
Z=C u \ /
Y = B (opposite angle).
Set the protractor with its base along q_.
DE, and its centre at B,
to cover the three angles X, Y, Z.
X + Y + Z = 180°.
.*. A + B + C= ?
' The three angles of a triangle always make up two right
angles.'
If ABC is equilateral the three angles are equal, and together
make up 180°.
But 3 X 60 = 180 ; and therefore
* Each angle of an equilateral triangle is 60°.'
INTR0D.1 EXPERIMENTAL GEOMETRY. 13
EXAMPLES— IV.
1. Construct an isosceles triangle, base AB 4 cm., sides AC, BC
each 3 cm.
2. Construct a point C, 3 cm. from A and from B, A and B being
4 cm. apart. Is the construction the same as that of Ex. 1 ? Why ?
How many such points do you find ?
3. Construct an isosceles triangle, equal sides 1^", their angle 40°.
Calculate the other angles.
4. Draw an equilateral triangle, side 1^". Measure one of its angles.
What ought it to be ? What angle can you thus construct ?
5. Two angles of a triangle are 33° and 106°. What is the third
angle ?
6. Take two points A, B, 5 cm. apart. Construct two points C, D,
each 3 cm. from A and 4 cm. from B. jNIeasure CD and mark its mid
point.
7. Find two points C and D each 3 cm. from A and B, when A and B
are 4 cm. apart. What distance is A from C and D ? Mark the mid
point of AB.
8. Can the three angles of a triangle be 72°, 94°, 30° ? Why ? If the
first two are correct, what ought the third to be ?
The Rhombus.
Fold over a piece of paper, and mark the fold AC. Fold over
again so that one part of the fold MC comes on the other MA, and
mark the second fold MD.
Cut across AD so as to cut out a fourfold \ a]^^^^^
triangle AMD; unfold and flatten out into " B ^_-^'''
the four-sided figure ABCD. I "^-^-vj^ U
What do you know about the sides of this i pj
figure 1
An equal-sided four -sided figure is a
rhombus. AC, BD are diagonals. ^.^^^'n.^
WTiat do you know about the angles ADB, ^i^— — l!2i^^_
CDB? About the angles at M ? About the ^^-^^-^
parts MA, MC of diagonal AC?
'The diagonals of a rhombus bisect its angles, and bisect
each other at right angles.'
14
EXPERIMENTAL GEOMETRY.
[iNTROD.
Ehombus and Bisectors.
It is easy to construct a rhombus when its side and one angle,
or its side and a diagonal, are given.
Ex. (i.). ' Construct a rhombus, angle 46'
3" '
side
draw arc of circle BC
Make angle A ==46°; with centre A, radius f"
to cut the sides of angle A in B, C.
With centres B, C, same radius (|"), draw arcs
cutting in D ; join BD, CD.
ABDC is the rhombus.
Also, since the diagonal AD bisects the angle
A of the rhombus, a similar construction enables
us to bisect a given angle A. Tims :
Draw a circle, centre A, any radius AB, cutting the sides AB, AC of
the angle in B, C ; with centres B, C, same radius, draw arcs D.
AD is the bisector of angle A.
Ex. (ii.). 'Construct a rhombus, one diagonal 1", side f'V
Make AB 1" ; with centres A, B, radius |", draw
arcs of circles cutting in C, D.
ACBD is the rhombus.
Also, since the diagonal CD bisects AB at right
angles, a similar construction enables us to draw a ^
line CD bisecting at right angles (at M) a given
line AB. Thus :
Draw circles of equal radii, centres A, B, to cut in
C, D ; the line CD bisects AB at right angles in M.
The Eight Bisector.
The line bisecting another line AB at right angles is the right
bisector (i.e. the perpendicular bisector) of AB.
If C (last figure) is any point equally distant from two points
A, B, the triangle CAB is isosceles, and folds over so that the
bisector CM of angle C is also the right bisector of AB. Thus :
* The locus of points equidistant from two fixed points A, B
is the right bisector of the line AB.'
By the aid of this locus we are able to draw circles to pass
accurately through two given points.
Ex. What kind of triangles are ABD, ACD in the upper figure?
ABC, ABD, ACD, BCD in the lower figure?
X 1
X 1
Yx
N. \
Ml y^
^\\
y
INTROD.
EXPERIMENTAL GEOMETRY.
15
Circle through Given Points.
Ex. (i.). *Draw a circle, radius h", to pass through two points
A, B, f " apart.
Draw and pioduce DM, the riglit bisector of AB
(as on last page).
With centre B, radius h", draw arc of circle to cut
DM in O.
Draw circle, centre O, radius ^", through A, B. A""
Can you do this more simply ?
Ex. (ii.). 'Draw a circle through the three
angular points A, B, C of a triangle.'
Draw and produce DM, the right bisector of AB ;
EN, „ M AC,
to meet DM in O.
^Yith centre O, radius OA, draw the circle
through A, B, C.
This circle is the circumcircle of triangle ABC.
EXAMPLES-V.
1. Draw a rhombus, ang. 56°, side 3 cm. Measure its diagonals.
2. Draw angles of 72°, 84°, 108°, 156°, and bisect them. Measure one
of the parts in each case.
3. Draw a rhombus, angle 66°, side 1^". Test with your set-square to
see if opposite sides are parallel. Are they ? Calculate the angles made
by the longer diagonal and the sides.
4. Draw a rhombus, diagonal 4-2 cm., side 3-5 cm. Calculate the sum
of its four angles. (Use the two triangles.)
5. Draw lines 1|", 2|", 2|" long, and draw their right bisectors.
Measure one part in each case.
6. Draw a rhombus ABCD, angle A 60°, side 1". Draw and measure
perpendiculars from A, B on the side CD (produced sufficiently).
7. Make a straiglit line AB, 6 cm. Draw a circle, radius 5 cm., to
pass through A, B. How far is its centre from the line AB ?
8. Draw a triangle, « = !", 6 = 1|'-', c=l^" ; and draw its circumcircle.
9. Make an angle BAC, 30°. Make AC = 2 cm., AD = 5 cm., along
AC ; and draw a circle to pass through C, D and have its centre in AB.
(Draw the right bisector of CD to meet AB in the centre of the circle.)
16 experimental geometry. [introd.
• Angle in a Semicircle — Angles in an Arc.
Draw a semicircle APB, centre O, radius 1", diameter AB.
Take a point P on the circle.
Draw PA, PB to T from P.
Measure the angle APB. How many-
degrees 1
Take a second point Q on the circle and
repeat. ^^.^ ^^^
The angle AQB is ? degrees.
Fix two pins (or divider points) at A, B, place the right angle of
your set-square on the semicircle so that its sides are close against
the pins at A, B ; the point of the right angle
is on the circle.
Repeat in several other positions ; the point
of the right angle is always on the circle.
* The angle in a semicircle is a right angle.'
Draw a circle, radius V\ centre O. Place in it a chord AB, If ;
take points P, Q as before, and measure the angles APB, AQB.
APB is 1 AQB is 1
Put the 60° angle of your set-square on the
circle APB, with its sides against pins at A, B.
The point of the angle comes on the circle.
Repeat in several positions. The point of the
angle is always on the circle.
Repeat the above process, using the 30° angle of your set-square,
a circle of V^ radius, and a chord of 1".
' Angles in the same arc of a circle are equal.'
This gives us another locus — viz. that of the point of a given
angle whose sides traverse fixed points. This locus is only^ar^
of a circle (or more strictly tioo parts of circles, one on each side
of AB).
In each case above measure the angle formed at the centre by radii
to the pin-points. Compare with that in the corresponding arc.
'An angle at the centre of a circle is double the angle at
the circumference on the same arc'
c-p^\- nr^
A_J___\ I lli
^ s
jntrod.] experimental geometry. 17
Points Equidistant from a Line.
There is still another locus which is very useful.
Draw a straight line AB, and a parallel CD about 1^" from it.
Bring a straight-edge S along AB,
so that the instrument lies on the
side of AB away from CD. Put one
side of the right angle of the set-
square on AB, and mark on the set-
square with a pencil the point P where the other side crosses
CD.
Slide the set-square along the straight-edge (which must be held
firm) ; P always comes on the parallel CD, so that the distance of
any point on CD from AB is always the distance PQ from P to the
corner Q of the set-square.
' The locus of all points equidistant from a fixed line is two
parallels to the line at that distance.'
Where is the second parallel ?
We can easily construct these parallels.
Ex. ' Construct a parallel to A B at a distance of 3 cm,'
Make AC perpendicular to AB, 3 cm. long.
Through C draw CD parallel to AB.
EXAMPLES— VI.
1. Draw 4 circles, radius 4 cm. Place chords 7 cm., 4 cm., 3 cm.,
2 cm. in the successive circles ; and measure an angle in the laiger
arc in each case.
2. Take AB, 4 cm. Find a point P such that angle APB is a right
angle, and distance AP = 2 cm. (Use semicircle.)
3. Construct a line AB, and a point C 1^" from it. Find a point P
1^" from C and 1" from the line BA.
4. One side of a set-square slides along a straight-edge. What is the
locus of the opposite point of the set-square ?
5. Construct an angle BAC of 72°. Construct a point P on the side
AB, li" from AC.
6. Construct a semicircle, centre O, diameter AB, radius 2 cm. Draw
OC perpendicular to AB, meeting the circle in C. Join AC, BC. What
angle is ACB ? Compare the sides AC, BC.
p. o. , B
18 EXPERIMENT AND THEORY. [iNTROD.
The experiments which show that all angles in certain arcs
are equal are excellent examples of the way in which new facts
may be discovered by experiment; but as we cannot try the
experiment in all possible cases, it is natural to see if we can
find some way to convince ourselves that the result must be
true in all cases. This is done by inventing a theory to account
for all the facts; the accuracy of our invented theory is
then put to the proof from time to time by testing experiment-
ally the truth of new results which it establishes. If any of
these are clearly not true, our theory must be erroneous or
incomplete.
Statements based on a theory are called theorems, and can
be proved by the original definitions, or by other theorems
previously established. Such proof is called formal; and the
succeeding chapters of this book constitute the more elemen-
tary part of formal as distinct from experimental geometry.
Practical methods of drawing and measuring are derived
from theorems and embodied in constructions. The proofs
of Theorems 2, 3, and 6 in Chapter I. are, however, experi-
mental, the theoretical proofs being given at the end of
Chapter IX.
The fundamental definitions of formal geometry are those of
the plane, straight line, angle, and direction or parallels ; in
addition to these we require notions of length, of extent of
surface and volume — i.e. of 'space' — of magnitude, of position,
of movement or change of position, and of shape or form. We
also assume — i.e. take it for granted — that a body or figure may
be moved without any change (of shape, &c.) except that of
position, and that any magnitude (length, area, &c.) may be so
multiplied as to exceed any other like magnitude (e.g. a mile
may be so multiplied as to exceed the distance from the earth
to the sun) ; and we also assume the general notions of equality
and inequality.
The theory of geometry is very old, dating back at least 2500
years to Thales ; it has been developed (to give a few names)
among the ancients by Pythagoras, Euclid, Archimedes, Ptolemy,
&c., and among the moderns by Pascal, Newton, Poncelet, Monge,
Carnot, Cremona, &c.
INTROD.] DIAGONAL SCALES. 19
We conclude this section with a brief account of scales,* because
of their connection with our instruments of measurement.
1 i •
1 ' 1
n U [T "
^! '
Qp
'^p
1
1
1
■" 11
1
1 1
M 1 1 M 1 1 n 1
tenths 8 6 4 2 / 2 units
The diagonal scale enables small fractions of an inch or half-
inch or centimetre to be measured more accurately than the plain
scale allows.
The base line is divided into a plain scale of units and, say,
tenths ; and perpendiculars are drawn through the unit divisions
0, 1, 2, &c. Ten parallels to the base line are drawn at equal
distances; and the tenth divisions of the base line are joined
diagonally to those of the top line.
This enables us to measure tenths of tenths — i.e. hundredths of
the unit.
'Measure a length PQ by an inch diagonal scale, showing
hundredths.'
Adjust the divider points to PQ, bring one of them along
a unit perpendicular and the other in the divided unit on a
parallel through the first.
Move the divider points, always parallel to the base line, until
the Q end is on a diagonal as well as a parallel, keeping P always
on a unit perpendicular.
In the example P is on the 1st unit, PQ = 1"
M Q II 5th diagonal, PQ = 1-5"
•• Q II 7th parallel, PQ= 1.57".
Similarly we can set the divider to a length of 1-57".
Note. If the base unit is divided into 12 parts, and there are 8
parallels, the diagonal scale shows eighths of twelfths— i.e. ninety-sixth
parts — and so on.
* The rest of this section may be postponed by beginners. The use of the
diagonal scale is required for some of the examples in Chapter III.
20 CONSTRUCTION OP SCALES. [iNTROD.
Scales are constructed by setting off the unit a suitable number
of times along a straight line, and dividing one unit by construc-
tion (p. 4) into the required number of parts, as in the figure
on p. 1.
In constructing a diagonal scale, a drawing-board should be
used; a T-square for the base line and its parallels, and a set-
square on the T-square for the perpendiculars. When the unit
on the base line has been divided, the corresponding points on the
top line can be marked from these by moving the set-square along
the T-square ; the top and bottom divisions must then be joined
diagonally.
Drawing ' to Scale ' and Comparative Scales.
In maps and plans, objects are rarely represented in their actual
size ; for drawing such plans a scale is made or used in which the
actual unit (1 mile, 1 yd., &c.) is replaced by a more suitable
length (1 in., | in., &c.) ; and the plan drawn by means of this
scale is said to be 'to a certain scale' — e.g. 1" to the mile, J" to
the yard, a scale of 1 : 20, &c.
The fraction acTuaiTnit ^^ called the representative fraction of
the scale. Thus, if f represent 1 yd.,
1 1
and for 1" to the mile, RF. = , ^^ —
1760 X 36 63360
Ex. Draw a scale of miles up to 4 miles to show furlongs, R.F. = ^-^u.
1 mile is represented by — — - mile
1760 X 36 . ,
= -84480-^"'^
Set off units of f inch, and divide one into eighths.
fjr. )8 ^ 10 I I
I I I I I, iy^ ^
- T
3 mi.
INTROD.] COMPARATIVE SCALES. 21
In measuring distances on foreign maps, it is sometimes neces-
sary to convert measures from one scale to another — e.g. from
kilometres to miles. In this case a comparative scale of kilo-
metres can be constructed from the scale of miles or vice versa.
Ex. Given a scale of \\" to the mile, construct a comparative scale of
kilometres, showing tenths, up to six kilometres.
1 kilometre = 39370", 1 mile = 63360".
3937
. •. 1 kilometre is represented by — _ - x 1^" = -93".
OoOO
Set off units of -93" and divide one into tenths.
EXAMPLES— VII.
Scales.
1. Make a scale of miles f " to the mile up to 6 miles, showing furlongs.
Indicate on it a length of 3 miles 7 fur.
2. Using scale 1, make a plan of a trapezium-shaped park, parallel
sides 2|, 3^ miles, their distance apart If miles, another side 2^ miles.
3. A French map is drawn 2 cm. to the kilometre. Make a compara-
tive scale of miles. (1 mile = 1-609 km. ; 1 km. =1000 metres = 100,000 cm.)
AVrite down KF.
4. Make a scale of yards, J" to the yard, up to 1 pole, showing feet.
Divide diagonally to show inches. Give R.F.
5. Make a scale of poles up to 1 chain, showing yards, R.F. y^r-*
6. From a scale of miles §" to the mile, up to 8 miles, construct
a scale of geographical miles (60 geographical miles = 69-1 miles),
showing tenths.
7. On a map a distance of 3 miles 2 fur. is represented by liV"- Find
the KF., and show a line representing 5 miles 5 fur.
8. Make a scale of Russian versts from a scale of miles 1-2" to the
mile. (1 verst = - 663 mile.) Calculate R.F.
9. Make a scale of furlongs up to 1 mile, R.F. Tro^- Divide diagonally
to show poles.
10. A yachting course is 9 km. E., 12 N., and 15 home. Calculate the
distance in miles, and represent on a map I" to the mile. (1 km. =§ mile.)
11. Make a scale of kilometres up to 5 km., 3 cm. to the km. ;
calculate the R.F., and represent a distance of 2-57 km.
12. Make a scale 25 cm. to the kilometre, up to \ km., showing tenths,
and show metres by diagonal division. (1 km. = 1000 metres.)
22 ABBREVIATIONS. [iNTROD.
ABBREVIATIONS.
These may be referred to as they occur in the text.
The following symbols are used :
. •. therefore ; '. * because ;
= is or are equal to, =i= is or are not equal to ;
^ II greater than, ^ n greater than ;
<; II less than, <j;^ n less than ;
= is greater than, equal to, or less than ;
^ is congruent to ; ||j is similar to ;
II is parallel to ; J. is perpendicular to ;
A area of triangle ; t — 7 area of parallelogram.
Obvious abbreviations of words are used :
str., straight; perp., perpendicular; pari., parallel; eql., equal
hyp., hypotenuse; alt., altitude, alternate; chd., chord;
tangt. , tangent ; bisr. , bisector ;
ang. , angle ; rt. ang. , right angle ;
compt., complement ; suppt., supplement;
fig., figure; poln., polygon; sq., square;
tr. , triangle ; quadl. , quadrilateral ;
parm., parallelogram; rect., rectangle;
circ, circle; semcle., semicircle.
opp., opposite; adj., adjacent;
propl., proportional; simr., similar.
simly., similarly; simde., similitude;
Def., definition; Th., theorem; Cor., corollary;
Constr., construction; Ex., example, &c. ;
line is used for straight line ;
a, b, c, sides of triangle ABC ;
A, B, C, angles of triangle opp. a, 6, c respectively ;
. ., a + b + c
s = semi-sum of sides = — ^ — ;
R, radius of circumcircle ;
r, II iiicircle ;
PLANE GEOMETRY,
CHAPTER I.
PRELIMINARY DEFINITIONS AND THEOREMS.
Geometry treats of the position, form, and size of bodies by
means of figures.
The fundamental assumption or postulate of geometry is that
figures or bodies may be supposed to be moved without changing
either form or size.
Definition 1. — A surface is the boundary of a body or part of
a body.
Ex. Top of a table, outside of a book.
Definition 2. — A line is the boundary of any part of a surface.
Ex. Edge of a table, of a page of a book.
Definition 3. — A point is the boundary of any part of a line.
Ex. Corner of a table, of a page of a book.
Points may be marked by a cross (x) or a dot (.).
Note. If a point moves, its path is a line, and it is said to generate
the line— e.g. a pen-point traces or generates a line on paper. A line
of this kind is often called a locus.
Similarly, a line moving generates a surface, as a hand of a clock
(considered as a line) ; a knife-edge cutting an apple generates the cut
surface.
Surfaces are recognised as of many kinds, rough, smooth, hard,
soft, &c. ; in geometry we are concerned with their shape and size
only, as round, flat, conical, &c.
Ex. Name bodies of round, flat, conical surfaces.
24 PLANE. [CH. I.
The most important surface is an ideally flat or level surface.
Many surfaces are nearly so — that of still water, of a well-finished
table, slate, or drawing-board; the best examples are ordinary
plate-glass, certain machine-made steel 'planes,' and 'optically
plane ' glass.
The Plane.*
Take a post-card, mark two points on one side, lay a straight-
edge against them, and with the point of a knife along the straight-
edge score the card along a straight line, taking the line right
across the card. Be careful not to cut the straight-edge, nor to
cut right through the card.
Now fold over the card so that one part fits closely on the other.
A surface which can be folded over in this manner about any two
points in it is a plane surface.
Take two post-cards, and place one on the other ; notice that you
can slide or turn one on the other, and that they coincide — i.e.
exactly fit — within their common boundary. This is always true
of two plane surfaces, or two parts of a plane surface.
Lay a post-card on a slate, a drawing-board, or a larger sheet
of cardboard ; its surface will coincide everywhere with that of
the other as it slides over itj hence there is a plane surface —
e.g. that of the board — greater than that of the post-card. Simi-
larly we could find a sheet of plate-glass with a plane sur-
face greater than that of the board, and so on ; and there is
every reason to believe that however great a plane surface is,
a greater one is possible : we say, then, that a plane surface
can be produced or extended indefinitely — or, briefly, that it is
infinite.!
Note. It is only the surface of the card that coincides with the
surface of the hoard or slieet of glass ; mentally, we picture the
surface as having position and as having sides — e.g. upper and under;
hut being a boundary only, it has no substance.
* A few post-cards and some tracing-paper will be useful in addition to ordi-
nary paper. Sheets of plate-glass would, of course, be better than post-cards
for showing coincidence of two planes.
1 1 think it possible that this fact is included in the folding definition of plane,
but I have not so far succeeded in proving this.
CH. I.] PLANE AND STRAIGHT LINE. 25
We have thus arrived at the following properties of a 'plane'
surface : it folds over about any two points, it is infinite, and it
can be placed so as to coincide with any other plane.*
The fold of a plane when folded over is the most important
kind of line — viz. a straight line. A well-made steel straight-edge
gives the best example of this.
The Straight Line.
Carefully fold a smooth sheet of paper,! and make the fold as
precise as possible. Note that the fold extends right across the
paper.
Unfold and lay the sheet on a flat
surface (board or slate), fix two points
A, B of the fold with divider points,
and hold down a third point C, not on
the fold, with a finger.
Now turn the free part P of the paper
in any way. Notice that the fold does
not move, but divides the free part of
the surface from the fixed part. Now
keep only the points A, B fixed and turn either or both parts of the
paper without pulling ; note that no part of the fold moves.
Fold a second sheet and fix two points A, B on both folds
simultaneously ; the two folds coincide, and neither moves as we
move the parts of either or both sheets. This is true of all
straight lines — viz. that when two points of one are fixed, the
straight line has one position only. Also, as the fold extends to
the boundary of the plane, by extending the plane we extend the
straight line ; hence a straight line is also infinite.
Now place your two folds with only one point, D say, common ;
note that they cross. Two straight lines (produced sufficiently)
with one point common always cross,
We now present these properties of plane and straight line in
the form of definitions and theorems.
* "We shall see presently that the last fact is a consequence of the others and
of the propei'ties of the fold,
f The paper should be fairly stiff.
26 PLANE AND STRAIGHT LINE. [CH. I.
Definition 4.^ A plane is a surface, infinite in extent, which can
be folded about any two points of the surface so that one part lies
entirely on the other.
Definition 5. — A straight line is the fold of a plane about any
two points in it, and is infinite in extent.
The limited straight line from a point A to a point B is the
join of A, B.
Theorem 1. — * A straight line can be drawn through any two
points in a plane, and lies entirely in the plane.'
Theorem 2. — * Any two straight lines coincide entirely when
two points of each coincide.' ^
These two theorems follow at once from the properties of a fold
shown above.
Theorem 3. — 'Every finite straight line has a mid point
about which it can be reversed so that the ends are inter-
changed.' *
Mark two points A, B on a fold of paper, f
Fold the paper over until one of them A comes
on the other B, and mark the new fold, crossing
the old fold at M.
Thus the part MA of the line folds over on to
MB. Similarly MB will fold on to MA; hence
MA = MB.
Thus the line can be reversed on itself so that A and B are
interchanged, the mid point M remaining unmoved.
It follows also that the straight line can be turned in the plane
about M so as to be reversed.
Note. The second fold in this case is perpendicular to the first (see
Def. 9 and 11), and makes right angles with it.
* Formal proof is given in a note after Chapter IX.
t Tracing-paper may be used for a straight line drawn in ink or pencil.
CH. I.] PLANE AND STRAIGHT LINE. 27
EXAMPLES— VIII.
1. Draw a straight line AB with the aid of a fold of paper. Test it by
placing the folded sheet first on one side and then on the other side of
the line.
2. Test in tlie same way all your straight edges (set-squares and
scales).
3. Mark two points in a straight line AB ; measure their distance with
divider and scale.
Interchange the divider points and measure again. Are both measures
the same ? Why ?
4. Stretch a piece of thread along the paper. Test with a straight-
edge. Is it straight ?
5. Bisect the edge of a piece of paper by folding over. Test by
measuring.
Properties of the Plane.
Lay any plane P (of tracing-paper) on another Q, and prick
through three points A, B, C not in one straight line, so that these
points are in both planes.
The complete straight lines AB, BC, CA
lie in both planes. (Thh. 1, 2.)
If now a straight edge* turns in the top
plane about a pin fixed at A, so as to sweep
out the space XX' in that plane, it will cross
some point of CB at every stage of the
process, and hence the complete line repre-
sented by the straight edge lies always in
both planes.
Thus the whole space XX' lies in both
planes.
Similarly YY' and ZZ' lie entirely in both planes, so that the
planes P, Q coincide altogether. Hence :
Theorem 4. — 'Any two planes coincide when three points of
each, not in one straight line, coincide.'
This theorem remains true when the two sides of either plane
are interchanged. Hence :
Theorem 5. — * A plane may be reversed upon itself/
* Supposed infinitely long.
28 ANGLE — PERPENDICULAR. [cH. I.
Plane Figures.
Definition 6. — Any group of points, lines, surfaces, is a figure.
Definition 7. — A plane figure is a figure whose points and lines
are all in one plane.
Definition 8. — An angle is a plane figure formed by two
straight lines which terminate at a common point.
If two lines cross they form four angles at
the point of crossing.
Theorem 6. — 'Every angle has a bisector about which it
can be reversed so that its sides are interchanged.' *
Draw an angle A on tracing-paper, fold over about A so that
one side AB comes on the other AC, and mark the
fold AD.
Thus the angle BAD folds on to CAD, about the
bisector AD.
Similarly CAD folds on to BAD ;
hence CAD = BAD.
Thus the augle BAC can be reversed on itself so ^j
that AB and AC are interchanged, the bisector AD
remaining unmoved.
Note. If AB, AC are opposite parts of one straight line, there is still
a bisector AD.
Definition 9. — A perpendicular to a straight line at any point
in it is the bisector of the angle formed by opposite parts of the
line at that point.
To obtain this, fold over a straight edge of paper about a point
A in it, so that one part AC of the line folds on to the other part
AB. The new fold is the perpendicular.
Definition 10. — The right bisector of a straight line is the
perpendicular to it at its mid point.
Ex. C
of paper.
* Formal proof is given iu a note after Chapter IX.
CH. T.] RIGHT, ACUTE, OBTUSE ANGLES. 29
Theorem 7. — * One only perpendicular exists to a straight line
at each point in the line.'
For there is one only bisector of the angle of opposite parts AB,
AC of the line at each point A of it (next fig.).
Definition 11. — A right angle is an angle formed by a straight
line and a perpendicular to it.
Theorem 8. — ' All right angles are equal.'
If AD ± BC, and EH _L FG,
place the figure FEHG on BADC, so that
FG comes on BC, and E on A.
/. the bisector EH comes on AD ; ''
.*. rt. ang. H EG coincides with rt. ang. DAC ;
i.e. rt. ang. HEG = rt. ang. DAC.
Similarly any two right angles are equal.
Definition 12. — An acute angle is less than a right angle.
An obtuse angle is greater than one, less than two right angles.
Definition 13. — The complement of an angle is its defect from
one right angle. (Compt. of A = 90° — A.)
The supplement of an angle is its defect from two right angles.
(Suppt. of A = 180° -A.)
EXAMPLES— IX.
1. Take a piece of paper with straight-cut edges. Fold over a corner
so that one edge comes on the other, and mark the fold. What is the
fold called ? Measure the angles thus formed.
2. Fold over a straight edge of paper so that one part of the edge
comes on the other, and mark the fold. What is the fold called ? The
two angles thus formed ? Measure them.
3. Draw angles of 20°, 54°, 90°, 135°, 167°. Which are acute, right,
obtuse ?
4. Write down the complements of 17°, 56°, 72°, and the supplements
of 36°, 72°, 153°.
5. Draw the right bisector of a straight line AB, 2" long. Compare
the distances from A, B of several points in it. (Bisect the line at M by
parallels as on p. 4, and draw a perp. as on p. 3, or use Ex. (ii.), p. 14.)
6. Through how many right angles does the minute-hand of a clock
turn in a day (24 hrs. ) ?
30 DIRECTION — PARALLELS. [CH. I.
Definition 14. — Two straight lines in a plane have the same or
different directions according as they malce
with any third line whatever equal or unequal
angles towards the same parts.
Thus AB and CD have the same direction
if X = Y, and then also Z - W ;*
and AB and CD have different directions
if X =1= Y, and then also Z 4= W.t
Definition 15. — Parallel straight lines are coplanar lines
having the same direction.
(Coplanar lines are lines in one plane.)
Ex. Two perpendiculars to a given straight Une are parallel.
Theorem 9. — * Two parallels either coincide or do not meet.*
Either (i.) the parallels AB, CD have a AC
common point E, as in fig. (i.) ;
then if EF is a third line in their plane,
ang. CEF = AEF, towards same parts;
.*. CE coincides with AE ;
i.e. CD coincides with AB ;
or (ii.), the parallels AB, CD have no
common point ;
i.e. they do not meet.
EXAMPLES— X.
1. If a set-square moves with one edge along a fixed straight line,
two positions of a second edge are parallel.
2. Draw two parallels, and a line cutting one at 70°; measure the
angles it makes with the second parallel. Which are equal ?
3. Draw two straight lines meeting, and a third line to cut both ;
measure the angles it makes with them. Are any equal ?
4. Draw two lines at an angle of 53°, and two parallels to these.
Measure the angle of the latter towards the same parts.
5. Make a four-sided figure of two sets of parallels ; compare its
angles, and its opposite sides.
6.t 'If «, h are parallels, and also c, d are parallels, show that the
angle a, c = angle 6, d towards the same parts.'
* This definition includes Euclid's parallel axiom, + Important.
CH. r.]
CIRCLE — TRIANGLE.
31
^rc
The Circle. — If a straight line turns in a plane about a fixed
point in the line, so as to sweep out the whole space of four right
angles round the point, any point of the straight line describes a
closed curve enclosing the fixed point.
Definition 16. — A circle is a closed pLane curve of which all
points are equally distant from a fixed point in its plane called the
centre.
A radius is a straight line from centre to curve.
A chord is the join of two points on the curve.
A diameter is a chord through the centre ;
it is double any radius, and divides the circle
into two semicircles.
An arc is a part of the curve.
A sector is a part of the circle bounded by two radii and the
intercepted arc.
Definition 17. — A triangle is a closed plane figure formed by the
joins of three points not in a straight line.
The three points A, B, C are vertices,*
their joins AB, BC, CA the sides, and
the three interior angles the angles, of
the triangle.
The angles may be denoted by the
capital letters A, B, C, and the sides
opposite these by the italic letters a, 6, c, when no confusion arises
from doing so.
An altitude f of a triangle is the length of a perpendicular from
a vertex to the opposite side, which is then called the base.
A median is a straight line from a vertex to the mid point of
the opposite side.
Definition 18. — An isosceles triangle has two equal sides.
An equilateral triangle has three equal sides.
A right triangle has one angle a right angle, and the side
opposite this is the hypotenuse.
An obtuse triangle has one angle obtuse.
An acute triangle has three acute angles.
* Singular, vertex ; plural, vertices or vertexes.
t Or perpendicular, when used of the direction only, without reference to
magnitude,
base
32
QUADRILATERAL — POLYGON.
[CH.
EXAMPLES— XI.
1. Draw a circle, 1" radius, and place in it a chord IJ" long. Cut off
an arc whose chord is 1|" long. Measure the longest chord of the circle.
2. Draw an isosceles triangle, given a = b = li", c = |". Measure the
altitude from C to AB.
3. Draw an isosceles triangle, given C = 34°, a = b = 2^". Measure the
altitude from C to AB.
4. Draw an equilateral triangle of If" side. Measure one angle.
5. Draw a right triangle, hypotenuse 2", one side 1". Measure the
median of the hypotenuse, and the larger acute angle.
6. Draw a triangle, altitude 1", base 1|", one side IJ". Measure the
third side.
7. Draw a triangle, « = 2|", B = 42°, C = 64°. Measure the altitude
from A to BC.
Definition 19. — A quadrilateral is a closed plane figure formed
by the joins of four points,* two and two, and two and two.
A rhombus is an equal-sided quadrilateral.
A parallelogram is a quadrilateral with opposite sides parallel,
two and two.
A rectangle is a right-angled parallelogram.
A square is an equal-sided rectangle.
A trapezium is a quadrilateral with two sides parallel. -
Definition 20. — Polygons are closed plane figures bounded by
straight lines, and are named from the number of sides and angles.
Sides and angles. Sides and angles.
Triangle 3 Octagon 8
Quadrilateral 4 Nonagon 9
Pentagon 5 Decagon 10
Hexagon 6 Dodecagon 12
Heptagon 7 Quindecagon 15
A regular polygon has all its sides equal, and all its angles equal.
Definition 21. — A diagonal of a polygon is a join of two
vertices, not a side of the polygon.
* No three coUinear, (This is always imderstood in definitions of polygons.)
CH. I.] SYMMETRY — CONGRUENCE. 33
EXAMPLES— XII.
1. Construct a rhombus, side 1^", one angle 60°. Measure the shorter
diagonal.
2. Construct a parallelogram, sides 1", 1^", angle 49". Measure the
shorter altitude.
3. Construct a rectangle, sides f", If". Measure diagonals.
4. Construct a square of 1^" side. Measure diagonal.
5. Construct a trapezium, pentagon, hexagon, heptagon.
6. Name in order the five figures given in Definition 19.
Symmetry and Congruence.
Definition 22. — A figure is symmetrical about a line when it
can be reversed about that line so as to coincide with its original
position.
EXAMPLES-XIIL*
1. A limited straight line is symmetrical about its right bisector.
2. An angle is symmetrical about its bisector.
3. A circle is symmetrical about any diameter.
4. The figure formed by two circles is symmetrical about the line of
their centres.
Definition 23. — Two figures are congruent when one can be
plnced on the other so as to coincide with it.
EXAMPLES— XIV.
1. Two planes or two sides of a plane, considered as infinite, are
congruent.
2. Two right angles are congruent.
3. The two parts of a bisected line, or a bisected angle, are congraent.
4. Two straight lines of the same length are congruent.
5. Two equal angles, or the two faces of an angle, are congruent.
6. Two circles with equal radii are congruent.
7. If two triangles ABC, DEF are congruent, in the order of the
letters, which sides and which angles are equal ?
8. Two rectangles of equal sides— e.g. rectangular post-cards— are
congruent. So also two squares of the same side.
* Tracing-paper may be used.
P. a C
34 projection — area. [ch. i.
Projection. *
Definition 24. — The projection (or right projection f) of a line
or curve AB on a straight line CD is the
length MN between perpendiculars to CD
from the ends of AB.
It is evident that if two lines are bounded
by the same perpendiculars to a given line,
their right projections on it are equal. ^
EXAMPLES-XV.
1. Draw a parallelogram, sides 1", 1^", angle 40°. Draw and measure
the projection of the first side on the opposite side, and on the adjacent
side. Which is greater ?
2. Draw an equilateral triangle, 1" side. Measure the projections of
two sides on the third.
3. Draw an isosceles triangle. Verify that the projections of the equal
sides on the third side are equal.
Area.
Triangles, polygons, circles, &c. enclose a certain amount of
the surface of a plane, which is not the same for all figures—
e.g. the floor spaces of two rooms may difi'er greatly in size.
Definition 25. — The area of a closed figure is the amount of
the surface which it encloses.
The unit of area is a square, generally on unit length — e.g.
square mile, square foot ; and the area of a figure is said to be so
many square miles, square feet, &c.
* Congruent figures have equal areas.'
Use a division of squared paper for 1 foot. Draw a plan of a
room 18 ft. by 15 ft., and find its area. How can you calculate it
without the plan 1
Draw several rectangles on squared paper. Verify that their
area (in terms of a square of the paper as unit) is the product of
lengths of their sides, or base x altitude.
Ex. How many square feet of carpet cover a floor 28 ft. by 18? If
the carpet is 2 ft. wide, how many strips are wanted ?
* The remainder of this chapter may be postponed if thought advisable.
t Generally called orthogonal projection.
CH. I.]
EXPERIMENTAL GEOMETRY — AREA.
35
Triangle — Parallelogram.
Take a rectangular post-card A BCD ;
cut in two along a diagonal AC.
Verify that the triangles are congruent.
What fraction of the whole area is that of
each triangle ?
Take another card ; take any point E in
AD ; draw EN perpendicular to BC, dividing
the card into two rectangles.
Join EB, EC.
What fraction of area AN is EBN ?
of DN is ENC ? of AC is EBC ?
'The area of a triangle is half the product of base and
altitude (J base x alt.).'
Cut out a parallelogram from a post-card,
as ABCD ; cut it across a diagonal AC.
Verify that trs. ABC, CDA are congruent.
What fraction of the parallelogram is
either triangle 1
What is the area of the parallelogram 1
' The area of a parallelogram is the product base x altitude,
and is that of a rectangle of the same base and altitude.'
Draw two parallelograms of the same base and altitude, but
with different angles. Show that they are equal in area. Eepeat
for two triangles.
EXAMPLES— XVI.
1. Move the triangle DEC (in the second figure above) to the left,
making DC coincitle with AB. The area of the whole figure is unaltered.
What is the new figure? What can you conclude?
2. Draw a parallelogram, sides 2", 3", angle 30°. Measure its area.
3. Draw an equilateral triangle, IJ" side. Measure its area.
4. Calculate the areas: triangle, parallelogram, base If", alt. 2";
trapezium, pari, sides 1", 1|", alt. 1^". (Divide into triangles.)
5. Draw a square of 2" side. Calculate and measure its area.
36
EXPERIMENTAL GEOMETRY AREA.
[CH. I.
Squares of Eight Triangle.
Theorem 10."^ — ' The square on the hypotenuse of a
triangle is equal to the sum of squares on the other
(Pythagoras' theorem.)
If B is the rt. ang. of tr. ABC,
make sq. ABDE, and CF eql. to BD ;
.*. DF = BC.
Make sq. DFGH, and HK eql. to AB ;
.-. EK=DH = DF = BC.
Join as in figure.
Then trs. 1, 2, 3, 4 are congruent.
.*. if 1 and 2 are taken from sqq.
ABDE, DFGH, and 3 and 4 added, the area is unchanged.
But the figure is then ACGK, the sq. on AC ;
/. sq. on hyp. AC = sum of sqq. on sides AB, BC.
Note. ACGK can be shown to be a square thus :
Produce KA, CB to meet in L ;
then ang. L=KAE (same parts) = GCF,
. •. KA II GC ; similarly KG || AC.
Also ang. KAC = KAE + EAC = BAC + EAC = a right angle ;
.*. fig. ACGK is an equal-sided right-angled parallelogram ;
i.e. ACGK is a square.
right
sides.'
Areas op Equiangular or Similar Triangles..
Divide a triangle ABC into small triangles by parls. dividing
the sides into, say, 6 equal parts.
Count up from A, and write down
(i.) on the left, the number of triangles
in successive trapeziums ;
(ii.) on the right, the total numher of
triangles in successive triangles, as AHK.
What do you notice about the two sets
of numbers 1
Give the ratio of areas ABC: AHK.
Compare with AB : AH.
* This is meant as an exercise in drawing ;
post-card and cut out.
the figure may he drawn on a
CH. I.] LENGTH AND AREA OP CIRCLE. 37
Fold a piece of paper round a coin or a cylinder, and mark the
length to fold just round. Measure this length and the diameter,
and calculate the ratio ^'^diameter ~' Thus :
Cycle Wheel* Cylinder.* Collar-box,*
Circumf. 7 ft. SJ" 2146 cm. 23//
Diam. 2 „ 3/ 6-83 cm. 7J/
Ratio TT 3.139... 3-142... 3-135...
Thus the ratio ^^'^;|;"-' denoted by tt, is a little greater
than 3. It has been calculated to several hundred decimal places,
and its value to 5 places is 3-14159. We may use 3-14 as
an approximation.
Area of a Circle.
If a circle is divided into very small sectors, each sector is
practically a triangle, and its area is half the pro-
duct of base (arc of sector) and altitude (radius). f
Hence the area of a circle is equal to that of
a triangle whose alt. is the radius and base the
circumference of the circle, t
The length I, and area A of a circle, radius r, are
I = 27rr, A = TT?--, where tt = 3- 1 4 . . .
The arc or area of a sector can be calculated, when its angle is
known, as a fraction of that of the whole circle.
EXAMPLES— XVII.
1. Write down the area of the second circle above.
2. Calculate lengths and areas of circles, radii 1-3", 2-7 cm., 5^ yd. (the
last area in square poles also).
3. Calculate arcs and areas of sectors of a circle, radius 2-4 cm., angles
45°, 60°, 90°.
4. Measure tt for a bicycle wheel or a drauglits-man.
5. The diameter of a carriage wheel is 44". How many turns does it
make in a mile ?
* These were meaaared with a tape. A set of draughts-men, unpolished,
serves for a class. The circumference can be measured by rolling on a diagonal
scale.
t This is strictly true, though the proof is incomplete. See Chapter "V.
38 EXAMPLES. fcH. I.
EXAMPLES— XVIII.
1. Draw a triangle, a = 2-4 cm., 6 =2^7 cm., c = 3'3 cm., and measure
the greatest angle.
2. Draw a triangle, A = 54°, B = 65°, c=l-8". Measure the altitude
from B to AC.
3. Draw a triangle, a=l-5", &=l-2", C = 53°, and measure the side c.
4. Two posts on one side of a river are 100 yd. apart. A post on the
opposite bank is 60 yd. from one post and 80 yd. from the other. Draw
a plan, 1 cm. to 10 yd., and find the width of the river.
5. A rope of a giant's stride 16 ft. high just reaches the ground.
When held by a boy 10 ft. from the post, it just reaches the top of his
head. Find his height. (Use 1 cm. for 2 ft. )
6. A man lying on the ground 30 ft. from a house can just see the
roof. If the slope of the roof is 50° with the ground, find the height of
the house wall. (Use 1" to 10 ft.)
7. Prick two points A, B on paper 5 cm. apart. Without pen or pencil
show the straight line AB, and bisect it.
8. Draw a diagram to show the four cardinal points (N., S., E., W.).
Show also the directions N.E., N.W., S.E., S.W.
9. Write down the values of the angles formed by the following
directions : N., S.E. ; E., N.E. ; E., S.W. ; S., N.E.
10. (Range finding.) The directions of a battery C make angles
A = 90°, B = 79° with a line AB, 1000 yd. long. Find the distance of the
battery from A. (Use 1" to 1000 yd.)
11. At a distance of 180 ft. from its foot, the direction of the top of a
tower makes with the ground an angle of 30° (called its angle of elevation,
or its elevation). Find the height of the tower.
12. A man walks 5 miles N., 4 miles N.W., 5 miles S., and then home.
Draw a plan, and state the whole length of the walk. What is the
figure of the walk ?
13. A man is 2 miles N. of another. They start walking at the same
time in the same direction, N.W., at 3 and 4 miles an hour respectively.
How far apart, and in what direction from each other, are they at the
end of 2 hrs. 50 min. ?
14. Draw an isosceles triangle, a = b = 2|", c = 2". Measure the angle C.
15. Draw an isosceles triangle, a=6 = 4 cm., C = 73°. Measure the
third side.
16. Draw a rhombus, side IJ", angle 67°. Draw an altitude.
17. Draw a parallelogram, sides g". If", angle 79°.
18. Draw a parallelogram, sides 3-2 cm., 2-7 cm., one diagonal
4-5 cm.
CH. I.] EXAMPLES. 39
19. Draw a circle, radius 1", centre O ; draw any radius OA. With
centre A, radius 1", draw a circle, catting the first circle in B. Complete
the triangle ABO. What kind of triangle is AOB? AVhat angle is
AOB ? How many such triangles will go just round the first circle?
20. Draw a circle, radius 3 cm. Step off points A, B, C, D, E, F round
the circle at equal distances of 3 cm. ; join the points in order. What
kind of hexagon is ABCDEF? What do you know about its sides? its
angles ?
21. Draw a circle of 1" radius. Mark off 6 points 1" apart on the
circle ; with these points as centres, radius I", draw six circles. With
the original centre, radius 1^''. draw a circle to touch all these.
22. Make an angle of 72" at the centre of a circle 3 cm. radius. What
fraction of the whole circumference does it cut out ? Complete a regular
pentagon in the ciif • ' Measure its side.
23. Draw a circle, radius 1-4". Draw two diameters at right angles.
Join their ends in order. What figure is formed ? Measure its side.
24. How can you use Ex. 23 to draw a square Avhen its diagonal is
given ? Draw a square of diagonal 7 cm. Measure its side.
25. Draw a circle, 1" radius. Set off three radii at angles of 120°, and
mark points A, B, C on them ^" from the centre. With centres A, B, C,
radii ^", draw three circles. With the original centre draw a circle
through the points where the three circles cut each other.
26. Divide a circle into eight equal sectors. What figure is formed by
joining the ends of the radii in order ?
27. If the angle of vision of an eye is 120°, what width of country can
the eye take in at a glance at a distance of 5 miles ? (Use 1 cm. to the
mile.)
28. Draw a regular dodecagon (twelve-sided polygon) in a circle,
radius 2". (Make angle of 30° at the centre. )
29. Fold over a sheet of paper, and mark the fold. Make a second
fold perpendicular to the first, cut across to make a fourfold triangle,
and unfold. What is the figure? What can you infer about its
diagonals ?
30. Repeat Ex. 29, but cut across so as to make the fourfold triangle
isosceles. What is the figure now ? And what additional fact do you
know about its diagonals ?
31. Describe a parallelogram, sides 2-7 cm,, 3*8 cm., angle 115°. Draw
perpendiculars to one side (prolonged sufficiently) from the ends of the
opposite side. State two facts about them.
32. Take a straight line AB, 4-8 cm. long. Draw circles, centres A, B,
radii 3 cm., to cut in C, D. What figure is ACBD? What line is CD
with reference to AB ?
4^ EXAMPLES. [CH. I.
33. Draw squares on sides of 2-3 cm. and 4-6 cm. Measure their
diagonals. How many of the first are contained in the second ?
34. Draw a square on diagonal 5-6 cm. Measure its side. (See Ex. 23.)
35. Draw a rectangle, one side 2-7 cm., diagonal 4*3 cm. Measure the
other side, and the other diagonal.
36. Draw a right triangle, hyp. 3|", one side 3". Measure the other
side. Does your construction resemble that of Ex. 35 ? Why ?
37. Draw a parallelogram, one side 1", one angle 136°, and the diagonal
through this angle 1^". How many can you draw ?
38. Draw a straight line AB, 1-7". At A, B make perpendiculars
AC, BD I" long. State an additional fact about AC, BD. What kind
of figure is ACDB ? State two facts about CD, AB.
39. Draw two parallel lines 1" apart. On one of them mark off AB
1|". Draw AC to the other 1|". Make CD on the second parallel 1".
Join BD. What kind of figure is ACDB ?
40. Draw a trapezium as in Ex. 39. Bisect the two non-parallel sides
AC, DB at E, F. Measure AB, CD, EF ; compare 2.EF with the sum
AB + CD.
41. Draw a parallelogram, sides 1-4", 2-3", one diagonal 3". Compare
its opposite sides and opposite angles. Can you prove opposite angles
equal? (Produce sides and mark angles of same parts.)
42. Draw a parallelogram, sides 3-2 cm., 4-7 cm., angle 137°. Write in
each of the other angles its number of degrees.
43. Make a parallelogram on squared paper, base 7 divisions, altitude
5, one angle 60°. Count the number of squares in it, and compare with
a rectangle of sides 7 divisions and 5.
44. Make two squares on squared paper, sides 3 divisions and 7 divi-
sions. Write down the ratio or fraction of their areas. Compare with
that of their sides.
45. Draw a triangle, a = 2-3 cm., 6 = 3-2 cm., 6=2-9 cm. Draw a line
parallel to BC, bisecting AC and AB ; show that the triangle cut off is
a quarter of the whole. (Draw parls. to the sides as on p. 36.)
46. Calculate the areas : rhombus, base 1-7", alt. 2-3" ; rectangle,. sides
132 yd., 79 yd. ; triangle, base 35 ft., alt. 27 ft.
47. One side of a rectangular field is 72 yd., and makes an angle of
30° with a diagonal. Draw a plan and calculate the area. (1 cm. to
10 yd.)
48. Calculate, without drawing, the length of the hypotenuse of the
right triangle whose sides are (i.) 2-7 cm., 3-9 cm.; (ii.) 3|", 2|" ;
(iii.) 3 miles, 4 miles ; (iv.) 127 yd., 322 yd.
49. Calculate, without measuring, the altitude of an equilateral triangle
of 2" side. Deduce from it those of equilateral triangles of 1", 3 cm.,
5 ft., 7 yd. side.
CH. I.] EXAMPLES. 41
50. The distance of two park gates cannot be measured because of a
lake ; but a third gate is 500 yd. from one, and 300 yd. from the other in
a line perpendicular to the join of the two gates. Find their distance.
51. A street is 100 ft. wide. Calculate and measure the length of a
string stretched from the bottom of a wall on one side to the top of
a 60 ft. Avail on the other.
52. Draw two squares, sides 2-3 cm. and 6-9 cm. How many of the
former are contained in the latter ? Why ?
53. Draw a triangle ABC, a =2-3", 6 = 1-9", c = 3-l". Divide the side
AB into 5 equal parts ; draw parallels to the sides as on p. 36. How
many small triangles are there? Make AH, AK f of AB, AC. Write
down the ratio or fraction of the areas AHK, ABC. Compare with that
of their sides.
54. How far does each of the following wheels go in one revolution :
(i.) driving wheel, 6 ft. diani. ; (ii.) cycle wheel, 28" diam. ; (iii.) carriage
wheel, 3 ft. 6 in. diam.?
55. How many revolutions do the wheels of Ex. 54 make in a mile ?
56. Calculate the area of each wheel in Ex. 54.
57. On what area can a donkey feed when tethered by a 20 ft. rope ?
58. Calculate the areas of sectors of a circle of 1" radius, angles 30°,
45°, 60°, 72°, 120°.
59. Draw a circle, 1" radius. Draw two diameters at right angles, and
form a square by joining their ends. What is the area of the square ? of
the part of the circle left after cutting out the square ?
60. Draw a square ABCD, side 4 cm. Draw diagonals AC, BD ; and
draw perpendiculars to these at the points A, B, C, D. What is the
new figure ? Find its area.
61. Draw a circle, 3 cm. radius. Draw two diameters AB, CD at right
angles, and perpendiculars to these at A, B, C, D. AVliat is the new
figure ? its area ?
62. Find the area left from the outside square in Ex. 61 after the circle
has been cut out.
63. Draw a circle, 1" radius. Mark off 6 points 1" apart on the circle,
and join three alternate (every other) points. What figure is formed ?
Measure its side and altitude, and calculate its area.
64. Can you calculate the altitude of the triangle of Ex. 63 without
measuring it ? What is its calculated value ?
42
CHAPTER 11.
ANGLES— PARALLELS— TRIANGLE— CIRCLE.
Theorem 11. — 'Two straight lines which meet have any
two adjacent angles supplements, and any two opposite angles
equal.' *
If AB, CD form at O the four angles X, Y, Z, W, and OP± AB ;
then (i.), X + Y = POA + ROB
= 2 rt. angs. ;
.-. X = 2 rt. angs. - Y
= suppt. of adj. ang. Y.
Similarly, Y = suppt. of Z,
Z of W, W of X.
Also (ii.), X + Y = 2 rt. angs. = Y + Z
.'. X = opp. ang. Z ; similarly, Y = W.
Theorem 12.— 'Two adjacent angles at a point which are
supplements and have a common side, have their other sides
in a straight line.'
If adj. angs. AOB, AOC are suppts.
and have side AO common,
produce BO to D ;
.*. ang. DOA = suppt. of BOA = COA ;
.'. CO coincides with DO ;
i.e. BO, CO are in a straight line.
Theorem 13. — 'Two parallels make equal alternate interior
angles with any third line ; and two straight lines making
equal alternate angles with a third are parallel.'
If AB, CD make with EF alt. angs. a
X, Y and angs. X, Z towards same parts;
then (i.), if AB || CD,
X = Z (same parts)
= Y, the alt. ang. E-
Also (ii.), if X = alt. ang. Y,
then Z = Y (opp. ang.)
= X, towards same parts ;
/. AB II CD.
* This should be proved practically by the protractor.
CH. II.]
ANGLES OF TRIANGLE POLYGON.
43
Theorem 14. — ' The sum of the three angles of a triangle is
two right angles, and an exterior angle is equal to the sum of
the two interior opposite angles.'
If ABC is a triangle, and CD pari, to AB makes
angles, Y with BC,
and X with AC produced ;
then ang. A = X, same parts,
and B = Y, alt. ang.,
.-. A + B + C = X + Y + C = 2 rt. angs. ;
and ext. ang. BCE = X + Y = A + B, int. opp. angs.
Cor. (i.). — 'An exterior angle of a triangle is greater than
an interior opposite angle.'
Cor. (ii.).— 'If one angle of a triangle is right or obtuse, each
of the others is acute.'
Theorem 15. — * The sum of the angles of a polygon is twice as
many right angles as the number, less two, of the sides.'
If one vertex A of the poln. ABCDEF
is joined to the others, the first and last
triangles have each two sides of the poln.,
and the rest one each ;
.'. there are as many triangles as the
number, less two, of the sides ;
.*. sum of angs. of poln.
= sum of angs. of triangles
= twice as many rt. angs. as the number,
less two, of the sides.
Note. If there are n sides,
sum of angs. = {2n - 4) rt. angs.
Cor (i.). — 'The sum of angles of a quadrilateral is 4 rt.
Cor. (ii.). — *The sum of exterior angles of a polygon is 4 rt.
angles.' *
If parallels are drawn at a point to the sides of poln.,
sum of ext. angs. U + V + . . . = 4 rt. angs.
* If there are any re-entrant angles, this is not true without modification.
44
SIDES AND ANGLES OF TRIANGLE.
[CH. II.
Theorem 16. — 'In an isosceles triangle, the angles opposite
the equal sides are equal ; and the bisector of angle of the
equal sides is the right bisector of the third side.'
If AB = AC in isosceles tr. ABC,
and AD bisects ang. A,
fold over the part DAC on to DAB ;
then AC comes on AB, */ ang. DAC = DAB ;
and point C comes on B, '.* AC = AB ;
.*. ang. C coincides with B ;
i.e. ang. C = B.
Also DC coincides with and = DB,
.'. D is mid point of BC ;
ang. ADC coincides with and = ADB,
.-. AD_LBC;
.*. AD is the right bisector of BC.
Cor. — 'An isosceles triangle is symmetrical about the bisector
of angle of the equal sides.' '^
Theorem 17. — 'In any triangle one angle is greater than,
less than, or equal to another, according as the opposite side
of the first is greater than, less than, or equal to that of the
other.'
If AC > AB in tr. ABC,
make AD eql. to AB, and CE pari, to BD ;
then ang. B > ABD
> ADB, V AB = AD,
> ACE, same parts,
>C;
i.e. (i.), if AC > AB, ang. B > C.
Similarly (ii.), if AC < AB, B < C;
and (iii.) if AC = AB, B = C (Th. 16).
Cor. — 'The hypotenuse is the longest side of a right
triangle.'
Ex. If one angle of a triangle > another, is the side opp. the greater
angle greater or less than that opp. the less angle ? Why ?
* This is a very impoi'tant principle.
CH. IT.] NON-PARALLELS. 45
Theorem 18. — 'If two angles of a triangle are equal, the
sides opposite the equal angles are equal.'
If the angle B = C in tr. ABC, reverse the
triangle so that points B, C are interchanged;
then the angles B, C are interchanged ;
.*. the point A has its original position,
.*. CA, BA are interchanged.
.-. CA = BA.
Ex. Prove this theorem by the previous theorem.
Theorem 19. — 'Any two non-parallel lines in a plane meet if
produced far enough.' *
If AB, CD are non-parallel,
make AE pari, to CD, and CF eql. to CA ;
.'. ang. CAF = CFA
= alt. ang. FAE.
.*. AF bisects ang. CAE.
Hence if any line AC through A meets CD, the bisector AF of
ang. CAE also meets CD.
Similarly, the bisector of FAE meets CD, and every successive
bisector, as long as the process is continued, meets CD.
But if the process is sufficiently continued, a bisector AD is
obtained making the ang. DAE less than BAE.
.'. AB lies inside the triangle CAD, and therefore meets CD if
produced far enough.
* This demonstration may be postponed by beginners. The result should be
learnt, being important for constructions.
46 SIDES OF TRIANGLE — RIGHT BISECTOR. [CH. II.
Theorem 20. — 'Any two sides of a triangle are together
greater than the third.'
If AD is perp."^ to BC in tr. ABC ;
then hyp. AB > BD, in tr. ABD.
Similarly, AC > DC ;
.-. AB + AC> BD + DC
> BC;
i.e. h + c> a;
similarly, a + h>c, G + a>b.
Cor. — 'Any side of a triangle is greater than the difference of
the other two.'
Ex. Draw the figures and prove for right or obtuse triangles.
Theorem 21. — 'The locus of a point in a plane equidistant
from two fixed points in it, is the right bisector of their
join.'
If P is any point equidistant from the fixed
points A, B, ^
then PA = PB in tr. APB ; \p
.'. NC bisecting ang. APB is the right bisector
of AB (Th. 16);
i.e. every point P on the locus is on the right
bisector of AB.
Also, if Q is any point on the right bisector CN,
the figure AQB is symmetrical about CN ;
.-. QA = QB;
i.e. every point of the right bisector CN is on the locus.
.*. the right bisector of A B is the locus.
Ex. Construct a right bisector and an isosceles triangle in paper
without ruler or compass.
* By definition a perp. to BC exists at B ; a pari, to this through A meets
BC by Th. 19, and is perp. to BC.
CH. II.]
CIRCLE THROUGH THREE POINTS.
47
Theorem 22. — 'One only circle can be drawn through any
three points not in a straight line.'
If A, B, C are three points not in a
straight line,
the right bisectors MO, NO of AB and
OB are non-parallel,
because AB, BO are non-parallel;
.'. MO, NO meet in a point O ; B
.*. O is equidistant from A, B, C, and a
circle, centre O, rad. OA, passes through
A, B, C.
Also, the centre of a circle through A, B, C is equidistant from
A, B and lies in MO, and similarly in NO ;
.'. O is the only centre, and there is one only circle.
Cor. — * Two circles cannot meet in more than two points.*
Theorem 23. — ' Two circles cut, in two only points, when the
join of their centres is greater than the difference, and less than
the sum, of their radii.' *
If A, B are centres of circs., rad. r, s,
?* ^ s, and AB meets the circle s
in H, L;
then if AB >?'-s and < r + 5,
(i.) AL = AB-t-BL
>r—s+s
>r,
.'. L is outside circle r.
(ii.) AH + BH <r + s,
.'. AH < r,
.*. H is inside circle r ;
.'. the circle s is partly outside and
partly inside the circle r, and cuts it
in two only points F, G.
Also, by symmetry about AB,
Cor. — 'The line of centres of two intersecting circles is the
right bisector of the common chord.'
* Beginners may use the first figure only.
48 ANGLE IN SEMICIRCLE. [CH. II.
Theorem 24. — ' The angle in a semicircle is a right angle ;
and the circumcircle of a right triangle has the hypotenuse as
diameter.'
(i.) If ABC is a semicircle, centre O,
join OB ;
then OA = OB in tr. AGS,
.'.ang. OBA = A.
Similarly, ang. OBC = C,
.*. ang. B = A + C in tr. ABC ;
.'. B is a right angle.
(ii.) If ABC is a rt. triangle, B the rt. ang.,
make ang. ABO eql. to A ;
.-. OB = OA.
Also, ang. OBC = compt. of OB A = compt. of A
= OCB, '.' B is a rt. ang. ;
.*. OB = OC.
.*. The circle, centre O, diameter AC, is the circumcircle of rt.
triangle ABC.
Cor. (i.). — 'The hypotenuse of a right triangle is double its
median.'
Cor. (ii.). — 'A triangle is right when one of the sides is
double its median.'
EXAMPLES-XIX.
1. Show that angle BOO = 2. B AC.
2. Any angle at the centre of a circle — i.e. having two radii
as its sides— is double the angle at the circumference on the same
arc.
3. Ang. BAC = compt. of BOA.
4. The right bisector of a chord of a circle bisects either arc of the
chord. (Use symmetry. )
5. The diagonals of a rectangle are equal and bisect each other.
CH. II.]
CONSTRUCTIONS.
49
Xc
Bisectors, Circumcircle.
Construction 1. — 'Draw the right bisector of a given line
AB ; or, bisect a given line at right angles.'
With centres A, B, any radius greater than
half AB,
draw equal circles cutting in C, D ;
join CD, cutting AB in M. ^""
C, D, equidistant from A, B, lie in the
right bisector of AB ;
.'. CD is rt. bisector of AB, and M the mid point.
Ex. Show also that AB is rt. bisector of CD. What figure is the
quadrilateral ACBD?
^o
Construction 2. — * Draw the circle through
three points A, B, C, not in a straight line.'
Draw MO, NO rt. bisectors of AB, BC ;
.'. O is equidistant from A, B, C.
Draw circle, centre O, radius OA,
passing through A, B, C.
(This circle is the circumcircle of tr. ABC.)
Construction 3. — ' Draw the bisector of a given angle.
If B is the angle, draw arc AC, centre B,
any radius, meeting the sides in A, C.
With centres A, C, same * radius,
draw arcs D, and join BD.
Then tr. ABC is isosceles ;
.*. bisector of ang. B is rt. bisr. of AC,
and therefore traverses D ;
.'. BD is bisector of ang. B.
Note. If AB, CB are produced to cross, BD produced bisects the
opp. ang. at B, and a perp. to BD at B bisects the two adjacent
angles.
Ex. Show that the bisectors of adjacent angles of two lines are
perpendicular.
* This is convenient, though not necessary.
P.O. D
50 PERPENDICULARS — ANGLE [CH. II.
Construction 4.— 'Draw a perpendicular to a straight line at
any point in it, by ruler and compass.' *
If P is the point in AB, - '^
with any point O as centre, draw arc APR, ^ X
join AO to meet arc in R, join RP. y'
Ang. RPA, in semicircle, is right; y^
.'. PR±AB. Ax^^^ ^ ^y^ B
Construction 5. — 'Draw a perpendicular to a straight line
from a point not on the line, by ruler and compass.' *
If P is the point, AB the line, with any two -^iF
points A, B on the line as centres,
draw arcs PQ, meeting in Q ; ^^^^^
join PQ. A
A, B are on right bisector of PQ ;
.*. PQ±AB. yf
Construction 6. — 'Draw an angle equal to a given angle.'
With the point B of the given angle as centre, any radius, draw
arc AC cutting its sides in A, C.
"With centre P in any line PQ, same
radius, draw arc QR.
With centre Q, radius CA, draw arc R ;
join PR.
If ang. B is placed on P so that BC
coincides with PQ, A comes on each of
the arcs at R, and coincides with R ;
/. ang. P = ang. B.
Scale of Chords. — This is formed by setting off the chords of
1°, 2°, 3°... at the centre of a given circle along a straight line
from a fixed point O.
To construct an angle, say 32°, from it, draw an arc QR, centre
P, radius did. of 60°; draw arc R, centre Q, rad. clid. of 32°;
.-. ang. P = 32°
* Of theoretical interest only. Construction by set-square is simpler.
CH. II.]
PARALLELS — ARC OF GIVEN ANGLE.
51
Construction 7.— 'Draw a parallel to a given line from a
given point, by ruler and compass.' *
Draw any str. line QA through the given
point P, meeting the given line AB in A ;
make ang. QPR eql. to PAB ; ^ ^^
PRIIAB.
"K
A ' B
Construction 8.— 'Draw a parallel to a given straight line
at a given distance from it.'
Set the hyp. of set-square along the
given line AB, fix a str.-edge along one
side, and interchange the sides of set-
square, so that hyp. comes into posn. PQ,
perp. to AB.
Prick off PQ from AB, along hyp.,
eql. to given line L ;
draw QR park to AB, at given dist. L.
R
/Iq
vA
,/ IP
B
\
Construction 9. — ' On a given straight line draw an arc of a
circle to contain a given angle.'
If A B is the line,
draw right bisector DNO,
make ang. DNE eql. to given ang. X,
draw AO park to EN.
With centre O, draw arc APB.
Ang. APB = JAOB (at centre)!
= AON=:DNE = X;
.'. arc APB contains "iven ansle X.
Note. If X is given in degrees, the angle NAO may be drawn at once
as the complement of tlie given angle X.
* Of theoretical interest only,
t See Ex. 2, Th, 24.
52
TRIANGLES.
[CH. II.
Construction 10. — ' Triangles.'
(Three conditions, one at least of which is not an angle,
determine a triangle : there are sometimes alternative solutions.)
State the number of solutions in each case.
Construct a triangle, having given
(i.) three sides a,h,C',
(ii.) two sides and their angle, a, b, C ;
(iii.) two sides and an angle opposite one, a, b, A;
(iv.) one side and two adjacent angles, a, B, C ;
(v.) one side, one adjacent, and the opposite angle, a, B, A.
If A, B are given in degrees, calculate C,
and use (iv.). Otherwise :
Make BC = a, CBA = B, ABD = A ;
make CA pari, to BD ;
.'. ang. CAB = alt. ang. ABD = A;
.'. ABC is the triangle.
(vi.) One side, one adjacent angle, and sum of the other (or of
the three) sides, a, B, b + c.^
Make BD = b + c, DBC = B, BC = «;
draw NA, rt. bisector of CD, join AC ;
.-. AC = AD, and AB + AC = BD = & + c ;
.'. ABC is the triangle.
(vii.) One side, the opposite angle, and
sum of the other sides, a. A, b + c*
Make BD
b + c, BDC=2' BC
draw NA, rt. bisector of CD, join AC.
Ang. BAC = ACD + ADC = 2ADC = A,
and AC = AD;
.*. BA + AC = b + c;
.*. ABC is the triangle,
(viii.) One side, its altitude, and opposite
angle, a, 7i, A.
Make BC = a, arc BAC containing A.
Draw a pari, to BC at distance h^
cutting the circle in A ;
.*. ABC is the triangle.
* Similar constructions serve when 6 - c is given instead oi b + c.
CH. II.] EXAMPLES. 53
EXAMPLES— XX.
Theorems.
1. When two lines meet, the bisectors of adjacent angles are perpen-
dicular.
2. If two straight lines, AB, CD, on the same side of AC make the
interior angles A, C supplements, then AB || CD.
3. If AB i! CD in trapezium ABCD, and AE is drawn making angle
BAE equal to CDA towards the same parts, show that AE, AD are in
a straight line.
4. If AB II CD, the angle BAD is either the supplement of ADC or
equal to it.
5. If two straight lines make two interior angles on the same side of a
third line supplements, the two first lines are parallel.
6. If two lines a, b are parallel respectively to c, d, the angles ab and
cd towards the same parts are equal.
7. One only perpendicular can be drawn from a given point to a given
line.
8. If ABCD is a trapezium, and AB || CD, ang. A = suppt. of D.
9. In an isosceles triangle a parallel to the unequal side cuts off equal
parts from the equal sides.
10. If a straight line cuts off equal parts from the equal sides of an
isosceles triangle, it is parallel to the third side.
11. The bisectors of the equal angles of an isosceles triangle form an
isosceles triangle with the unequal side.
12. Show that equilateral triangles have all their angles equal.
What is the value of each ?
13. If any angle of an isosceles triangle is 60°, the triangle is equi-
lateral.
14. The exterior angle of the equal sides of an isosceles triangle is
equal to twice one of the equal angles.
15. If the straight line bisecting an exterior angle of a triangle is
parallel to the opposite side, the triangle is isosceles.
16. If a bisector of angle of a tiiangle is perpendicular to the opposite
side, the triangle is isosceles.
17. If BD, CE bisecting the equal angles B, C of a triangle meet AC,
AB in D, E, then BE = CD.
18. Show also in Ex. 17 that DE = BE and || BC.
19. If D, E are points on the equal sides AC, AB of an isosceles
triangle such that BE = CD, then DE || BC.
54 EXAMPLES. [CH. II.
20. If in Ex. 19, DE also =BE, then BD bisects angle B.
21. The perpendicular is the shortest straight line from a given point
to a given straight line.
22. If lines OP, OQ, &c. are drawn from a point O to points P, Q,
&c., in a straight line, they increase with their angular distance from
the perpendicular ON.
23. Any line from a vertex of a triangle to the opposite side is less
than the greater of the two other sides.
24. If P is a point inside a triangle ABC, then BP + PC is less than
BA + AC, and angle BPC is greater than BAC.
25. Show also that PA + PB + PC in Ex. 24 is less than the sum, and
greater than half the sum, of the sides.
26. The sides of a quadrilateral are together greater than the sum of
the diagonals.
27. The sum of the diagonals of a quadrilateral is greater than half
the sum of the sides.
28. If two straight lines are right bisectors of each other, the joins of
their ends form a rhombus.
29. If from any point on the bisector of an angle parallels are drawn
to the sides, a rhombus is formed.
30. A rhombus is a parallelogram.
31. The diagonals of a rhombus bisect its angles, and bisect each other
at right angles.
32. If a diagonal of a parallelogram bisects its angles, the figure is a
rhombus.
33. The diagonals of a square are equal, bisect each other at right
angles, bisect the angles, and make angles of 45° with the sides.
34. The right bisectors of sides of a square pass through the intersec-
tion of diagonals.
35. If AB, AC are equal sides of an isosceles triangle, and O is the
circumcentre, OA bisects angle A.
36. If circles are drawn with the ends of a straight line as centres, and
equal radii greater than half the line, show that they cut twice.
37. The circumcentre of an equilateral triangle is on each bisector of
angle. •
38. If O is the circumcentre of an equilateral triangle ABC, the angle
BOO is 120°.
39. If the angle A of the equal sides of an isosceles triangle ABC is
50°, and O is the circumcentre, show that angle BOO is 100°.
40. If angles B + C in a triangle = A, the triangle is right-angled.
41. If the median AD of a triangle ABC is half BC, the triangle is
right-angled.
CH. II.] EXAMPLES. 55
42. In ail isosceles right triangle the perpendicular and median
through the right angle coincide.
43. The right bisectors of the right-angled sides of a right triangle
meet on the hypotenuse.
44. If one line AB meets a given line AC, any parallel to AB
meets AC.
45. If CM, ON are perpendicular respectively to two parallels, then
DM, ON are in a straight line.
46. If OP, OQ make equal angles respectively, towards the same
parts, with two parallels, then OP, OQ are in a straight line.
47. Straight lines bisecting two consecutive angles of a parallelogram
are at right angles.
48. By producing two sides of an angle of a parallelogram, show that
opposite angles of a parallelogram are equal.
49. If four straight lines at a point make opposite angles equal in
pairs, they are two and two in a straight line.
50. A straight line NDE, perpendicular at N to the side BC of an
isosceles triangle, meets the equal sides AB, AC in D, E ; show that
the triangle ADE is isosceles.
51. If a right triangle has one acute angle double the other, the
hypotenuse is douljle the shortest side.
52. A trapezium ABCD has AB parallel to CD, and angle C equal
to D. If AD, BC meet in E, show that triangle EAB is isosceles.
53. In a quadrilateral ABCD, angle C = D, side AD = BC; show that
AB II CD. (Produce sides to meet.)
54. On the side BC of a triangle as diameter a circle is drawn cutting
AB, AC in D, E ; show that BDC, BEC are right triangles, and have
equal medians from D, E.
55. Two isosceles triangles have the angles of their equal sides equal ;
show that their other angles are also equal.
56. If DE parallel to the unequal side BC of an isosceles triangle cuts
the equal sides in D, E, the right bisector of BC is also the right bisector
of DE.
57. If the sides AB, AC of a triangle are produced to D, E, and the
angle DBC is equal to ECB, the triangle is isosceles.
58. On AB, one of the equal sides AB, AC of an isosceles triangle, a
point D is taken. Show that the angle ADC> ACD.
59. MP, MQ are drawn perpendicular to a given line AB on opposite
sides of it, from a point M on the line; show that MP, MQ are in a
straight line.
60. If also MP = MQ, in Ex. 59, Avliat do you know about the sides of
the figure APBQ ? about its diagonal AB ?
56 EXAMPLES. [CH. II.
EXAMPLES— XXI.
Constructions.
1. Calculate the sum of angles of polj^gons of 7, 8, 10 sides. Also the
sum of exterior angles.
2. Calculate each angle of regular polygons of 5, 6, 7 sides.
3. Draw a triangle, a — I", b = ll", C = 40°. Bisect C and measure its
parts.
4. Draw an isosceles triangle, a = b = ll", C = 46°. Draw the right
bisector of c ; produce it as far as C.
5. Draw a right triangle, hyp. a = 3-2 cm., 6 = 1«6 cm. Measure C.
6. Draw a triangle, a = i", B = 52°, C = 67°. What is the 3rd angle?
Draw a perpendicular AD on BC, produce so that DE = AD. Measure
angles EBC, ECB and compare with B, C.
7. Draw a triangle, « = If", & = 2|", c = 2J" ; bisect B, draw CD per-
pendicular to the bisector and produce to meet BA in E. Measure BE.
8. Draw any straight line, and mark two points A, B off the line ; find
a point P in the line equidistant from A, B.
9. Construct a point P on a given circle equidistant from two points
A, B on the circle.
10. Calculate the angle of an equilateral triangle, and construct the
angle.
11. Construct a triangle, one side 3-7 cm., two angles 60°. What kind
of triangle is it ?
12. Trisect a right angle by ruler and compass.
13. If one angle of a triangle A is four times each of the others,
calculate the angles ; construct such a triangle, « = 2-6 cm.
14. Draw a triangle, a — 2-S cm., 6 ==3-5 cm., c = 4-2 cm. ; construct the
circumcircle and measure its radius.
15. Construct a triangle, a = b = S-2 cm., altitude from AB, 2-4 cm.
16. Construct an isosceles triangle, « = &, c-l-S", C = 56°. (Calc. angs.
A, B.)
17. Find the locus of mid points of chords of a circle, parallel to a
given diameter.
18. Construct a triangle, a = 3-3 cm., B — 43°, R (circumradius) = l-9 cm.
(Begin with circumcircle.)
19. Construct an isosceles triangle, a = 6=: 1-7", R^l-l".
20. Construct a right triangle, a=l-8 cm., R=:2-3 cm.
21. Construct a square on a side of 3-3 cm.
22. Draw any straight line AB, find the locus of a point P which is
always 1" from AB.
CH. II.] EXAMPLES. 57
23. Find a point equidistant from a given point and a given line.
(Take 2-2 cm. from each.)
24. Draw an angle of 58°. Construct a point 2 cm. from each of its
sides.
25. Construct the locus of i)oints equidistant from the sides of an
angle of 46°.
26. Construct a right triangle, hyp. « = 2", 6 + c=2|".
27. Construct a triangle, a = Z-5 cm., 6 + c = 5-3 cm., A = 78°.
28. Construct a triangle, a + b + c = S", B = 38°, C = 72°.
29. Construct a parallelogram, sides If", 2^", one diag. 3^".
30. Draw a straight line AB, take two points C, D on opposite sides
of AB, construct a point P in AB so that AB bisects angle CPD.
31. Construct a circle, radius 1-3", to pass through two points A, B,
2" apart. Can it be done if AB = 3" ?
32. Construct a point equidistant from two parallel lines 1^" apart.
33. Construct the locus of a point equidistant from two parallels.
34. Draw two parallels, and take two points, one between the
parallels and one outside them; construct a point equidistant from
the parallels, and also equidistant from the two points.
35. Construct a point in one side of an angle whose (perp. ) distance
from the other is a given length.
36. Through a given point P draw a straight line to make equal
angles with the two sides of a given angle. How many such lines can
be drawn ?
37. Through a given point A draw a straight line making a given
angle with a given straight line BC.
38. Through two given points P, Q dra'sv straight lines to form with
a given line BC an equilateral triangle ABC. (Use Ex. 37, and
angles of 60°. )
39. Draw two straight lines at an angle of 108° ; construct a point 1"
from one line, 1|" from the other.
40. From a ship at sea, a second ship is 3 miles E., and a third is
5 miles N.W. Find by a diagram the distances between the last two.
41. Could you do Ex. 40 in the same way if the distances were 3000
miles E. and 5000 miles N.W. ? Why ?
42. Draw a circle, 1|" radius. Draw a chord in it whose arc has an
angle of 76°.
43. Draw a triangle, « = 4.7 cm., A = 67°, altitude from BC, 3-2 cm.
44. A ship is 5 miles from shore, and the directions from the ship of
two lighthouses on the shore, 8 miles apart, form an angle of 70°. Draw
a plan, and measure the distances of the ship from the lighthouses.
58 EXAMPLES. [CH. II.
45. A boatman heading for shore sees a tower, ch)se to the shore, 45°
to his right ; after making two miles he sees it 60° to his right. What
Avas his distance from shore ?
46. Find the locus of mid points of all chords of a circle through a
given point P on the circle.
47. Construct a square, diagonal 2-83".
48. Construct a rhombus, diagonal 6 cm., opp. ang. 98°.
49. Construct a rhombus, diagonals 3-7 cm., 2-9 cm.
50. The sides of a jointed parallelogram ABCD are AB, 2-5 cm., BC,
3 cm. If BC is fixed, find the locus of the points A, D as the parallelo-
gram moves.
51. Construct a rectangle, sides 2-7" and 1-9", and draw the right
bisector of the first side. Show by symmetry that it is the right bisector
of the opposite side.
52. Draw two diameters at an angle of 60°, in a circle of 1" radius.
Show that their ends form a rectangle. Measure the sides.
53. Find the greatest line that can be drawn in a parallelogram from
a vertex to one of the sides. Give proof.
54. Construct a triangle, given a = 4 cm., b-c — l cm., C = 54°.
55. Find the greatest chord of a circle. Give proof.
56. Construct a trapezium, AB i| CD, ang. A=:B = 60°, AB = 5 cm.,
AD = 3 cm. Measure BC, BD.
57. One angle of an isosceles triangle is half each of the others.
Calculate the angles.
58. On a l)ase of 3-4 cm. describe an isosceles triangle, having base
angles each double of the third angle. (See Ex. 57.) Measure the equal
sides.
59. On a base BC of 1" draw an isosceles triangle having ang.
B = C = 2A. Draw CD to AB, bisecting angle C. Measure CD, AD.
What kind of triangle is CD A ? Calculate its angles.
60. Construct the triangle ABC and the bisector CD of Ex. 59. On
AB, AC outwards, describe isosceles triangles AEB, AFC, equal sides
equal to BC^ What is the figure ADCF? Calculate the angles A, F, C
of the pentagon AEBCF. What kind of pentagon is it?
UNW^P'
CHAPTER III.
CONGRUENT TRIANGLES— SIMILAR FIGURES— AREAS.
(The symbol ^ is used for ' is congruent to.')
Theorem 25. — ' Two triangles are congruent which have one
side and two angles of one equal respectively to one side and
two corresponding* angles of the other.'
If the triangles ABC, DEF have
any two angles A = D, B = E,
then also third angle C -• F.
If also one side BC = EF,
place tr. DEF on ABC, reversing if necessary,
so that EF coincides with BC, ang. E with B,
and ang. F with C ;
.'. sides ED, FD fall on BA, CA ;
.'. point D coincides with A ;
.*. triangle DEF ^ ABC.
Note. The corresponding sides are equal —
viz, AB = DE, opp. angles C, F;
AC = DF, M B, E.
Theorem 26. — * Two triangles are congruent which have two
sides and their angle of one equal respectively to a
two sides and their angle of the other.'
If the triangles ABC, DEF have
AB = DE, BC = EF,
and their ang. B = ang, E ;
place tr. DEF on ABC, so that ang. E coincides
with B, and side EF with BC ;
.*. D coincides with A, '.' ED = BA ;
.*. triangle DEF ^ ABC.
Ex. State what sides and angles are equal in con-
sequence.
* In congruent triangles those sides correspond which are opposite to equal
angles ; and angles correspond which are opposite to equal sides.
60 CONGRUENT TRIANGLES. [CH. III.
Theorem 27. — 'Two triangles are congruent which have
three sides of one equal respectively to three sides of the
other.'
If the triangles ABC, DEF have
AB = DE, BC = EF, CA = FD ;
draw arc AG, centre B,
and arc AH, centre C.
These circles meet again, once only, on the
other side of the line of centres BC.
Place tr. DEF on ABC, on the same side
of BC, so that EF coincides with BC ;
then D falls on arc AG, '.• ED == BA ;
and D also falls on arc AH, *.' FD = CA;
.'. D coincides with A ;
.*. triangle DEF ^ ABC.
Ex. (i.). State what angles are equal.
Ex. (ii. ). A diagonal divides a ihonibus into congruent triangles.
Theorem 28. — 'Two right triangles are congruent which have
the hypotenuse and one side of one equal respectively to the
hypotenuse and one side of the other.'
If the right triangles ABC, DEF have
hyp. AB = DE, side AC = DF ;
draw semicircle ACB, diam. AB,
and arc CG, centre A. / / /^
These circles meet again, once only, on the ^
other side of the line of centres AB.
Place tr. DEF on ABC on the same side of AB,
so that DE coincides with AB ;
then F faUs on semcle. ACB, ".• ang. F is right ;
and F falls also on arc CG, '.' DF = AC ;
.'. F coincides with C ;
.'. triangle DEF ^ ABC.
Ex. (i.). State what sides and angles are equal.
Ex. (ii.). If a quadrilateral ABCD has A, C right angles, and side
AB = CB, the triangles ABD, CBD are congruent.
CH. III.]
BISECTOR OF ANGLE.
61
Theorem 29. — 'If a triangle has two sides equal respectively
to two sides of another, but their angle greater than that of
the equal sides of the other, its third side is also greater than
that of the other.'
If the triangles ABC, DEF have
sides AB - DE, BC = EF,
but their ang. B > E ;
place tr. DEF on ABC in position GBC,
so that EF coincides with BC.
Draw BM bisecting ang. ABG, join MG.
Then by symmetry about BM,
AM = MG, •/ BA = BG;
.'. AC, i.e. AM + MC = MG + MC
>GC;
i.e. AC>DF.
Ex. Prove the converse of this theorem.
Theorem 30. — 'The locus of a point in a plane equidistant
from two lines in the plane is the bisectors of their angles.'
If P is a point equidistant from the lines AB, AC,
perp. PM = PN ; *
then in rt. trs. PMA, PNA,
PM = PN, hyp. PA is common,
.'. ang. PAM = PAN ;
i.e. any point on the locus is on a bisector of
angle of the lines.
Also, if Q is on a bisector of angle ; ^
then in rt. trs. QKA, QLA,
ang. K = L, ang. QAK = QAL, M
QA is common ; j|
.'. perp. QK = QL ;
i.e. any point on a bisector of angle is on the
locus ;
.*. the locus is the bisectors of angles of AB, AC.
Ex. State the theorem when the two lines are parallel.
* Tlie distance from a point to a straight line generally means the shortest—
i.e. perpendicular— distance.
62
THE PARALLELOGRAM.
[CH.
Theorem 31. — 'Opposite sides and angles of a parallelogram
are equal ; and a diagonal divides it into congruent triangles.'
If AC is a diag. of parm. ABCD ;
then in trs. ABC, CDA, AC is common,
ang. BAC = alt. ang. DCA,
ang. BCA = alt. ang. DAC ;
.•. triangle ABC ^ CDA,
and side AB = opp. side DC, BC = AD ;
also ang. B = D ; similarly, A - C.
Cor. — 'The perpendicular distance between two parallels is
everywhere the same.'
Theorem 32. — * The joins towards the same parts of the ends
of two equal and parallel straight lines are equal and parallel.'
If AB = and || CD, and AD, BC are joins towards ^
same parts ;
then in trs. ABC, CDA,
BA = DC, AC is common,
ang. BAC = alt. ang. DCA ;
.'. BC = AD; i.e. the joins are equal;
and ang. ACB = CAD, the alt. ang. ; B
.*. BC II CD ; i.e. the joins are parallel.
Cor. — 'The locus of a point at a given dis-
tance from a straight line is two parallels at that distance.'
Ex. Whatkindof figure is ABCD?
Theorem 33. — ' The diagonals of a parallelogram bisect each
other.'
If the diags. AC, BD of parm. ABCD
meet in O ;
then in trs. ABO, CDO,
AB = CD, ang. CAB = alt. ang. OCD,
ang. OBA = alt. ang. CDC;
.-. OB = CD, and OC = OA ;
i.e. AC and BD bisect each other at O.
Ex. If two straight Unes bisect each other, the joins of their ends
form a parallelogram.
CH. III.
PROPORTIONAL DIVISION.
63
Theorem 34. — 'A system of parallels wMch cuts off equal
parts from one straight line cuts off equal parts from any other
which it meets.'
If a system of parls. cuts off equal parts
AB, CD from one line AD,
and cuts off parts PQ, RS from another PS ;
draw AH, CK pari, to PS.
Then in trs. ABH, CDK,
AB = CD, ang. A = C (same parts),
ang. B = D;
.-. AH = CK.
Also PQ = AH in parm. AQ,
and RS = CK in parm. CS,
/. PQ=RS.
Theorem 35. — 'A parallel to one side of a triangle cuts off
the same fractional part from the other two sides ; and a
straight line which cuts off the same fractional part from two
sides of a triangle is parallel to the third.'
If ABC is a triangle, and DE pari, to BC
cuts off 4, say, of AB ;
divide AB into 7 equal parts so that
AD contains 4 of them ;
then a system of parls. through the points
of division divides AC into 7 equal parts,
of which AE contains 4 ;
i.e. DE cuts off 4 of AB and AC.
(ii.) If DF cuts off the same fractional part,
4 say, from the sides AB, AC of the triangle ;
make DE pari, to BC ;
.-. AE = ^AC = AF.
.*. E coincides with F ;
i.e. DF II BC.
Prove similarly for any other fraction.
Ex. Show that DE=:fBC.
64 MEASURE. [CIT. III.
J Note on Measure, Eatio, and Proportion.
This may be taken, as required, with the theorems on propor-
tion in this chapter.
Measurement.
Definition of Measure. — The measure of any magnitude is the
number expressing how much of the unit magnitude it contains.
This number is defined as greater or less according as the
magnitude it measures is greater or less.
Q> Q'
vP
10 5 '^ '
MM MM
A '-
>
2
The measures of magnitudes can sometimes be expressed (as
PQ = 2 J'', above) as arithmetical fractions, and are then rational ;
but just as often measures cannot be so expressed (as RS, the
diag. of a square, unit side, above), and are then irrational.
Each kind of measure can, however, be expressed by means of
an infinitely continued decimal, thus :
Make a scale, and measure PQ or RS by setting P or R on a
unit division, and dividing the unit containing Q or S decimally
into tenths, hundredths, and so on indefinitely.
Then Q will be found between the 3rd and 4th tenths, 3rd and
4th hundredtlis, and so on.
Thus, measure of PQ is 2-33S... ad inf.
Similarly, measure of RS is 1-414213 *...ad inf.
If PQ' is any length less than PQ, some of the scale divisions
will come between Q and Q' ; hence :
(i.) 'Unequal magnitudes are measured by diflferent decimals ;
(ii.) Magnitudes measured by the same decimal are equal ;
(iii.) Each decimal represents one only number.'
These decimals can be multiplied, &c., as in ordinary arithmetic,
working from the left.\
* Irrational numbers do not gfive recurring decimals.
f This is proved in note on ratio at the end of Chapter V.
CH. III.] RATIO AND PROPORTION. 65
(Capital letters denote magnitudes, Greek letters numbers.)
Definition of Eatio. — The ratio of a magnitude to another of
the same kind ^ is the number expressing how much of the second
is contained in the first.
The first is the antecedent or numerator, the second the con-
sequent or denominator, of the ratio.
X
The ratio of X to Y is written X : Y, or — ; and if /a is the number
X
we write, X : Y = /x, ^ = 1^^ or X = {jY.
Construction of a ratio X : Y = /x :
Find the measure /x of X on a decimal scale, unit Y ;
.-. X:Y = /x.
Thus the ratio PQ : 1 inch, on the last page, is 2-S ; and
diagonal of square : side of square is 1414213,..( =\/2).
Proportion consists in the equality of ratios.
Four magnitudes are in proportion when the ratio of the 1st to
the 2nd is equal to that of the 3rd to the 4th ;
i.e. X, Y, Z, W are in proportion when X : Y = Z : W.
X, W are the extremes, Y, Z the means of the proportion ; W is
the fourth proportional of X, Y, Z.
Three magnitudes are in proportion when the ratio of the 1st to
the 2nd is equal to that of the 2nd to the 3rd ;
i.e. X, Y, Z are in proportion when X : Y = Y : Z.
Y is the mean proportional of X, Z.
Z is the third proportional of X, Y.
The mean part of a magnitude is the mean proportional of the
whole magnitude and the other part.
A straight line is divided harmonically in four points A, B, C, D
when two points B, D divide the line AC between the others,
internally and externally in the same ratio.
Note. The numerical definition of ratio is required only for comparing
magnitudes of one kind with those of another kind. For geometry of
figure, ratio may be treated entirely geometrically. (See Ch. VII.)
* Two magnitudes are of the same kind when one is greater than, equal to, or
less than the other,
P.O. B
66 RATIO BY ALGEBRAIC FORM. [CH. III.
If Z is any unit, and X = AZ, Y = vZ, and ratio X : Y = />t ;
then XZ = X = [iY = /xvZ ;
/, A = /xv. (Dei. of equal number.)
Hence /x is the ratio X:v; i.e. the fraction* or quotient -
expressing how much of the number v is contained in A ;
i.e. the ratio X : Y = the fraction A/v.
Hence ratios X : Y, &c., may be treated by the ordinary rules of
fractions in algebra, when the results are intelligible.
Unit Theorem. — ' Two magnitudes are equal (i.) which have
the same ratio to a third ; (ii.) to which a third has the same
ratio.'
X = Y(i.)if| = |or(ii.)if| = |-
Product Theorem. — * If four magnitudes are proportional, the
product of extremes is equal to the product of means.'
If ^ = ,^. then XW = YZ.
Y W
Note. If X, Y, Z, W are lengths, their products represent rectangles.
Thus, ' If four straight lines are proportional, the rectangle of extremes
is equal to the rectangle of means.' t
Alternando. — *If four magnitudes of the same kind are
proportional, the second and third terms can be interchanged.'
^„ X Z ,, X Y
Ify^W'^^^^Z^W
XXX
Summation. — 'If -^. — ^i — ?. &c., are equal ratios of one
kind of magnitude, each ratio gi __, X^ + X, + X3 + &c. ^ X^^,,
Yj Yj + Yg + Y3 + &c. Yj - Yg
and so on.'
Other theorems may be assumed from algebra as required.
These theorems are formally proved in Ch. V., note.
* Since a. and v are numbers.
f A strict geometrical proof of this is given in Th. 52, below.
CH. III.]
PROPORTIONAL DIVISION.
67
Theorem 36.* — 'A system of parallels cuts off proportional
parts from any two straight lines which it meets.'
If a system of parls. cuts off AB, CD, and
PQ, RS from the two lines ;
make a scale, unit AB, from C along CD,
divide decimally the unit containing D into
tenths, hundredths, &c., draw parls. to AP,
&c., through the points of division ;
these form a similar scale from R, unit PQ.
Also, since DS || system of parls. AP, &c.,
the points D and S come between the same
divisions respectively of the scales of AB,
PQ;
.'. ratio -— - = — — . since both are represented ^
AB PQ ^ D-
by the same decimal.
Ex. Name all the sets of propoitional parts.
Theorem 37.* — *A parallel to one side of a triangle cuts off
proportional parts from the other two ; and a straight line
which cuts off proportional parts from two sides of a triangle
is parallel to the third.'
If ABC is a triangle, and DE || BC ;
make AF pari, to BC ;
then the system of parls. AF, DE, BC cuts off
propl. parts from AB, AC ;
. AD AE „
AB AC
(ii.) If DG cuts off propl. parts from AB, AC,
AG AD
so that — - = -— -. make DE pari, to BC :
AC AB ^
AG AD AE,
•• AC~AB~AC'
.*. AG = AE, and G coincides with E ;
i.e. DG||BC.
Ex. Name all sets of proportional parts.
* These may be postponed by beginners. They are, however, general, and
apply to rational and irrational numbers alike.
68 SIMILAR FIGURES. [CH. III.
(Tlie symbol ||| is used for 'is similar to.')
Definition 26. — Two polygons of the same number of sides
are similar whose angles in the same order are respectively equal,
and whose corresponding sides are proportional.
Cor. — 'Regular polygons of tlie same number of sides are
similar.'
Theorem 38. — 'Two triangles are similar which have two
angles of one equal respectively to two angles of the other.'
If the trs. ABC, DEF have
ang. A = D, B = E, and /. C = F ; a
place tr. DEF on ABC in j)osition AGH, / \
so that ang. D coincides with A, q^/_ \ j_j
and DE, DF with AG, AH ; / \
.•. ang. G = B, and GH II BC; /;^ A^
AGAH
•• AB~AC' S
.-. tr. DEF III ABC.
Note. The sides opp, the eql. angs. are propl. and are called corre-
sponding sides.
Theorem 39. — 'Two triangles are similar which have two
sides of one proportional to two of the other, and the angles of
these sides equal.'
If the trs. ABC, DEF have — - = -— i
A r^ A AB AC
and ang. D = A ;
place tr. DEF on ABC in position AGH,
so that ang. D coincides with A,
and DE, DF with AG, AH ;
agdedfah
^^^''ab-ab-ac-ac'
.*. GH||BC;
.'. ang. B = G (same parts) = E, and C = F ;
.'. tr. DEF III ABC.
Ex. If two triangles Ijave corresponding sides parallel, slio\y tUa,t
they are similar.
CH. III.]
SIMILAR TRIANGLES.
69
Theorem 40. — 'Two triangles are similar which have three
sides of one proportional respectively to three sides of the
other.'
If the trs. ABC, DEF have
DE EFDF
AB~'BC~AC'
make AG eql. to DE, and GH pari, to BC ;
AHAGDE DF
•• AC~AB~AB~AC'
.-. AH = DF; similarly, GH = EF ;
also AG = DE in trs. AGH, DEF ;
.'. tr. DEF = AGH III ABC.
Theorem 41. — * If two triangles have one angle of one equal
to one of the other, and the sides of a second angle of each
proportional ; then either (i.) the third angles are equal, and
the triangles similar, or (ii.) the third angles are supple-
ments.'
If trs. ABC, DEF have ang. B = E,
and sides of angs. A, D proportional,
AB
viz. — — = — — : then
AC
DE DF
either (i.) ang. A = D ; but B = E,
.-. C = F, and tr. DEF ||| ABC ;
or (ii.) ang. A =f D ;
make BAG eql. to D ;
.-. BGA = F (•.• B = E);
.-. tr. DEF III ABG ;
. ^_AB_AC
• • DF ~ DE ~ DF *"
.-. AG = AC;
.*. ang. C = AGC = suppt. of AGB
= suppt. of F.
Cor. — 'Two right triangles are similar which have the
sides of one acute angle of each proportional.'
(Make B, E rt. angs.)
Ex, Write the theorem when AB, AC = DE, DF respectively.
E- 'F
70
RIGHT TRIANGLE — BISECTED ANGLE.
CH. III.
Theorem 42. — 'A right triangle is divided into similar tri-
angles by the perpendicular from the right angle.'
If A is the right angle in rt. tr. ABC,
and AD the perp. from A ;
then ang. DAC = compt. of BAD = B,
and rt. ang. D = A ;
.-. tr. DAC III ABC.
Similarly, tr. DBA ||| ABC ||| DAC.
Note. Corresponding sides are opposite
equal angles ; B"
e.g. j^ (short sides) = ^ (long sides) = g^ (hypO-
Ex. Show that BA is the mean propl. of BD, BC.
Theorem 43. — ' The bisector of an angle or exterior angle of
a triangle divides the opposite side in the ratio of the other
sides; also (ii.), a straight line through an angle of a triangle
dividing the opposite side in the ratio of the other two is a
bisector of the angle.'
and
If AD bisects ang. A of tr. ABC,
make CF pari, to AD ;
.*. ang. AFC = BAD, same parts,
= DAC
= ACF, alt. ang. ;
.-. AC = AF,
BDBABA
DC~AF~AC'
Also (ii.), if AE divides BC so that
BEBA
EC" AC'
make CF pari, to bisr. AD ;
BE BA .. -_ .-
then— - = -— , . AF = AC;
EC AF
.'. AE II CF, and coincides with bisector AD.
Ex. Prove the theorem for the exterior ang. at A ; and show that the
two bisectors divide BC harmonically.
CH. III.]
MEAN BASE TRIANGLE — SIMILAR POLYGONS.
71
Theorem 44. — 'An isosceles triangle whose unequal side is
the mean part* of the equal sides has the equal angles each
double the third angle.'
If isosceles tr. ABC has BC the mean part of eql. sides AB, AC ;
make AD eql. to BC, join CD ;
.'. AD is the mean part of AB ;
BDDABC.
•• da^ba'ca'
.*. DC bisects ang. BCA in tr. ACB ;
also, ang. B is common to trs. DBC, CBA,
BC
BA'
tr. DBC III BCA :
and ratio of sides -— - = -— -
BC DA
anor. A = DCB
JBCA = ? = |
Definition 27. — Polygons are similarly situated whose corre-
sponding sides are parallel, and whose joins of corresponding
vertices pass all through a fixed point.
Theorem 45. — ' Two similar polygons can be placed so as to
be similarly situated about any chosen point.'
If FGHKL III ABCDE, in order of letters ;
take any point O, join OA, OB, &c. ;
make OX pari, to AB, eql. to FG,
XM pari, to OB ;
make MN, NP, PQ, &c., pari, to
AB, BC, CD, e^c.
Then ang. M N P = ang. B = G ;
and MN = OX, in parin. XN, = FG ;
FG GH.
AB"BC'
/. NP = GH.
Similarly, ang. NPQ = C = H, and PQ = HK, and so on ;
.*. fig. FGHKL ^ MNPQR, and may be placed in position MNPQR,
similarly situated to ABCDE about O.
Ex. If corresponding sides of similar polygons are parallel, or joins of
corresponding points concurrent, the polygons are similarly situated.
* See Def., p. 65, near bottom of page.
, NP ON MN
also zrzz = j-=:= -rw
BC OB AB
72
SIMILAR DIVISION.
[CH. III.
Theorem 46. — 'The perimeters of two similar polygons are
proportional to corresponding sides.'
If ABCDE, FGHKL are similar iigures, in the order of the
letters (see last fig.),
ABBC CD DEEA
FG~GH~HK~KL~LF
AB + BC + CD + DE + EA
FG + GH + HK + KL+LF
perimeter of ABCDE
perimeter of FGHKU'
(summation)
Note. Tliis is true of similar triangles. Deduce a construction for a
triangle, angles and sum of sides given.
Theorem 47. — 'Concurrent straight lines divide similarly
any two parallels which they meet, and (ii.) the joins of cor-
responding points of similarly divided parallels are concur-
rent.'
(i.) If concurrent lines through O cut the parallels in A, B, C, D
and E, F, G, H respectively ;
AB OB BC
then
EF OF FG
= (similarly)
CD
GH"
EFGH
are similarly
(ii.) If ABCD
divided parallels ;
draw EA, FB to meet in O,
and OC to meet EH in K ;
•• FK~ EF~FG'
.*. FK = FG, and G coincides with K ;
.*. GC passes through O ;
similarly, DH passes through O ;
i.e. the joins EA, FB, GC, HD are concurrent.
Note. The proof is general, and applies to any number of points
satisfying the given conditions. Also, the lines are similarly situated,
and this is a particular case of Th. 45.
CH. III.] AREA OP PARALLELOGRAM, ETC. 73
(The symbols A, O are used for area of triangle, parallelogram.)
Theorem 48. — (i.) 'Two rectangles or two parallelograms of
given angle, which, have the sides of one equal respectively to
those of the other, are congruent and have equal areas.'
(ii.) 'Two rectangles of equal areas and bases have equal
altitudes.'
(iii.) ' Two squares of equal areas have equal sides.'
rjL
One figure can be placed to coincide with the other.
Note. We may denote a rectangle by its base and altitude — thus,
rect. AB . KL — since these completely determine it.
Theorem 49. — 'A triangle has half the area of a rectangle
or parallelogram of the same base and altitude.'
D A
If tr. ABC and rect. or parni. DBCE have
same base BC, and same altitude AN ;
then A lies in DE pari, to BC.
Make AF pari, to BD and CE ;
then A ABF = Jz:7DF,
and AACF = 1oEF;
.-. sum or difFce. A ABC = | O DC.
Cor.— A ABC = 1 rect. BC . AN.
Theorem 50. — *A parallelogram has the area of a rectangle
of the same base and altitude.'
If ABCD is a parm., base BC, alt. AN ;
O AC = 2 A ABC = rect. BC.AN of same
base and altitude.
Ex. A trapezium has the area of the rectangle
of the altitude of its parallel sides and half their sum. (Divide intx>
triangles.)
74
AREA OP PARALLELOGRAM, ETC.
[CH. III.
Theorem 51. — (i.) 'Two triangles or two parallelograms of
equal bases and altitudes have equal areas.'
(ii.) 'Two triangles or two parallelograms of equal areas and
bases have equal altitudes.'
(i.) If trs. ABC, DEF or parms. AC, DF have
base BC = EF, and alt. AM = DN ;
then A ABC = J rect. BC . AM = J rect. EF . DN
= ADEF;
and O AC = rect. BC . AM - rect. EF . DN
- O DF.
(ii.) If trs. ABC, DEF or parms. AC, DF
have equal areas, and base BC == EF ;
then rect. BC . AM = 2 A ABC = 2 A DEF
= rect. EF.DN;
.-. alt. AM = DN ;
and rect. BC . AM = O AC = O DF
= rect. EF . DN ;
• AM = DN.
Theorem 52. — ' The rectangles of the extremes and means of
four straight lines in proportion are equal.'
If a, h, c, d are four straight lines in propn., a'.h = c:d\
make OA eql. to a, OB eql. to 6,
along one line, ^
OC eql. to c, CD eql. to c?,
along a perp. ;
complete rects. AD, BC, join AC, &c.
Then AC || BD, •.* OA : OB = OC : CD ;
.'. A ACD = A ACB, same base and alt. ;
.*. whole A OAD = A OBC ;
.'. rect. AD ^ rect. BC ;
i.e. rect. a.d, of extremes = rect. h . c, of means.
Cor. — 'If three straight lines are proportional, the rectangle
of extremes is equal to the square of the mean.'
Ex. Prove the theorem for two parallelograms or two triangles of
given angle.
CH. III.] EQUAL RECTANGLES — PYTHAGORAS* THEOREM.
75
Theorem 53. — 'Two equal rectangles have their sides the
extremes and means of a proportion.'
Place the equal rects. AD, BC to have an angle
common at O ;
then A OAD = A OBC ; take away OAC ;
.*. A ACD = A ACB, of same base AC ;
.*. trs. ACD, ACB have same alt. ;
.-. BD II AC, and OA : OB = OC : CD ;
i.e. OA, CD are extremes, and OB, OC means, of a proportion.
\
On
Theorem 54. — (i.) 'The square on the hypotenuse of a right
triangle is equal to the sum of squares on the other sides.' *
(ii.) 'A triangle is right which has the square on one side
equal to the sum of squares on the other sides.'
(i.) If AD, BE, OF are squares on
the hyp. and sides of rt. tr. ABC ;
draw CMN perp. to AB, pari, to BD.
Then tr. CBM ||i ABC ;
. BMBC,
•• BC"BA'
.*. sq. on BC = rect. BM . BA = rect. BN.
Similarly, sq. on AC = rect. AM . AB
= rect. AN ;
.*. sq. on BC + sq. on AC
= rect. BN+rect. AN = fig. AD
= sq. on AB.
(ii.) If triangle PQR has
sq. on PQ = sq. on QR + sq. on PR ;
make BCA a right angle,
BC = QR, AC = PR;
.*. sq. on PQ = sq. on QR + sq. on PR = sq. on BC + sq. on AC
= sq. on AB ;
.*. PQ = AB ; and in trs. PQR, ABC,
PQ = AB, QR = BC, PR = AC ;
.'. ang. R = C = a right angle.
Ex. The sum of squares on the sides of a rectangle is equal to the sum
of squares on the diagonals.
* For proof by dissection, see p. 3G, Ch. I.
76
CONSTRUCTIONS.
[CH. 111.
Proportionals.
Construction 11. — 'Construct a straight line having
ratio /x to a given line' (i.e. given a line X and
number /x, construct /xX).*
If AB is the line, set off a unit AC at an
angle to AB, calculate ju- as a decimal a-a^ag-'-i
make fKQ = a-a^a^... units, to as many
decimal places as possible or required ;
make QP pari, to CB ;
.-. AP:AB = AQ:AC = /x;
i.e. AP lias the ratio /x to AB.
Note. If the ratio is given as that of two lines I, m
AP can be found by the following construction.
Construction 12. — 'Construct the fourth proportional W of
three straight lines X, Y, Z.'
From a point O set off OA, OB equal to X, Y
along one line,! and OC eql. to Z along another ;
make BD pari, to AC.
Then X : Y = OA : OB = OC : CD = Z : CD ;
.-. CD is the 4th propl. W of X, Y, Z.
Similarly, if X, Z, W are given, Y is
constructed by making DB pari, to AC.
Thus any term in a proportion of lines can be
constructed when three are given.
Ex. 1. Construct the third proportional Z of X, Y.
Ex. 2. Given 1st, 2nd, and 4th terms X, Y, W, construct the 3rd Z.
Construction 13, — 'Construct the mean proportional Z of
two lines X, Y.'
Make AB, BC eql. to X, Y, in one line ;
draw semicircle ADC, and BD perp. to AC.
Then rt. tr. ABD ||| DBC ;
;. AB:BD = BD:BC;
i.e. BD is the mean propl. Z of X, Y.
Note. Make X a unit, Y n units ;
.♦. 7. = \/n units. We thus construct '\/n. ^
Construct thus \/3, \/5, and measure. "" '
* The theoretical construction is given at the end of Ch. V.
f A and B need not be on opp. sides of O.
CH. III.]
SIMILAR POLYGONS — MEAN SECTION.
77
Construction 14. — 'Construct a polygon on a given side,
similar to a given polygon.' ^
If ABCDE is the polygon, FG the side
coiTesponding to AB, M^
join AC, AD ; c'
make AM = FG ; MN, NP, PQ pari, to sides
of polygon ;
on FG make tr. FGH = AMN ;
„ FH M FHK = ANP,
and so on ;
.'. fig. FGHKL^ AMNPQ ||| ABCDE. C^
"Id
Ac
Construction 15. — ' Construct the line of which a given line
is the mean part;* and divide a line in mean section! (i.e. so
that one part is the mean part of the line).'
If AB is the line, make BC half AB and perp. to AB ;
draw semicircle DBE, centre C, to cut AC in
A.
D, E, so that DE = AB ; draw DF pari, to BE.
Then tr. ADB ||| ABE, *.* ang. A is common,
and ang. ABD = compt. of CBD
= CBE (semicircle) = AEB (•/ CE^CB).
Also DE = AB;
AD AD AB DE,
• DE~AB~AE~AE'
i.e. AE is the line whose mean part
is DE or AB ;
and AFB is divided similarly to ADE ;
.*. BF is the mean part of AB,
and AB is divided in mean at F.
Note, (i.) AD is also the mean part of AB ;
hence AD = BF ; thus we may construct by draw- G
ing arc DH, centre A, to meet AB in H.
(ii.) If EGIIDB, AG is divided in mean at B; i.e. AB is divided
externally so that one part BG is the mean between the line AB and
the other part AG.
Ex. Divide a straight line so that the square on one part is equal to
the rectangle of the whole and the other part.
* See definition at foot of p. 65.
t Also called * in medial section ' and * in extreroe and mean ratio,'
78
MEAN BASE TRIANGLE — SQUARE AREA.
CH. III.
Construction 16. — 'Construct an isosceles triangle whose
base is the mean part of the equal sides, and whose equal
angles are each double the third angle.' (Angles of 72° and
36°, 54°, 18°.)
If BC is the base, construct BD of which
BC is the mean part ;
with centres B, C, radius BD, draw arcs A ;
ABC is the triangle.
By Th. 44, ang. B = C = 2A.
Also A + B + C:=5A = 180°;
/. A = 36°, B = 72°.
Draw BE perp. to AC ;
.*. ang. EBC = compt. of C = 18°,
and EBA = 54°.
Note. We can construct by ruler and compass "p
only,
(i.) 90°, 45°, by right angle and bisection.
(ii. ) 60°, 30°, 15°, by equil. triangle and bisection,
(iii.) 72°, 54°, 36°, 27°, 18°, 9°, as above and by bisection,
(iv.) Any angle obtained from these, as sum or difference.
These are all the angles in whole numbers of degrees that can be
drawn in this manner.
Ex. Show that the complete series 3°, 6°, 9°... (adding 3° each time)
can be obtained.
B-^
Construction 17. — * Construct a square equal in area to a
given rectangle of sides X, Y.' * ^
Make AB = X, BC = Y, in one line ;
draw semcle. ADC, make BD perp. to AC,
and draw sq. on BD.
Then BD is mean propl. of X, Y ;
i.e. X:BD = BD:Y;
.*. sq. on BD == rect. X . Y. u <^^ ^"q
Note. The next constructions show how to make a rectangle equal
to a given triangle or polygon ; thus a square can be constructed equal
in area to any polygon.
* Reduces to construction of roean propl.
CH.
ni.]
PARALLELOGRAMS OF GIVEN AREA.
79
Construction 18. — 'Draw a rectangle, or parallelogram of
given angle, equal in area to a given parallelogram or triangle.'
If ABCD is a given parm., make CBE the
given aiig., draw parm. EBCF ;
.*. O EC = O AC, same base and alt.
If ABC is a given tr., make BM half BC,
ang. CBE the given ang., draw parm. EM ;
,*. OEM = 2 AABM = AABC.
Construction 19. — 'Draw a triangle, rectangle, parallelogram
of given angle, or square equal in area to a polygon.'
If ABCDE is the polygon,
produce a side BC, take an end diag. CE,
draw DF pari, to CE ;
.'. ACEF = ACED;
.*. quadl. ABFE = poln. ABCDE in area. ^
Thus, without altering the area, the number
of sides is reduced by unity.
Similarly, take an end diag. AF of
quadl. ABFE ;
draw EG pari, to AF ;
.-. AABG = quadl. ABFE = poln. ABCDE.
(ii.) On a side AB of tr. ABG, make AH a
rect., or parm. of given angle, eql. to A ABG ;
.*. OAH = poln. ABCDE.
Similarly for a polygon of any number of sides,
be made eql. to rect. AH — i.e. to given polygon.
A square can
Construction 20. — 'On a given side construct
a rectangle, or parallelogram or triangle of
given angle, equal in area to a given polygon.' *
On a side BC of poln. P construct rect., or
parm. or tr. of given ang., AC, eql. to P.
Make BE the given side along BA,
draw AF pari, to EC, draw parm. EF ;
.*. O EF = O AC = P. (Th. 52.)
Orif AABC = P, AEBF = ABC = P.
* This reduces to Constrs. 19 and 12.
\
\
s
\
\
^'\
1
=e^-.=
80 SOLUTION OP PROBLEMS. [CH. IIL
In order to solve a problem, read the statement carefully,
making a rough drawing of each part of the figure as soon as you
understand the statement sufficiently. Compare different parts of
the figure to find equal sides, angles, triangles, similar triangles,
&c., and note the results of the comparison.
If this is not sufficient, draw auxiliary lines, circles, perpendi-
culars, bisectors of angles or sides, parallels, to see if any fresh
relations can thus be discovered amongst the parts of the original
figure ; and search your text-book for theorems or constructions
which may throw light on the problem.
In construction problems, start with the finished construction,
and proceed as above, until you discover the key.
Do not make a triangle isosceles or equilateral, or a parallelogram
rectangular or equal-sided, unless it is so given in the statement.
Appended are a few examples :
Theorem 55. — 'The joins of mid points of sides of a triangle
are parallel to the sides, and divide the whole triangle into
four congruent triangles.' (Propl. division.)
Theorem 56. — 'The sum of parallel sides of a trapezium is
double the line bisecting the other sides.'
Theorem 57. — * The medians of a triangle are concurrent, and
trisect each other.' (Simr. triangles.)
They meet in the centroid of the triangle.
Theorem 58. — 'The bisectors of angles of a triangle are
concurrent.'
They meet in the incentre of the triangle. (Th. 30.)
' The bisectors of exterior angles of a triangle are concurrent,
two and two, with one of the bisectors of an interior angle.'
They meet in the three ecentres of the triangle.
Theorem 59. — 'The three perpendiculars of a triangle are
concurrent.' (Ang. in semicircle and in same arc, or simr,
triangles.)
They meet in the orthocentre of the triangle.
The joins of their feet form the pedal triangle.
CH. III.]
PROBLEMS.
81
Theorem 60. — 'If one angle B of a triangle ABC is greater
than another C, the bisector of B is less than that of C
If BD, CE are the bisectors,
^ BCH = B.
2'
BCE.
make ang. ABF = CBH =
Show triangle CBH
Compare BG and CE.
Show triangle ABF ||| ACE.
Compare BF and CE. ^
BD<the greater of BG and BF. (Ex. XX. 23.)
Compare BD and CE.
Construction 21. — ' If A, B, C are three points in order in a
straight line, find a point P in it such that PB is a mean
proportional between PA, PC
If P is the point, and PF, PG on another line = PB, PC,
.U rsr- ^« 1 PG PC PB
then GF = CB, and zr= = zrzz = ^^-r ',
PF PB PA
.*. GB||AF.
Also, if BE II CD II PF,
CD CB GF
PF -pB ~PF'
.*. CD = GF = BE, and CDEB is a rhombus.
Hence the construction.
Make a rhombus BCDE, join BD, AE to F,
draw FP pari, to BE or CD.
(Show that PF = PB, GF = CB.)
Pythagoras' Theorem Generalised. — 'The sum Q + R of the
areas of two similar figures on the right-angled sides of a
right triangle is equal to that of the similar figure P on the
third side.'
If P, Q, R are first similarly situated, and
then reduced to equal squares P", Q', R' by
Constr. 19, corresp. points and sides being used
throughout the construction ; then
side of P' : side of Q' : side of R'
= side of P : corresp. side of Q : side of R ;
.*. the sides of sqq. P', Q', R' form a rt. tr. A'B'C, simr. to ABC ;
.'. P + Q = P' + Q'=R' = R.
P, G, F
82
PROBLEMS.
[CH. III.
^^-'"D
Construction 22. — * Divide a given line AB internally and
externally in a given ratio /x.'
(i.) If the ratio is given as that of two
lines Z, 7n, and P is the internal point,
and parls. AC, BD, BE eql. to I, 7n, m are
drawn from A, B ;
then tr. APC ||| BPD ;
.*. C, P, D are collinear. Hence the constr.
Make AC = l, BD || AC and = m,
join CD to cut AB in P ;
.*. AP: PB = Z:m = /x.
Similarly, make BE || AC and = m,
join CE to cut AB in Q ;
.'. AQ: BQ = l:m = fx.
(ii.) If the ratio is given numerically as /a, make BD, BE a unit
and AC /a units, and proceed as before.
Note. Since P must be on CD, which cuts AB in one only point, it
is clear that there is one only internal, and similarly one only external
point.
Theorem 61. — 'The locus of a point P whose distances from
two fixed points A, B have a given ratio PA : PB = /x is the circle
whose diameter divides AB in this ratio,'
If CD is the diamr. of this circle,
and CA : BC = /A = DA : DB ;
then (i.) if P is on the locus,
PA : PB = CA : BC = DA : DB ;
.'. PC, PD are bisectors of angs. APB, BPH ;
.'. CPD is a rt. ang., and P is on the circle.
(ii.) If Q is a point on the circle,
and CQ bisects ang. AQE,
then DQ_LCQ, and bisects ang. EQK ;
.-. DA : DE = QA : QE = CA : EC ;
.'. Alternando, DA : CA = DE : EC ;
i.e. DE : EC = DA : CA = DB : BC ;
.*. E coincides with B (see note, Constr. 22) ;
/. QA : QB = CA ; BC, and Q is on the locus.
CH.
in.]
LOCI STRAIGHT LINE AND CIRCLE.
83
Construction problems reduce generally to finding certain points
by means of the loci on which they lie. The following summary
will be useful.
A. Locus a straight line.
(i.) Point whose join to a fixed point has fixed direction.
(ii.) Point equidistant from two fixed points.
Locus of vertex of isosceles triangle, given base; of centre of
circle through two points ; right bisector.
(iii.) Point equidistant from two lines; or whose distances in
fixed directions from two lines have a given ratio.
Bisector of angle. Locus of vertex of triangle of
given form (simr. to given triangle), its base resting
on two given lines and pari, to fixed direction.
This is the general case of similarly situated
figures or multiplication.
(iv.) Point dividing in given ratio any intercept
of two parls., or any intercept of fixed direction between two lines.
(v.) Point at given distance from a line, or whose distance in
fixed direction from a line is constant.
Locus of vertex of triangle or parm., given alt. and base line, or
area, base, and base line.
This is the general case of translation.
(vi.) Point the sum of whose distances from two lines is constant.
(Locus is sides of a rect., pari, to bisectors of angles of lines.)
Locus of vertex of quadl., given sum of distances from two sides
of given angle.
B. Locus a circle.
(i.) Point at given distance from fixed point.
Locus of any point of a figure turning about fixed point.
Simple rotation.
Theorem 62. — 'Any straight line AB can be rotated to
coincide with an equal line PQ ; and any
polygon AB... to coincide with a congruent
polygon PQ... of the same aspect.*
Rt. bisectors of AP, BQ determine the centre
of rotation O.
.No
This is the general case of congruence.
Note. A polygon of opp. aspect may be first reversed.
84 LOCI STRAIGHT LINE AND CIRCLE. [CH. III.
B. Locus a circle.
(ii.) Point whose joins to two fixed points form a given angle.
Locus of vertex of triangle, given base and opp. angle, of rt.
triangle of given hypotenuse. (Constr. 9, 10 (viii.), Ch. II., or
A, i., ii.)
(iii.) Mid point of chord of circle of given length, or through
fixed point.
(iv.) Point whose distances from two fixed points have given
ratio. (Th. 61.)
A, ii. is a limiting case of this.
EXAMPLES— XXII.
1. Construct a point P in a line PQ, whose join to a point A makes a
given angle with PQ.
2. Construct an isosceles triangle, or circumcircle of triangle, given
base and opp. ang. (A, i., ii., or B, ii.)
3. Inscribe in a triangle an equilateral triangle, also a triangle similar
to a given triangle, given one side pari, to fixed direction. (A, iii. )
4. Construct a triangle, given base, area, median from one end of base.
(A, iv., v., B, i.)
5. Find locus of diagonal point of a trapezium, one parallel side fixed,
the other and its alt. given in magnitude. (A, iv. )
6. Construct a triangle, given one angle in position, and sum of
distances from sides of foOt of perpendicular from vertex of angle on
opposite side. (A, vi., B, ii.)
7. Construct quadl. ABCD, given angle C in position, triangle ABD
in form, direction of BD, sum of distances of A from sides of ang. C.
(A, iii., vi.)
8. Construct a triangle, given base, opp. ang., and alt. (A, v., B, ii.) ;
given base, opp. ang., and area (A, v., B, ii.) ; given base, alt., ratio of
sides (A, v., B, iv.) ; given base, opp. ang., ratio of sides (B, ii., B, iv.).
9. Inscribe a triangle in a circle, given one side in length, opp. vertex,
and distance of another side from centre (B, i., iii.) ; given one side in
length, a point on another side, and distance of this side from centre
(B, i.,iii.).
10. Inscribe a triangle in a circle, given one side fixed in position and
the ratio of the other sides. (B, iv.)
Suggest a simpler alternative construction.
11. If one end of a straight line of given length and direction moves in
any path, the other end describes a congruent path. (A, v.)
CH. III.] EXAMPLES. 85
EXAMPLES— XXIII.
Theorems.
1. A diagonal of a quadrilateral bisects the augle between two equal
sides ; show that the other sides are equal.
2. Equilateral triangles ABD, BCE are formed externally on two sides
of a triangle ABC ; show that DC = AE.
3. In a regular hexagon the triangle formed by joining three alternate
vertices is equilateral.
4. In a regular pentagon ABODE, show that the triangle ACD has
the angles C, D each double of A. (Calc. in degrees ang. of pentagon
and hence of triangles ABC, ACD.)
5. If the bisector of angle A of a triangle coincides with the median,
the triangle is isosceles.
6. Every rhombus is a parallelogram.
7. A quadrilateral is a parallelogram when (i.) its diagonals bisect
each other, (ii.) opposite angles are equal, (iii.) opposite sides are equal.
8. A parallelogram whose diagonals bisect its angles is a rhombus. Is
this true of any quadrilateral ?
9. How can you distinguish parallelogram, rhombus, rectangle, square
by their diagonals ?
10. The total height of a staircase is the sum of heights of the separate
steps.
11. If AB, CD are equal chords of a circle (the order on the circle being
ABDC), the triangles ABD, CDB are congruent. (Use symmetry.)
12. Squares ABDE, ACFG are described on two sides of a triangle ;
show that the median AX of the triangle is perpendicular to and the
half of EG.
13. Show also in Ex. 12 that BG is perpendicular and equal to CE.
14. If the non-parallel sides of a trapezium are equal, two opposite
angles are supplementary.
15. Medians AX, BY of a triangle are produced to double their lengths
at F, G ; show that FG passes through C.
16. One straight line AB from a point A to a circle, centre O, is
greater than another AC, if the angle AOB at the centre is greater
than AOC.
17. If two sides of a parallelogram are given in magnitude, their
diagonal increases as their angle diminishes. What are its greatest and
least values ?
18. If a straight line bounded by two parallels is bisected at a point
O, any other straight line through O bounded by them is bisected at O.
86 EXAMPLES. [CH. III.
19. The joins of mid points of sides of a quadrilateral form a parallelo-
gram.
20. Parallels to the equal sides of an isosceles triangle are drawn from
a point in the third side ; show that the perimeter of the parallelogram
thus formed is the sum of the equal sides.
21. M is the mid point of a line AB. The sum of perpendiculars from
A and B to a line cutting AB produced is twice the perpendicular
from M.
22. The diagonals of a trapezium divide each other proportionally.
23. A parallel to the parallel sides of a trapezium divides the other
sides proportionally.
24. If two diagonals of a quadrilateral divide each other proportionally,
the figure is a trapezium.
25. AD is the bisector of angle A of a triangle ABC, M the mid point
of BC, and BE, CF perpendiculars to AD. Show that ME=MF.
26. The lines BE, DP drawn to the mid points E, F of the sides AD,
BC of a parallelogram, trisect the diagonal AC.
27. If DE parallel to BC in triangle ABC meets AB, AC in D, E,
then BE and CD divide each other proportionally.
28. In Ex. 27 any line through A divides DE, BC proportionally.
29. If parallels APB, CQD are divided proportionally at P, Q, the
joins AC, PQ, BD are concurrent.
30. The non-parallel sides of a trapezium and the line through the mid
points of the parallel sides are concurrent.
31. The projections of two parallel lines on a given line are proportional
to the lines.
32. If two triangles have their angles respectively equal, the sides of
one are either all greater than, or all less than, or all equal to the
corresponding sides of the other.
33. The circumradii of similar triangles are proportional to the sides.
34. If three straight lines CAB, OCD, OEF are similarly divided, the
triangles ACE, BDF are similar.
35. If two similar triangles are placed with corresponding sides
parallel, the joins of corresponding points are concurrent. Is this true
of similar polygons ?
36. If ABCD, EFGH are similarly divided lines, the triangles of
their parts are similar.
37. A parallelogram having an angle A common with a parallelogram
ABCD, and the opposite vertex E on the diagonal AC, is similar to it.
38. If two corresponding sides of two similar parallelograms having
a common angle are in a straight line, their diagonals through this
angle coincide.
CH. III.] EXAMPLES. 87
39. It* the pairs of opposite sides of a parallelogram cut proportion-
ally the parallels to the other sides through a point P, then P is on
a diagonal.
40. If Pi Q divide the sides AD, BC of a parallelogram so that
PA : PD = QC : QB, then PQ bisects the diagonals. (Use alternando.)
41. If DE parallel to BC meets AB, AC in D, E, and if BE, CD
meet in O, and OF parallel to BC meets AB in F; then ADFB is
divided harmonically.
42. If P is any point in a diameter AB of a circle, PC any line to the
circle, Q a point on AB produced such that ang. ACQ = ACP, show that
QAPB is divided harmonically.
43. The lines joining the ends of parallel diameters of two circles
divide the join of centres harmonically, in the ratio of the radii.
44. If M is the mid point of a line AB, divided harmonically at P, Q,
show that MP. MQ=MB'^. (Treat as algebraic products.)
45. If in a right triangle ABC, B the right angle, AB is half BC, then
AC - AB is the mean part of BC. (Constr. 15, Note i.)
46. If in Th. 44 CD is produced to F so that CF = CA, show that
AF = FB = BC, and BF || AC.
47. The sum of perpendiculars to the sides of an equilateral triangle
from a point inside it is equal to an altitude.
48. In a square ABCD, Q is the mid point of CD, and AP meeting CD
in P is equal to CP + CB. Show that the angle BAP is double of QAD.
49. A triangle is isosceles when (i.) two altitudes are equal, (ii.) two
medians are equal, (iii.) two bisectors of angle are equal.
50. One side of a triangle is greater than another when (i.) its altitude
is less, (ii. ) its median is less, (iii. ) its bisector of angle is less than that
of the other.
51. The four triangles formed by the half diagonals of a parallelogram
with the sides are equal in area.
52. Any straight line through the diagonal point of a parallelogram
bisects its area.
53. If triangles ABC, A'B'C have a=a', b = b\ C = supplement of C,
their areas are equal.
54. If straight lines AB, CD intersecting at E make the triangles
AEC, BED equal in area, then BC || AD.
55. If G is the centroid of the triangle ABC, the areas GAB, GBC,
GCA are equal.
56. If ABC, DBC are triangles equal in area on opposite sides of BC,
then AD is bisected by BC.
57. If G is a point in a triangle such that areas GAB, GBC, GCA are
equal, G is the centroid.
68 EXAMPLES. [cH. III.
58. If tlie diagonal AC of a quadrilateral ABCD is bisected by BD,
the triangles ABD, CBD are equal in area.
59. A straight line through a vertex D of a parallelogram ABCD cuts
BC, AB in F, G ; show that ABF, CFG are equal in area.
60^ If AD is the perpendicular on BC in triangle ABC, H the ortho-
Centre, show that rect. DA. DH = rect. DB. DC.
61. In a trapezium the rectangle of two of the parts into which the
diagonals divide each other is equal to that of the other two.
Q2. If AD, BE are perpendiculars of a triangle ABC, show that
iect. CD.CB = rect. CE.CA. If X, Y are mid points of BC, CA,
show also that CX . CD = CY . CE.
63. The rectangle contained by one of the equal sides of an isosceles
triangle, and the projection on it of the third side, is half the square of
this side.
64. If BD, CE are drawn from the ends of the hypotenuse of a right
triangle to the sides, BD'-^ + CEs^BC^+DE^.
65. If a point P is joined to the points of a rectangle ABCD,
then PA2 + PC2=PB2+PD2.
66. If BC is the hypotenuse of a right triangle, BC^ is four-tifths of
the sum of squares of medians from B, C.
67. A square has a greater area than a rectangle of the same perimeter.
(Constr. square equal to rectangle. )
68. A square has a less perimeter than a rectangle of the same area.
69. Of all parallelograms of given area, the square has the least
perimeter.
70. Of all parallelograms of given perimeter, the square has the
greatest area.
71. Can you prove Exx. 69 and 70 if quadrilateral is written for
parallelogram ?
72. A square ABCD and rectangle AEFG have a common angle A,
and the point E is on AB. If their areas are equal, show that ED || BG.
73. Conversely, if in Ex. 72 ED || BG, sIioav that the areas are equal.
74. Assuming that angles in the same arc of a circle are equal,
show that if AB, CD are two chords of a circle meeting in P,
the rect. PA . PB = rect. PC. PD.
75. Using Ex. 74, show that if a chord CD of a circle is bisected
at M by a diameter AB, then rect. AM . MB = MC2.
76. Show also in Ex. 75 that AC2 = AIVI . AB.
77. If P is any point on a chord of a circle, centre O, bisected at M,
show that P02 = PM2 + M02.
78. Show that the sum of squares on the right-angled sides of a right
triangle is four times the square on the median of the hypotenuse.
CH. III.] EXAMPLES. 89
EXAMPLES— XXIV.
Constructions.
1. On two sides of an angle A find points B, C such that BC is equal
in length to one line D, and parallel to another E.
2. Find a point P in the side BC of a triangle such that if PQ, PR
are perpendicular to AB, AC, then AQ=:AR.
3. From a given point P draw lengths PQ, PR to two parallel straight
lines so that PQ and PR are equal and perpendicular to each other.
4. Construct a triangle, given in position the mid points of sides.
5. Construct a triangle, given in position the feet of perpendiculars.
6. Measure the angle of elevation of the sun when a post 12 ft. high
throws a shadow of 24 ft.
7. Draw an elevation of a staircase, height of each step 9", depth 12",
to show 10 steps, half an inch to the foot.
8. A tower stands on the edge of a cliff 100 ft. high. From a point
below the cliff the elevation of the base of the tower is 45°, and of the top
60°. Find by diagram the height of the tower.
9. Two posts A, B on one bank of a river are 100 ft. apart. Two
others C, D on the same side, each 200 ft. from the bank, are 500 ft.
apart. The joins AC, BD just cover a post P on the opposite bank.
Find the width of the river.
10. Inscribe a rhombus in a parallelogram, having one vertex at a fixed
point on one side of the parallelogram.
11. Construct a quadrilateral A BCD, diagonal BD = 2-3 cm., AB = AD
= 2-7 cm., CB = CD = 3-2 cm. Draw the right bisector of BD.
12. Draw a parallelogram, sides 1^", 2|", angle 60°. Is the diagonal
through this angle greater or less than that of the rectangle of the same
sides ? Why ? Measure the two.
13. Given a finite straight line AB, construct a continuation of it CD,
separated from A B by an obstacle.
14. Construct a fourth proportional to 1", 2-07", 2-44", and measure
its length.
15. Sliow that the result in Ex. 14 represents the product of 2-07 and
2-44. Hence deduce a method of lepresenting a product geometrically.
16. Two finite straight lines ABCD and PQ, not parallel, are given
in position. Divide PQ similarly to ABCD.
17. Draw a line between two sides of an angle to pass through a fixed
point and to be divided in a given ratio at that point.
18. Take any point P inside an equilateral triangle, 3 cm. side. Draw
a straight line terminated by the sides and divided at P in the ratio 3 : 7.
90 EXAMPLES. [CH. III.
19. Construct the locus of a point P whose distances from two given
lines OA, OB at an angle of 63° are as 4 : 5.
20. Draw a parallelogram, sides 2-8 cm., 1-9 cm., angle 54°. Construct
a similar figure, with sides greater in the ratio 12 : 7.
21. Construct a regular pentagon (use protractor) in a circle of 2 cm.
radius. Construct a similar figure whose diagonals are half as long
again as those of the first.
22. Construct the locus of a point dividing a line between two parallels
li" apart in the ratio 3 : 5.
23. AB, CD are straight lines which meet if produced outside the
paper. Through a given point P draw a straight line to their point of
intersection. (Use parallel lines similarly divided, or similarly situated
triangles. )
24. Draw also the bisector of angle of two such lines as AB, CD in
Ex. 23.
25. Construct a triangle the sum of whose sides is 3", and two of whose
angles are 52°, 48°.
26. Draw a square whose side is the mean proportional of 1-8 cm. and
3-2 cm. How does it compare in area with the rectangle of these sides ?
27. Give a geometrical construction to find \J1 to two decimal places.
Is your value correct ? (Use ^-inch diagonal scale.)
28. Take AB = |", BC = ^" in a straight line, and find P in BC pro-
duced so that PB is mean proportional of PC, PA. Measure PC.
(Constr. 21.)
29. Take AB = |", BC = i", as in Ex. 28, and construct AD the
mean proportional of AB, AC. Compare DC with PC in Ex. 28.
30. Divide a straight line 2-6" long in the ratio 9:4, and find the
mean proportional of the parts.
31. Find points D, E in the sides BC, CA of a triangle so that DE
parallel to AB is the mean proportional of BD, DC.
32. Draw a straight line ABC, AB = 2-7 cm., BC = 1.8 cm. ; find a
point D so tiiat ABCD is divided harmonically.
33. Draw an isosceles triangle, a=h = %Q cm., C = 30'. Draw CD
the bisector of angle C. Can you find a point E so that ADBE is
divided harmonically ? Why ? How do you interpret the result ?
34. If AP, AQ, AR are drawn from A to the opposite side BC of
a parallelogram, bisecting BC, BP, BQ respectively and meeting the
diagonal BD in H, K, L ; find the ratio BL : BD.
35. Divide AB, =1.4", harmonically with internal parts in the
ratio 3 : 4.
36. Construct and calculate numerically the mean part of a line
1" long.
CH. III.] EXAMPLES. 91
37. Calculate from Construction 15 the ratio of a line to its mean part
(i.) as a surd, (ii.) as a decimal.
38. If in a right triangle ABC, B the right angle, AB = 2BC, and BD
is perpendicular to AC, find the ratio AD : DC.
39. Construct an isosceles triangle ABC such that the side BC, 1"
long, is the mean part of the equal sides AB, AC. Bisect angles B, C
by BD, CE each equal to AB. Prove that AEBCD is a regular pentagon.
40. Find the locus of a point P dividing in a given ratio, say 5 : 3,
the join of a fixed point V to a point Q moving on a given circle.
41. Construct the locus of P when PA : PB = 3 : 8, AB = 3.3 cm., A and
B being fixed points.
42. Construct the locus of the point A of a triangle when a=2'3 cm.,
b:c = S:5.
43. Construct a triangle, « = 2-5 cm., b:c = S:2, C = 30'. Can it be
done if C = 80°?
44. Construct by ruler and compass only an angle of 27°. (Constr. 16.
Bisect ang. A twice.) Test by protractor.
45. Draw a triangle, « = 1.12", 6 = 1-28", c = 2.08". Inscribe in it an
isosceles right triangle with the hypotenuse parallel to the bisector of
angle A.
46. Calculate the areas: (i.) triangle, base 2-7 cm., alt. 3-2 cm.;
(ii.) trapezium, pari, sides 1-9 cm., 2-9 cm., alt. 1«3 cm. ; (iii.) right
triangle, perp. sides 12'7 cm. and 33-8 cm.
47. Calculate the sides of the equivalent squares in Ex. 46.
48. Draw a regular hexagon on a side of 2-5 cm., and draw a triangle
of equal area. AVhat is the area?
49. Construct a rhombus equal in area to a given parallelogram.
50. Construct a square equal to the-(i.) sum, (ii.) difference of squares
on sides of 2-5 cm. and 1-8 cm.
51. A pentagon ABCDE represents an estate (V to the mile). AC
= li"; altitudes of B, D, E from AC, 1", 1^", 1^"; projections of AB, AE,
AD on AC, h", I", 11". Draw a plan and calculate the area. (This is
the ordinary surveyor's method of measuring the area of a field.)
52. ABCD (4" to the mile) represents a field ; B, D are right angles ;
AC = 3-2"; altitudes of B, D from AC, 1-3", M". Draw a plan and
calculate the area.
53. Calculate the areas : (i.) equil. triangle, side 3-2 cm. ; (ii.) parm.,
base 2-2 cm., alt. 3-8 cm. ; (iii.) isosceles rt. triangle, hyp. 1-41".
54. Construct a quadl, AC = 2.7 cm., AB = BC = 2.3 cm., CD = DA
=2-9 cm. Describe an equal isosceles right triangle.
55. On a scale of t^^" to the yard represent a square field of 10 acres.
56. Make a square equal to a rectangle of sides 1-8", 6".
92 EXAMPLES. [CH. III.
57. Divide a straight line of 2" into two parts whose rectangle is
equal to the square on |" side.
58. Given one side 2«74" of a rectangle equal to a square on 1-54" side,
construct the other side.
59. Construct the side of a square of 3-76 sq. in. Measure.
60. Construct an equilateral triangle of 3-7 cm. side, and construct a
square of the same area.
61. Construct a mean base triangle (Constr. 16), base H", and make a
square equal to it.
62. Inscribe a given square in a given square. (Constr. 10, vii.)
63. Calculate the diagonal of a square of 5-96 sq. in. area.
64. Construct a triangle, a=l-32", 6 = 2-75", 6*= 1-96". On the other
side of a construct a triangle of equal area, having one of its sides 2".
65. Construct a rectangle, diagonal 2", equal to the triangle of Ex. 64.
66. Draw a rectangle, sides 3-6 cm., 4-8 cm., and construct an equal
rectangle, one side 3-2 cm.
67. Given a line AB, 1", and a point P in BA produced, y from A;
find points D, D' such that the square on PD or PD' is equal to the
rectangle PA . PB.
68. Given a line AB, 4 cm., find the points C, D dividing AB internally
and externally in the ratio 5 : 3.
Find also the point P such that PC is the mean proportional of
PA, PB.
69. Set off a line AB, 1". Make AP perpendicular and equal to AB.
Along AB from A set off in succession lengths AC equal to PB, AD to
PC, AE to PD, and so on.
Show that AB, AC,... represent the complete series of quadratic surds
of whole numbers \/l> \/2, \/3...
Which of this series of surds would be the simplest to pick out as a
check on the accuracy of your Avork ?
70. Take a line AB, 1". Make BC, perpendicular to AB, 1" ; CD,
perpendicular to AC, 1" ; DE, perpendicular to AD, 1", and so on.
Show that the series of lines AB, AC, AD, AE repiesents the same
series of surds as those of Ex. 69, and that the two constructions are
really the same.
71. Construct an isosceles right triangle, side 1". On the hypotenuse
construct a second, and on its hypotenuse a third, and so on, until a
hypotenuse is constructed in line with the original base produced.
Measure this.
72. On a base of 1 cm. construct a mean base triangle (Constr. 16).
On its side construct another, .and so on until the side of the fifth
triangle is obtained along the original base. Measure this side.
93
PART II.
CHAPTER IV.
THE CIRCLE— CHORD, TANGENT, ANGLE, AND RECT-
ANGLE PROPERTIES — EXPERIMENTAL SOLID
GEOMETRY.
Definition 28. — A secant or transversal of a circle through any
point in its plane is a straight line through the point cutting the
circle.
The tangent of a circle at a point A on it is
the secant or transversal AP through A whose
second point P on the circle coincides with A.
(Turn a straight-edge about a pin fixed at A,
until P is as near A as possible.)
In Th. 64 it is shoAvn that there is one only-
tangent at each point of the circle.
Note. The letigth QA from a point Q on a tangent
to the point of contact A is often called the tangent from Q to the circle.
Definition 29. — Two circles touch which have a common
tangent at a point where they meet.
Definition 30. — An angle in a circle is an angle whose point
is on the circle, and whose sides cut the circle. A
It stands on the arc cut off the circle by its /^ y\\
sides ; and it is in the arc or segment containing / / \ \
the point of the angle and terminated by the sides. \/ \\
The angle A stands on arc BDC, and is in arc \^ J
or segment BAG. d
Definition 31. — A segment of a circle is a part of a circle
bounded by a chord and one of its arcs.
The two arcs or segments into which a chord divides a circle
are opposite or alternate.
Definition 32. — A polygon whose vertices are on a circle is
inscribed in the circle ; and the circle is then circumscribed to
the polygon.
Definition 33. — A polygon whose sides touch a circle is circum-
scribed to the circle ; and the circle is then inscribed in the polygon.
94
CENTRE — CHORD — TANGENT.
[CH. IV.
Theorem 63. — 'The centre of a circle lies in any right
bisector of a chord; and a chord meets the circle in two
only points.'
If NC is the rt. bisector of chd. AB of a circle, centre O ;
(i.) O is equidistant from AB ;
.'. O lies in rt. bisector NC.
(ii.) If G is any third point in AB,
the rt. bisector of AG || NO,
and therefore does not traverse O ;
.'. G is not on the circle, centre O ;
.'. chd. AB meets the circle in two only points
A. B.
Ex. Show that two chords of a circle, not both diameters, cannot
bisect each other.
Theorem 64. — 'The tangent at any point on a circle is per-
pendicular to the radius of the point, and meets the circle
at its point of contact only.'
If A is a point on a circle, centre O,
P any other point on the circle ;
draw ON, rt. bisector of A P.
Then (i.) if P moves along circle to coin-
cidence with A,
the rt. bisector ON coincides with OA,
and sect. AP coincides with tangt. AT ;
.'. tangt. ATj.rad. OA.
, (ii.) Hyp. OT > OA in rt. triangle OAT ;
.'. T is outside the circle ;
.*. A is the only point of the tangent on the circle.
Note. The tangent is entirely outside the circle except at A ; hence
no tangent can be drawn from a point inside a circle.
Ex. 1. Show that the circle on OA as diameter touches AT and also
the first circle.
Ex. 2. If TO meets the circle on OA as diameter in M, show that
TA is the mean proportional of TN, TO.
CH. IV.]
CHORD — TANGENTS.
95
Theorem 65. — 'Every straight line, except a tangent, in the
plane of a circle and meeting it, cuts the circle twice.'
If a str. line AN meets a circle, centre O,
at A;
make ON perp. to AN, and NB eql. to NA ;
.'. NO is rt. bisector of AB ;
.*. OB = OA, and B is on the circle.
But ON <OA in rt. triangle OAN ;
.'. N is inside the circle, and AB passes from
inside to outside in crossing the tangents at A, B.
Cor. — 'A chord of a circle lies entirely inside the circle.
Theorem 66. — * Two only tangents can he drawn to a circle
from an outside point in its plane ; these are equal in length,
and symmetrical about the join of the point to the centre.'
If P is a point outside a circle, centre O,
and PA is tangt. at A ;
then PAO is a rt. ang.
.'. A is on circle, diam. OP, which cuts
first circle in two only points A, B ;
.'. there are two only tangts. PA, PB ;
and the whole figure is symmetrical about PO ;
.*, the tangts. PA, PB are equal,
and symmetrical about PO.
Theorem 67. — 'The locus of the centre of a circle touching
two straight lines is the bisectors of their angles.'
If O is the centre of a circle touching AB, AC
at M, N,
perp. ON = OM ;
.'. O is on a bisector of BAG. (Th. 30.)
And any point on the bisectors is equidistant
from the lines, and may be a centre.
Ex. Show that A is on the bisector of ang. NOM.
Ex. Construct a centre of a circle to touch three straight lines,
96
TANGENT CIRCLES ANGLE IN CIRCLE.
[CH. IV.
Theorem 68. — 'Two circles which touch meet on their line
of centres, and meet in one only point.'
If two circles, centres O, Q, touch at A,
and AT is the common tangt.,
OA and QA each ± AT ;
.*. OA, AQ are in a straight line.
Also, the common chd. from A _L OQ,
and coincides with AT, which meets the
circles at A only ;
,'. the two circles meet at A only.
Note. The join of centres of two circles whicli touch is the sum or
difference of their radii.
Ex. Show that two circles do not naeet at all if the join of their
centres is greater than the sum or less than the difference of the
radii.
Theorem 69. — 'An angle in a circle is half the angle at its
centre on the same arc'
If ang. AOC at centre O of a circle is on
same arc ADC as ang. ABC in circle,
produce BO to E ;
then ang. OBA = CAB, •/ OA ^^ OB ;
.*. ang. AOE = suppt. of AOB
= OBA + CAB, in tr. AOB,
= 20BA.
Similarly, ang. COE = 20BC ;
.'. sum or diffce. ang. AOC = 2ABC.
Ex. Show by this method that the angle in a
semicircle is a right angle.
Theorem 70. — 'Two angles in a circle in the same arc or
segment are equal.' ^ — -^ a
If ABC, ADC are angs. in same arc or segment
of a circle, centre O,
ang. ABC = JAOC, on same arc AEC,
=:ADC.
CH. TV.] ANGLE IN CIRCLE OF TANGENT AND CHORD.
97
Theorem 71. — 'The angle of a tangent to a circle and a
chord through the point of contact is equal to the angle in
the opposite arc, or alternate segment.' *
If TAB is an angle of a tangt. AT and
clid. AB of a circle, and C, P are points in
the opposite arc ;
move P round the arc to A ;
then as P moves to coince. with A,
PA II II tangt. AT,
ang. BPA n m ang. BAT.
But ang. P always = C, same arc ;
.*. ang. TAB = ACB in opp. arc.
Theorem 72. — 'Angles in opposite arcs of a circle are sup-
plements ; or, opposite angles of a cyclic quadrilateral are
supplements.' X
If B, D are angs. in opp. arcs ABC, ADC
of a circle, draw tangt. SAT ;
.'. ang. TAC = B in opp. arc ;
similarly, ang. SAC = D m
.*. ang. B = TAC = suppt. of SAC
= suppt. of D.
Ex. Show this also by means of angles at centre.
Theorem 73. — ' If equal angles C, D stand on the same line
AB, on the same side, the points A, B, C, D are concyclic'
Draw circ. ABC, cutting BD in E ; ^
.'. ang. AEB = C, same arc, = ADB ;
.*. AE II AD and coincides with it ;
.*. point D coincides with E ;
i.e. D is on circle ACB.
c- "D-
Cor. — ' The locus of the vertex of a given angle, whose sides
traverse fixed points, is two equal arcs of circles.'
Theorem 74. — 'If angles C, D standing on the same line AB,
on opposite sides, are supplements. A, B, C, D are concyclic'
Draw circle ACB cutting BD in E ; prove as in Th. 73.
* Illustrate by angle of set-square at P, and pins at A, B,
P. G. G
98
RECTANGLE OP CHORDS AND SECANTS.
[CH. IV.
Theorem 75. — *A tangent from a point to a circle is the
mean proportional of the parts of any secant from the point ;
and a line from a point to a circle, mean proportional of the
parts of a secant from the point, is a tangent.'
(i.) If PT is a tangt., PAB a secant from point P to a circle ;
then ang. PTA = PBT, in opp. arc,
and ang. P is common to trs. PTA, PBT ;
.-. tr. PTA III PBT ;
.-. PA : PT = PT : PB ;
i.e. tangt. PT is mean propl. of PA, PB,
parts of secant PAB.
(ii.) If PS is mean propl. of PA, PB ;
then PA : PS - PS : PB,
and ang. P is common to trs. PSA, PBS ;
;. tr. PSA III PBS ;
.*. ang. PSA = PBS, in opp. arc ;
.'. SP coincides with the tangt. at S.
Note. Sq. on PT^rect. PA. PB = sq. on PS.
Theorem 76. — 'The rectangle of the two parts of any chord
or secant from a point to a circle is constant, and equal to the
square on the tangent from the point.' ^ p
If PCD is a fixed secant or did. from P to a
circle, PAB any chd. or secant whatever,
PT tangt. from P ; *
then in trs. PBC, PDA,
ang. P is common, ang. B = D, same arc ;
.-. tr. PBC III PDA ;
.-. PB : PD = PC : PA ;
.•. rect. PA . PB = PC . PD = const.
= PT^, if PAB moves to
coincidence with PT.
Ex. Ptolemy's Theorem.— The smn oi rectangles of
opposite sides of a cyclic quadrilateral is equal to the rectangle of
diagonals.
* When the point is outside.
CH. IV.]
CENTRES OF SIMILITUDE.
99
Definition 34. — The centres of similitude of two circles are the
two points dividing the join of centres in the ratio of the radii.
They are constructed (Constr. 22, Ch. III.) by joining ends of
parallel diameters of the circles.
If a secant through a centre of similitude of two circles cuts
them in points A, B and C, D, two nearest or two farthest points
A, C or B, D are corresponding points, and one nearest and one
farthest, A, D or B, C, inverse points. (See fig. below.)
Theorem 77. — 'The radii from corresponding points of a
secant througli a centre of similitude of two circles are
parallel.'
If SABCD is a secant from centre of simde. S
of circles, centres O, Q ; \
make QE pari, to OA, to meet SD in E ;
.-. QE : OA = QS : OS = QC : OA ;
.'. QE =: QC, and E is on circle CD ;
i.e. E coincides with C ;
.-. QC II OA ; similarly, QD || OB.
Cor. — *The common tangents of two circles
pass through the centres of similitude.'
For if SAD turns about S until the points
A, B coincide at T, the points C, D will also
coincide at U.
Theorem 78. — 'The rectangle of the parts of a secant of two
circles from a centre of similitude to two inverse points is
constant.'
In the above figure, if STU is a common tangent ;
then SB : SD = SO : SQ = ST : SU ;
.*. tr. SBT III SDU, and ang. SBT = SDU.
But ang. STA = SBT, opp. arc, '.• ST is tangt.;
.*. tr. STA III SDU ;
.-. SA : ST = SU : SD ;
.'. rect. SA . SD = ST . SU = const.
Note. TB I! UD, and we can prove in like manner that any two
correspond in jr chords of the circles are parallel. Thus the circles are
similarly situated with respect to their centres of similitude.
100
CHORD AND DISTANCE FROM CENTRE.
[CH. IV.
f y
'\P M
\
)
(T
^3^^?^
4
Theorem 79. — 'A greater chord of a circle is nearer the
centre, and subtends a greater angle at the centre, than a less
chord ; and equal chords are equidistant from the centre and
subtend equal angles at it.'
If AB, CD are chcls. of a circle, centre O,
and OM, ON bisect angs. AOB, COD,
and therefore also are right bisectors of
chds. AB, CD ;
turn fig. OCND round in the plane into
the position OELA.
Then if ang. AOB > COD, i.e. > AOE,
the half AOM> AOL;
.'. the compt. OAM<compt. OAL;
.*. OL passes first inside, and then
outside the triangle CAM,
and cuts AM in a point P ;
.'. OM<hyp. OP, in rt. tr. OPM,
<OL
<ON.
Similarly, AM > AP > AL ;
.*. the double AB>AE
>CD;
i.e. (i.) if ang. AOB > COD,
chd. AB>chd. CD, and dist. OM<dist. ON.
Similarly (ii.), if ang. AOB < COD,
chd. AB<chd. CD, and dist. OM>dist. ON.
Also (iii.), if ang. AOB = COD = AOE,
E coincides with B, and L with M ;
.-. chd. AB = AE = chd. CD, and dist. OM = OL = dist. ON.
Cor. — 'Equal chords from a point on a circle, on the same
side of the diameter through the point, coincide.'
Ex. 1. Show by congruent right triangles that equal chords are
equidistant from the centre ; and conversely, that chords equidistant
from the centre are equal.
Ex. 2. Find the greatest chord of a circle. "What angle does it
subtend at the centre ?
CH. IV.]
EQUAL CHORDS, ARCS, AND ANGLES.
101
Theorem 80. — * The arcs and chords in a circle of equal
angles at centre or circumference are equal ; the angles and
chords of equal arcs are equal ; the angles and arcs of equal
chords are equal.'
If AOB, COD are angles at the centre O of a circle,
AB, CD their chds. and arcs ;
turn sector COD about O in the plane so
that OC coincides with OA, P
and arc CD lies along AB.
Then (i.), if ang. AOB = ang. COD,
CD coincides with OB, and D with B ;
.'. arc AB = arc CD, and chd. AB = chd. CD.
Also, if ang. APB = CQD,
the double AOB = COD, &c., as before.
(ii.) If arc AB = arc CD,
D coincides with B, and CD with OB ;
.". ang. AOB = ang. COD, and chd. AB ^ chd. CD,
and half ang. APB = half ang. CQD.
(iii.) If chd. AB - chd. CD,
ang. AOB = ang. COD. (Th. 79.)
.*. also half ang. APB = half ang. CQD,
and arc AB = arc CD.
Cor. (i.). — 'Two sectors AOB, COD in a circle are congruent
which have equal angles, equal arcs, or equal chords.'
Cor. (ii.). — 'If angles in equal circles are equal, their arcs,
chords, and sectors are equal, and similarly for arcs, chords, or
sectors.'
Ex, 1. Show by congruent triangles that equal chords of a circle
subtend equal angles at the centre, and hence that they cut off equal
arcs.
Ex. 2. Show that AD, BC are parallel. Also that AB, DC intersect
on the line bisecting AD, BC.
Ex. 3. If AB and DC meet in E, show that EB = EC.
Ex. 4.* The bisector of an angle in a circle bisects the opposite
arc.
Ex. 5. The bisectors of angles in a given arc of a circle are con-
current.
* Very important.
102
NINE-POINT CIRCLE — SIMSON LINE.
[CH. IV.
Theorem 81. — ' The circle through the mid points of sides of
a triangle passes also through the feet of perpendiculars and
bisects the joins of the orthocentre to the vertices.' (N -circle,
or nine-point circle.)
If X, Y, Z are mid points of sides of a triangle ABC, H the
orthocentre, D, E, F feet of perps.,
K, L, M mid points of HA, HB, HC;
then XM bisects BC, HC,
and II BE ;
similarly, MK || AC;
.'. ang. XMK = BEA = a rt. ang. ;
.'. the circle on diam. XK
passes through M, and simly. ^' d X
through L, and it also passes through D ;
.*. the circle through K, L, M passes through X and D,
similarly through Y and E, and Z and F ;
i.e. the circle through X, Y, Z passes through D, E, F and K, L, M.
Note. The angle XYD in the arc on XD= B - C. (B>C.)
For ext. ang. CXY = int. opp. angs. XYD + YDC, in tr. XYD;
i.e. ang. B = XYD + C, since medn. YD^YC, in rt. tr. ADC;
.-. XYD=:B-C.
Ex. Find in the same way the values of angles YZE, ZXF.
and
Theorem 82. — 'The feet of perpendiculars to the sides of a
triangle from a point on its circumcircle are collinear,' (Simson
line.)
If PQ, PR, PS are perps. to the sides of ^/^
triangle ABC from P on circumcircle ;
PSAR and PRQC are cyclic quadls. ;
.*. ang. PRS = PAS, in same arc,
= SQppt. of PAB
= PCB (circumcircle)
= suppt. of PRQ, opp. arc ;
.*. SR, RQ are in a straight line.
Ex. If from a point P on the circumcircle of a triangle, lines PQ, PR,
PS to the sides make equal angles with these in order, then Q, R, S are
collinear.
CH. IV.] TANGENTS OP IN- AND E-CIRCLES — ANGLE IN CIRCLE. 103
Theorem 83. — ' The tangents from the vertices A, B, C of a
triangle to the incircle and the ecircle of angle A are s-a*
s - b, 8 - c, and s, s-c^ s- h, respectively.
If P, Q, R and P', Q', R' are
points of contact of the sides, then
(i.) the six tangts. of incle.
= sum of sides = 2s ;
.'. AR + BP + CP = AQ + BR + CQ
= s\
and BP + CP = a;
.*. AR = 6' - a = (similarly) AQ.
Similarly, BP = BR = s-6,
CQ = CP = 6'-c.
(ii.) AR' = AB + BR' = AB + BP',
AQ' = AC + CQ' = AC + CP';
.-. AR' + AQ' = AB + AC + BC = 26' ;
/. AR' = AQ' = s,
and BP' = BR' = AR'-AB = s-c,
CP' = CQ' = s-h.
Ex. 1. Show that the mid point of BC bisects PR'.
Ex. 2. Write down the lengtlis of tangents from A, B, C to the
ecircles in angles B, C.
Theorem 84. — 'Two angles at the centre of a circle, their
arcs, and the areas of their sectors are proportional.'
If AOB, COD are angles at the centre
O of a circle,
make a scale of arcs, unit AB, from O
along CD, divide decimally the unit con-
taining D, join points of divisn. to O.
These joins form a similar scale of
angles or sectors, unit AOB ;
and D, CD come between the same
divisions of the two scales ;
• ^"g- CQD _ ''^rc CD _ sector COD
arc AB sector AOB
*« stands for — ^ — ; i.e. semi-sum of sides.
104 STRAIGHT LINE A MINIMUM PATH. [CH. IV.
Theorem 85. — 'The straight line is the shortest path in a
plane between two points in the plane.' *
If ADB is a path from A to B not coinciding with str. line AB
— e.g. not passing through C ; -.
draw circles CD, CE, centres A, B. ^^^--^^r-<p
Then part of the path, DE, is outside /-- -^X,^^
the circles ; f ^ cl A
turn the parts AD, BE round A, B in the plane into the positions
AGO, BFC ;
.*. path AGCFB<ADB;
.'. not all paths are equal, and there must be some shortest path
or paths ;
and no path not coinciding with AB is the shortest ;
.*. the str. line AB is the shortest path from A to B.
Cor. — Any arc of a curve is greater than its chord.
Theorem 86. — ' The sum of the tangents from a point to a
circle is greater than their enclosed arc' (Legendre.)
If TA, TB are two tangents enclosing d
(with their chord) the arc ACB ; then if ^^^ ^"^X
ADB is any path enclosing ACB, and not / y\ \
coinciding with this arc, / /^ ^s^ \
draw tangt. ECF ; \/^ ^ ^^/^
.-. path AECFB<ADB. (Th. 85.) B A
Hence no path not coinciding with ACB is the shortest enclosing
path, and there must be some shortest ;
.'. arc ACB < any enclosing path < sum of tangts. TA, TB.
Theorem 87. — 'The circumference of a circle is greater than
the perimeter of any inpolygon, and less than that of any
circumpolygon.'
This follows at once from the previous theorems.
Ex. Show that circumfnce. of circle, rad. r, >6r, but<8r.
* Adapted from Fmncis Newman's Difficulties of Elementary Geometry.
t Because these are on opp. sides of the tangt. at C.
CH. IV.]
CONSTRUCTIONS.
105
Construction 23.
outside point.'
Tangents.
'Draw two tangents to a circle from an
If P is the point, O centre of circle ;
bisect OP in Q, draw arcs of circle A, B,
centre Q, diameter OP ;
join PA, PB.
Then PA, PB_LOA, OB respectively
(ang. in semcle.) ;
.*. PA, PB are tangts. from P to circle.
Construction 24. — * Draw common tangents to two circles.'
If O, Q are centres of the circles
DH, AG,
with centre Q of the larger circle,
draw two new circles BE, OF,
radii the dilFce. and sum of radii of given
circles.
Bisect OQ at M, and draw circle,
centre M, diameter OQ,
cutting the new circles in B, C, E, F.
Join QB, QC to meet the circle AG in
A, G;
make OD || QA, OH |I QC ;
then DA and HG are two common tangts.
For AB = and || OD ( •/ QB = QA - OD);
.-. DA II OBj_QA and OD ;
.'. DA, and similarly GH, is a common tangt.
In the same way two other common tangents are derived from
the points E, F.
Ex. Discuss the cases (i.) wlien the circles touch, (ii.) Avhen the circles
cut. v»
Note. The common tangents are more simply drawn as tangents from
the centres of simihtude to either circle as a theoretical construction ;
hut it is often inconvenient, as the outer centre of similitude may be
outside the limits of the paper. They may be easily drawn by eye.
106
TANGENT CIRCLES.
[CH. IV.
Construction 25. — 'Draw a circle to touch three given
straight lines ; or
Draw the incircle and ecircles of a triangle.'
(i.) The incircle.
If the three lines form a tr. ABC
bisect angles B, C by Bl, CI,
draw ID, IE, IF perp. to BC, CA, AB ;
then I is equidistant from BC, CA, AB.
Draw circle, centre I, radius ID ;
this touches BC, CA, AB at D, E, F.
(ii.) The ecircles.
If the ext. angles at B, C, &c.
are bisected by BE^, CEj, &c.,
circles with centres E^, E.2, Eg can
be drawn to touch BC and AB, AC
produced, &c.
These are the three ecircles of
the triangle ABC.
Thus there are four circles touch-
ing three straight lines.
Ex. Discuss the cases when (i.) two
lines are parallel, (iL) three lines are
parallel, (iii.) three lines are concurrent.
Construction 26. — ' Draw a circle to touch two given straight
lines and pass through a given point.'
If AB, AC are the lines, P the point,
draw AC bisecting ang. A ;
with any point Q in AG as centre,
draw a circle to touch AB, AC in D, F,
and to cut AP in E ;
draw PC pari, to EQ, and
draw circle PB, centre O, touching the lines.
By similar triangles, if OB_LAB,
OP : QE = AC : AQ = OB : QD ;
.-. OB = OP.
Ex. Show that there is a second circle.
CH. IV.]
TANGENT CIRCLBS.
107
Construction 27. — 'Draw a circle to touch a given circle and
two given straight lines.'
Draw parallels to the given lines at the distance of the given
radius. The centre of a circle touching these parallels and passing
through the centre of the given circle is centre of the required
circle. (Constr. 26.)
Construction 28. — 'Draw a circle through two given points
to touch a given line.' *
If A, B are the points, CD the line, \q
draw QD the rt. bisector of AB ; \\
from H, any point on it, as centre, \ ^v
draw circle GEF touching CD at G, \ ^^
and cutting AD at E, F.
Draw AC pari, to HE,
and ON perp. to CD.
Then ON : HG = OD : HD-OA : HE ;
.*. ON = OA = OB.
Draw circle, centre O, rad. ON or
OA touching CD at N, and passing
through A, B.
If AQ II HF, Q is another centre.
Ex. Can the problem always be solved*!
Construction 29. — 'Draw a circle through two given points
to touch a given circle.'
Draw a circle through given
A, B cutting the given circle,
O, in C, D ;
join CD to meet AB in P.
Construct S, T, points of contact
tangts. from P to given circle.
Draw circles ABT, ABS.
These touch the given circle at T, S.
For PT2 = PC.PD = PA.PB;
.*. PT is mean propl. of PA, PB, and
touches circle ABT ;
.'. circle ABT, and similarly ABS, touches given circle.
* This construction has been substituted for the traditional one because of its
analogy with Constr. 26.
points
centre
108
TANGENT CIRCLES.
CH. IV.
Construction 30. — 'Construct a circle to touch two circles
and pass through a given point.'
Construct a centre of simde. S of the given
circles, centres O, Q, draw a secant SAB
(A, B inverse points), find point R in SP
such that rect. SP.SR = SA.SB."^
Draw circle, centre G, through R, P
to touch circle QB in C. (Constr. 29.)
This also touches circle AD.
If SC cuts circle AD in E, D,
rect. SD.SC = SA.SB = SP.SR;
.*. D is on circle RPC ;
and OE |1 QC (Th. 77) ;
.'. ang. GDC = GCD = OED (alt. ang.) = ODE;
.". GDO is a str. line, and circle RPC touches ADE.f
Note. There is a second circle (Coristi-. 29) through RP touching circle
BC, and therefore also touching AD ; and two other circles through P
touching the two circles, derived from the other centre of similitude.
Construction 31. — ' Construct a circle to touch three circles.'
If O, Q, R are centres of circles X, Y, Z in order of magnitude,
diminish the radii of the greater X, Y by that of the least Z, draw
circles X', Y', centres O, Q, with the diminished radii.
The centre P of a circle through R touching X^ Y' is centre of
the required circle.
Construction 32. — ' Construct a circle to touch a given circle,
and a given line at a given point.'
If T is the point in line PT, O the
centre of given circle AB, make TQ perp.
to PT. Make OA pari, to TQ, join AT to
cut given circle in B, join OB to meet TQ
in Q. Then
ang. QTB = alt. ang. OAB = OBA = QBT;
.'. QB = QT, and circle BT, centre Q,
touches PT and circle AB.
Ex. Find the centre of the other circle.
* SR is 4th propl. of SP, SA, SB.
t The points of contact C, D are centres of simiUtude of circles G, Q and G, O,
and are collinear with a centre of simde. of O, Q.
CH. IV.]
REGULAR POLYGONS.
109
Construction 33. — 'Construct a regular polygon of n sides on
a given side.'
(Same construction as arc to contain
given angle.)
If AB is the side, draw the rt. bisector
DNO ; and make ang. DNC = the wth part
of tioo rt. angs. ; "'^
draw AO pari, to CN,
draw circle AB, centre O ;
.-. ang. AOB = 2 . AON = 2 . CND
= ?zth part of 4 rt. angs. ;
.'. arc AB = wth part of circle.
Step off the arc AB round the circle and
complete polygon as in figure.
Note, (i.) For a hexagon, make AOB an eql. triangle,
(ii.) For a pentagon, f construct an isosceles triangle AFB wliose
base AB is the mean part of sides FA, FB.
With A, B, F centres, radius AB, draw arcs meeting in G, H ; AGFHB
is the pentagon.
Construction 34. — (i.) 'Inscribe a regular polygon in a given
circle.'
Take any radius OA of the circle ;
make ang. AOB = the wth part oi four rt. angs. ;
.*. arc f^B — nih. part of circle ; complete as before.
Note, (i.) For a hexagon, chd. AB = AO.
(ii.) For a pen tagon,t construct on OA an isosceles triangle FOA of
which OA is the mean part of sides OF, FA ;
FO meets the circle in B. Complete as before.
(ii.) * Circumscribe a regular polygon to a given circle.*
Divide the circle into n parts as in last constr.
Draw tangents at the points of division.
Ex. Inscribe and circumscribe a regular octagon to a circle.
♦In the example n = 7, DNC=18077=26° nearly,
t If the exact geometrical construction is required.
110
CIRCUM- AND IN-CIRCLES INSCRIBED SQUARE. [CH.
IV.
Construction 35. — 'Draw the circumcircle and incircle of a
given regular polygon.'
Draw rt. bisectors of two consecutive
sides of the polygon, meeting in O ;
or draw bisectors of two consecutive
angles, meeting in the same point O.
O is the centre of both circles.
This is easily made evident by rotation
of the figure about O.
Construction 36. — ' Inscribe a square in any figure which is
symmetrical about a bisector of angle.'
(This includes a sector of circle, and any
regular polygon.)
If fig. ABCD is symmetrical about AP,
the bisector of ang. A, so that AB =.AD ;
set off DE perp. and eql. to BD.
Join AE to meet the figure in F ;
draw parls. to BD, DE, starting
from F, forming the square FGKH.
_ KH AH HF
For ^^ = v^=,^, and BD = DE;
BD AD
.'. KH = HF;
.*. FGKH is an
i.e. a square.
DE'
equal-sided rectangle-
MULTIPLICATION."^
This construction and Constrs. 26 and 28 are examples of the
general method of multiplication. We construct a figure similarly
situated to that required, satisfying all the required conditions but
one ; we then enlarge or reduce — i.e. multiply — to obtain the figure
satisfying this remaining condition (Ch. III. A (iii.), p. 83).
Thus, above, BD, DE are sides of a square satisfying all condi-
tions except having a vertex on CD, so that the required vertex of
the similarly situated square lies on AE ; we thus obtain F. Simi-
larly in Constrs. 26, 28, the circles DEF, EFG satisfy all the con-
ditions except passing through P, A respectively.
A few examples of the method are given on the next page.
* A very good account of this and other general methods of construction is
given in Pedersen's Mithodes et TMories.
CH. IV.]
MULTIPLICATION.
Ill
Ex. (i.). 'Inscribe in a given triangle ABC a triangle whose
sides have given directions.'
Make rq in one of the given directions,
and rp, qp in the other two directions.
Join Ap to P on BC ;
and make PR, PQ pari, to j^r, j)q-
Then AQ : Aq^AP : Ap = AR : Ar ;
.-. QR II qr, and PQR is reqd. triangle.
Ex. (ii.). 'Inscribe in a semicircle (with two vertices on the
diameter) a ctuadrilateral of given form (similar to a given
quadrilateral JW^)- '
p^r
Q ^ .0
Tin
ZQ
Draw rt. bisector of ps to O in qr ; then jyqrs is
inscribed in a semcle. , centre O.
Draw semcle. PS, centre O, diam. along qr, eql. to
given semcle.
From O multiply ^g-rs into PQRS ;
copy PQRS into A BCD inscribed in the given
semicircle AD.
EXAMPLES-XXV.
1. Inscribe in a triangle ABC an isosceles right triangle having its
hypotenuse parallel to BC.
2. Inscribe a square in a given segment of a circle. (Two vertices on
the base of the segment.)
3. Place a chord in a circle so as to be trisected by two radii whose
angle is 30°.
4. Place a chord in a circle in a given direction, and divided in a given
ratio by a fixed diameter.
5. Inscribe a triangle PQR in a given triangle ABC, having given
P in position on BC, the angle P, and direction of QR.
6. Draw a straight line through a given point P to cut two given
circles at A, B, so that PA : PB = ratio of corresponding radii.
7. Inscribe a parallelogram, sides 2:1, angle 60°, in a semicircle.
8. Draw a straight line in a given direction to divide the opposite
sides AB, DC of a quadrilateral in the same ratio.
9. Inscribe a square in a triangle. (Two vertices of the square on one
side of the triangle.)
10. Inscribe a square in a sector of a circle.
11. Inscribe a parallelogram in a segment of a circle, chd. 1", rad. f"
(two vertices on base), one angle 50°, ratio of sides 2 : 3.
We add an example of the method as applied to the proof of theorems.
112
MULTIPLICATION.
[CH. IV.
Ex. (iii.). 'The centroid, circumcentre, orthocentre, and nine-
point centre of a triangle are coUinear.'
If G, O, H, N are the above points in order in a triangle ABC ;
the median triangle XYZ can be derived from
ABC by multiplication from G by the ratio
GX:AG = 1:2.
Hence any point of tr. XYZ is collinear
through G with the corresponding point
of ABC.
But O, the circumcentre of ABC, is the
orthocentre of XYZ,
•.XO, YO±ZY, ZX;
.*. corresponding orthocentres H, O are collinear with G.
Similarly O, N, the circumcentres of ABC, XYZ, are collinear with G
i.e. G, O, H, N are collinear.
Ex. Find the ratios of parts of HO.
The following example does not illustrate multiplication, but is an
interesting property of the triangle, which we require for a property of
the parabola in Ch. VIII.
Ex. (iv.). 'The Simson line of a point on the circumcircle of a
triangle bisects the join of the point to the orthocentre of the
triangle.'*
If QR is Simson line of point P on cmcirc. of tr. ABC, and AHDK
the perp. to BC through orthocentre H ;
make RL eql. to PR in line PR.
Then ang. KBC=:KAC, same arc,
= compt. of C^HBD;
.-. KD = DH; alsoPR=RL;
.'. trapm. LHKP is symml. about RD ;
.-. ang. HLR-RPK = suppt. of PKH
C.PRIIKD)
= suppt. of PBQ, same arc,
= QRP, opp. arc;
.-. QR II HL, and bisects HP.
* Casey's Sequel to Euclid, Bk. III.
CH. IV.] EXAMPLES. 113
EXAMPLES— XXVI.
Theorems.
1. The locus of mid points of a system of parallel chords of a circle is a
diameter of the circle.
2. The joins of the ends of two parallel chords of a circle meet in P, Q.
Show that PQ passes through the centre. (Use symmetry.)
3. Perpendiculars on a chord of a circle from the ends of a diameter
cut off equal lengths from the line of the chord, measured from its two
ends.
4. The centre of any circle is the centre of the circle which touches
any three equal chords,
5. Straight lines from a point to a circle increase in length as the
angles they subtend at the centre increase from zero to two nght
angles.
6. Two circles do not meet at all if the join of their centres is greater
than the sum or less than the difference of their radii.
7. When do circles (i.) cut, (ii.) touch, (iii.) not meet at all?
8. The sum of two opposite sides of a quadrilateral circumscribing a
circle is equal to the sum of the other sides.
9. A parallelogram circumscribing a circle is a rhombus.
10. A tangent PQ of a circle, centre O, cuts two fixed tangents in PQ ;
show that the angle POQ is constant. What is its value when the fixed
tangents are parallel ?
11. The chord of contact of two tangents forms with them an isosceles
triangle.
12. The chord of contact of two parallel tangents is a diameter.
13. Each side of a rhombus circumscribing a circle subtends a right
angle at the centre.
14. The bisector of an angle in an arc of a circle bisects the opposite
arc.
15. If AP, AQ are diametei-s of two circles cutting in A, B, show that
BP, BQ are in a straight line.
16. The arcs between two parallel chords of a circle are equal.
17. A parallelogram inscribed in a circle is a rectangle.
18. The joins of ends of two equal arcs of a circle are either equal or
parallel.
19. The joins towards the same parts of the ends of two equal chords
of a circle complete a trapezium.
20. A circle cuts off equal chords from the two sides PA, PB of an
angle P. Show that the lengths PA, PB outside the circle are equal.
114 EXAMPLES. [CH. IV.
21. Chords BAC, DAE drawn through a common point A of two
circles, and equally inclined to the line of centres, are equal.
22. A triangle is equilateral whose incentre and circumcentre coincide.
23. Two arcs stand on a common chord AB. AP, BP meeting on one
arc cut the other in Q, R. Show that the chord QR has constant length.
24. The sides AB, DC of a quadrilateral in a circle meet at P, and
AD, BC at Q ; show that the bisectors of angles P, Q are perpendicular.
25. If two triangles have a side and opposite angle of one equal
respectively to a side and opposite angle of the other, their circumradii
are equal.
26. If DE parallel to the unequal side BC of an isosceles triangle ABC
cuts the sides in D, E, the i)oints D, E, C, B are concyclic.
27. If the ends of two chords AB, CD of a circle are cross joined, the
two triangles formed are similar.
28. The bisector of the angle PTA of a tangent PT and chord TA also
bisects the arc TA.
29. If the bisector of the angle ACB in an arc of a circle is parallel to
the tangent AP meeting BC produced in P, then AP = AB.
30. If two circles, centres O, Q, touch at T, and ATB is any cliord
through T, then AC, BQ are parallel, and arcs AT, BT are similar.*
31. If A, B, C are points in order in a straight line, similar aics on
AB, AC, on the same side of the line, touch at A ; and similar arcs
on AB, BC, on opposite sides, touch at B.
32. If a circle touches a straight line AB at P, and a circle through
AB touches the first circle in Q, the angles AQP, BQP are either equal
or supplementary.
33. The radius of the nine-point circle of a triangle is half that of the
circumciicle.
34. The internal common tangent of two circles which touch externally
bisects the other common tangents.
35. If PT, PS are tangents to a circle, centre O, and OT meets PS in
Q, show that QP.QS = QT.QO.
36. If a straight line cuts two intersecting circles in A, B and C, D,
and their common chord in P, then PA. PB = PC. PD.
37. The tangents to two intersecting circles from a point on their
common chord are equal.
38. Enunciate Ex. 37 when two circles touch.
39. The line of any chord AP through a fixed point A on a circle cuts in Q
a fixed line parallel to the tangent at A. Show that AP . AQ is constant.
40. A tangent PT of a circle at T is e^ual to the radius, and TA is
the diameter through T ; show that the circle cuts off four-iiftlis of PA.
* That is, their angles are equal.
CH. IV.] EXAMPLES. 115
41. CD is the bisector of angle C of a triangle whose angles B, C are
each double of A. Show that BC touches the circumcircle of ADC.
42. If I, E are incenbre and ecentre on bisector AP of angle A of a
triangle, AIRE is harmonically divided.
43. If I is the incentre of a triangle, and Al meets BC in P, show that
PI: PA=a:a + b + c,
44. The point of contact of two circles which touch is a centre of
similitude of the circles.
45. If AP is a bisector of angle of a triangle, A, P are centres of
similitude of the in circle and an ecircle.
46. The centroid of a triangle is a centre of similitude of the circum-
circle and nine-point circle.
47. A straight line through a vertex A of a triangle cuts the incircle
and ecircle in inverse points P, Q ; show that AP . AQ=s{s - a). (A is a
centre of simde. )
48. A tangent TA at T of a circle, centre O, is equal to a diameter
and is bisected at M. Show that MO is equal to the tangent from A to
circle MTO.
49. If AD, BE are perpendiculars of a triangle, orthocentre H, show
that CA.CE^CB.CD; also that DH. DA=DB. DC.
50. If two circles cut at right angles, the square of their join of centres
is the sum of squares of their radii.
51. If two circles are each touched by a third, the join of points of
contact passes through a centra of similitude of the two circles.
Can you connect this fact with the construction for drawing a circle to
touch two given circles ?
52. Two circles, centres O, Q, cut in A, B. OC, QD are parallel
radii of their respective circles in opposite directions, and CD meets OQ
in S. Show that
(i.) S is a centre of similitude ;
(ii.) the orthocentres of triangles SOC, SQD are collinear with S.
53. Show that in Ex. 52, S must be inside each circle, and hence that
two circles which cut in two points cannot have any internal common
tangents.
54. If the points of contact of the external common tangents of two
circles are collinear, two and two, with the internal centre of similitude,
show that the circles are equal.
What is the internal centre of similitude of two circles which
touch ?
116 EXAMPLES. [CH. IV.
EXAMPLES— XXVII.
Constructions.
1. Draw a circle, with given centre, to cut off a given length from a
given line.
2. Construct the locus of mid points of equal chords of a circle.
3. Draw a straight line through a fixed point so that a fixed circle
cuts off a given length from it.
4. Draw a circle, 2-2 cm. radius ; take a point P, 2-8 cm. from the
centre ; draw a circle, centre P, to pass through points of contact of
tangents from P to the first circle.
5. Draw a circle, 1-35" radius ; take a point 1-08" from the centre, and
construct the shortest chord through this point. Measure the chord.
6. Through a point A common to two circles draw a chord BAC
hisected at A. Can this be done if the circles touch ?
7. Draw a circle, radius 1-65", place in it a chord of 1-32" parallel to a
fixed direction.
8. Find the longest chord BAC through the common point A of two
circles.
9. Construct a square of 1-47" side, and circumscribe a circle.
10. Draw a circle, radius 1.73" ; cut off an arc whose chord is 1" and
bisect the arc.
11. Draw a circle, radius |", to touch two straight lines forming an
angle of 40°.
12. Draw a circle through the centre and two fixed points of a given
circle.
13. Inscribe a circle in a square of 2-37" side.
14. Inscribe a square in a circle of 5-6 cm. diameter, and measure the
side.
15. Construct a rectangle, sides 2-4 cm. and 3-2 cm, ; circumscribe a
circle, and measure the radius.
16. A ladder 30 ft. long is placed horizontally, perpendicular to and
just touching a wall. It is then raised with one end always touching
the wall. Construct the locus of its mid point (1" to 10 ft.).
17. Construct a point in a fixed line such that two fixed points off the
line subtend at it a given angle. How many solutions?
18. Draw a circle, radius 2-7 cm., and a chord 3-2 cm. Find a point
on the circumference equidistant from the ends of the chord.
19. Construct the locns of a point such that the sum of its greatest
and least distances from a fixed circle is constant.
CH. IV.] EXAMPLES. 117
20. Inscribe a circle in a sector of a circle, angle 60°, radius 2".
Measure the radius of the incircle.
21. Inscribe a regular heptagon in a circle, radius 2 cm.
22. Describe a regular hexagon about a circle, radius 1-8 cm.
23. Construct the locus of points whose tangents to a given circle have
a given length.
24. Draw two tangents to a circle to contain a given angle.
25. Draw a rhombus, angle 87°, to circumscribe a circle.
26. Draw a circle through the mid point of one side of a triangle to
touch the other two sides. How if the triangle is isosceles ?
27. Draw a semicircle, radius 2-3 cm. Inscribe in it a circle, radius
1 cm.
28. On two sides of an angle A take equal lengths BC, PQ, 1^" long,
h" and I" from A respectively. Verify that the circles APC, ABQ and
the right bisectors of CP, BQ are concurrent.
29. Construct a regular pentagon, not using the protractor, on a side
of 1 inch. Inscribe a circle.
30. Find in a given straight line a point whose tangent to a given
circle has a given length.
31. Given any two points A, B, construct the length of the tangent
from a point P in AB produced to any circle through A, B.
32. Draw a circle passing through the mid point of one side of a
triangle, the foot of the perpendicular of another side, and touching the
third side.
33. Find a point P in a given line AB such that the rectangle
PA . PB = «^, where « is a given length,
34. Construct the locus of a point whose distances from two fixed
points 1^" apart are in the ratio 3 : 2.
35. Draw a circle, radius 2-5 cm., cut off an arc whose chord is 3-1 cm.,
and divide the arc into two parts whose chords have the ratio 3 : 4.
36. Draw an isosceles triangle, angle of equal sides 50°, third side IJ".
Draw two equal circles each to touch the other, one of the equal sides,
and the third side of the triangle,
37. Inscribe in an equilateral triangle three equal circles to touch each
other,
38. Construct three equal circles to touch each other and to touch
internally a circle of 2-8 cm. radius.
39. Draw three circles, with the points A, B, C of a triangle as centres,
to touch each other.
40. Draw a sector of radius If", angle 120°. Inscribe a square in it.
41. Inscribe in a square of 1-63" side a parallelogram of angle 78°,
having one vertex ^" from a vertex of tlie square.
118 EXAMPLES. [CH. IV.
42. Construct a cyclic quadrilateral, ^dven diagonals AD and BC, the
angle A, and the ratio AB : AC.
43. Construct a quadrilateral, given diagonals AC = li", BD = 1", their
angle 108° ; A =60°, = 71°. (Translate tr. BAD from A to C.)
44. Construct a quadrilateral, given the sides AB, BC, CD and the
angles CAD, CBD.
45. Circumscribe a rhombus of 7-2 cm. side about a circle of 2-7 cm.
radius.
46. Construct a point P in the side BC of a triangle such that the
square on PA is equal to the rectangle PB . PC.
47. If P is any point in the side BC produced of a triangle, construct a
transversal PDE such that rect. PD. PE = PB. PC.
48. Describe an octagon on a side of 2-3 cm. Give a general con-
struction for an inscribed square Avith a vertex at any point on a side of
the octagon.
49. Construct a regular pentagon, diagonal 5-7 cm. (The side is the
mean part of the diagonal. )
50. Given the angle A of a triangle, construct the locus of the mid
point of BA when vertices B, C are fixed.
51. The parallel sides BA, CD of a trapezium are given in magnitude,
and B, C are fixed points. Find the locus of intersection of bisectors of
angles B, C, also of the mid point of AD.
52. Draw the nine-point circle of a triangle whose sides are 4 cm.,
5 cm., 6 cm. Test it at the feet of perpendiculars.
53. Draw a circle, radius 1". Inscribe a triangle, two sides 1|", If".
Construct its orthocentre H ; measure the parts into which the Simson
line to the triangle of any point P on the circle divides PH.
54. Construct the absolutely shortest path from a given point to a
given circle. Indicate the proof. Assuming that maxima and minima
occur alternately, what is the longest path ?
55. Draw a circle to touch two lines at an angle of 60°, and a circle of
^" radius, whose centre is distant f" and 1|" from these lines.
56. Draw a triangle, a =4= cm., 6 = 6 cm., c = 7 cm. With centre C,
rad. 3 cm., draw a circle. Draw a circle through A, B to touch this
circle.
57. Construct a mean base triangle (Constr. 16, Ch. III.), and in it
inscribe an isosceles right triangle, whose hypotenuse is parallel to an
equal side of the first triangle.
58. Draw a sector whose angle is 60°, radius 1". Inscribe in the sector
a triangle whose sides are as 2 : 3 : 4, the greatest side parallel to chord
of sector.
59. Inscribe in a semicircle, 3 cm. radius, a parallelogram of angle 70°
and sides 1 : 3 (two vertices on arc, two on diam.).
CH. IV.]
EXPERIMENTAL SOLID GEOMETRY.
119
c c
b i c \~~b I c
I I
I I <
a \ \a a\ a
b \ c \ b I c
c c
Solid Figures — Volume — Cube — Cuboid.
Figures whose points, lines, surfaces are not all in one plane
are solid figures. When closed they include a certain amount of
space called their volume or cubic content, the unit being the
cube whose edge is unit length. (Cubic inch, &c.)
Draw and cut out the adjoining
figure of rectangles (thick paper or
thin board). ^ Fold along the dotted
lines to form the second figure ; fasten
the edges with gummed paper. This
makes a rectangular solid or cuboid.
The lines a, &, c, sides of the rect-
angles, are called the edges, and the
rectangles the faces of the cuboid.
Make a second figure, in squares instead of
rectangles. This makes a cube, and a side of
a square is the edge of the cube. Thus,
A cube is a solid formed by six equal
squares.
A cuboid is a solid formed by six rect-
angles.
Volume of a Cuboid.
If the edges of the cuboid are divided into unit lengths, 5, 4, 3
units, say;
then we can make 5 cuboids, edges 1, 4, 3 ;
and each of the last makes 4 cuboids, edges 1, 1, 3 ;
and each of the last makes 3 cuboids, edges 1, 1, 1 — i.e. 3 unit
cubes.
Thus we have 5x4x3 unit cubes. Hence,
* The volume of a cuboid is the product of its three edges, or
of its height, length, and breadth.'!
^
Ex. 1. Find the volume of a box 30" by 18" by 15" ; and of a room
25 ft. by 20 by 12.
Ex. 2. Find the volumes of cubes of edge 7", 13 ft., 2-4 cm.
* A post-card serves well.
t The complete proof is given below, Ch. IX.
120
EXPERIMENTAL SOLID GEOMETRY.
[CH.
IV.
Wedge.
Make and cut out the adjoining
figure of three rectangles and two
triangles, Avhose sides are those of
the rects., fold along dotted lines, and
fasten the edges, making a wedge,
A (right) wedge is a solid formed
by three rectangles and two triangles.
The volume of a wedge is half that of a
cuboid whose rectangular base is twice the
triangular base of the wedge, and which has
the same altitude ; thus,
triangular base x alti-
* Volume of wedge
tude.'
Cylinder.
Cut out a rectangle and fold it
round"*^ so that two opposite sides coin-
cide and the others form equal circles.
This with the circular ends is a
(right) cylinder. The join of the
centres of the ends is its axis, and
either end is the base.
Area of a right cylinder, radius of base r, altitude h.
The area of each base is irr'^.
Area of curved surface is the rectangle, altitude
X length of base circle {h x ^ttv).
' Area of curved surface of cylinder = height x cir-
cumference of base circle.'
^::^
^kl^
Volume of a Cylinder.
If planes through the axis cut out small sectors from the ends,
they cut out practically wedges from the cylinder, whose volume
is height x area of sector, f Thus,
* The volume of a cylinder = height x area of base.' f
Ex. 1. Calculate volume of wedge, alt. 2-3 cm., base area 5 sq. cm.
Ex. 2. Calculate area and volume of cylinder, alt. 6 cm. , rad. 1 -2 cm.
* It can be first folded round a bar or tube, and cut as a rectangle afterwards.
t This is strictly true, though the proof is not complete.
CH. IV.]
EXPERIMENTAL SOLID GEOMETRY.
121
Tetrahedron — Pyramid
Cut out the adjoiuing figure of four
triangles, adjacent sides of any two outer
triangles being equal, fold along dotted
lines, and fasten the edges, making a tetra-
hedron.
A tetrahedron is a solid formed by four
triangles.
If instead of the triangle ahc we take a
polygon, and make triangles on its sides as
before, the solid figure is a pyramid.
A pyramid is a solid formed by a polygon
and triangles based on its sides, and having
a common vertex.
* The volume of a tetrahedron or pjnramid
is a third of the product, height x base.' *
Cone.
Cut out a sector of a circle, and fold round
so that the sides coincide, and the arc forms a
circle. This with the plane circular end or base
is a (right) cone.
The centre of the sector is the vertex,
the join of vertex to centre of base the
axis and altitude or height ; and the
join of vertex to any point on the base
circle is the slant height of the cone.
The area of the curved surface of the cone is
that of the sector which forms it. Thus,
'The area of curved surface of a cone is half the product,
slant height x length of base circle.'
' The volume of a cone (like that of a pyramid) is a third of
product, height x base.' *
Ex. 1. Calculate the volume of a pyramid, height 5", base a square of
2.7" side.
Ex. 2. Calculate surface and volume of a cone, height 4 cm., slant
height 5 cm., radius of base 3 cm. How could you calculate the slant
height ?
* The proof is given below, Ch. IX.
122 EXPERIMENTAL SOLID GEOMETRY. [cH. IV.
Sphere.
A sphere is a closed surface whose points
are all equidistant from a fixed point, its
centre. The distance from centre to surface
is the radius, and from surface to surface
through the centre the diameter of the sphere.
A billiard ball is a good example.
Any plane cutting the sphere cuts it in a circle — e.g. a line of
latitude on a globe of the world.
Any plane through the centre cuts the sphere in a great circle
— e.g. line of longitude.
Great circles can be moved about on a sphere just as straight
lines on a plane ; and one only can be drawn between two points,
unless these are opposite ends of a diameter of the sphere.
'The area of a sphere is four times the area of a great
circle ; ' "' i.e. area = ^irr'^.
Volume op a Sphere.
If the sphere is divided into small pyramids
by planes through the centre, the volume of a
pyramid is J area of base x altitude (radius).
And the volume of the sphere — i.e. the sum of
the pyramids — is that of a pyramid whose base
is the surface of the sphere, and altitude the
radius ; thus —
' The volume of a sphere is one-third the product of radius
47r7
and surface ; ' i.e. vol. of sphere =
Ex. 1. Calculate the surface and volume of spheres of radii 2 cm.,
1-6", 5 ft.
Ex. 2. Taking a degree as 69-12 miles, calculate the earth's circum-
ference (length of a great circle).
Ex. 3. Taking the earth's radius as 3960 miles, calculate its circum-
ference, and its area.
* The proof is given below, Ch. IX.
CH. IV.] EXAMPLES. 123
EXAMPLES— XXYIII.
1. Calculate the volume of a room 24 ft. long by 18 ft. wide and 12 ft.
high.
2. Calculate the area of a box 33" by 16" by 18".
3. Calculate the volume of a pile of books 100 deep, each book 8^" by
6" by li".
4. How much water is required to fill a stretch of canal 3 miles long,
6 ft. deep, 24 ft. wide ? Answer in cubic yards.
5. A wedge has a triangular base of sides 3", 4", 5", and an altitude
of 6". Calculate its volume.
6. A cuboid block of wood 2 ft. by 1 ft. by 3" is sawn across the
diagonals of its greatest faces so as to form four wedges. Calculate the
volume of each.
7. How many cubic inches of milk does a can 1 ft. high, 1 ft. diameter,
hold?
8. If 1 gallon contains 277 cubic inches, how many gallons does the
can of No. 7 hold ?
9. A bullet tied to one end of a string is swung round in a circle,
the otiier end of the string being fixed ; what surface does the string
generate ? If the diameter of the circle is 10", and length of string 21",
what is the area of the curved surface ?
10. Taking the height of the Great Pyramid as 480 ft., and the side of
the square base as 762 ft., calculate the volume in cubic yards.
11. A conical funnel has a diameter of 4" and a depth of 6". AVhat
volume of water will just fill it?
12. A conical church spire is 16 yd. high, on a base of 8 yd. diameter.
Calculate its volume, and the area of its curved surface.
13. A sphere can just be got into a cylindrical vessel 2 ft. high, 2 ft.
diameter. Calculate the volume of the sphere.
14. What volume of water would just fill the cylinder in Ex. 13 with
the spherCj supposed solid, inside ?
15. The area of the curved surface of a cylinder is 94-2 sq. in. If the
height is 10", what is the diameter of the base ?
16. How many gallons of water does a cask hold, 6 ft. long, average
diameter 3 ft. ? (1 cub. ft. =6-25 gallons.)
17. Show that 1" of rainfall is equivalent to 101 tons per acre. (1 cub.
ft. of water weighs 1000 oz.)
18. Calculate the area of a triangular face of the Great Pyramid of
Ex. 10.
124
CHAPTER V.
RECTANGLES BY ALGEBRAIC FORM— MEASURE AND
RATIO OF AREAS— RECTANGLE PROPERTIES OF
THE TRIANGLE— MISCELLANEOUS.
Rectangles by Algebraic Form.
The area of a rectangle being determined by its sides X, Y may
be written XY as an algebraic product. The area of the square on
side X may be similarly written XX or X^.
Theorem 88.— (i.) 'The rectangle whose sides are a line X,
and the sum of two lines Y + Z, is equal to the sum of rect-
angles XY + xz;
If AB is X, and AC, CD in a perp. line
are Y, Z, complete rects. AE, AF.
Then X(Y + Z) = rect. AE = AF + CE
= XY + XZ (i.).
C Z D
(ii.) 'The rectangle whose sides are a line X, and the diflfer-
ence of two lines Y - Z, is equal to the difference of rectangles
XY-XZ.'
If AB is X, and AC, CD in a perp. line
are Y, Z, complete rects. AE, AF.
Then X(Y - Z) = rect. AE = AF - CE
= XY-XZ (ii.).
We thus derive a very important principle :
'Rectangles whose sides are the sum or difference of lines
can be treated by the rules of algebraic products.'
For the two forms (i.) and (ii.) are the fundamental forms of
products x{y + z) and x{y - z) in algebra.
A few examples of the method are given. With due care as to
sign ( + and - ) results are readily obtained.
CH. v.] RECTANGLE BY ALGEBRAIC FORM. 125
Theorem 89.—' If X and Y are any two straight lines '—
(i.) (X + Y)2 = X2 + Y2 + 2XY; ---^^'"'^ ^ ^
(ii.) (X-Yf = X2 + Y^-2XY; ^^-^-^ x~^
(iii.) X2-Y2 = (X + Y)(X-Y); ^
(iv.) (X + Y)2 + (X-Y)-^ = 2X2 + 2Y2. -" "^ J"^
These follow exactly as in algebra. The fol- ^y-^^y-^
lowing diagrams illustrate them geometrically.
XY
XY
(X-Y)_
XY
2 XY
2X2 i2Y2
(X+Y)!
X2-hY2
X2_Y2
(X+Y)2 + (X-Y)
In the third and fourth figures the rectangle P must be moved
into the position of the congruent rectangle P'.
Theorem 90. — 'If a straight line AB is bisected at M,
and P is any other point in the line, PA^- PB2 = 2 . MP. AB.'
(PA>PB.)
By Th. 89 (iii.) PA^ - PB^ = (AP + PB)(AP - PB).
But when P is between A and B,
(AP + PB) = AB,
(AP - PB) = MP + MB - PB = 2MP ; ^ "^ f °
.-. PA2-PB2 = 2MP.AB.
Similarly, when P is outside A and B, ^ » •
(AP + BP) = 2MP, AP - BP = AB ;
• PA2-PB2 = 2MP. AB.
Cor. — 'There is one only point P in a straight line the
difference of whose squares of distances from two fixed points
in the line in a given order has a given area.'
For if the given area S = rect. X . AB, then X is a fixed length ;
and 2MP. AB = PA2-PB2 = S = X. AB;
.*. MP = X/2, a fixed length, and P is a fixed point.
126 SQUARES ON SIDES OF TRIANGLE RECTANGLE OF SECANT. [CH. V.
Theorem 91. — ' The sum of squares on two sides of a triangle
is twice the sum of squares on half the third side and its
median.'
If AD is the median of BC, in triangle ABC,
make AN perp. to BC ;
then AB2 + AC^ = BN2 + AN^ + CN^ + AN^
= BN^ + CN2 + 2AN'^;
but BN2 + CN2 = (BD-ND)2 + (DC + ND)2
= (BD - ND)2 + (BD + ND)^
•/ BD = DC,
= 2BD-^ + 2ND2 (Xh. 89, iv.);
.-. AB- + AC" = 2BD2 + 2ND2 + 2AN2
-2BD2 + 2AD2.
Ex. Prove the theorem when the angle B is obtuse,
theorem become when B is a right angle ?
What does the
Theorem 92. — ' The square on one side of a triangle is greater
or less than the sum of squares on the other two sides by twice
the rectangle of either of these sides and the projection on it of
the other, according as the angle opposite the first side is obtuse
or acute.'
This can be proved in a similar manner to the last theorem.
It is interesting only as completing Pythagoras' theorem.
Theorem 93. — 'The rectangle PA.PB
of the parts of any chord or secant of a
circle, centre O, is equal to the difference
of squares OP^ - OA^.'
Draw NO, the rt. bisector of AB ;
then, if P is external,
PA . PB = (NP - NA)(NP + BN)
= (NP-NA)(NP + NA)
= NP2-NA2 (Th. 89, iii.)
= OP"^ - OA^ (by Pythagoras' theorem).
Similarly, if P is internal,
AP . PB = OA2 - OP2.
CH. v.]
RADICAL AXIS.
127
Definition 35. — The radical axis of two circles is the locus of
points whose tangents to the circle are equal.
A coaxial system of circles is one of which all pairs have the
same radical axis.
Theorem 94. — ' The radical axis of two circles is a straight
line perpendicular to the line of centres.'
If TP, SP are equal tangents to
two circles, centres O, Q, and TO >^QS,
then P is a point on the radical axis.
Make PN perp. to OQ. Then
N02 - NQ2 = (P02 - PN2) - (PQ2 - PN2)
= P02-PQ2
= (PT2 + OT2) - (PS2 + QS2)
= 0T2 - QS2 = constant ;
.-. (Th. 90, Cor.) N is a fixed point,
and NP a fixed line.
It is readily seen that every point outside the circles on the line
NP has equal tangents to the two circles.'^
Note, (i.) If the circles cut in A, B, the line AB is the radical axis.
For if RK, RL are tangents from any point R in AB,
RK2=RA.RB = RL2; .-. RK = RL.
Note, (ii.) If two circles touch, the radical axis is the tangent at their
point of contact.
Theorem 95. — ' The three radical axes of three circles meet
in a point.' (The radical centre of the
circles.)
If the radical axes of circles A, B and
A, C meet in V, and VR, VS, VT are
tangents to A, B, C ;
then VT = VR = VS;
.'. V is on the radical axis of B, C ;
i.e. the three radical axes are concurrent.
Note. If V is inside the circles, prove
hy the rectangle property of chords.
Cor. — * If three circles touch each other, the common tangents
at their points of contact are concurrent.'
If a point on the line is inside one circle, it is also inside the other.
128
RATIO OP AREAS.
[CH.
Theorem 96. — ' The areas of two rectangles of equal altitudes
are proportional to their bases.'
If rects. ABC, ABD are placed with
their eql. alts, on AB, and their bases
BC, BD along BD ;
make a scale, unit BC, along BD,
divide decimally the unit containing D,
draw parls. to AB through points of division.
These form a scale of rectangles, unit AC ;
and point D and side DE come between the same divisions of the
two scales ;
rect. AD base BD
rect. AC base BC
Theorem 97. — ' The areas of two triangles or parallelograms
of equal altitudes are proportional to their bases.' ^
If ABC, DEF are triangles or parms. of
eql. alts. AM, DN ;
A ABC _ J rect. AM . BC _ base BC
rect. DN . EF "base EF '
rect. AM . BC base BC
then
and
A DEF
QAC
ODF
rect. DN . EF base EF
N E F
Theorem 98. — 'The ratio of two rectangles, or of two
parallelograms or triangles of given angle, is the product of
ratios of the sides of the given angle of the first to those
of the second.'
Set the two rects., parms., or trs. AB, CD with
the given angle at O, and the sides of this angle
along OA, OB, and complete as in fig.
Then if OA = /xOC, and OB = vOD,
£7 -^AB = juCB, and O CB = vCD ; o'
.'. OAB = /xvCD;
i.e. O AB : O CD = /^v = product of ratios of sides.
Cor. — * The ratio of two squares is the square of the ratio of
their sides.'
*OrAAOB, &c.
CH. v.]
MEASURE AND RATIO OF AREAS.
129
The problem of measuring an area is to find how much of unit
square is required to make up the area — i.e. to find the ratio of the
given area to unit area.
From Theorem 98, if AB is any rectangle and CD unit square,
/x, V are the measures of the sides, and [xv the measure of the
rectangle ; hence,
' The measure of a rectangle is the product of measures of its
sides, and of a square the square of measure of its side.' *
Other areas may be measured by measuring the equivalent
rectangle. Thus :
* The measure of a parallelogram is the product, and of a
triangle half the product, base x altitude.'
Theorem 99. — 'The areas of two similar triangles or two
similar polygons are proportional to the squares of corre-
sponding sides.'
(i.) If tr. ABC III DEF,
they have a given ang. B = E ;
AABC AB BC /AB\2
• • A DEF ~ DE "" EF ~ [deJ '
i.e. A ABC : A DEF = AB2: DE^.
(ii.) If the similar polns. AB..., FG...
are placed to be similarly situated about O,
so that FG II AB, &c. ;
AOAB 0B2_ AOBC
A OFG ~ OG^
A,
then
AOGH
. . ., . . A OCD .
= (Bimdarly)^^^ = &c.
sum of As CAB + OBC + &c. , , . ,
^ sumofAsOFG + OGH + &c. (^^^^^^^^o^) ^
.-. area ABCDE : area FGHKL = A CAB : A OFG
= AB^':FG2.
Cor. — 'Corresponding sides of similar polygons ara propor-
tional to the square roots of their areas.'
This is important for constructing similar figures of given areas.
Thus if areas of similar pentagons are as 3:1, the sides are as
* The proof of this suggested in Ch. I. applies to rational numbers only.
P.O. I
130
INRADIUS — ERADIUS — AREA OF TRIANGLE.
[CH. V.
Theorem 100. — 'If r, r^ are radii of incircle and ecircle in
angle A of a triangle, centres I, E^, then
„. A .. A
s-a
rra = {s-b)(s-c),
A = Js{s - a){s - h){s - c), Al . AE^ = be'
If P, Q, R and P', Q', R' are points of contact of sides ;
then —
(i.) A = AIB + BIC + CIA
= J(?'c + ?'a + rb) = 7's ;
A
.*. r =
s
(ii.)
EiR'll
IR;
' ' r
AR'
AR
s
s-a'
:. ra =
rs
s — a
A
s-a
(iii.) The bisector E^B^IB;
.-. ang. EiBP' = compt. of IBP = BIP;
E P' BP'
.-. triangle E^BP' |||BIP, and -^= -,p-;
.-. rva = BP . BP' = {s- b)(s - c).
(iv.) A- = ^^ • ra{s -a) = s{s - a){s - b)(s - c) by (i.), (ii.), (iii-) ;
i.e. A= Js{s - a){s - b)(s — c).
(v.) Ang. IBEi = rt. ang. = ICEj,
.'. quadl. IBE^C is cyclic;
.*. ang. IEjC = IBC (same arc) = IBA,
and ang E^AC = BAI, '.• E^A bisects ang. A ;
/. tr. E,AC|||BAI,and^ = ^;
.-. AI.AEi = AB.AC = &c.
Ex. 1. Write down corresponding forms for rj, Vc.
Ex. 2. Calculate the radii r, ?-«, n, Vc, and area of a triangle of sides
4 cm., 5 cm., 7 cm. Verify by drawing and measuring.
CH. v.]
CIRCUJIRADIUS VALUE OF OI^.
131
Theorem 101. — 'If R is the circumradius of a triangle,
„ abc ,
If NO is the rt. bisector of BC,
O the circumcentre,
make BE perp. to AC ;
then ang. BON = J BOG = A (at circumf.) ;
.-. rt. tr. BON ||| BAE ;
OB_BN_ BC. AC _ a&
•• AB~BE~2BE.AC~4A'
„ abc
■■ '' = iA-
Ex. Calculate R for the triangle of sides 4 cm., 5 cm., 7 cm., and
verify by drawing.
Theorem 102. — 'If O, I are circumcentre and incentre of a
triangle, 0|2=R2-2Rr.'
If the incircle touches AB at N, and
Al bisects arc BC of circumcircle at M,
and MO meets this circle in D ;
then ang. MIB = IBA + lAB, int. opp. angs.,
= IBC + IAC
= IBC.+ MBC, same arc,
= MBI;
/. MB = MI.
Also, ang. IAN = BDM, same arc ;
.'. rt. tr. IAN ||| MDB ;
" MD~MB~Mi'
.*. IA.Mi = r. MD = 2R7\
But lA . Ml = R2 - 0|2, in circumcircle (Th. 93) ;
.*. R^-0|2 = 2Rr;
i.e. 0|2 = R2-2R?\
Ex. 1. Prove similarly, if Ej is ecentre in angle A,
OEi2=R2 + 2Rr„.
Ex. 2. Measure Ol for the triangle of sides 4 cm., 5 cm., 7 cm,, and
compare with the value calculated from Exx. of Thh. 100, 101.
132 LENGTH AND AREA OF A CIRCLE. [CH. V.
Theorem 103. — (i.) * A circle is the limit of a regular in- or
circum-polygon when the number of sides is infinite.'
By definition of tangent (Cli. IV., Def. 28), if one angular point
of a regular polygon moves up to coincidence with its neighbour,
the number n of its sides increasing infinitely, the inpolygon coin-
cides with the circumpolygon, and therefore also with the circle
which always coupes between them.
We derive two important consequences :
/.. \ ,m-L ^. circumference _ . , / x . ^ ^ ,
(u.) ' The ratio — — t of a circle (tt) is constant.
diameter
If AB, A'B' are sides of reg. inpolns. of n
sides in circles, centres O, O', diameters D, D',
then trs. OAB, O'A'B' are similar ;
perimeter of poln. AB ?i , AB n. A'B'
A'
2. OB 2.0'B'
_ perimeter of poln. A'B'
- ^,
And each circle is the limit of its inpolygon when the number
of sides is infinite ;
circumf. of AB circumf. of A'B' , ,
.'. ^r^ = —, = constant.
D D
Value of tt. — The perimeters of in- and circum-poln. of 16384
sides of diameter 1" are 3.1415925"... and 3-1415927"...,
and TT (nearly half-way between these numbers) = 3-1415926...
(iii.) ' The area of a circle is half the product of circumference
and radius.' (Area = irr'^.)
The area of the reg. circumpoln. of n sides AB is \ ^
?^ A0AB=:1 71. AB.ON
= 1^ perimeter x rad. of circle ;
.*. area of circle = limit of circumpoln. _
= J circumference x radius. /
^^i
Ex. Calculate the circumference and area of a circle of radius 2-9 cm.
Also the arcs and areas of sectors of 32°, 72°, 225°, 304° of the same
circle.
Some examples are given at the end of Ch. I.
CH. V
MENELAUS' AND CEVa's THEOREMS.
133
Theorem 104. — 'A transversal cutting the sides a, &, c of
AR BP CO
a triangle in P, Q, R makes ^^- ;=^- ^= +'^''* (Menelaus'
BK OP Ay
theorem.)
Draw BD pari, to AC.
Then by similar triangles,
AR AQ BP BD
BR~BD' CP~CQ'
. AR BP CQ AQ BD CQ
* • BR ■ CP ■ AQ ~ BD ' CQ ■ AQ
= +1.
Conversely, 'If P, Q, R satisfy this condition, and PR meets
AC in Q', we can show CQ' : AQ' = CQ : AQ ;
.*. Q coincides with Q', and hence P, Q, R are collinear.'
Theorem 105. — 'Three concurrent lines OA, OB, OC from the
vertices of a triangle cutting the opposite sides in P, Q, R make
AR BP CQ ^
BR"CP"AQ~
(Ceva's theorem.)
Draw DBE pari, to AC. Then
the parls, DBE, CQA are similarly divided ;
.*. — ^ = — ; and by similar triangles,
AR AC BPBE
BR ~ BD' CP ~ CA '
. AR BP CQ AC BE DB
••BR'CPAQ~BDCAEB
- -1.
Conversely, 'If P, Q, R satisfy this condition, and AP, BQ',
CR are concurrent, we can show CQ' : AQ' = CQ : AQ ;
.'. Q coincides with Q', and AP, BQ, CR are concurrent.'
Note. These are simple tests to apply for the coUinearity of three
points on the sides, or the concurrency of three rays through the
vertices, of a triangle.
Ex. Practise these on the ' Solution of Problems ' section of Cli. III.
* The ratio AR : BR is positive if the directions from A and B to R are the
same, and negative if these directions are opposite.
134
CONSTRUCTIONS.
[CH. V,
\
>E
Square op Mean Section — Eadical Axis — Polygon of
Given Area and Form.
Construction 37. — 'Divide a straight line so that the rect-
angle of the whole and one part is equal to the square on the
other part.' *
If AB is the line, make BC half AB
and perp. to it,
draw circle DBE, centre C, to cut AC
in D, E ; make AF eql. to AD ;
then AB touches circ. DBE, and DE = AB;
.-. AF2 = AD2 = AD(AE-DE)
= AD.AE-AD.DE = AB2-AF,AB
= AB.FB.
Hence AB is divided as required at F.
Construction 38. — * Construct the radical axis of two circles.'
If the circles cut or touch, the common
chord or tangent is the radical axis.
If the circles do not meet :
Draw a third circle to cut the given circles,
centres O, Q, in A, B and C, D.
Draw AB, CD to meet in P, and PR, the
radical axis, perp. to OQ.
If PT, PS are tangents to the circles,
PT2 = PA . PB = PC . PD = PS2 ;
.'. P is on radical axis, which is therefore PR.
Construction 39. — * Construct a polygon similar to a polygon
P and equal in area to another Q.'
On side AB of P make rect. BC eql. to P,
on side DE eql. to AC make rect. EF
eql. to Q,
make BG in AB produced eql. to DF,
draw semcle. AHG, make BK eql. to BH.
Construct R on BK similar to P ;
.-. R : P = BK2 : BA2 = BH2 : BA^ = BA . BG : BA^
= BG : BA = DF : BA = Q : P ;
.-. R = Q, and R ||| P.
* Alternatively, construct the mean part BG of AB ;
.-. BG2=AG.AB.
-nG
CH. v.] DIVISION OF TRIANGLES. 135
Construction 40. — 'Cut off the iith. part of a triangle by a
straight line through a given point in one side.'
If P is the point in side AC of tr. ABC,
make CD pari, to PB ; A
.-. Apbd = pbc, ^
and A PAD = ABC.
Make AE the nth part of AD ;
• APAE = ^''^ = ^^55. B. ^\P
n n /
Note. This construction can be extended D'^ "*C
to divide a triangle or polygon into n equal
parts.
Construction 41. — (i.) 'Cut off the nth. part of a triangle by
a parallel to one side ; (ii.) divide a triangle into n equal parts
by parallels to a side.'
If HK pari, to BC makes AHK the wth
part of ABC, then AB = j7iAH, vP, _ _ _ P
•.• triangle ABC III AHK. ^'^ ^"-.^^^D
(i.) On any line DAE, / ^^/\
make AE = 7i. AD on opp. sides of A ; ' ^^^ / \
make DF pari, to BE; .*. AB = 7iAF. \^ ,-^^ h/ V
On diameter BF make semicircle FGB, \ \ v \*
, \ m/ \Q
make AG perp. to BF, \^/^ AR
.*. AG = mean propl. of AB, AF; B q
.*. AB:AG = AG:AF = -v/7i.
Make AH eql. to AG; /. AB = JnAG = shiAH.
Make HK pari, to BC ; Z. triangle ABC ||| AHK ;
.*. AABC: AAHK = AB2:AH2 = w.
Hence AHK is the nth part of ABC.
(ii.) Make in succession along AB,AL = GH,AM = GL, AN = GM,''*"
and so on; and draw LP, MQ, NR, &c. pari, to BC.
Then AL2 = GH2 = 2AH2, AM2 = GL2 = 3AH2, &c. ;
.-. AALP = 2AHK, AAMQ = 3AHK,
and so on.
Thus HK, LP, MQ, &c. divide the triangle into n equal parts.
* K the work is correct, GN = AB.
136 EXAMPLES. [CH. V.
EXAMPLES— XXIX.
Theorems.
1. If P is a point in a straight line AB, Q in AB produced, then
AP2 + PB2 + 2AP . PB = AQ2 + BQ^ - 2AQ . BQ.
2. If M is the mid point of AB in Ex. 1, and AP, AQ>PB, BQ, then
AP.PB = MB2-MP2; AQ . BQ-MQ^- MB2.
3. Show also that PA^- PB2 = 4MP. MB. And write down the corre-
sponding form for the point Q.
4. The sum of squares on the diagonals of a parallelogram is equal to
the sum of squares on the four sides.
6. The sum of squares on the diagonals of a quadrilateral is less than
the sum of squares on the sides.
6. The sum of squares on the diagonals of a trapezium is equal to
the sum of squares on the non -parallel sides and twice the rectangle
of the parallel sides.
7. Show that Ex. 4 is a particular case of Ex. 6.
8. If AB is a diameter and CD a chord of a circle, the sum of squares
on AC, AD, BC, BD is constant and equal to 2AB2.
9. If P is a fixed point inside a circle and AB a chord parallel to the
diameter through P, then PA^ + PB^ is constant.
10. A square has for its side the sum of perpendicular sides of a right
triangle ; show that it exceeds the square on the hypotenuse by four
times the area of the triangle.
11. If M, D are mid point and foot of perpendicular on the side BC of
a triangle ABC, and AB> AC, then AB^- AC2 = 2BC . MD.
12. The locus of a point whose difference of squares of its distances
from two fixed points is constant is a straight line.
13. Circles, centres B, C, are drawn through the orthocentre of a
triangle ABC. Show that the tangents from A to the two circles are
equal. Show also that the circles meet on the circumcircle.
14. The mid points of the four common tangents of two circles are
col linear.
15. A circle whose centre is on the axis of a coaxial system of circles
cuts one of them at right angles ; show that it cuts all of them at right
angles.
16. A straight line cuts the axis of a coaxial system of circles in V, and
the circles in PP', QQ', &c. ; show that VP . VP' = VQ . VQ', &c.
17. If a point P is joined to the vertices of a parallelogram ABCD,
and AC>BD, then PA^-j-PC^-lPB^-f PD2) = ^(AC2- BD^).
CH. v.] EXAMPLES. 137
18. A jointed rhombus ABCD moves so that one diagonal AC is
always a chord of a fixed circle, centre O. Show that BD always
passes through O, and that OB . OD is constant.
19. The greatest rectangle of the two parts into which a given straight
line can be divided internally is the square on half the line.
20. A square has the greatest area of all rectangles of a given
perimeter.
21. A square has the least perimeter of all rectangles of a given area.
22. Are Ex. 20 and Ex. 21 true if parallelogram is written for rectangle?
23. If ABCD are four points in order in a line, and AC is the mean
proportional of AB, AD, then AB : AD = AB^ : AC^.
24. If four straight lines are proportional, similar polygons on the first
two are proportional to similar polygons on the last two.
25. Show that Pythagoras' theorem is true if similar polygons are used
instead of squares. (See Ch. III.)
26. A diameter AB of a circle is divided into two parts at C, and
semicircles described on AC, CB as diameters, on opposite sides of AB ;
show that the sum of arcs AC + CB is half the length of the tirstcircle.
27. Show also that the two areas into which the arcs AC, CB divide
the whole circle are proportional to the diameters AC, CB.
28. Deduce from Ex. 27 a construction for dividing a circle by semi-
circular arcs into n equal parts.
29. Simihxr sectors of circles are proportional to the squares of their
radii or chords.
30. Similar segments of circles are proportional to the squares of their
radii or chords. (Treat as the sums or differences of similar sectors and
similar triangles.)
31. Show that Pythagoras' theorem is true if similar segments of
circles are used instead of squares.
32. Show that the centres of similitude of three circles lie three by
three on straight lines. What figure is formed by the straight lines ?
33. Tangents to the ciicumcircle of a triangle at its vertices meet the
opposite sides in collinear points.
34. If P, Q, R are collinear points on the sides BC, CA, AB of a
triangle, and P, P' divide BC harmonically, show that AP', BQ, CR are
concurrent.
35. If lines AP, BQ, CR to the sides of a triangle ABC are con-
current, and P, P' divide BC harmonically, show that P', Q, R are
collinear.
36. If a circle cuts the sides BC, CA, AB of a triangle in pairs of
points P, P'; Q, Q'; R, R'; then if AP, BQ, CR are concurrent, so also
are AP', BQ', CR'.
138 EXAMPLES. [CH. V.
EXAMPLES— XXX.
Constructions.
1. Divide a straight line 3-24" so that the rectangle of the whole line
and one part is equal to the square on the other. Measure.
2. Give a construction for lines of \/2", V^", V^", &c. from a line of 1".
3. Construct a line whose ratio to a given line is V^ + l.
4. Calculate the altitude of an equilateral triangle in terms of a side.
Calculate the area of an equilateral triangle of 1-73" side. Calculate the
area of an isosceles triangle, a = b=5 cm., c — 6 cm.
5. Construct a triangle, a = l-S", 6 = 2-4", c — 2'7". Calculate and
measure its area.
6. Calculate the area and radii of the circles of a triangle, a =12 ft.,
6 = 16 ft., c=lS ft. (Tabulate s, s -a, s-b, s-c, Thh. 100, 1.)
7. Construct the radical axis of two circles, radii f ", 1|", distance of
centres 3".
8. Construct a point whose tangents to two circles, radii 3-2 cm.,
2-5 cm., join of centres 4 cm., are 2 cm. long. How many points?
9. Construct the side of a square equal to the difference of squares on
3-5 cm. and 2-8 cm.
10. Construct the locus of a point whose difference of squares of its
distances from two fixed points 5 cm. apart is 4 sq. cm.
11. Find a point in a straight line 3" long whose difference of squares
of its distances from the ends is 1 sq. in.
12. Construct the locus of a point P whose sum of squares of its
distances from fixed points A, B is constant. (See Th. 91.)
13. Find a point in a line 6 cm. long such that the sum of squares of
its distances from the ends is 26 sq. cm.
14. Construct the locus of all points the sum of whose squares from
the ends of the line in Ex. 13 is 26 sq. cm.
15. How far can you see out to sea from a height (of the eyes) of
(i.) 10 ft., (ii.) 1000 ft., earth's radius 3960 miles?
16. Construct the side of a square of 3-76 sq. in. Measure.
17. Construct a square equal to (i. ) 3 times, (ii. ) J of a given square.
18. Construct a regular pentagon of 1" side ; and a similar pentagon
of double the area. (Calc. the new side, Th. 99, Cor.)
19. Construct a regular hexagon having an area of 1 sq. in. (Make
any hexagon, and adapt Constr. 39.)
20. Draw an isosceles triangle, « = 6 = 1.73", = 51°; take a point on
CA, 1" from C, and draw a line through it bisecting the triangle.
CH. v.] EXAMPLES. 139
21. Bisect a quadrilateral by a line through a vertex. Write down the
construction for cutting off a sixth part.
22. Construct an equilateral triangle whose area is 12 sq. cm. Verify-
by measuring. (Adapt Constr. 39. )
23. Divide a straight line into two parts whose squares have a given
ratio, say 3 : 5. (See Constr. 13, Note, for constr. of \/3, V5.)
24. Divide a straight line so that the sum of squares on the whole and
one part is three times the square on the other part. (Mean section. )
25. Draw an equilateral triangle, side 2-7 cm., and divide into four
equal parts by parallels to one side. Show that Constniction 41 can be
shortened in this case.
26. Draw a triangle, a = l-72", 6 = 1.31", c = 1.56". Take AP on AB
= 1", draw a line through P to cut off one-fifth of the triangle.
27. Construct a triangle, rt = 3-5 cm., & = 4-2 cm., = 2-8 cm., and cut off
a fifth part by a parallel to the side 6. Which side gives the easiest
construction ?
28. Calculate the area of a regular hexagon on a side of 1". What is
that of a regular hexagon of 2" side ?
29. Find the ratio of the areas of the circumscribed square and
inscribed hexagon of a circle.
30. Divide the arc of a semicircle into two parts, the squares of whose
chords have the ratio 2 : 3.
31. Divide a straight line 3-8 cm. long into two parts whose rectangle
is the greatest possible. What is the area of this rectangle ?
32. Make a square equal to five times the square of Ex. 31.
33. Show how to derive in succession the complete series of squares of
areas 1, 2, 3, 4... What series of numbers do their sides represent?
34. Construct the radical axis of two non-intersecting circles, centres
O, Q, to meet the line of centres in N.
Construct a third circle coaxial with the first two, and passing through
a fixed point A on OQ.
35. In the construction of Ex. 34, determine the two points D, D^ on
OQ such that ND = NDi = length of tangent from N to the circles.
36. Construct the circles through the points D, Dj of Ex. 35, coaxial
with the original circles. W^hat do you find? (These circles are the
limiting circles or foci of the system of coaxial circles.)
37. Can you construct the limiting circles of a coaxial system when
the circles cut in two points A, B ? Why ?
38. Two regular pentagons have sides 3 cm., 5 cm. Construct a square
whose area is the sum of their areas.
140
MAXIMA AND MINIMA.
[CH. V.
The greatest and least values of a varying quantity can some-
times be found readily by a simple application of a known
inequality; the most general method is that of Thh. 85, 86,
Ch. IV., in which some variation of a given form or value in-
creases or diminishes the quantity in question. We give a few
examples; others will be found in Ex. XXXIV., 114-133.
Construction 42. — 'Find the shortest path between two
points A, B to touch at a given line PQ.'
If P is the point where the shortest path BPA touches PQ, then
BP and AP are straight, as otherwise the path may be shortened.
Make ANC perp. to PQ, NC eql. to AN,
join BC to cut PQ in P, join AP.
Then NP is rt. bisector of AC ; B
/. PC = PA, and path BPA = BC.
If BQA is any other path,
BQ-l-QC, i.e. BQ-j-QA>BC, in triangle BQC
i.e. path BQA > BPA;
.*. BPA is the shortest path.
Note. If A, B are on opposite sides of the line PQ, then AB is the
shortest path.
Theorem 106. — 'If all the sides of a polygon ABCDE are
given except one AE, the greatest polygon is that in which
the last side AE subtends right angles at the other vertices
B, C, D.'
If ACE is not a right angle,
rotate triangle CDE about C to make
CE perp. to AC ;
.*. alt. of E from CA increases,
base CA remains unaltered, and area
of tr. ACE and of the poln. increases ;
.*. if angle ACE is not a rt. ang.,
area of poln. ABCDE can be increased ;
.*. the maximum polygon has ang. ACE, and similarly ABE and
ADE, each a right angle.
Ex. The maximum quadrilateral with three equal sides a given, is
hah the regular hexagon on side a.
D'
0^1
-^.E'
CH. V,
MAXIMA AND MINIMA.
141
Theorem 107. — 'Of all polygons of given sides, the greatest is
that which can be inscribed in a circle.'
If ABCD is inscribed, and A'B'C'D' having the
same sides cannot be inscribed, in a circle, draw ^
diameter AE ;
make triangle D'E'C congruent to DEC.
Then the circle on diam. A'E' does not pass
through all the points B', C', D' ; not through C', say.
Also, quadls. ABCE, A'B'C'E' having three sides
the same, ABCE in a semicircle > A'B'C'E'.
Similarly ADE^A'D'E';
.-. poln. ABCED>A'B'C'E'D', and ABCD > A'B'C'D'.
P "v \ yR
Construction 43. — * If A, B are given points, PQ a given line,
find the points in PQ at which AB subtends (i.) maximum,
(ii.) minimum angles.'
If AB meets PQ in C, and A, B are on the same side of PQ,
and if circles through A, B touch PQ
in P and Q respectively ;
take R any other point in CP, and
draw circle ARB cutting AP in D ;
.*. ang. APB>ADB, int. opp. ang.,
>ARB, same arc; /
.'. APB is a maximum on one side Q s c
of C.
Similarly, AQB is a maximum on the Other side ;
and ACB ( = zero) and the angle of the parallels to PQ from A, B
( = zero) are minima.
This is a good example of maxima and minima occurring
alternately.
Discuss the case when A, B are on opp. sides of PQ.
Ex. 1. Find the point P within a triangle ABC such that
PA + PB + PC is a minimum.
Ex. 2. An isosceles triangle has a less perimeter than any other
triangle of the same base and area.
Ex. 3. An equilateral triangle has a less perimeter than any triangle
of equal area.
142
QUADRILATERAL IN OR ABOUT QUADRILATERAL. [ciI. V.
Construction 44. — ' Circumscribe a quadrilateral of given
form* to a given quadrilateral PQRS.'
If a/3y8 has the given form, on PQ, QR
describe circles whose arcs PAQ, QBR have
angles a, /?, and which meet in O.
Draw chd. A'QB', make B'RC' such that
B'C':B'A' = ^y:/?a;
.*. triangle A'B'C ||| a/3y.
Make circ. ORG', and ang. ROX = suppt.
of 7 ; draw C'X, A'P to D'.
.'. C' = y, hence D' = S.
Draw in order SXC, CRB, BQA, APD.
Then tr. OAB ||| OA'B', OBC ||| OB'C;
.-. tr. ABC III A'B'C III a/?y. (Th.39,Ch.III.)
And ang. DCA = y- BCA = 8ya, DAC = Say ;
:. tr. DCA III Sya ;
.'. quadl. ABCD ||| a/5yS, and circumscribes PQRS.
A very simple construction serves for a parni., rect., or sq.
If QX' makes ang. of parm. with PR, and QX': RP = ratio of
sides, SX' fixes one side.
Note. A'B'C'D' III ABCD, and can be rotated about O to be simi-
larly situated to ABCD, and then multiplied (by OA : OA') into
coincidence.
Construction 45. — 'Inscribe a quadrilateral of given form in
a given quadrilateral ABCD.'
lipqrs is the given form, circumscribe a^yS similar to ABCD;
divide the sides of ABCD in the same propn. at P, Q, R, S.
Theorem 108. — ' If three sides of a polygon of given form
pass through fixed points P, Q, R, any fourth side traverses a
fixed point X.'
The fourth side CD completes a quadl. ABCD of given form.
Draw circs. PAQ, QBR meeting in O, ORC meeting CD in X ;
.-. X is a fixed point; and if A'B'C'D' ||| ABCD,
C is on circ. ORC, and ang. BC'X = C = C;
.*. CD' coincides with C'X, and traverses fixed point X.
* That is, similar to a given quadl.
CH. v.]
TRIANGLE ON THREE CURVES.
143
C N
Construction 46. — 'Construct a triangle ABC of given form*
so that, A being given, B, C are on given curves.'
If apy has the given form, and A is a
vertex given, e.g. on some third curve :
(i.) If one curve is a straight line P,
make AN perp. to P, tr. AMN simr. to a^y,
MB perp. to AM, to meet the curve R at B ;
make ang. BAG eql. to a, to meet P at C.
Then rt. tr. BAM ||| CAN,
*.• ang. BAM = a - MAC = CAN ;
.-. tr. BAC III MAN i|| a/?y. (Th. 39.)
(ii.) If P is a circle, centre O, rad. r ;
make tr. AQO simr. to a^y, draw arc B,
cent. Q, rad. r x — , to cut curve R in B :
ay '
make ang. BAC = a.
Then tr. ABQ ||| ACQ f (Th. 41, i.) ;
.-. tr. ABC III AQO ||| a/5y. (Th. 39.)
In each case the curve P is rotated through
a about A and then multiplied by the ratio
a/3 : ay into a similar curve cutting R in the point B.
The method applies to any curve P.
EXAMPLES-XXXI.
1. Inscribe a triangle of given form in a semicircle. (Choose A some
point on the circumference, and use the diameter as the curve P and the
circle as R.)
2. Describe a triangle, A = 60°, 6 : c=2 : 1, with its vertices on (i.) three
parallel lines, (ii. ) three concurrent lines, (iii. ) three sides of a triangle.
(Choose A on one line. )
3. Describe an isosceles right triangle with the right-angled vertex
fixed and the others on given lines.
4. Describe an equilateral triangle (i.) with two vertices on a given
circle, and one on a given tangent; (ii.) with its vertices on three
concentric circles.
5. Circumscribe to a square a parm. , ang. 60°, sides 8 : 7.
* That is, similar to a given triangle.
t If the right point C is chosen on the line AC.
144
ROTATION LOCI.
[CH. V.
The principle underlying Construction 46 is useful for construct-
ing certain loci.
Theorem 109. — 'If a triangle of given form is rotated about
a fixed vertex A, and a second vertex B describes any curve,
the third vertex C describes a similar curve.'
If ABC, AB'C' are two posns. of the
triangle, BB', CC are corresponding chds.
of their curves, and triangle ABB' ||| ACC ;
/. chd. CC' and curve of C can be rotated
through ang. A into posn. yy', similar and
similarly situated to BB' and the curve of B.
Ex. * On chords AB through a fixed point A of a circle, equi-
lateral triangles ABC are drawn ; find the locus of a point
dividing BC in a given ratio.' (Take 2:1.)
In triangle A BP, BP :AB = BP : BC = const.,
and ang. B = 60°;
. •. triangle ABP has a given form ;
and it rotates about A, B describing a circle ;
. • . P describes a similar circle.
Make triangle ACQ simr. to ABP ;
.-. tr. APQIIIABO;
.-. QP : OB ==QA:OA = const.;
. • . Q is centre of circle.
Plotting Locl
Loci not straight lines or circles can be drawn by plotting any
number of points, and drawing a freehand curve through them.
Ex. The ends of a rod AB move round two circles,
centres O, Q ; plot the locus of a point P on AB.
Take OA 1", QB f", OQ 3^, AB 3", AP 1".
Set the divider to AB (3"), mark two points A, B on
the circles, bring a straight-edge to them ; prick P, 1"
from A.
Take a fresh position and repeat, going carefully
round one circle, until the form of the locus is clear.
The curve is like a distorted 8. But for different values of the radii
and length of rod, and for different positions of P, the curve may have
other forms.
CH. v.] LOCI — CONICS. 145
o
In plotting curves where several pairs of lines in a fixed ratio /x
are required, this construction is useful : o
Make OM : ON = the given ratio fi ;
parls. M P, NQ give OP : OQ = />t.
Definition 36. — A conic is the locus of a point whose distance
from a fixed point has a constant ratio to its distance from a
fixed line.
The fixed point is the focus, the line the directrix, and the ratio
the eccentricity of the conic.
A conic is a parabola, ellipse, or hyperbola,
according as eccentricity = 1, < 1, > 1.
EXAMPLES-XXXII.
1. Plot a parabola ; an ellipse, ecc. 2:3; a hyperbola, ecc. 3:2;
focus— directrix, 1". (Use squared paper.)
2. Plot* the locus of P when the sum of distances PA + PB from fixed
points A, B is constant. (An ellipse, foci A and B. It is a closed oval
curve. )
3. Plot the locus of P when PA - PB is constant. (A hyperbola, foci
A, B.) p
4. Draw a circle, draw perpendiculars PN to a
diameter AA' ; multiply these by a constant ratio
QN : PN =At (say 2 : 3). Plot the locus of Q.
5. In Ex. 4, show that QN^ : AN . NA'^/x^.
This is an ellipse whose axes are AA' and yuAA'.
It represents the jilan of a circle tilted up from the
horizontal.
Envelopes.
If a straight line moves according to some law — e.g. the side
BO of the triangle in Th. 109 — a curve, the envelope of the
line, exists to which the straight line is a tangent. If a number
of positions are drawn sufiftciently near, the form of the envelope
is shown ; and the curve can sometimes be found by our knowledge
of geometry.
On the next two pages are given examples of methods of
finding the point of contact of an envelope with its generating
line in any position.
* This should also be drawn with divider, a loop of cotton, and pencil,
P. G. J
146
ENVELOPES.
[CH. V.
Ex. PJot the envelope of one side of a riglit angle whose vertex
moves on a circle, and whose other side traverses a fixed point.
The form of the curve is readily recognised from
the figure as an oval. It is an ellipse with the
fixed point A as one focus, and H at the same
distance from the centre on AO for the other.
If A is external, the envelope is a hyperbola.
Note. If BC, B'C are two near positions of the
moving line, meeting in D, and touching the en-
velope at P, R ;
then, as the tangt. B'C moves to coince. with BC,
R and D move to coince. with P.
Thus the point of contact P on BC has the limiting
posn. of the intersection of two coincident tangents.
The envelope is found as the locus of this point.
Thus, in the above example,
angs. B, B' in two near positions are equal ;
.'. A, B, B', D (2nd fig.), are concyclic ; and as
B'C' moves to coince. with BC, BB' becomes
tangt. to the first circle, and to circ. ABP ;
these therefore touch.
Hence the centre of circle ABP is Q on OB
(3rd fig.), and on the rt. bisector of AB ;
and AQP is the diameter of circ. ABP.
Also, OQ II HP, •/ OQ bisects AP, AH ;
;. PA + PH = 2QA + 20Q = 2QB + 20Q = 20B
= constant ;
.'. locus of point of contact P is an ellipse, foci A,
If A is outside the circle,
is a hyperbola.
PA -- PH is constant, and the envelope
Note. A line BC of given length (2nd fig. ) can be moved into posi-
tion B'C by turning about a centre of rotation C (intersection of
rt. bisectors of BB', CC, or of circs. BB'D, CCD) ;
then CD bisects ang. BDC.
If now B'C turns about C to coincidence witli BC,
D moves to P, point of contact of BC with envelope,
and CD to CP, perp. to tangt. BP.
Hence the point of contact of a moving line in any position Avith its
envelope is the foot of perpendicular on the line fo'om the centre of
rotation.
CH. v.] ENVELOPES INSTANTANEOUS CENTRE. 147
This centre of rotation is not in general fixed, but varies with the
position of the line ; it is therefore called the instantaneous centre
of rotation for the particular position of the moving line.
In fig. 3, last page, if E is the point of the side BA of the right angle,
which momentarily coincides with A ;
then as B begins to move on circle O,
E moves at first along direction AB ;
.'. inst. centre O' is intersection of BO', AO', perp. to BT, AB.
And O'P, perp. to BC, determines pt. of contact P.
The method is useful for a figure given in magnitude and form.
By the aid of Theorem 109 Ave can show that
' The envelope of one side of an angle whose vertex describes
a fixed circle, and whose other side traverses a fixed point, is an
ellipse or hyperbola.'
c
For if ABC is the angle, A the fixed point,
andANj_BC;
the triangle ANB has a given form, and turns about
one vertex A, whilst another vertex B describes a u/
circle ; ^■
.*. the third vertex N describes a circle (Th. 109);
.'. envelope of NB — i.e. of BC — is an ellipse or hyperbola.
EXAMPLES-XXXIII.
1. The envelope of a line at a fixed distance from a given point is a
circle.
2. If the sides AB, AC of a given triangle traverse fixed points
X, Y, the locus of A is a circle, and the envelope of BC another
circle. (Bordillier. )
If AP is a diameter of the circle XAY, and PM perp. to
BC meets this circle in Q ; (P is inst. centre) ;
Q is a fixed point, and the centre of the envelope.
3. The envelope of one side of a right angle whose
vertex describes a straight line, and whose other side
traverses a fixed point, is a parabola.
4. Plot the envelope of Ex. 3. Distance focus — line ^".
5. The vertex of an angle describes a straight line, and one side passes
through a fixed point ; show that the other side envelopes a parabola.
(Use Ex. 3 and Th. 109, as for ellipse above.)
6. Plot the envelope of the side BC of an equilateral triangle of which
A is fixed and B moves on a circle.
AVhat is the envelope when A is on the circle ?
148 EXAMPLES. [CH. V.
EXAMPLES— XXXiy.
General.
1. A straight line is determined in position by one point and the angle
it makes with a given direction.
2. Take a point P one inch from a given line, and draw through it a
straight line to make an angle of 54° with the line.
3. The right bisector of one side of a rectangle is the right bisector of
the opposite side.
4. If opposite sides of a parallelogram have the same right bisector,
the angles are right angles.
5. Draw a square of 3-7 cm. side. Without pen or pencil, mark the
bisectors of angles and right bisectors of sides.
6. Two straight lines in a plane parallel to a third are parallel to one
another.
7. A straight line perpendicular to one side of a parallelogram is also
perpendicular to the opposite side.
8. The angle A of a parallelogram ABCD is 63° ; a straight line EF
cuts AB at an angle of 97°, towards the same part as A ; what angle does
EF make with AD ? with DC ? with BC ?
9. Draw a parallelogram ABCD, AB = 2 cm., AD = 3 cm., ang. A = 70°.
At C outside the parm. make ang. BCE = 110°. Show that DC, CE are
in a straight line.
10. A straight line OP through a fixed point O meets a fixed straight
line ABC in A. It is then turned in the plane through an angle X, into
the position OQ, cutting ABC in B. If C is in AB produced, show that
X = PAC-QBC.
11. Draw two lines not meeting on the paper. Measure the angle
between them. (Draw a line AB to cut them and apply Ex. 10.)
12. Through a point P on one of the two lines of Ex. 11 draw a
parallel to the other, and measure the angle thus formed. What do you
notice ?
13. Show that the angle of two lines X, Y may be measured bv the
angle of X, Z, if Z || Y.
14. A straight line makes with two straight lines cutting it angles of
95° and 87° towards the same parts. Calculate their angle.
15. Calculate the exterior angles of a hexagon having five angles 135°,
147°, 161°, 108°, 83°. Sum them.
16. Draw the pentagon ABCDE, given AB = 1", B = LS5°, BC = 1-41",
C = I05°, CD = 3" = DE, EA = 1-41". Measure the other angles.
CH. v.] EXAMPLES. 149
17. Construct a triangle, « = 3-72", B = 99°, C = 41°; draw a line DEF
cutting AB, AC, BC in D, E, F, making ang. ADE = 60°. Calculate
angles AED, BFD ; and test by measuring.
18. The bisectors of consecutive angles of a quadrilateral meet at E ;
show that their angle is half the sum of the other angles of the quadl.
19. Construct a triangle, given a =2-7 cm., exterior angles at B and C,
108° and 152°.
20. Straight lines AD, A E to the side BC of an isosceles triangle make
equal internal angles with the equal sides AB, AC. Show that ADE is
isosceles.
21. If BE, CD make equal angles BEC, CDB with equal sides AB, AC
of a triangle, show that BE = CD.
22. Construct a triangle, a = 3-72", 6=2-59", c = 2-23". Draw the right
bisector of AC, meeting BC in D. Prove that BD=:a- AD.
23. ABC is a triangle having 6 = 4-6 cm., = 3-7 cm., A = 150°. The
light bisectors of 6, c meet a in D, E. Show that a = sum of sides
of ADE.
24. One only parallel can be drawn to a straight line through a given
point.
25. If a right triangle is turned in a plane so that its right-angled sides
are interchanged in position, the two positions of the hypotenuse are
perpendicular.
26. Deduce from Ex. 25 a construction for a perpendicular by set-
square.
27. If two lines a, h are perpendicular respectively to two lines c, dy
the angle ah is equal to the angle cd.
28. Given the diagonal 2-12" of a square, construct the square.
29. The projection of any straight line on a non-parallel line is less
than the original line.
30. If A is the greatest angle of a triangle ABC, a line DE, cutting
the sides AB, AC in D, E, is less than BC.
31. Construct a rhombus, diagonals 3-2" and 2-8". Calculate the
length of a side.
32. If M is the mid point of a straight line AB, P any point on it, MP
is half the sum or difference of MA and MB, according as P is external
or internal.
33. Perpendiculars from two vertices of a triangle on the opposite
sides form an angle equal to the third angle of the triangle.
34. Construct a parallelogram, sides 2-32" and 1-76", one diagonal
1-97". Draw a parallel to one side through the diagonal point. Verify
that the two parallelograms formed are congruent.
35. Show that a square is divided into eight congruent triangles by
bisectors of angles and right bisectors of sides.
150 EXAMPLES. [CH. V.
36. Show that if straight lines PQ, RS bounded by the two pairs of
opposite sides of a square are perpendicular, they are also equal.
37. Construct a quadrilateral with equal diagonals 2-7 cm. long per-
pendicular to and trisecting each other ; draw a rectangle whose sides
pass through its vertices.
38. Show that if the diagonals of a quadrilateral are equal and perpen-
dicular, every circumscribing rectangle is a square.
39. Construct a square to circumscribe a given quadrilateral.
40. If PQR is a triangle entirely inside another ABC, show that
p + q + r<:a + b + c.
41. If PQR... is a polygon, with no re-entrant angle, entirely inside
another ABC..., the perimeter of PQR... is less than that of ABC...
42. Draw any pentagon, and copy it by ruler and compass.
43. The sum of two sides of a triangle is greater than twice their
bisector of angle.
44. If in a trapezium the angles at the ends of one of the parallel
sides are equal, the non-parallel sides are equal ; and if the non-
parallel sides are equal, the angles at the ends of either parallel side
are equal ; also in either case the diagonals are equal.
45. If the diagonals of a trapezium are equal, the non-parallel sides are
equal.
46. If the diagonals or non-parallel sides of a trapezium are equal, the
trapezium is cyclic; and the median line of the parallel sides passes
through the centre of the circumcircle.
47. Divide a straight line AB at a point P so that PA : PB = the ratio
of any two given lines X : Y.
48. Show that there is one only internal point in AB which can divide
AB as in Ex. 47 ; and similarly one only external point.
49. Given a = 4 cm,, A = 70°, b:c—5:S, construct the triangle.
50. All straight lines through a point P and cutting two parallels in
A, B are divided at P in the same ratio PA : PB.
51. Draw two parallels 5 cm. apart, and construct the locus of a point
P dividing in the ratio 3 : 4 any line bounded by the parallels.
52. If A' is the mid point of BC in triangle ABC, and CD is perpen-
dicular to the bisector of angle A, DA' is parallel to AB, and is equal to
half tlie difference of AB and AC.
53. If AD, AP are perpendicular and bisector of angle A of a triangle,
the angle DAP is half the difference of B, C.
54. If one of the parallel sides of a trapezium is fixed, and the altitude
and magnitude of the opposite side are given, construct the locus of the
diagonal point.
55. The locus of a point the sum of whose distances from two fixed
lines is constant is four parallels to the bisectors of angle of the lines.
CH. v.] EXAMPLES. 151
56. The locus of a point whose distances from two sides of an angle
have a fixed ratio X : Y is two straight lines through the angle.
57. Construct a point in a triangle whose distances from the sides are
proportional to those sides (« = 5 cm., 6 = 6 cm., c = 8 cm.).
58. The mid points of sides of a triangle form a similar triangle,
similarly situated to the original triangle.
59. Given in position an angle A of a triangle and the mid point of the
opposite side, construct the triangle.
60. Construct a straight line through a given point such that two
given pairs of parallels cut off equal parts from it.
61. If the equal parts of Ex. 60 are PP'.QQ' in order, find the locus of
the mid point of PQ' for all positions of the given point.
62. A parallel to a side of a parallelogram cuts the diagonals in P, Q ;
show that the join of the mid point of PQ to the diagonal point is parallel
to another side.
63. If BE is the fourth part of diagonal BD of parallelogram ABCD,
and AE meets CD in F, then FC = 2CD.
64. Draw a parallelogram ; and inscribe in it a rhombus, given the
length of one diagonal.
65. If three vertices of a parallelogram whose sides are parallel to
fixed directions move on three fixed straight lines, the fourth vertex
moves on a straight line.
66. Inscribe in a given quadrilateral a parallelogram, given the
directions of sides. (Use Ex. 65.)
67. Construct a parallelogram, two opposite vertices being fixed, and
the other two lying on a fixed circle.
68. Similar segments of circles on equal bases are congruent ; or,
similar arcs of circles on equal chords are congruent. (Similar segments
or arcs have equal angles. )
69. Two triangles ABC, DEF have BC=EF, ang. A=D; show that
the vertices of DEF can be placed on the circumcircle of ABC.
70. Given a, B, and tlie distance of the circumcentre from BC, con-
struct the triangle.
71. Construct a point at which the three sides of a triangle subtend
equal angles.
72. If a jointed quadrilateral ABCD has AB fixed, find the loci of
C, D and of the mid points of AC, BD.
73. Inscribe a given square in a given square.
74. If H is the orthocentre of ABC, then A, B, C are orthocentres of
HBC, HCA, HAB.
75. The circumcentre of a triangle is the orthocentre of the median
triangle ; and the triangles have a common centroid.
152 EXAMPLES. [CH. V.
76. If O, H are circumcentre and oitliocentie of a triangle, and X the
mid point of BC, show that OX^^AH. And if AH meets BC in D and
the cireumcircle in K, then DK=DH.
77. Construct a triangle, given in position the circumcentre, ortho-
centre, and an angular point.
78. If G, O, H, N are centroid and circum-, ortho-, N-centres of ABC,
and X, Y, Z, L, M, N mid points of sides and of AH, BH, CH ; show
(i.) Triangles XYZ, LMN are similarly situated to ABC,
about G, H.
(ii.) LX is a diameter of the N -circle and i| AC.
(iii.) If AC meets BC in P, the circle, diam. AP, touches the
N -circle.
79. The centroid, circum-, ortho-, and N-centres are collinear,
80. Construct a triangle, given the N -circle, circumcentre, and direc-
tion of one side.
81. If straight lines AD, BE to the opposite sides of a triangle trisect
each other, they are medians.
82. Two straight lines from two vertices of a triangle to the opposite
side cannot each bisect the other.
83. Construct a triangle, given angle C and the medians through A, B.
84. The incentre and ecentres of a triangle are each the orthocentre of
the triangle of the other three.
85. The orthocentre and angular points of a triangle are the incentre
and ecentres of the pedal triangle.
86. Given the ecentres, or two ecentres and incentre, construct the
triangle.
87. The cireumcircle of a triangle bisects the join of an in- and
e-centre.
88. Construct a triangle, given cireumcircle, incentre, and a vertex.
89. If AP is the bisector of angle A of a triangle, I the incentre, show
that A\ :\P = b + c:a, and hence construct a point S in Al such that
AS:SI:IP = &:c:a.
90. If two altitudes of a triangle are given, show that the ratio of the
corresponding sides is given.
Construct a triangle, given the altitudes.
91. There is one only point P in a triangle such that angle PAB = PBC
= PCA, and one only point P' such that P'BA = P'CB = P'AC. (P and
P' are positive and negative Brocard points.)
92. If P and P' are Brocard ])oints of a triangle, show that PAB = P'BA,
&c. (Brocard angle.)
93. Construct the Brocard point P of an isosceles right triangle ABC,
right-angled at A, and show that CPA is a right angle.
CH. v.] EXAMPLES. 153
94. If P, P' are Brocard points of an isosceles triangle whose angles
B, C are each double of A, show that
(i.) Ang. PBA = PCB, ang. APB = 108° = CPB.
(ii.) PC, PB are mean parts of PB, PA, and hence PB + PC = PA.
95. If a is the mean part of h, and h of c, then a + b = c.
96. If AC is the mean part of AB, then AB2+ BC-^SAC^.
97. The circles circumscribing the four triangles of a complete
quadrilateral are concurrent.
98. Construct four concurrent circles to meet, two and two, in four
given points (e.g. circles A, B meet in P ; B, C in Q ; C, D in R ;
D, A in S).
99. If M is the mid point of arc AB of a circle, and a straight line MP
cuts the chord in P and the circle in Q, then MA is tangent to circle
APQ.
100. If the tangent AC at the end of a diameter AB of a circle is
equal to the diagonal of the square on AB, the centroid of ABC is
on the circle.
101. If AB is a diameter of a circle, PQ perpendicular to AB from a
point P on the circle, and R is taken on AB so that AR = AP ; show that
(i.) PR bisects angle QPB ;
(ii.) QR is radius of circle touching PQ, QB and arc PB. '
102. Inscribe a circle in a part of a semicircle cut off by a pei"pen-
dicularto the diameter. (Use Ex. 101, ii.)
103. How many revolutions does a 28" bicycle-wheel make in a ten-
mile ride? (7r = 3-1416.)
104. If ABCD is a cyclic quadrilateral, the rectangle of diagonals is
equal to the sum of rectangles of opposite sides. (Ptolemy's theorem.
Diaw AE to diag. BD, making ang. BAE eql. to CAD. Triangle
BAE|I|CAD, DAE III CAB.)
105. If ABCD is a non-cyclic quadrilateral, the rectangle of diagonals
is less than the sum of rectangles of opposite sides.
106. Construct tangents to a circle from a point without using the
centre of the circle.
107. If P is a point on one of two circles, centres O, Q cutting in A, B,
PN perpendicular to AB, and PT tangent to the other circle cuts AB in
C and the first circle in R, then PC(PT+ RT) = 2PN . OQ.
108. Draw a circle to touch a given circle, a tangent to it, and a line
through the point of contact.
109. Find the point in a straight line at which two given points not on
the line subtend the greatest angle.
110. The sum of squares on the medians of a triangle is | of the sum
of squares on the sides.
154 EXAMPLES. [CH. V.
111. The sum of the two liines formed by semicircles on three sides of
a right triangle, on tlie same side of the hypotenuse, is equal in area
to the triangle. (Apply Pythagoras' theorem.)
112. Cut off the nth part of a quadrilateral by a line through a point
in one side.
113. Draw a circle of given radius to cut two given circles ortho-
gonally.
114. The greatest distance from the centre of a circle of a point
P, from which lines PQ, PR drawn to the circle are perpendicular, is
^^2 X radius.
115. The maximum length cut by two circles from lines having a
given direction is that which traverses a centre of similitude.
116. Find the greatest triangle inscribed in a given circle.
117. Find the greatest and least distances from a point P, IJ" from the
centre of a circle of f" radius, to the circle.
118. Find a point P on a circle such that the sum of squares of its
distances from two fixed points is (i.) minimum, (ii.) maximum.
119. Plot a graph of some varying quantity — e.g. the join of a
point on a circle to some fixed point, as the angle which this join
subtends at the centre of the circle varies — and show that maxima
and minima occur alternately.
120. If A, B are fixed points on a circle, find P on the circle so that
PA + PB is a maximum.
121. Find P in a straight line so that if A, B are fixed points on
opposite sides of the line, PA - PB is a minimum.
122. On a given base and with given sum of sides the maximum
triangle is isosceles.
123. The maximum polygon of 7i sides of given perimeter is regular.
124. Of two regular polygons of the same perimeter, that which has
the greater number of sides is greater ; and a circle is the greatest figure
of given perimeter.
125. Of right triangles on a given hypotenuse, the isosceles has the
greatest perimeter.
126. Construct a point P on a circle the rectangle of whose perpen-
diculars on two fixed tangents is a maximum.
127. A tangent to a circle, centre O, meets two fixed tangents in
P, Q ; show that PQ and the triangle PQO are minima when the point
of contact bisects the arc between the fixed tangents.
128. Find a point P in a straight line the sum of whose squares of
distances from two fixed points A, B is a minimum. "What does this
become when AB is the straight line?
129. The minimum triangle formed with two fixed lines by a line
through a fixed point is that whose side is bisected at this point.
CH. V.J EXAMPLES. 155
130. Find a point P in a triangle the sum of whose squares of dis-
tances from the sides is a minimum. (The symmedian point, whose
perps. are propl. to sides.)
131. Inscribe in a triangle the triangle of minimum perimeter. (The
pedal triangle. )
132. Construct a tangent to a circle such that the product of perpen-
diculars on it from two fixed points on the circle is a maximum.
133. Of all triangles of given form whose sides pass through three
given points the maxinium is that whose perpendiculars to sides through
those points are concurrent. (Use circles as in Constr. 43.)
134. If O, A, B are three points in order in a line, find C so that
OC is (i.) arithmetic mean, (ii.) geometric mean, (iii. ) harmonic mean of
OA, OB. (If O, C divide AB harmonically, OC is H.M. of OA, OB.)
135. If a figure ABC... is moved in a plane into a position A'B'C...
so that A'B' is parallel to its old direction AB, then every side CD' is
parallel to its old position CD. (This process is called translation.)
136. Show that any figure ABC... can be converted into any congruent
figure A'B'C... of like aspect by translation and simple rotation.
137. A straight line PQ travels with one end P on a given figure, and
keeps always parallel to a fixed direction ; show that Q describes a
congruent figure.
138. Construct a triangle, given the three medians AX, BY, CZ.
(Translate BC to AD, CA to AE, then sides of tr. BDE are double
medians, and A is its centroid.)
139. Construct a quadrilateral ABCD, given diagonals AC, BD and
their angle, angle A, and (i.) side BC, (ii.) angle C. (Translate triangle
ABD, to bring point A on C, and the constr. is obvious.)
140. If the end A of a straight line AB moving parallel to itself moves
on a circle, then B describes a circle.
141. A straight line PQ is moved into any other position P'Q' ; show
that the angle between these is the difference of the angles, measured in
the same sense, made by them with any third line AB.
142. Any straight line AB can be converted into any other CD by
rotation and multiplication. (If AB, CD meet at K, circles ACK, BDK
meet at centre of rotation O.) Interpret when AB |! CD.
143. Any figure ABC... can be converted into any similar figure
A'B'C... of like aspect by rotation and multiplication.
144. If a rhombus is inscribed in a similar rhombus, each is a square.
145. A rhombus circumscribed by a square is a square.
146. Draw a square circumscribing a parallelogram, sides 1", 1^",
angle 80°. (Constr. 44, note.)
147. If PR, QS between opposite sides AD, BC and AB, CD of a paral-
lelogram make an angle POQ = suppt. of A, then PR : QS = AB : BC.
156 EXAMPLES. [CH. V.
148. Draw a rectangle, sides 2 : 3, circumscribing a parallelogram,
sides 3 cm., 5 cm., angle 75°. (Constr. 44, note.)
149. A jjarallelogram (other than a square) cannot be circumscribed to
a similar parallelogram. (In Constr. 44, if PQRS||j A'B'C'D', S is on
circle ROX, and the construction gives the original parm.)
150. Arcs on the sides AB, BC of a parallelogram contain angles A, B,
and meet in O when produced. Show that O is on AC.
151. Show that in Constr. 44 the points O, P, D, X are concyclic.
152. All triangles of given form inscribed in a given triangle have a
common centre of rotation.
153. Draw a triangle, sides 1", 1§", If". Inscribe an equilateral
triangle with a vertex 1" from one end of side If". (Use Constr. 46.)
154. Find the centre of rotation of all equilateral triangles of Ex. 153.
155. Inscribe an equilateral triangle of side 1" in the triangle of
Ex. 153. (Multiply the first eql. triangle from the centre of rotation to
make its sides 1", rotate a vertex on to one side. )
156. Inscribe a square in a parallelogram. (Construct an isosceles
right triangle with vertex at diag. point, the others on sides of parm.)
157. The locus of points whose tangents to two circles have the ratio
of the respective radii is a circle.
158. One end A of a straight line AB which traverses a fixed point
traces a circle ; plot the locus of a point on AB. (Piston of oscillating
cylinder.)
159. One end A of a straight line AB of given length describes a circle,
the other end B a straight line through its centre. Plot the locus of a
point on AB. (Connecting-rod of crank and piston.)
160. Plot the locus of a point P on a straight line AB of given length
moving with its ends on two perpendicular lines OX, OY. (Trammel.)
161. If A, A' are the extreme positions of P on OX in Ex. 160, and
PN is perpendicular to OX, show that PN^ : AN . A'N is constant, and
hence (see Cli. V. p. 145, Ex. 5) that the locus is an ellipse.
162. Plot the locus of a point on a wheel as it rolls along a straight
line. What is the locus of the centre ?
163. Plot the envelope of a line AB of given length whose ends move
on two fixed lines. Show that the envelope touches the fixed lines, and
find the locus of its instantaneous centre.
164. Plot the envelope of a diameter fixed in a circle, as the circle rolls
along a straight line. Indicate the instantaneous centre for one position
of the circle and diameter.
165. Plot the envelope of a line AB of constant length whose ends
move on two fixed circles. Discuss the case when AB has the length of
a common tangent.
CH. v.] EXAMPLES. 157
Antiparallels and Symmedians.
Definition 37.— A line in a triangle, making with two sides
the angles of the triangle opposite to these sides, is an anti-
parallel of the triangle to the third side.
166. The sides of the pedal triangle are antiparallels to the sides of
the triangle.
167. The locus of mid points of antiparallels to a side of a triangle is a
straight line through the opposite vertex (a symmedian).
168. Perpendiculai-s on the two sides of a triangle from a point on the
symmedian of their vertex are proportional to the sides.
169. The three symmedians of a triangle meet in a point (the sym-
median point or syntroid).
170. Tangents to the eircumcircle of a triangle at the vertices are
antiparallels to the sides.
171. A diameter of the eircumcircle of a triangle through a vertex is
perpendicular to the antiparallels of the opposite side.
172. The joins of the vertices of a triangle to the opposite vertices
of the triangle formed by tangents to the eircumcircle of the original
triangle at its vertices, meet in the syntroid.
173. A bisector of angle of a triangle bisects the angle of the median
and symmedian through its vertex.
174. An antiparallel to one side of a triangle forms a cyclic quadri-
lateral with the three sides.
175. Construct the syntroid of a triangle ABC.
176. Establish the following theorems :
(i.) If a triangle AiBjCi is similarly situated to any triangle
ABC about the syntroid of ABC, its sides cut those of ABC in
six coney clic points. (Tucker circle. )
(ii.) Parallels to the sides of a triangle through the syntroid
cut the sides in six concyclic points. (Lemoine circle.)
(iii.) Antiparallels to the sides of a triangle through the syn-
troid meet the other sides in six concyclic points. (Cosen circle.)
(iv.) The lines bisecting the sides of the pedal triangle meet
those of the triangle in six concyclic points. (Taylor circle.)
(v.) Show that (ii.), (iii.), (iv.) are Tucker circles.
177. The Brocard points, orthocentre, and syntroid are concyclic.
(Brocard circle. The demonstration is much more difficult than that of
the theorems of Ex. 176.)
158 NOTE ON NUMBER, RATIO, AND PROrORTION. [CH. V.
(Tlie symbol >> denotes 'is the least quantity greater than.')
As elementary algebra gives no proof of the laws of operation of
irrational numbers (e.g. ^2, tt), I propose here to establish their
laws by means of decimals. I shall assume that terminating
decimals, which can be reduced to fractions ^9/^, can be added or
multiplied by the ordinary rules of arithmetic.
We may define algebraically that a number ju, is represented by
a decimal a-a^a^-.-an... ad inf., when
/x .> all approximations fXn ;
where fXn denotes the termng. decimal a'ajag'-'^n-
I. 'A decimal a-a^a^... ad inf. represents one only ratio
number.'
I I 1 Iii,i i Ji i i t|
O Y I Q P
Take OJ = Y as unit, make a scale to a units, «j tenths,
^2 hundredths, &c., in succession;
thus every nth approximation fin is represented ;
also (a + l)Y > any /z^Y (since adding unity to the last digit an
cannot make fin exceed a-\-l).
Take OP, = X, > a}\ finY ]
.'. X:Y, = fi say, > all fin (Del of > or < number) ;
i.e. the decimal a-a-^a^... represents fi.
Again, fiY is a magnitude X or OP,
and if V is a number < /x,
vY is a magnitude OQ or Z < X (Def. of less number) ;
but some scale divisions fall between Q and P ; *
.'. V < some fin-
.'. a different decimal represents a less number v, and
similarly n it n greater n A;
.'. a-a^a2... represents one only number fi.
* The fundamental axiom of measure is Archimedes' axiom :
'Any magnitude A, however small, when multiplied by a large enough integer m,
can he made to exceed any other B of the same kind.'
i.e. mQP> OJ, m large enough ;
. * . QP> OJ/?Ji> OJ/10">some whole element of the scale.
CH. v.] NmiBER — RATIO. 159
Cor. We have sliown incidentally how to construct a number /x
from its decimal, and to construct the magnitude /xY from /x and Y.
This is the general solution of the problem of finding a fourth term
in a proportion.
II. 'The sum of the decimals of two numbers /x, v is the
decimal of their sum /x + v.'
If the decimals of /x, v are added, from the , 7^^/ ^
left, the resulting decimal c-c^Cg--- is that of a i 2 s-'-
number A (by I). ^•^i^2^3--
Then any less number k < some A„ < A^ say, and by the process
Ap</^,i + v„, n large enough ;
.*. any less number k < some (ja„ + v^) ;
but A > all ja„ + v„ ;
.-. A •> all fin + Vnl
i.e. A = /x + v = v + /x (similarly).
Thus we can add and subtract numbers by their decimals.
The fact fi+v = v + fiisB. commutative law.
III. 'The product of the decimals of two numbers is the
decimal of their product.*
If the decimals of /x, v are multiplied, from . ^ - ^"'
the left, the resulting decimal is that of a — I_2_3i:-
number A. 1 2 3--*
And any less number k <Xp say, < /XnV„, n large enough,*
and A > all jtx„v„ ;
.'. A •> all fMnVn ;
i.e. X = fxv = vfi, similarly (commutative law).
Thus we can multiply and divide numbers by their decimals ;
and we can operate fractions of ratio numbers by the ordinary
rules of arithmetic, defining the fraction A/v as /x, where /xv = A.
(Kouse Ball's Algebra.)
Note. This can he extended to include the complete theory of
indices and logarithms.
* This is a consequence of the process. See the corresponding stage in the
proof of II. above.
160 PROPORTION. [CH. V,
The laws which we have established in I., II., III. cover all the
operations of 'real' number, which may thus be defiued as all
number which can be represented by continuous decimals.
We can now establish the theorems of Ch. III., p. 66.
Altemando. (Exchange of second and third terms of a propn.)
If X : Y = Z : W = />t, and Y : Z = V ;
X = />tY = /xi/Z ; and Y = vZ - v/xW;
.-. X:Z = /xi/ = i//x = Y:W.
Inversion, (Inverting each ratio in a propn.)
If X:Y = Z:W = ^, and Y:X = v;
X = fxY = fjLvX ; .'. 1 = ju,v = v/x ;
/. W = v/xW = vZ ;
i.e. W:Z = v = Y:X.
Unit Theorems.
(i.) If X : Y = Z : Y, then X = Z.
If X:Y = /x = Z:Y,
X > all /x^Y = Z.
(ii.) If X : Y = X : Z, then Y = Z.
By inversion, Y : X = Z : X, .*. Y = Z, by (i.).
XXX
Summation. — If — 1==— 2 = — ?, &c., the magnitudes being all of
''l ^2 ^B . »
one kind,*
then each ratio = — ^^ — J^ ^yf /" '
For if each ratio = /x,
X^ •> all /x^Yp Xg •> all /x„Y2, &c. ;
.-. X1 + X2 + X3+... >all/x„(Yi + Y, + Y3+...)
= KYi + Y2 + Y;+...).
Note. This theorem is the general form of the theorem
^xX+^lY=^l[X + y).
Product Theorem. — If k, A, /x, v are measures of X, Y, Z, W in
proportion,
K : A = X : Y = Z : W = /x : V ;
and by ordinary algebra, kv - A/x.
* Two magnitudes are of the same kind when one is equal to, greater than, or
less than the other.
CH. V.
ANGLE LINE — CONTINUOUS MAGNITUDE.
161
The hand of a clock may be regarded as a straight line turning
in a plane, and may be said to generate the angle
between two positions ; in doing so it generates every
possible less angle, so that an angle may be con-
ceived as growing from zero, where the hand started,
lip to any given value.
And as the hand may complete 1, 2, or more
turns, angles may be generated of more than 4, 8...
right angles — i.e. of any magnitude whatever.
As the hand may turn either way, an angle may
be diminished as well as increased ; we thus get the
notion of positive ( + ) and negative ( — ) angles, the positive
way of turning being earthwise,"*" and tlie negative sunwise f or
clockwise.
In the same way, length may be generated by a point moving
along a line, lengths in opposite directions being
positive and negative.
For straight lines N and E are positive, S and
W negative directions.
A quantity of this type, which can be increased
or diminished continuously in either sense, is real
quantity of the most general kind. And the
'range of number (positive and negative) representing the measure
of continuous real quantity is the whole range of real number of
algebra. It may be defined algebraically as all number, positive
and negative, that can be represented by decimals.
In the next chapter we shall assume that angles and lines have
sign as well as magnitude, according to the above rules. Figures
will be drawn on the assumption that the line due E is the zero
of angular rotation ; angles in any other position may be supposed
to be moved so that one of their sides coincides with this position ;
and angles of figures Avill generally be considered positive. But in
general theorems of angles, negative angles are included as well aa
positive.
* The way the earth turns round its axis, regarded from IS[. Pole,
t The way the sun appears to move in the sky.
w^
s
p.o,
162
CHAPTER YI.
ELEMENTARY TRIGONOMETRY.
TRIGONOMETRICAL RATIOS— THE TRIANGLE.
If PM, QN are perpendiculars from one side of an angle a to the
other,
then tr. POM ||| QON ;
hence the ratio MP : OP ( = NQ : OQ) has a given
value for each angle a ; and for any greater or
less angle /? the ratio has a greater or less value ;
and similarly for ratios OM : OP, MP : OM, &c.
These ratios therefore serve to distinguish
angles, though they are not directly proportional O*
to the angles.
Definition 1. — If from a point P on one side of an angle
AOP a perpendicular PN is drawn to the other, forming the
right triangle PON, then the ratio
perp. . NP . ^, . i? *^r,
T—^i I.e. — -, IS the sine of AOP ;
hyp.'
^•" OP'
base
ON
hyp.'
" OP'
perp.
base '
NP
" ON'
base
ON
perp.'
" NP'
hyp.
base'
OP
" ON'
hyp.
OP
perp.
NP'
II cosine
II tangent
II cotangent
II secant
M cosecant
Note. In writing the ratios, always make the vertex of the angle the
first, and the other end of the hypotenuse the last letter ; the ratios then
have always the correct sign. The last three ratios are the reciprocals
of the first three, and are not much used.
CH. VI.]
TRIGONOMETRICAL RATIOS — SIGN.
163
The ratios are written —
sin AOP, cos AOP, tan AOP, cot AOP, sec AOP, cosec AOP;
or if a is the angle AOP —
sin a, cos a, tan a, cot a, sec a, cosec a.
Ex. Write down all ratios for the triangle QON, last page.
The cosine of an angle AOP is the ratio of the projection (ON)
of OP on OA to OP itself. Hence an important theorem —
Theorem 1. — ' The projection of a straight line X on a line
forming an angle a with it is X cos a,'
K
A
N/
r 1
i +
N
1
1
-1
N i
Algebraic Sign op the Ratios.
The projection ON of OP on OA is positive or negative according as
it falls along OA or along AO produced ;
NP is positive for all angles up to 180°,
OP (the hypotenuse) is always positive.
Hence (i.), when a is an acute angle,
sin a is + ^^ cos a + "^^ tan a + ^""^ &c.
(ii.) When a is ohtuse,
sin a is + ^'®, cos a — ^'®, tan a — ^^^ &c.
For angles greater than 180°, less
than 360°, NP falls on the other side of OA, and is then negative;
OP is positive, and ON positive or negative as before.
Thus signs as well as magnitudes of the ratios are distinguished ;
but we shall not in general concern ourselves with angles greater
than 180°.
An angle of any given figure — e.g. triangle or polygon — must
be supposed to be moved so that one side lies along OA. Such
angles are usually considered as positive.
Ex. 1. Find by drawing and measuring the first three trigonometrical
ratios of 20°, 75°, 112°, 150°. Compare with tables.
Ex. 2. In each case of Ex. 1, find the sum of squares of sin a, cos a.
Ex. 3. In each case of Ex. 1, compare ^ Avith tan a.
^ cos a
Ex.4. Draw a triangle, a = 1-5", 6 = 1.9", = 2-7". Calculate the
values of a/sin A, 6/sin B, c/sin C, and measure the circumradius.
Ex. 5. In Ex. 4, calculate sin A cos B + cos A sin B ; compare sin C.
164
RATIOS OF 30°, 60°, 45°
[CH. VI.
The values of sines, cosines, &c. of angles can be found from
tables, or by construction. The latter method is not applicable
where great accuracy is required, as in ordnance survey. The
ratios of angles which can be constructed by ruler and compass
are easily found as surds.
Construction 1. — (i.) ' Construct ratios of 60°, 30
If POA is an equilateral triangle, each angle is 60°
and PN, bisector of ang. P, is the rt. bisector of OA.
Hence, if each side is 2 units,
ON = 1, OP = 2, NP2 = 0P2 - ON2 = 3;
:. NP= J3;
V3.
sm
60° =
sin AOP =
OP
2 '
cos
60° =
OP 2^
tan 60°
NP
ON
Also,
ang
NPO = ^
= 30°;
• ono NO , _-.^ PN
sm 30 = — - = h; cos 30 = — -
PO ^ PO
V3
; tan 30° =
V3'
(ii.) * Construct ratios of 45°.'
If PON is an isosceles right triangle,
ang. AOP = NPO = -J rt. ang. = 45°.
And if ON = 1, NP = 1, OP2 = 1 + 1 = 2 ;
/. OP= ^2;
. • AKO NP 1
••«^^^^-6p = 72^
... ON 1 / J2\ . ,.^ NP ,
For calculation use J2 = 1414, ^3 = 1-732.
Ex. 1. Calculate the above ratios to four decimal places, and compare
with the tables.
Ex. 2. Calculate the ratio of a shorter diagonal of a regular hexagon
to its side.
Ex. 3. Calculate side of insquare of circle, rad. 3 cm.
Ex. 4. Calculate length of shadow of a tower 150 ft. high, when the
sun's altitude (angle of elevation) is 60°.
CH. VI.]
RATIOS OP 0% 90°, 180°.
165
Construction 2. — * Construct ratios of 0°, 90°, 180°.*
If AOP is an acute angle, PA an arc, centre O,
and PN±OA;
then if P moves to coincidence with A,
the ang. AOP, arc AP, perp. NP, all become zero,
and OP and ON coincide with OA.
0^
OA
OA
NA
/ NP\
*. Sin AOP ( = p— j becomes
cos AOP ( = °^) ,
: i.e. sin = 0.
T„AOP( = ^)
OA
OA
Similarly, if AOB = 90°, AOP < 90°,
PB an arc, centre O, and PN_LOA;
then if P moves to coincidence with B,
ang. AOP becomes 90°, ON becomes zero,
OP and NP coincide with OB.
i.e. cos = 1.
i.e. tan = 0.
OB
Sin AOP becomes -— -
OB
i.e. sin 90° = 1.
Cos AOP .. — r : i.e. cos 90° = 0,
Tan AOP
0^
OB
OB
i.e. tan 90° = oo
Prove similarly :
Sin 180° = 0, cos 180°= -1, tan 180=
0.
Ex. 1. Find the ratios of 270° and 360°.
£x. 2. For any triangle ABC, write down the values of
(i.)sin(A + B + C), cos (A + B + C), tan (A + B + C);
(ii.) sin
A + B + C A + B + C
tan
A + B + C
2 ' 2 ' — 2
Ex. 3. In a right triangle, A the right angle, R the circumradius,
show that 2R =
a
sin A'
c
sinC
sin B'
Ex. 4. Can you prove Ex. 3 for an acute triangle ?
Ex. 5. If a<90°, show that tan a : sin a>l.
Ex. 6. What does the ratio tan a : sin a become when a is zero?
Ex. 7. Show that cos a, sin a never exceed unity, but that tan a can
be used to represent any real number.
166
RELATIONS OP RATIOS.
[CH. VI.
A few theorems follow readily from the definitions. They should
be learnt by heart.
Theorem 2. — * If a is any angle ' —
1
,. . sm a
(1.) = tan a- ^ .
^ ' cos a cot a
(ii.) sin^a + cos^a = 1 ;
(iii.) 1 + tan^a = — n- = sec^a.
^ ' cos-a
(i.) results from writing down the ratios.
.... CI- 9 9 NP2 ON2 NP2 + ON2
(n.) Sin2a + cos2a = ^^, + ^^ = _^^p2—
_OP2
~OP2~ •
..... -, ^ 9 , NP2 ON2+NP2 OP2
(m.) l+tan2a=l+-— ,
ON2
ON2
ON^
cos^a
1. Prove that cos a= , sin a= ,
sec a cosec a
and reciprocally.
Ex. 2. Given sin a- -816, calculate cos a, tan a; given cos a=«387,
calculate sin a, tan a ; given tan a = l'35, calculate cos a, sin a.
Ex. 3. If C is a right angle in a triangle, sin^A + sin^B^l.
When the sine, cosine, or tangent (or any ratio) of an angle is
given, the other ratios may easily be found in terms of the given
ratio by construction.
Construction 3. — * Determine the other ratios of an angle in
terms of the given sine, cosine, or tangent.'
If sin a is given = s, say,
make the hypotenuse OP unity; .*. perp. NP = s;
:, base ON = ^1 - s^ ;
,', cos a = Jl—s'^= Jl- sin^a ;
tan a =
sm a
z, &c.
Jl-s^ ;^l-sin^a
Similarly, if cos a is given = c, say,
make hyp. OP unity ; .*. base ON = c, &c., as before.
And if tan a is given = t, say,
make base ON unity ; .'. perp. NP = ^ ;
.*. hyp. OP = JT+^, &c., as before.
Ex. Given sin a=x, cos ^=y, tan y=z, find by construction the
other two ratios of a, j8, y.
CH. VI.]
Theorem 3.-
RELATrONS OF RATIOS.
167
The sine of an angle = cosine of its complement
M A
the cosine u =sine
the tangent m = cotangent
If AOP is a, and PNJ.OA,
make AOQ = 90° - a, and QM _L OA ;
.-. tr. OPN III QOM ;
NP OM .-.^o .
Similarly, cos a =... = ...= sin (90° - a) ;
tan a =... = ... = cot (90° - a).
Note. These theorems are true in niagnitude and sign for all angles ;
a similar proof, with due care as to sign, applies to all cases.
Theorem 4. — ' Sin (90° + a) = cos a ; and cos (90^ + a) = - sin a."
With due care as to sign, this may be proved after the same
manner as Th. 3. It also is true for all angles.
Ex. 1. Fill up the following equalities, and verify from the tables:
sin 60° = cos... ; sin 150° = cos... ; cos 54° = sin... ; cos 120°= - sin... ;
tan 48° = cot... : cot 37° = tan...
T. n T * • 1 • C A + B
Ex. 2. In any triangle sin -^ = cos — ^—
cos
B
sin
A + C
Theorem 5. — ' The sine of an angle = sine of its supplement ;
the cosine n = - cosine ti ;
the tangent n = -tangent u .*
If AOP is a, and PN_LOA,
make AOQ = 180° - a, and QM ± OA ;
.-. ang. QOM = PON, and tr. OQM ||| OPN.
. CI- NP MQ . ,.,^,^0
.-. Sm a = -— = + — 5 = sm (180°
OP OQ ^
ON
cos a = — - =
OP
OM
OQ
NP
tana = ^^ =
MQ
OM
-tan (180° -a).
The theorem can be proved in a similar manner, with due care
as to sign, for all values of a.
Ex. 1. By the aid of the tables, M-rite down the numerical values of
the three first ratios of 172°, 158°, 110°, 137°.
Ex. 2. Fill up the equalities, sin 25° = sin... ; cos 74°= -cos...;
tan 57°= -tan...
168
EXAMPLES. [CH. VI.
EXAMPLES— XXXy.
(In the following examples write -. , , or -?^^ resnec-
sin a cos a tan a sin a ^
tively for cosec a, sec a, cot a where they occur. Also tt stands for
an angle of two right angles. )
1. If a = 60°, /3 = 45°, 7 = 30°,
(i. ) cos'-^a + 3 cos a - sin^a = 1 ;
(ii.) cos a sin 7 + cos /3 sin ^ + cos 7 sin a = li ;
..... cos a + cos /3 + cos 7 ,
(ill.) -. — ^ -. — ^ = 1.
sin a + sin jS + sin 7
2. Find the value of (i.) cos 60° sin 30° + tan 60° cot 30° ;
(ii. ) sec 45° + cosec 45° ;
tan - + tan - + tan 7:
.... . o 4 d
(ill.
cot ^ + cot 7 + cot ^
O 4 o
3. Find the value of sin a + cos a + tan a, when
(i.) a = |, (ii.) a = 45°, (iii.) a = ^.
4. Show that (i.) -— — — =sin a cos a ;
cot a + tan a
... 1 - tan a _ cot a - 1 ^
1 + tan a cot a + 1 '
.... , tan a + tan /3 , ,0
(ill. ) — 7 7-^ = tan a tan B.
cota + cotjS ^
5. Show that (i. ) (sin a + cos a)^ = 1 + 2 sin a cos a ;
(ii. ) sin^^ + cos''^ = 1-2 sin^^ cos^^ ;
(iii.) sin^a + cos^a = 1-3 sin^a + 3 sin^a ;
,. . l+tan2^ , „.
(iv.) :^ = tan2^.
l+cot-(9
6. Find the values of sin a cos ^ + cos a sin j8, and compare with the
values of sin (a+j8) (found from tables) in each case, when
(i.) a = 30°, ^ = 45°; (ii.) a = 30°, |S = 60°; (iii.) a = 60°, ^ = 45°.
(You may assume that sin 105° = sin 75° for the last case.)
7. Solve the equations :
13 1
(i.)cos^ = r -; (11.) 4sin^ = -; — -; (iii.) 3 tan ^ = - -;
2 cos ^ ' ' sin ^ ' ' ' tan ^
3
(iv.) 4 sin3^ = 3 sin d ; (v.) sec ^-cos ^ = ^-
8. (i.) If sin a= ,. , find cos a, tan a.
■Wa^ + b'^
(ii.) If cos a= . =, find sin a, cot a.
CH. VI.] EXAMPLES. 169
8. (iii.) If tail a =^^, lind sin a, cos a.
9. In any triangle (i.) cos ^— =sm -^ ; (u-) tan —^ =cot - ;
(iii.) sin (B + C)=:sin A ; (iv.) cos (A + B)= -cos C.
10. Write down from the tables the sines and cosines of 63°, 42°,
31° ; and make a table showing what other angles less than 180° have
the same numerical values of sine or cosine; stating in each case the
algebraical sign of the ratio.
11. Find, by comparing with the corresponding acute angle, the ratios
sin 150°, cos 135°, tan ^, sin 225°, cos ^, tan 330°.
12. If sin a=k, cos p=l, find the value of
(i. ) Sin a cos /3 + cos a sin ^.
(ii.) Cos (TT-a), sin (7r + /3), tan (7r-/3), cos {ir+a).
(Use tables in the following examples.)
13. When the angle of elevation of the sun is 35°, a house throws its
shadow just across a street 100 ft. wide. Calculate the height of the
house.
14. From the top of a cliff 200 ft. high the angle of depression or dip —
i.e. angle l)elow the horizontal — of a ship is 15°. Calculate its distance
from the foot of the clitf.
15. From a boat at sea the angles of elevation of the top and bottom of
a lighthouse are 36° and 30°. If the cliff on which the house stands is
180 ft. high, what is the height of the house ?
16. A hill has a slope of 1 in 15 (1 ft. rise in every 15 ft. along the
slope). Half a mile up is a coal-shaft whose foot is on a level with that
of the hill. Find the slope of the hill, and the depth in feet of the
shaft.
17. From two boats at sea in line with a headland, and half a mile
apart, the angles of elevation of the headland are 14° 29 and 7° 15'
respectively. Find the height of the headland.
18. From two points 100 yd. apart on the bank of a river the directions
from the river line of a post on the opposite bank are 26° 34' and 14° 2'
respectively. Find the width of the river.
19. From a ship sailing due north for the mouth of a harbour a church
spire near the shore is seen in a direction of 55° E. of N. After another
mile its direction is 65°. If the coast runs E. and W. , find the distance
of the church from the harbour mouth.
20. A church tower 150 ft. high stands on a hill 300 ft. high. Find
the angle which it subtends at a point on the flat at the foot of the hill
300 yd. distant in direct line from the foot of the tower.
170
FORMULA OF A TRIANGLE.
[CH. VI.
Theorem 6. — 'In any triangle '-
,. . a ^ _ ^ _
^ '' sill A sin B sin C
(ii.) a = ^ cos C + c cos B,
2R. (Sine formula.)
(Projection formula.)
(i.) Draw the circumcle., and diam. BA' ;
then either ang. A' = A, same arc,
or ang. A' = suppt. of A, opp. arc ;
also ang. BCA' = 90° (ang. in semcle.) ;
CB a
.*. in each case, sin A = sin A' =
A'B 2R '
I.e.
2R = (similarly)
sin A "" ^" '^ ' sin B sin C
(ii.) If AN A. BC, BC = BN + NC, fig. (i.),
and BC = BN - CN, fig. (ii.).
But NC = & cos C, in fig. (i.),
and CN = & cos ACN = & cos (180°-C)
= - 5 cos C, in fig. (ii.) ;
and BN = c cos B in each case ;
.*, a = BC = & cos C + c cos B.
Ex. Write down the formulae for b and c.
Ex. 1. Show that a sin B = 6 sin A, & sin C = c sin B.
Ex. 2. Show that (sin A + sin B)-^sin C = {a + b)-^c.
Ex. 3. Show that a cos B = c~b cos A ;
a cos C = b-c cos A.
Ex. 4. Calculate R and the elements — i.e. sides and angles — of a
triangle, given :
(i.) « = 530yd., A = 50°, B = 75°; (ii.) 6 = 12-3 cm., A = 87°, C = 48°.
Note. Formulae (i.) and (ii.) are really particular cases of projection
formuloe, useful in mechanics and co-ordinate geometry.
If a path b in direction ^ from a given line,
+ a path c II II y n the m ,
are equivalent to
a path a in n a h u m ;
then a cos a ( = ON) = & cos /? + c cos y,*
a sin a ( = NP) = & sin /3 + c sin y.
In formulae (i.) and (ii.) a = 0, ^ = B, y = - C.
* These formulae are true for all values of the angles «, /3, y when these are
measured in the same sense.
CH. VI.]
FORMULAE OF A TRIANGLE.
171
Theorem 6. — (iii.) ci^ = b^ + c^- 21)6 cos A ; cos A =
^2 +
(Cosine formula.)
(iv.)Sinf = ^{
26c
tan
r=V(
(s-b)(s-c)
be
b){s
A*
is -a)
c)
|. (Half angle formulae.)
s{s — a)
(iii.) From formula (i.) a sin B = h sin A ;
II 11 (ii.) a cos B == c - 6 cos A.
Square and add ;
.-. a2(sin2B + cos^B) = h^sin^A + cos^A) + c^ - 2bc cos A ;
but sin2B + cos^B = 1 = sin^A + cos^A (Th. 2, ii.) ;
/. rj = b- + c2 - 2bc cos A.
Hence, also, 26c cos A = 6^ + c^ - a^ ;
62 + c2-a2
i.e. cos A =
2bc
Ex. Write down the formulae for b"^, c^, cos B, cos Q
(iv.) If R, R' are points of contact of in- and
e-circles, centres I, E^, on AB ;
then {^-b){s-c) = r.ra (Th. 100, Ch. V.),
and 6c = AI.AEj n ;
. . 2A_RI R'E^_ r.ra _ (s-b)(s-c)
••''''" 2 "AI'AE^ ~ '
be
be
2A_AR AR'_ g(g-a)
"^""^ 2" Ar-AEi~~6^^
tW^--?" R^E,_ (.-6)(.-c)
^^^ 2~AR' AR' 4s -a)
B C
Ex, Write down the formulae for ^, — •
Formula (i.) is used when a side and two angles are given ;
formula (iii.) or (iv.) is used when three sides are given.
For logarithms always use (iv.) and not (iii.) ; without logarithms
use (iii.) and not (iv.).
Ex. 1. Calculate the angles, given (1.) a=5l0 yd., 6=750, c=660;
(ii.)a = 5cra., 6 = 6 cm., c = 7 cm. ; (iii.) a = 53-71 ft, 6 = 68-72 ft, c= 91.34 ft.
Ex. 2. In formula (ii.), c = a cos B + 6 cos A, write 2R sin A, &c., for
«, 6, c; hence prove sin (A + B) ( = sin C) = sin A cos B + cos A sin B.
A
* The sign of the sq. root is +,
is acute.
172
Theorem 6.— (v.)
FORMULA OP A TRIANGLE.
A-B
^ _ 7. *
(Tan. formula.)
[CH. VI.
tan
a-h*
tan
A+B a+h
(vi.) A= J{s{s-a){s-h){s-c)}=lhc&m f< = &c,
abc
(vii.) R = j^ = 2
&C.
sm A
(viii.) r = — = (s-a) tan ^, &c.
s Ji
Va = = s tan -^, &c.
s-a 2
(v.) In BC make CD = CE = CA ;
C is centre of semcle. EAD ;
BN, perp. to AD, || EA.
Now ang. D = JC, at cent, of semcle. ;
C A + B
.-. NBD = compt. of D = 90°-
and NBA = NBD-B = ^'^--B
2 '
A-B
2 '
tan
A-B
NA ND NA |E(.,BN||EA)
A + B BN BN ND BD
tan
Write similarly the forms for
a — b
a + b'
A-C C-B
Ex. Ifa=7, 6 = 5,C = 60°
A-B
then tan — 5— = 2-886;
.-. ^^ = 16°.l, and ^^=60% .'. A = 76°.l, B = 43°.9;
whence, by sine form., c = 6-25.
(vi.) The first form is Th. 100, Ch. V. ; the second, \ base x
(vii.) Th. 101, Ch. v., and 6 (i.) above,
(viii.) Th. 100, Ch. V. The second form is tan ^- — =
in 6 (iv.) above.
Ex. 1. Find the elements of a triangle, given :
(i.) & = 15, c=12, A = 50°; (ii.) a = 10-5, 0=13-2, B = 84''.6.
Ex. 2. Show that sin B = 'i\J{s{s - a){s - b){s - c)} + ac.
Ex. 3. Find A, R, r, n, given (i.) a=12, 6 = 15, c = 18;
(ii.) 6 = 25, B = 60°, s=36.
* This is more convenient than the old form, tan „ ~ ~— Tl cos g*
alt.
R^
AR'
CH. VI.]
SOLUTION OF TRIANGLES.
173
By means of the formulae just established, any required element
(side, angle, &c.) of a triangle can be calculated, by tables, with
greater accuracy than is possible by geometric construction.
Construction 4. — ' Trigonometrical solution of a triangle.'
(i.) * Two angles and one side given, A, B, c' (Sine F.)
Calculate C = 180° - A - B.
a c
Then - — - = - — — , whence a, and similarly b.
sm A sin C "^
(ii.) 'Three sides given, a, b, c' (Half ang. F.)
Calculate s = ^r , s - a, s-b, s-c.
Then tan ^ = .W^^ — / _ v , whence A, and similarly B; then
C=180°-A-B.
(iii.) 'Two sides and their angle given, a, b, C (Tan F.)
Calculate ^^ = 90° -|.
_,. , A-B a-b . A + B , A-
Then tan — h— = =■ x tan — ^~ , whence — ~
2 a+b 2 2
whence A and B by addition and subtraction.
c a
B
Also,
whence c.
sin C sm A
(iv.) The ambiguous case.
' Two sides and the angle opposite one given, a, b, A.
(Sine F.)
Sin B = - sin A, whence B.
a
Then C = 180° - A - B, and c = a x
sin C
sin A
But the value of sin B does not show whether
B is acute or obtuse, since sin (180°- B) has the b'
same value as sin B. (Th. 5.)
If, however, (i.) b'<a, then B<"A ;
.*. B is acute, and there is only one solution ;
but if (ii.) b>a, there are two possible solutions, the
two values of B being supplementary.
There are then also two values of C and of c.
(v.) ' The right triangle.'
This reduces to the definitions of ratios.
174 EXAMPLES. [CH. VI.
EXAMPLES— XXXVI.
(Angles may be found to the nearest minute, sides to four figures.)
In a triangle —
1. a = 5 ft., & = 7 ft., A = 45°; find B.
2. & = 21-6", 6':=: 144", C = 34° 25'; find B.
3. « = 1320 yd., c = l760 yd., C = 76° 30'; find A.
4. « = 7 ft., 6=5 ft., c = 4 ft. ; find B and C. What is A?
5. « = 22 ft., 6 = 44 ft., c=55 ft. ; lind A and C.
6. a = 3", 6 = 4", C = 51° 19' ; find c.
7. « = 10ft., c = 12ft., B = 60°; find 6.
8. A = 82% B = 68°, c=100 yd.; find 6.
9. A =43°, C = 77°, 6 = 150 yd. ; find c.
(In the following examples four-figure logarithms should be used.)
10. a = 379, B = 86°, C = 52°; find 6.
11. c = 4765, A = 118°, B = 27°; find a.
12. 6 = 391.2, C=102°, A = 34°; find c and a.
13. « = 143.1, 6 = 241-2, c=157-3; find A.
14. a = 31.2, 6 = 21-3, c=27-9; find B.
15. a = 415.7, 6 = 347-9, c = 521.6; find C and A.
16. 6 = 354-7, c = 426-8, A = 49° 16'; find C.
17. c= 12-36, a = 21-87, B = 73° 37'; find A.
18. a =2087, 6=3075, C = 103° 15' ; find B and c.
19. a = 3547, 6=4690, A = 37° 45'; find B.
20. 6 = 5665, c = 6006, B = 39° 30'; find C.
21. c = 824-3, a = 958-2, C=45° 43' ; find A and 6.
22. Find tlie greatest angle of the triangle which has the least side 5",
the sum of its sides 21", and the three sides in arithmetic progression.
23. Find the circumradii of the triangles in Exx. 12, 13.
24. Find the r and r„ of the triangles in Exx. 13, 14.
25. Find the areas of the triangles in Exx. 15, 16.
26. Find in terms of x and a the third side of a triangle whose sides
are x sin a, x cos a, and included angle 60°. Find the value of this side
when a = 45°. What kind of triangle is it in this case?
27. If a = 26, and C = 120°, find the ratio a : c.
28. If the second greatest angle of a triangle is 45° 10', and the sides
are in geometrical progression, find (as a surd) the common ratio.
29. Calculate the other sides and angles of the triangle of Ex. 28
when the second greatest side is 1".
30. Solve the mean base triangle (Constr. 16, Cli. III.) when the base
is 5 cm.
CH. vr.]
TRIANGULATION.
175
I
^-
r
For purposes of mapping and survey, a theodolite is used to
measure angles.
It consists of a telescope TS mounted
on an axis so as to turn about a divided
circle showing degrees and (with the aid
of a vernier) minutes * of angle.
The eye-piece has a spider line by
which the telescope can be accurately
focussed on to a given point. With the
best instruments differences of a quarter
minute, or 15 seconds,* can be observed.
The circle and telescope can be ad-
justed horizontally by means of screws
and levels ; and the elements of a
triangle ABC in a horizontal plane
measured as follows.
A base line AB is measured as accu-
rately as possible ; the theodolite placed
at one end B of this line, the telescope focussed on A, and its
position on the circle noted.
It is then turned through the arc TT' until focussed on C, and
the new position T'S' on tlie circle noted.
Thus the angle ABC ( = SBS') is measured by means of the
circle ; similarly, angles ABD, ABE, CBD, DBE to other points
D, E, &c. are measured.
The instrument is then moved to A, and corresponding angles
at A measured.
Thus triangle ABC is calculated from AB, angs. B, A ; similarly
trs. ABD, ABE, &c. are calculated. Thus CD, DE, &c. are known.
DE, for example, may now be used as a new base line, its
calculated value being used. Thus a whole district or country
can be mapped out very accurately by means of triangles.
This process is called triangulation. The calculations are made
by the sine formula.
Range-finding. — This is the snme process. A distance AB and
angles CAB, CBA are measured, and distance AC calculated.
*One degree=60 minutes, 1 ininute=60 seconds.
Thus 52° 17' 26" reads 52 degrees 17 min. 26 sec.
176 MEASURE OP HEIGHTS. [CH. VI.
For measuring heights the telescope is mounted so as to turn
about a second divided circle, perpendicular
to the first, and which itself turns about the
centre of the first circle."^
Thus, if the first circle is horizontal, the
second is vertical, and the telescope turns in a
vertical plane about the second circle, and in
a horizontal plane along with the second
circle.
Construction 5. — 'Measure the height of an inaccessible
peak.'
Measure the distance BA of two points ;^P
in a horizontal plane as base line ; (when AB
is not horizontal a corresponding correction is
required) ;
then if P is the peak, PC perp. to the
horizontal through AB ;
measure the elevation PAC at A, by the
vertical circle. A
Measure angle CAB by turning the vertical circle until the
telescope focusses on B.
Similarly measure angle CBA.
Calculate CA, in triangle BAC, given c. A, B ;
then height PC = CA tan PAC, and is thus known.
EXAMPLES— XXXVII. (Use 4-figure tables. )
1. At a point A on a straight road the corner of a house bears 58°
to the left. At a point B, 200 yd. farther on, it bears 82°. Find the
distance of the house from A and from the road.
2. The elevation of the top of the roof of a church is 25°, and of the top
of the spire 35°, from a certain point. If the roof -top is 80 ft. from the
ground, what is the height of the spire ?
3. At a point A the peak of a mountain bears due west, and has an
elevation of 22° ; at a point B, 400 yd. N. from A, it hears W.S.W. If
A is 2000 ft. above sea-level, find tlie height of the mountain.
4. A horizontal base line AB is 1235 yd. long ; a mountain peak,
altitude 18° 32' at A, has directions of 82° 31' and 88° 27' from A, B,
measured in the same sense. Find height of mountain.
* Fpr terrestrial work only half the vertical circle is constructed.
CH. VI.]
RATIOS OF a + 13, a — 13.
177
Theorem 7. — ' If a, /3 are any two angles,
sin (a + f3) = sin a cos /3 + cos a sin 13 ;
sin (a - /3) = sin a cos ^ — cos a sin y8.'
(i.) Turn a str. line AA' through a at A
into posn. AB ; turn it through ^ at B into
posn. BB', forming tr. ABC ; then
ang. A'CB' = a + ,/5 (by pari' to AB at C) ;
.'. sin (a + 13) = sin A'CB' = sin C
_ c a cos B + b cos A
~ 2R ~ 2R '
in tr. ABC,
= sin A cos B + cos A sin B
= sin a cos /3 + cos a sin 13.
(ii.) Turn a str. line AA' through a at A into posn. AB ;
turn it through -/3 at B into posn. BB',
forming tr. ABC ;
then ang. A'CB' = a-/3 ;
.'. sin (a -13) = sin A'CB' = sin C \
= sin A cos B + cos A sin B
= sin a cos ^ - cos a sin ^ ^
(since cos a = — cos A).
Note.* The method is readily adapted to angles of any sign and
magnitude, with due care as to sign. Thus,
in fig. (i.), a>270°, ^>90°;
ang. A'CB' = a + i3 = 360° + C;
.-.sin (a + /3) = sin (360° + C) = sin C
= sin A cos B + cos A sin B.
And sin a= -sin A, cos a = cos A,
sin /3 = sin B, cos ^= - cos B ;
.*. sin (a + |8) = sin a cos /3 + cos a sin /3.
In fig. (ii.), a<90°, i3>180°;
ang. A'CB' = a-/3 = C-180° ('.• j3>a);
.*. sin (a-j3)= -sin C
= - sin A cos B - cos A sin B.
And sin a = sin A, cos a= - cos A,
sin p= - sin B, cos /3= - cos B ;
.*. sin (a-/3) = sin a cos j3-cos a sin j3.
* Beginners may postpone the rest of this page.
P.O. L*
178 RATIOS OF a + /3, a-/3. [CH. VI.
Theorem 8. — ' If a, f^ are any two angles,
cos (a + ^) = cos a COS P- sin a sin ^ ;
cos {a- P) = cos a cos /? + sin a sin p.'
These are most easily derived from the sine forms.
Since cos 6 = sin (90° + 6), whatever ang. 6 is, (Th. 4) ;
Cos (a + ^) = sin(90° + a + /?)
= sin (90° + a) cos P + cos (90° + a) sin p
= cos a cos /5 — sin a sin p.
Cos{a-P)=sin{W + a-P) W
= sin (90° + a) cos p - cos (90° + a) sin yS" '
= cos a cos P + sin a sin p.
These results may be obtained geometrically by using (90° + a)
instead of a in the process of Th. 7.
Note. The forms for sin {a-p), cos (a-/3) may be derived from those
of sin (a + /3), cos {a + j3) by changing the sign of jS, since the original
process for sin (a + /3) is quite general.
Theorem 9. — ' If a, p are any two angles,
. , r,\ tan a + tan P , , n\ tan a - tan /? j
tan {a + P)= — ^ ; tan (a-P)= ^ ^-^•
^ '^^ 1 - tan a tan ^ ' ^ ^^ 1 + tan a tan p
■T:^ j_ , o\ sin (a + P) sin a cos ^ + cos a sin P
For tan {a + P) = j—roi = i -■ ^"S
^ cos (a + P) cos a cos p - sin a sm p
_ tan a + tan P
1 - tan a tan ^
(dividing each term by cos a cos P).
Prove similarly for tan (a - P).
EXAMPLES— XXXVIII.
1. Find sin 15°, cos 105°, sin 195°, cos ( - 15°) as surds.
(Use 60° - 45°, 60° + 45°, 45° + 150°, 45° - 60°.)
„ T) sin (a + |3) + sin (a-j3) ,
2. Prove ; . ^, ] ^ = tan a.
, cos (a + p) + cos (a-/3)
„ T> tana + taniS sin (a + S)
3. Prove 7 ^—- — ) ^.
tan a - tan ^ sm (a - p)
. p_ 1 + tan a tan (3 _ cos (a - ^)
1 - tan a tan ^ cos {a + p)
5. Find tan 165°, tan 105° as surds.
6. In any triangle tan A + tan B + tan C = tan A tan B tan C.
(Use tan C = - tan (A + B). )
rr r> 4. /^co A^ sin A + cos A
7. Prove tan (45 + A) = - - — .- ^•
sm A - cos A
CH. VI.] SUM AND DIFFERENCE OF SINES AND COSINES. 179
Theorem 10. — *If a, p are any two angles,
sin (a + j8) + sin (a - yS) = 2 sin a cos P
sin (a + /?) - sin (a — ^) = 2 cos a sin /?
cos (a + ^) + cos (a — ^) = 2 cos a COS /?
cos (a + p) — cos (a — /3) = - 2 sin a sin ^.' *
These result at once from the forms of Thh. 7 and 8.
Also, by multiplication, we obtain :
Sin(a + y8)sin (a-^)
= (sin a cos P + cos a sin ^)(sin a cos P — cos a sin ^)
= sin^a cos^P — cos^a sin^/3
= sin2a(l - sin^/?) - (1 - sin2a) sin2^
= sin^a - sin^^.
Cos (a + P) cos {a- P) = cos-a — sin^^ (similarly).
Theorem 11. — * If a, /? are any two angles,
• o o • OL + p a-P
Sin a + sin p = 2 sin — ^ cos — ~- ;
■ o o • «-/^ a + i^.
sm a - sin p = 2 sm — ^ cos — ^ ;
^ _ a+P a-P
cos a + COS p = 2 cos ' cos *^ ;
COS a - COS p = - 2 sm — ^ sm — ^.
Make y = ~2 ' ^ "^ ^^ ^ *'• 7 + ^ = "' ^^^ y-^ = Pf
.*. sin a + sin /3 = sin (y + 5) + sin (y - S)
= 2 sin y cos 5 (Th. 10)
= 2 sin — ^ COS .
Similarly the other forms may be derived.
EXAMPLES— XXXIX.
1. Express sin 60° + sin 72° as a product ; and sin 60° sin 72' as a
difference of ratios.
2. If 2 cos 54° cos a; =008 150° + cos 42°, find x.
- _, sin a + sin ^ , a + /3
3. Prove ^=tan ^
cos a + COS j3 2
. „ COS a - cos B ^ a- 8
4. Prove — -. — ^= -tan —zr^'
sm a + sm ^ 2
5. Prove sin2 75°- sin^ 37° = sin 112° sin 38°.
* It is important to remember the minus sign.
180 RATIOS OF MULTIPLES 2a, |. [CH. VI.
Theorem 12. — ' If a is any angle,
sin 2a = 2 sin a cos a ;
cos 2a = cos^a - sin^a
= 2 cos^a - 1
= 1-2 sin^a ;
2 tan a
tan 2a =
^ . a a
sm
a = 2 sm X cos - ;
A A
cos
„ a . „ a
a = cos^ ^ - sm^ -
J A
= 2cos2^-l
= l-2sin2^
tan
2tan^ ,
a= •
l-tan^l
1 - tan^a '
Make)8 = | /. a = 2^. Then,
(i.) Sin 2a = sin (a + a) = sin a cos a + cos a sin a
= 2 sin a cos a ;
and sin a = sin 2^ = 2 sin /3 cos /?
^ . a a
= 2 sm^cos -.
(ii.) Cos 2a = cos (a + a) = COS a cos a - sin a sin a
= cos^a - sin^a
= cos^a — (1 — cos^a)
= 2 cos^a - 1 ;
or cos 2a = cos^a - sin^a = 1 — sin^a — sin^a
= 1-2 sin^a ;
and cos a = cos 2/3 = cos^/? - sin^^
= cos2 ^-sin2 ^ = 2 cos2 ^- 1 = 1 -2 sin2 ^.
_ , . tan a + tan a 2 tan a
(iii.) Tan 2a=tan (« + «) = ^_^^^ ^ ^^^ a^l—^^'
^ 2 tan ^
, _^ 2 tan P 2
tan a = tan 2p = , — r — oo =
Ex. 1. K sin |=p, find cos a, sin a ; if cos ^ = ^, find cos 9, tan ^,
Ex. 2. Prove sin 3^ + sin = 4(sin ^-sin^^).
« « T^ i. tt si" '^
Ex. 3. Prove tan 5 = 7— -r—r*
*••*•• ^'^ 2 1+ cos a
CH. VI.]
EQUATIONS.
181
sm" ^ + cos
The solution of equations involving trigonometrical ratios is of
great importance in mechanics. It is effected by reduction to a
known type, as a quadratic ; by reduction to factors ; and in the
case of simultaneous equations by elimination, often a very difficult
process. We give a few examples.
Construction 6. — (i.) Solve the equation a cos 6 + b sin d = c.
Calculate a, where tan a = &/a, and r = >JaP- + 6^.
Then a = r cos a, b = r sin a ;
/»
.*. cos (a - ^) = cos a cos ^ + sin a sin ^ = -.
Find a-d from table, and subtract from a
.-. e=a-(a-e).
This may also be done as a quadratic :
a( cos^ ^ - sm^ ^ j + 2& cos - sm k — '"' ^^^
.'. (a + c) sin^ ■x-2b sin - cos ^ + (c - a) cos'^ « = ^ i
a n
dividing by cos^ - gives a quadratic in tan r,
z z
9 ft
{a + c) tan^ - - 2& tan ^ + c - a = 0.
(ii.) Solve the equation sin 5^ + sin 3^ = cos 6.
Then 2 sin id .cos 6 = cos 6 ;
:. either cos (9 = 0, (9 = 90°; or sin 4^ = 1 = sin 30° or sin 150°,
z
(9 = 7° 30' or 37° 30'.
(iii.) Solve the equations sin {6 + cf)) + sin (9 -<}>) = a,
b cos <fi + c sin 6 = d.
Then 2 sin ^ . cos <f> = a, and cos <^ = ;
.*. 2 sin 6{d - c sin 6) = ab, a quadratic in sin d,
EXAMPLES— XL.
Solve the equations :
1. 2 cos2^ - 5 cos - 3=0.
2. Sin^ + -^=2i
sin
3
3. Sin 6^ -sin 4^ = 2^^" ^'
4. 2 cos
1
2 cos 6
5. Sin ^ = 2 cos d.
6. Sin ^=4 cos e-l.
+ 1=0.
182 INVERSE FUNCTIONS — CmcULAR MEASURE. [cH. VI.
It is sometimes convenient to speak of an angle in terms of its
sine, cosine, and tangent. For this purpose inverse functions,
sin~^a;, cos~^x', tan'^x, &c., are used. These mean
* The angle whose sine, cosine, or tangent, &c., is x'
Ex. The slope of a hill is tan'^ — • How mucli does it rise in 10 miles,
measured up the hill ?
Circular Measure.
The general, as distinct from the merely geometrical, theory of
trigonometrical ratios is an important part of the general theory
of number. For this part of the subject the most convenient
measure of angle is the ratio of two lines, since an angle may
then be treated as the same kind of magnitude as its sine, cosine,
or tangent.
Definition 2. — The circular measure of an angle is the ratio
of the intercepted arc of a circle whose centre is its vertex, to
the radius of that circle.
If PQ, RS are the arcs of two such circles for an angle a,
and r, / their radii,
and p the measure of two right angles j _ _ ^ p
,, a PQ RS
then - = -
/) semicircle r semicircle r' '
p arc . 1
a = - X —^. — m each case.
TT radius
L..^S^.
Hence, by making tt the measure of two right angles, the
measure of a is — j-. — , which is the circular measure of a.
radius
Angles in circular measure are generally expressed in terms
TT IT
of tt; thus TT, -, ^ are 2, 1, § right angles.
2
The actual unit is a radian, and is equal to - right angles.
2
(For TT radians = 2 right angles, .*, 1 radn. = - rt. angs.)
Ex. The sun's disc subtends at tlie earth an angle of 32', and its
distance from the earth is 91,000,000 miles. Calculate its diameter.
(Treat the sun's diameter as an arc of a circle, rad. 91,000,000 mi.)
CH. VI.] CIRCULAR MEASURE — SINE CURVE. 183
TT, the circular measure of two right angles, is a very convenient
symbol for that angle ; but it must be remembered that in such
expressions as tt + a, a must be expressed in circular measure and
not in degrees. It is easy to convert angles from one system of
measure to the other by the following principle :
'An angle is the same fraction^ of two right angles in all
systems of measurement.'
Hence, if d is the number of degrees, c of radians — i.e. the
circular measure — of an angle,
d _c
180 "tt*
Ex. (i.). Express 77° in circular measure (as fraction of ir).
If c is the circular measure,
77
^ = "^180-
Ex. (ii). Express in degrees the unit of circular measure.
Here c=l.
180°
.-. unitof c.m. = — - = 180°-f3-141592...
TT
= 57°-295779...
= 57° 17' 45".
EXAMPLES-XLI.
1. Express in circular measure (fractions of tt) 60°, 75°, 108°, 135°.
2. Express in degrees y ; - ; y ; j^-
3. Find the length of a degree of longitude in latitude 60°. (Earth's
rad. 3960 mi.)
Sine Curve — Graphs of Ratios.
The graph of sin x (y = sin x), when x is the circular measure of
the angle, is the sine curve ; corresponding graphs give the cosine
curve and the tan. curve.
If graphs of sin x, cos x, tan x are made on tenth-inch squared
paper, one inch denoting unit number (sin - or sin 90°) ; then if
one tenth-inch denotes 6°, 1-5" denotes 90°; and in the true
sine, &c., curves, 1-57"... denotes ^, the difference being less than
5 per cent. These graphs, therefore, represent very nearly the
true sine, &c., curves.
* Used in the general algebraical sense.
184 EXAMPLES. [CH. VI,
EXAMPLES— XLII.
Trigonometry.
1. In a right triangle, if A = 90° and a—], the measures of sin B, sin C
are b, c. What are the measures of cos B, cos C ?
2. By Ex. 1 give a construction for measuring the sine and cosine of a
given angle.
3. In a right triangle, if A = 90° and c = l, the measure of tan B is b.
What is the measure of cot C ?
4. By Ex. 3 give a construction for measuring the tangent of a given
angle.
5. Construct angles of 32°, 49°, 127°, 152°, and tabulate their ratios by
measuring. (Make the hyp. the unit for sines and cosines, and the base
the unit for tangents. )
6. If A = 90° in a triangle ABC, show that b = a cos C = a sin B ; and
b=c tan B.
7. If « is a chord of a circle, radius r, show that aJ2r is the sine of the
angle in the larger arc of a.
8. Find by trigonometry the radius of the circumcircle of a regular
pentagon of 1" side.
9. Find the radius of the incircle of a regular hexagon of 3 cm. side.
10. Find the radii of circum- and in-circles of a regular polygon of n
sides, each side a.
11. Fiom the triangle whose base is the mean part of the other sides
derive sin 18°, and compare with the table.
12. How far must the side BC, 2 ft. long, of a square ABCD be
produced to P to make the angle BAP 60° ?
13. Calculate the chord of a sector of a circle of radius r, angle a.
What is the distance of the chord from the centre ?
14. Show that the cosine and sine of an angle can have any numerical
value between - 1 and + 1, but none > 1 or < - 1.
15. Show that any real number ±/i may be represented by the tangent
of some angle.
16. Given that sin a = -73, cos ^=-58, tan 7 = 1-62, construct the angles
a, B, y. Measure them.
A
17. In a triangle ABC, incentre I, show that inradius r=Al.sin g-,
A T
and tan — =
2 s-a
18. Given the sine of an angle, e.g. sin a = s, show how to construct
cos a, tan a. How can these be calculated Avithout construction ?
19. Given tan a = 3-5, calculate sin a, cos a.
CH. VI.] EXAMPLES. 185
2
20. Given cos a = ^, calculate sin a, tan a.
21. Show that cos^a^ 1 - sin'^a, and sin'^a = l - cos^a.
22. Show that sin^a + cos% = (sin a + cos a)(l - sin a cos a). Write down
the corresponding form for sin^a - cos^a.
23. Show that 1 + 2 sin a cos a = (sin a + cos of. Deduce a value for
1-2 sin a cos a.
24. Show that (sec a -tan a)^^- ;
^ 1 + sin a
25. Trace the changes in magnitude of cos a, sin a, tan a, as a increases
from 0° to 90°.
26. Write the follo\\ing in a line : sin 177°, cos 228°, tan 296° ; under-
neath write the sign of each ; underneath again write the acute angle
whose sine, cosine, tangent respectively has the same numerical value as
these.
27. By the aid of the tables (which are given up to 90°), and your
knowledge of the signs, write down the values of sin 250°, cos 140°,
tan 260°.
28. Solve, with the aid of tables, the equations
7 sin ^ = 3-06, 9 cos ^==6-02, 5 tan ^ = 1-457.
29. Solve (i. ) 3 sin^^ - 4 sin ^ + 2 = cos^^ ;
(ii.)sin2^=l-2cos3^;
(iii.) tan2^ + cot20 = 3J.
30. Trace the changes in magnitude and sign of sin a, as a increases
from 0° to 180°.
31. Repeat Ex. 30, for cos a and tan a successively instead of sin a.
32. Show that sin (-a)=-sin a; cos (-a) = cos a; and tan (-a) =
- tan a, where a is an acute angle.
33. If a, is some angle ^q^o- show that there is some angle j3<45° such
that cos ^ = sin a.
34. How could you from a table of sines find the values of cos 38°,
cos 47°, cos 75° ?
35. Show that if tables of sines and cosines are both printed for all
angles up to 45°, the sines and cosines of all angles up to 90° can be found
from them.
36. Prove cos a = sin (90° -a) when a^^^°;
and sin a = cos (90° -a) when a^i5?°.
<2/0
37. Prove cos a = sin (90° + a) > 90°
sin a = cos (90° + a) '^'^" "<180°'
'38. Prove cos a= -cos (180° -a) . >180'*
sin a = sin ( 180° - a) ^^^^^" "^ < 270-*
186 EXAMPLES. [CH. VI.
39. Find the sun's altitude — i.e. angle of elevation — when a post 18 ft.
high throws a shadow 10 ft. long.
40. A man 6 ft. higli standing 8 ft. from the base of a lamp-post has a
shadow of 10 ft. ; find the height of the post, and the angle it subtends
at the end of the shadow.
41. An iceberg has an elevation of 15° from a ship. At ^V "lile nearer
the elevation is 43^°. What is its height ?
42. If the earth's radius is 3960 miles, calculate the radius and circum-
ference of the parallel of latitude 60°. How much faster is a point on
the equator moving than a point in this latitude ?
43. At a point on one diagonal of a square fort, 73-2 yd. from the
nearest corner, the other diagonal subtends an angle of 60°. Find the
dimensions of the fort.
44. Find the angle which a chord of 1" subtends at the circumference
of a circle of -75" radius. (Draw a diameter.)
45. A trapezium ABCD has angles A, B right angles, and angle ABD
equal to BCD ; if also AD . BC = AB . CD, find this angle.
46. Verify your result in Ex. 45 by making AB 1", and measuring the
other sides.
(In Exx. 47 to 57, a, b, c. A, B, C, &c. are sides, angles, &c. of a
triangle ABC.)
47. {b + c) cos A + {c + a) cos B + {a + b) cos C = a + b + c.
48. c(sin A+p sin B) = {a+pb) sin C.
49. (6 + c)(l-cos A) = a(cos B + cos C).
50. If 25=2-75", a = l", A = 50°, calculates, R, a.
51. If ADj.BC, calculate AD, CD, given a=l-3", & = 1.7", B = 78°.
52. {c-b cos A) tan B = 6 sin A.
53. Tan A : tan B = (c^ + a' - 6^) : (6^ + c^ - a^).
54. Given a=157, 6=215, c=193, find the greatest angle.
55. Given A = 51° 36', B = 79° 23', c = 200 yd., find the other sides.
56. Given a =1570 m,, 6 = 2396 m., C = lll°, find c and a.
57. Given 6 = 51° 51', 6 = 213 yd., c^250 yd., find C and a.
58. In a quadrilateral ABCD, AB is a measured base line of 1000 yd.,
angs. DAC, CAB = 15° 18', 46° 42', and angs. CBD, DBA = 2r 12',
36° 30' ; calculate the sides and angles of the quadrilateral.
59. If two men start from the corner of two straight roads to walk
along them at 5 and 5^ miles an hour respectively, and are 30 miles apart
at the end of four hours, find the angle of the roads.
60. From the end A of a horizontal base line AB, 1000 m. long, the
altitude of a peak P is 21° 42' ; and the angles A, B which AB makes
with the horizontal directions of P at these points are 87° 24' and 72° 18'.
Find the height of the peak.
CH. VI.] EXAMPLES. 187
61. (i.) Sin (a + j8) cos a -cos {a + p) sin a = sin j3.
... , m a + /S, a-B cosa-cosj3
(11.) Tan— ^ tan — -^= ^.
^ ■* 2 2 cosa + cos)8
62. (i. ) Sin 3a = 3 sin a - 4 sin^a.
(ii.) Cos 3a = 4 cos^a-3 cos a.
63. Express sin {a + ^ + y), cos (a + /3 + 7) in terms of sines and cosines of
a, /3, 7; and find their values (i.) when a + ^ + 7 = 180°, (ii.) when a = /3 = 7.
64. (i.) Cos {n + l)d cos ^ + sin {7i + l)d sin ^ = cos nd.
(ii.) Sin 5a = 16 sin^a-20 sin^a + S sin a.
2. o ^ r>. o 1 - tan — tan —
nr r x • i ^ a tan B + tan C ^ B 2 2
65. In any triangle tan A = - 5— ^ — , ; tan —= --=- ^^ •
^ ° tan B tan C - 1 ' 2 , C A
tan — + tan
«« r,. IT sin3^ + sino^ cos2a-cos2j3
66. Simplify
2 ' •^" 2
sin 2 + ^°® 9) 5 write down the corresponding
value 01 1 - sin u.
68. Show that with proper signs to the roots,
2sin-=\/(l + sin ^) + v'(l-sin^). ,
/»rt CI- ^^£. cosa + sina cosa-sina
69. Simplify -. -.
cos a - sin a cos a + sm a
nn a' vr sin (a - /3) sin (/3 - 7) sin(7-a)
70. Simplify • ^ ^ + ^ — ^-'- + ^-^ '-»
cos a cos ;3 cos /3 cos 7 cos 7 cos a
71. In any triangle,
,. , sin B-sin C ^ A
(!• ) B ?^ = ~ ta,n 17 J
cos B - cos C 2
(ii. ) c2 = (a - 6)2 cos2 ~ + {a+ bf sin^ ^.
72. In any triangle,
A
cos —
^'•^ ^»-c~ . B-C
sm-^-
(ii.) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
73. A man sees a balloon due north at an elevation of 60°. Another
man half a mile to the west sees it N.N.E. What is its height and its
angle of elevation at the second point?
74. A house and a railway station are on opposite sides of a rect-
angular wood, sides 1 mile and 2 miles. Their directions fiom the
nearest corners of the wood are i)arallel, at 108° to the short sides
of the rectangle ; and their distances from these corners are \ mile
and f mile. Find the shortest road that can be made from house to
station without going through the wood.
188 EXAMPLES. [CH. VI.
75. Solve the equations :
(i.) cos 2^ = cos2^;
(ii.) tan 6. tan 2^ = 1 ;
(iii. ) 2 cos ^ cos 2d - cos 3^ = 1 ;
(iv. ) 1 - sin=*^ = cos=^^ - 2 sin ^ cos d.
2 3
76. If sin a = -, cos ^ — j, find the values of sin (a+j3) and cos (a-j3).
77. (i. ) (1 + cot ^ + tan ^)(sin 6 - cos 6) = ^^^"^^ ^^^^^
cos 6 sin 6
(ii. ) Cos 4^ = 8 cos^^ - 8 cos26i + 1.
78. (i.) Sin(a-i3) + sin(i3-7) + sin(7-a)= -4sin^^.sin^^sin^^^-
z z J
(ii.) Sin (a + /3 + 7) sin a = sin (a + /3) sin (a + 7) -sin j3 sin 7.
_. _. , , ^ V 1 - tan a . tan 3 - tan 8 tan 7 - tan 7 tan a
79. Prove tan a + /3 + 7 =- --^- ^ '— — ^ ;
tan a + tan /3 + tan 7 - tan a . tan /3 . tan 7
and deduce a value for tan 3a in terms of tan a.
-^ . . Sin g + 2 sin 3a + sin 5a _ sin 3a
Sin 3a + 2 sin 5a + sin 7a sin 5a
(ii. ) Tan a + tan (60° + a) + tan ( 1 20° + a) = 3 tan 3a.
81. {Sin (a + ^) + cos (a - i8)}{cos (a + ^3) + sin (a - p)} = cos 2/3 (cos a + sin a)^.
82. If a+j8 + 7=180°, prove the following equalities :
(i.) sin a + sin /3 + sin 7=4 cos | cos | cos ^;
..... . ^ . ^ . a . B 7
(11.) sin a + sm jS-sm 7—4 sm ^ sin | cos ~ ;
(iii.) cos 2a + cos 2)8 + cos 27= -4 cos a cos /3 cos 7-I ;
,. , a 3 7 ^ a + jS /3 + 7 7 + a
(iv.) cos p: + cos ^ + cos ^=4 cos —-.--- cos '-—J- cos --7— ;
2 2 2 4 4 4
(v. ) sin^a + sin2|3 + sin27 = 2 + 2 cos a cos ]3 cos 7.
4 12 '^4
83. If sin a=-=, sin p=j^, sin 7 = ^, and the angles are all acute, find
o ±0 ^o
the value of cos (a+/3 + 7) as a fraction. (See Ex. 63.)
84. If sin ^ + sin2^ = l, find sin 6, and show that cos2^ + cos4^=l.
85. If sin a + cos a=p, and tan a + cot a = 2-, show that q{p^-l) = 2.
86. If tan^a = 1+2 tan^/S, show that cos 2j3 = 1 + 2 cos 2a.
87. Solve the equations :
(i.) sin 11^ + sin 5^ = sin 8^;
(ii.) cos 5^ + cos 3^ + cos = 0;
(iii. ) cos20 + sin 6-1=0;
(iv.) tan20 + cot20 = 2;
(v.) (cos 0- sin 0) = cos a -sin a ;
(vi.) 5 sin = 4 sin {6 + <p), 7 sin = 4 sin 0;
(vii.) 8 cos + 12 cos = 13, 12 sin (0 + 0) = 13 sin 0.
CH. VI.
EXAMPLES. 189
88. If tan d = ~, and a=r cos 6, show that r=\/{a?-\-b^).
89. Plot the graphs y = &m x, 2/ = cos x from 0° to 360° ; and the graph
2/= tan X from 0° to 180°. (Use inch-squared paper, one horizontal
ten til inch for 6°, and one vertical inch as unity. )
90. Show from the form of your graphs that cos (90° + a) = - sin a;
sin (90° + a) = cos a.
f) /) f) /J
91. Sin ^ = 2" sin — cos -cos 2^ cos 2^. (Euler's theorem.)
92. Calculate the distance of the visible horizon from a point (i. ) 20 ft. ,
(ii.) 200 ft., (iii.) 2000 ft. above the sea-level. (Treat the earth as a
circle, and calculate the lengths of tangents from the points. Take
for earth's radius 3960 miles. )
93. Calculate the length of a degree at the equator.
94. If a chord of 1" is placed in a circle of 1-5" radius, what is the
length of its arc ? (Calculate the sine of half the angle at the centre.)
95. An arc of 5" subtends an angle of 40° at the centre of a circle of
radius 7-16". What angle does an arc of 5" subtend when the radius
is 3" ?
96. Calculate 30°, 45°, 150°, 225° in radians. (Use w.)
2ir 7 IT IT 2ir . ,
" -5 ' 18' 6' ¥ ^" '^'^'^^^-
97. If is the circular measure of an angle, tan 6>d> sin 6.
98. If a is any chord of a circle, radius r, the length of its arc is
2?' sin"^ p— • (Put sin 6=-^, then sin"^ 77- = ^, and 6 is the half angle
2r 2r 2r ' *=•
of arc. )
99. Calculate the radius of a circle at whose centre a length of 1 cm.
of arc subtends an angle of 1°.
100. If n is an integer, and a measured in radians, show that
cos (2w7r+a) = cos a; sin (?i7r-a) = sin a, n odd, and sin (W7r + a) = sin a,
n even ; tan (wx-f a) = tan a.
101. Give six solutions of each of the equations :
(i.) Sing = ^ ^ •
^ ' 4 cos
(ii.) Sin = 3 cos d.
(ilL) 4 COS20 + 4 cos 0-3=0.
190
PART III.
CHAPTEK VII.
MODERN GEOMETRY.
INVERSION— HARMONIC AND POLAR PROPERTIES-
CROSS RATIOS— INVOLUTION— PERSPECTIVE.
Definition 1. — The sign of a ratio PA : PB, in which any line
AB is divided by a point P, is positive or negative according as
the directions PA and PB are the same or opposite — i.e. as P is
external or internal to AB.
Definition 2. — Any arrangement of points in a straight line
forms a range, and any arrangement of concurrent lines a pencil.
Definition 3. — A transversal is a straight line cutting any
system of lines (curved or straight).
Definition 4. — The cross ratio of a four-point range A BCD is
AB CD
the ratio , ^' ^^ , written (ABCD).
AD.CB ^ '
This may be regarded as the product — x — — ,
or as the ratio AB . CD : AD . CB.
The 1st and 3rd points A, C are conjugates,
and the 2nd and 4th points B, D are conjugates.
Note. To write down the ratio (ABCD), set the points in order round
a circle, and start from the first in the order of the letters for numerator,
and in the reverse order for denominator.
Definition 5. — The cross ratio of a four- way pencil is that of a
transversal range of the pencil.
Definition 6. — A range or pencil is harmonic when it is
divided harmonically; in this case (ABCD)= -1, and A, C are
harmonic conjugates of B, D, and conversely.
Definition 7. — The polar of a point to a circle or conic is the
locus of its harmonic conjugate with respect to the points in
which any transversal through it cuts the circle.
The point is the polQ (Of its polar,
,f
CH. VII.]
INVERSION.
191
Definition 8. — Two points P, P' are inverse when PP' passes
through a fixed point I, the centre of inversion, and IP.IP' = ^-,
a constant in magnitude and sign, the constant or factor of
inversion.
If k = a^^ a is the radius of inversion. In this case Ic must be
positive.
Two figures are inverse when the points P, P', in which any
ray through a fixed point I cuts them, are inverse.
Ex. Two circles are inverae figures with respect to a centre of
simiUtude.
Theorem 1. — 'The inverse of
through the centre of inversion.'
a straight line is a circle
If I is centre, k constant of invn.,
IN perp. to given line PN, and N', P'
inverse points of N, P ;
then IN.IN' = /i; = IP.IP';
and tr. IP'N' ||| INP,
'.' IN': IP = IP': IN ; ang. I common:
.'. ang. IP'N' = INP, a rt. ang.;
.•. locus of P' is the circle on diam. IN'.
Note. If I is on PN, the inverse is that line ; and the inverse of the
centre I on a figure is the point at infinity on the tangent to the figure
at I.
Theorem 2. — 'The inverse of a circle is in general a circle,
the centre of inversion being a centre of similitude.'
The proof is similar to that of Th. 78, Ch. IV.
Note. If the constant of inversion is the square of the tangent to a
circle from the centre of inversion, the inverse of the circle is the same
circle.
Ex. 1. A pencil of straight lines inverts into a system of coaxial
circles.
Ex. 2. A tangent to a circle inverts into a circle touching the fii^st
circle.
Ex. 3. When is the inverse of a straight line a straight line?
Ex. 4. When is the inverse of a circle not a circle ? AVhat is the
inverse in this case ?
192
INVERSION — PEAUCELLIER's CELL.
[CH.
VII.
Construction 1. — * Inscribe a triangle, given the area and
one angle, on given curves.'
By inversion, we derive this construction from that for triangles
of given form. (Ch. V., p. 143.)
Thus, if A ife the given ang., A the given area,
of the triangle ABC,
make AC'. AC == ^, a constant ;
then, if C moves on any curve, C describes its
inverse, which is therefore a given curve.
Also
AC' AC. AC sin A Ic sin A
= constant ;
AB AB . AC sin A 2 A
hence the triangle ABC' has a given form, and can be inscribed on
the curves B, C'.
.'. the triangle ABC of given area can be inscribed on the given
curves.
Ex. Inscribe a triangle, one angle 30°, area 1 sq. in., one vertex given,
in an equilateral triangle of 2" side.
Theorem 3. — ' A BCD is a jointed rhombus ; opposite points
A, C are joined by equal rods to a fixed point I ; then, if B
describes any curve, D describes its inverse.' (Peaucellier.)
If AC meets BD in M,
I, B, D, each equidistant from A, C,
lie on MB, the rt. bisector of AC ;
and M is the mid point of DB ;
.-. rect. IB. ID = IM2-BM2
= IC2-BC2 = constant;
i.e. D is the inverse of B, constant IB ..ID.
If this system is constructed in wood or
metal, the inverse of any curve can be traced.
Note. If B moves on a circle, centre E, passing through I, the invei'se
of this circle is a straight line (Th. 1), so that in this case D describes a
straight line.
We thus have a mechanical construction for a straight line, by making
an additional rod EB, equal to El, turning about a fixed point E.
(This construction is not, however, independent of the planCy since
the circle on which B moves is a plane figure ; if B is not constrained
to move in a plane, it will describe any number of points on a sphere,
centre E, and its inverse D will describe a number of points in a
plane.)
CH. VII.] HARMONIC PROPERTIES OF QUADRILATERAL.
193
Theorem 4. — 'Any side of a quadrilateral is harmonically
divided by the opposite side and the join of the intersection
of the other two sides to that of the two diagonals.'
In the quadl. ABCD the side CD is cut
in*E, K by AB and FG. Then,
in tr. FCD, FK, CA, DB are concurrent,
and EBA is a transversal ;
ck da fb__cedafb
•• dk'.fa'cb" ~ de'fa'cb'
.*. ck:dk= -ce:de;
.*. (ECKD), and similarly (EBLA), is a harmonic range.
Note. FADCEB is called a complete quadl., with three diagonals
AC, BD, EF. Remember that the pencils at the additional points
E, F, G are harmonic.
Theorem 5. — * If M is the mid point of a pair of conjugates
A, C in a harmonic range ABCD, then' —
(i.) IV!B.MD = MC2 = MA2;
(ii.) MB.BD = AB.BC; MD . BD = AD . CD;
(B in each term) ; (D in each term) ;
(iii.) MB:MD = AB2:AD2 = BC2:CD2.
A ., .,., B C — - D
(i.) AB.CD = AD.BC, '.• AB:AD=BC:CD;
.*. (MC + MB)(MD - MC) = (MC + MD)(MC - MB) ;
.-, 2MB.MD = 2MC2 = 2MA2.
(ii.) MB . BD= MB(MD - MB) =-- MC^ - MB^
= (MC + MB)(MC - MB) = AB . BC.
Similarly, MD . BD = MD^- MC2 = AD. CD.
'MB\ MB . BD _ AB . BC _ AB^ _ BC^
B D ~ AD. CD ~ AD2 ~ CD^'
(iii.)
MB\ MB
Mb/ md"
Cor. — ' If MB . MD = MC^ = MA^, then A, C are harmonic con-
jugates of B, D.'
For make D^ harm. conj. of B to A, C ;
.*. MB . MDj = MC2= MB . MD, and D coincides with Dj.
P.O. M
194
POLE AND POLAR.
[CH. vir.
Theorem 6. — * The polar of a point to a circle is a straight
line, and is perpendicular to the central ray of the point.'
If PAB is a transvl. of P, and PCD
a transvl. through centre O of a circle ;
divide PCD, PAB harmly. at N, Q, so
that N, Q are on polar of P.
Then NC : CP = DN : DP = jtA, say;
.'. the circle (diam. CD) is locus of
points whose distances from N, P have
ratio [1. (Th. 61, Ch. III.)
.'. AN:AP = />t=BN:BP;
i.e. AN : BN = AP : BP = QA : BQ ;
.*. PN, QN bisect angs. N in tr. ANB,
(Th. 43, Ch. III.) ;
.'. QN ± PN ; also N is a fixed point ;
.*. locus of Q — i.e. polar of P — is a straight line, perp. to OP.
Note. Calling N, the point on the central ray of P, the foot of the
polar ; then since ON . OP^OC^,
Cor. (i.).— *A point and the foot of its polar to a circle are
inverse points to the centre, constant the square of radius.'
Cor. (ii.).— ' The polar of a point on a circle is its tangent.*
Theorem 7. — * If Q is on the polar of P to a circle, P is on the
polar of Q.'
Make PM perp. to OQ ; .'. tr. OMP ||| ONQ.
.*. OM:ON = OP:OQ;
.'. OM.OQ = ON.OP = OC2;
.*. M is foot of polar, and PM the polar, of Q.
Theorem 8. — 'The polar of any point to a circle passes
through the points of contact of tangents from the point ; and
through the intersection of the two tangents of any transversal
of the point.'
If the polar of P meets the circle at S ;
polar of S — i.e. the tangt. at S — passes through P.
If TE, TF are tangents at the ends of transversal PEF,
EF through points of contact is polar of T, and traverses P ;
.*. polar of P passes through T.
Definition 9. — A self-polar triangle of a circle or conic has
each side the polar of the opposite vertex (as PQR above).
CH. VII.] CONSTRUCTION OP POLAR — INFINITY. 195
Construction 2. — 'Construct the polar of a point P to a circle
or conic' "^
P Q
Draw two transversals PAB, PDC, * '
join AC, DB and AD, BC to meet in R, Q ; D^
.•, QR divides PAB, PCD liarmly. in the
quad!. ABCD ;
.'. QR is the polar of P.
Note. PQR is a self -polar triangle, and if P, Q, two vertices of a self-
polar triangle, are joined to a point A on the curve, and produced to
D, B, and PD produced to C ; then CB meets DA on polar of P — i.e.
in Q ; and DB passes through the 3rd vertex R. Hence :
Cor. — ' Transversals through a point on a circle or conic and
through two vertices of a self-polar triangle determine a chord
through the third vertex.'
Ex. Construct the pole of a line. (Draw polars of two points.)
Point at Infinity.
If P, P' are inverse points in a line, centre I, constant ^, to
every P corresponds one P'; and if IP is small enough, IP' is
greater than a given length ; hence, if P coincides with I, IP' is
greater than every length — i.e. infinite ; and we say,
* A straight line has one point at infinity.'
Examples. In Th. 5, if B coincides with M, its harm. conj. D is at
infinity, and DA : DC = BA : CB = I ; hence,
* The point at infinity on AB bisects AB externally.'
If ABCD is a harm, range, M the mid
point of AC, PQ pari, to AB ;
then if B coincides with M, MD is infinite,
and PD coincides with the pari. PQ since
any other line PE cuts AB. Hence,
'Two parallels meet in one point
at infinity.*
IfHK,Th.9,meetsVDatinf.inl;then(HBKI) = ^^j^ = j^ = (ABCD).
Hence Th. 9 is true when the transversal || one ray.
Ex. *The limit of a circl4 through a fixed point A, whose
centre O moves to infinity alwig a fixed line AC, is a perpen-
dicular line.' \
* In Ch. YIIL we show that conies bave the polar properties of a circle.
196 CROSS RATIOS. [CH. VII.
Theorem 9. — 'The cross ratio of a fourway pencil VA, VB,
VC, VD, in a given order, is constant.'
If ABCD is a transversal,
draw HBK pari, to VD ;
. AB_HB, CD VO/^
• • AD ~ VD' CB ~ KB '
. /A«^rv\ AB.CD HB
(ABCD) = =
"^ ' AD.CB KB
Similarly, any transvl. tbrongh B has cross
ratio HB : KB ; and any other transvl. || one of
these, and has the same cross ratio ;
.*. the cross ratio of all transversals is the same.
Theorem 10. — (i.) * The value /a of a cross ratio is unaltered
if the order of points is reversed, or if the points of each pair of
conjugates are interchanged.'
(ii.) 'ft is changed into - if the points of one pair of conjugates
are interchanged.' ^
For (i.) if (ABCD) = /z, reverse the order ;
^u /r^^oAX DC.BA AB.CD
then (DCBA) --- ^, „^ = ,^ ^ — = ii.
^ ^ DA . BC AD . CB ^
Interchange the points A, C and B, D of conjugates ;
CD AB
then (CDAB) = ' = ja = (BADC), reversing.
UB . AD
Note. The 24 possible orders of points A, B, C, D give only 6 different
cross ratios, since they are equal in fours.
(ii.) Interchange A, C or B, D only;
OB AD 1
•*• (^^^^) = 5d7aB = ]. = similarly (ADCB).
Note. The six values due to different orders of the points can be
derived from HBK above. Make KB = 1, HB=/i; then we have
fi {i.e. (ABCD)}, and I/ai, dividing HK at B,
\-li, {(ACBD)}, and 1/1-At, .. HB at K,
ixjix-l, {(ABDC)}, and (/*-1)/ai, m BK at H.
These are easily obtained by using the point at infinity on the pari,
transvl.
* Care must be taken to make all equalities true in sign.
CH. VII.]
CROSS
( UNI\
197
Theorem 11. — * A range or pencil is harmonic when, and only
when, its cross ratio is unaltered by interchange of one pair of
conjugates.' *
If {ABCD)=fi, and is unaltered by interchanging A, C ;
IM = (CBAD) = -, and /a^ = 1.
.*. /^= ±1-
But no cross ratio can have value + 1, since then
AB : AD = CB : CD, and either A, C or B, D coincide.
,', fjL= - 1, and the range is harmonic.
And if the range is harmonic, fi— —I ;
:. (CBAD) = - = - 1 = (ABCD) ;
i.e. a pair of conjugates A, C can be interchanged without altering
the value of the cross ratio.
Definition 10. — Two ranges or pencils A, B, C..., A', B', C'...
are homographic which have the cross ratio of any four elements
A, B, C, D of one equal to that of the corresponding elements
A', B', C, D' of the other.
Theorem 12. — 'Two homographic ranges which have one pair
of corresponding points coincident, have the joins of all other
pairs of corresponding points concurrent.'
If homographic ranges ABCD, AB'C'D have
A common, and (ABCD) = (AB'C'D') ;
join BB', CC to V, and VD to meet AB' in D^.
/. (AB'C'Dj) = (ABCD) = (AB'C'D') ;
.-. C'Di : ADj = CD' : AD' ;
.*. Dj coincides with D', and D'D traverses V.
Theorem 13. — * Two homographic pencils which have one pair
of corresponding rays coincident, have the intersections of all
other pairs of corresponding rays collinear.'
If homg. pencils V(ABCD), W(ABCD') have
ray VAW common, and V(ABCD) = W(ABCD') ;
draw BCA cutting VD, WD' in D, D'.
.'. (ABCD') = (ABCD), and D' coincides with D ;
i.e. the pair VD, WD' intersect on BC.
* This theorem gives a very simple proof of Th. 4, above.
198 CONSTRUCTIONS OF CROSS RATIOS — CROSS RATIO CURVE. [cH. VII.
>
Construction 3. — * Given three elements in a range or pencil
of given cross ratio /x, determine the fourth.'
If A, B, C are three points in a range,
draw HBK, making HB : KB eql. to fx;
draw AH, KC to V, and VD pari, to HK.
Then if I is the point at inf. on HK,
V(ABCD) = V(HBK.) = ^ = ^
If VA, VB, VC are three rays in a pencil, determine as above the
fourth point D on any transversal ; then VD is the fourth ray.
Ex. Show that there is only one solution.
Construction 4. — 'Construct the locus of a point whose
pencil at four fixed points, no three collinear, has a given cross
ratio /z.'
If A, B, C, D are the points, draw any
ray AP to A' on BC ; make (A'BCD') eql.
to fx, and join D'D to P on A A'.
.-. P(ABCD) = (A'BCD') = fx.
,'. P is a point on the curve.
And any third point Q on AP has a
different cross ratio Q(ABCD) ;
hence one only point (other tlian A) can
be found on every ray through A ;
and the curve can be plotted by finding a number of points.*
We may call this curve a cross ratio curve of the points.
Theorem 14. — * A cross ratio curve of four fixed points, no
three collinear, passes through each point.'
If A, B, C, D (last fig.) are the fixed points, P a point on the
curve, draw AD to D^, make (A^BCDj) eql. to /a=P(ABCD).
Then, if AP turns about A to coincidence with A^A,
A', D' coincide with A^, Dj,
D'P coincides with DjA, and P with A.
Hence A is a point on the curve.
Any transversal through A cuts the curve in two only points,
and AAj through two coincident points at A is the tangent at A.
We show in the next chapter that this curve is a conic.
* Pascal's theorem gives a simple construction for the point P.
CH. VII.]
PERSPECTIVE.
199
^S-^
Theorem 15. — 'If the joins A A', BB', CC of vertices of two
triangles are concurrent, the intersections P, Q, R of correspond-
ing sides BC, B'C', &c. are collinear; and conversely.' (Desargues.)
If BB', CC meet in V, and AA'
cuts BC, PQ, B'C in H, K, L ; then
(i.) If AA' passes through V,
pencil V(PBAC) is cut by BC, B'C,
.-. (PBHC) = (PB'LC);
.-. A(PBHC) = A'(PB'LC).
These pencils have HL common,
.*. the intersections of corresp. rays
are collinear ;
i.e. P, R, Q are collinear.
(ii.) If R is on PQ,
A(PRKQ) = A'(PRKQ);
.-. (PBHC) = (PB'LC) on transvls. BC, B'C
These ranges have P common,
.*. joins of corresp. points are concurrent ;
i.e. BB', AA', CC are concurrent.
This is the fundamental theorem in perspective.
Definition 11. — Two figures in a plane are in perspective which
have the joins of corresponding points concurrenV in ilie centre of
perspective, and the intersections of corresponding sides collinear
on the axis of perspective. ^
^Ti
Construction 5. — 'Construct the perspective of a given figure
ABC..., to centre V, axis PQ.'
Choose A' on VA to correspond to A, produce AB to meet the
^axis in R,
join RA' to meet VB in B' ;
.*. B' is the point corresp. to B,
and A'B' n side n AB.
Similarly any number of points, sides, or chords may be
constructed.
Ex. Show that the triangles RBB', CQC in the above figure are in
peispective. Hence derive (ii.) of Th. 15 from (i.).
200
PERSPECTIVE OF A FIGURE — PROJECTION. [cH. VII.
The process of deriving a figure as the perspective of another in
a plane is called plane projection, or simply projection.
Orthogonal or right projection is a particular case of this, in
which all rays through the vertex are parallel (the vertex being at
infinity), and perpendicular to the axis of perspective.
* Any polygon projects into a polygon, a curve into a curve,
a chord into a chord, a transversal into a transversal, a tan-
gent (limiting transversal) into a tangent, and a pencil into
a pencil.'
Theorem 16. — 'The intersections of corresponding pairs of
chords or tangents of two figures in perspective are correspond-
ing points in perspective.'
If PQ, PR a.iid pq, ]pr are corresp. pairs
of chds. or tangts. in persp., vertex V,
axis HK, meeting at P,p',
then QR, qr are also corresp. chds., and
intersect on the axis HK.
.*. trs. PQR, pqr are in persp.,
and Pp, cone, with Q^, Rr, traverses V;
i.e. P, p are corresp. points in perspective.
Theorem 17. — 'Corresponding ranges or pencils in perspective
are homographic'
For two ranges in persp. are formed on
transversals of a common pencil ;
and two pencils in persp. have a common
axial transversal.
Ex. 'The polar of a point to any per-
spective of a circle is a straight line.'
Note. We show in the next chapter that every conic is the" perspective
of a circle, and conversely ; we are thus able to deduce many properties
of the conic— e.g. its polar and cross-ratio properties — from those of the
circle.
The properties of the line at infinity are most important in
projection. The line of one figure which projects to infinity in
its perspective is called the vanishing line of the first figure,
because it does not appear in the second.
CH.
VII.]
PROJECTION — LINE AT INFINITY.
201
Theorem 18. — 'Points at infinity in a figure project into a
straight line parallel to the axis in its perspective.' -^
If V, HK are vertex and axis of persp.,
a the persp. of A, and K, A, a fixed points,
and HA any line through A in fig. A ;
then parls. V/, HA and Vj, KA meet in
points I, J at inf. in fig. A ;
hence i, j on Ha, Ka are the projns. of
points at inf. I, J in fig. A.
Also, j is fixed, *.' KA, Ka, V; are fixed ;
and ia : aH = Va : aA =ja : aK ;
,*. z}' II HK ; i.e. locus of i is a fixed pari, to HK.
Cor. — ' Points at infinity in a figure form a straight line.' *
(The straight line at infinity.)
Note, ij is the vanishing Une of fig. a.
Ex. ' A parallelogram projects into a complete quadrilateral.'
Construction 6. — 'Project a figure so that a given straight
line may be projected to infinity, and any two angles into given
angles.'
If I J in fig. A' is to project to inf. ;
choose any vertex V, any point a corresp. to A,
and nialie aH, aK pari, to VI, VJ ;
.*. IH:HA = Va:aA = JK:KA;
.*. HK, axis of persp., || I J.
Also, I in fig. a is on aH, VI ;
i.e. i is at inf. and is projn. of I ;
sinily., j is at inf. and is projn. of J ;
.*. IJ in fig. A projects to inf. in fig. a.
(ii.) Ang. HaK = IVJ. Hence, by choosing V on an arc IVJ of
given ang., the ang. HaK = given ang. Thus the ang. lAJ can be
projected into any given angle.
Also, by making two arcs IVJ, LVM, of given angles, to meet
in V, we can project any two angles I A J, LAM at A into given
angles at a.
Note. IJ is the vanishing line of fig. A.
Ex. Project the angles formed by a given angle and its bisectors into
two right angles. Also project any quadrilateral into a rectangle.
* They also form circles and other curves.
202
DIAGONALS OF QUADRILATERAL.
[CH. VII.
Also — — = -— =
(simr. trs.) j
Theorem 19. — 'The mid points of diagonals of a complete
quadrilateral are coUinear.'
If L, M, N are mid points of diags. AB, CD,
EF of a quadl. ; make APH, CK pari, to BE,
and DQK, BH pari, to CE.
Then M, L are diag. points of parms. EK, EH.
FBFDFQ
BP"'DA~QC'
FBBPBH
• FQ~QC~QK
.'. KHF is a str. line ;
also M, L, N bisect EH, EK, EF;
.*. L, M, N are collinear.
The following generalisation by projection is a good illustration
of the process, and also of the importance of the line at infinity.
Theorem 20. — * If three points on the diagonals of a complete
quadrilateral are collinear, their harmonic conjugates are also
collinear.'
If X, Y, Z are collinear points on diags.
AB, CD, EF of a quadl.,
and X', Y', Z' their harm, conjs. ;
project the figure so that XYZ projects
to infinity ;
.*. X', Y', Z' project into mid points
of diags. L, M, N of the new fig. (see
above), which are collinear j
,*. X', Y', iJ are collinear.
Perspective op a Circle.
In any conic or circle, every central chord is bisected at the
centre; hence the harm. conj. of the centre bisects each chord
externally and is at infinity ; the polar of the centre is therefore
entirely at infinity, i.e. —
'The centre of a conic or circle is the pole of the line at
infinity.'
We may define the centre of a perspective of a circle as the
pole of the line at infinity, and diameters as central chords.
CH. VII.]
CROSS RATIO IN CIRCLE — PASCAL.
203
Theorem 21. — * The cross ratio of the pencil from any point
on a circle or conic to four fixed points on it is constant.'
If A, B, C, D are the fixed points, and if
P is on arc AD, Q on another arc AB, and
AQ is produced to H ;
then ang._HQB = suppt. of AQB = APB ;
ang. BQC = BPC; CQD = CPD;
/. pencil Q(ABCD) = P(ABCD).
Thus, wherever P is in AD, and simly.
in any other arc,
P(ABCD) = Q(ABCD) = constant.
Also, if Q moves up to coincidence with A,
AQ becomes the tangent AT ;
.-. A(ABCD) = A(TBCD) = P(ABCD).
Thus the cross ratio has the same value when the fifth point
coincides with one of the four.
Note. By projection this theorem is true for any perspective of a
circle.
Theorem 22. — Pascal's theorem : ' The intersections of oppo-
site sides of an inscribed hexagon of a circle or conic are
coUinear.*
If the opp. sides of ABCDEF m
circ. meet, AB, DE in L ;
BC, EF in M ; CD, FA in N ;
and BC, EF meet AF, AB in H, K ;
then C(ABDF) = E(ABDF) ;
.*. on transversals AN, AL,
(AHNF) = (ABLK), with A common;
.*. HB, NL, FK are concurrent;
i.e. L, M, N are collinear.
Note. This theorem remains true when one of the points moves to
coincidence with the next. Thus the quadl. ABCD with the tangents
BT, DT form a Pascal hexagon ABBCDD ; and the intersections of
BB, DD ; AB, CD ; BC, AD, are concurrent, BB, DD being the tangents
at B, D, and T their point of intersection.
The polar properties of a circle or conic can be thus derived. (See last
page of chapter.)
204
CROSS RATIO OF TANGENTS — BRIANCHON.
[CH. VII.
Theorem 23. — ' The cross ratio of the range formed on any
tangent of a circle or conic by four fixed tangents is constant.'
If A A', BB', CC'; DD' are fixed tangents of
a circle, centre O, TA, T'A' any two tangents ;
ang. AOA' = lTOr (Ex. XXVI. 10),
= BOB' (similarly) ;
.-. ang. AOB = A'OB'; BOC = B'OC', &c. ;
.'. pencil O(ABCD) = O(A'B'C'D') ;
.*. (ABCD) = (A'B'C'D') = constant.
Note. By projection this theorem is true for
any perspective of a circle.
Theorem 24. — Brianchon's theorem : ' The joins of opposite
vertices of a circumscribed hexagon of a circle or conic are
concurrent.'
If ABCDEF is a circumhexagon of a
circle, and AD, BE meet in P ;
the ranges in which tangts. AF, BC meet
the other four are homographic ;
/. E(BCD'F') = D(AC'E'F),
and these pencils have ray D'E' common ;
.•. intersections of corresponding rays are
coUinear ;
i.e. P, C, F are collinear ;
i.e. AD, BE, CF are concurrent.
Note. By the aid of pole and polar this can be derived from Pascal's
theorem.
If tangents at A, B, C..., in Th. 22 above, form a hexagon UVWXYZ,
UX is the polar of L ; VY of M ; WZ of N ; these all traverse the pole
of LMN, and are therefore concurrent.
Also, Thh. 22, 24 are true for any orders of the letters. There are
60 different Pascal lines, obtained by arranging the vertices in all
possible orders. An interesting account of these lines is given in
Salmon's Conies, p. 260 (fifth edition).
In order to apply the theorems, write down the points or sides in
suitable order— e.g. A_X P^^L^R Z ; * opposite sides or vertices are
AX, LR ; XP, RZ ; and PL, ZA.
* Or round a circle.
CH. VII.] INVOLUTION. 205
Definition 12. — An involution range or pencil is a system of
pairs AA', BB', CC... of points or rays sucli that any four are
homographic with their conjugates.
AA', BB', CC are pairs or conjugates of the involution.
The centre of an involution range is the conjugate of the point
at infinity.
Cor. — 'Any transversal of a pencil in involution is cut in
involution.'
An involution is separate when the points or rays of one pair
are one internal and one external to those of any other pair ; and
continuous when the points or rays of one pair are both internal
or both external to those of another pair.
Theorem 25. — (i.) 'Inverse points in a line to a given centre
form an involution range.'
(ii.) 'Pairs of an involution range are inverse points to the
centre of the range.'
(i.) If AA', BB', CC' are inverse points
to centre I, const. A",
draw a circle AA'V, and through V the
circle VBB' cutting VI in W ;
.-. IW. IV = IB.IB'=IA.IA';
hence W is also on circle VAA';
and similarly on circ. VCC, &c.;
.*. ang. A'VB' = VB'I-VA'B'
= IWB - IWA = AWB.
Similarly, B'VC'=BWC, &c.;
.-. pencil V(A'B'C'...) = W(ABC...),
.'. (A'B'C'D') = (ABCD) &c., and (AA', BB'...) is in involn.
(ii.) If AA', BB', CC',... are pairs in involution,
draw circles AA'V, BB'V to meet in W, draw VW to I ;
/. IA.IA' = IV.IW-IB.IB'.
Find Cj so that IC . ICi = lA . IA' = IB . IB';
.*. AA', BB', CCj are pairs in involution, by (i.) ;
.-. (AA'BCj) = (A'AB'C) = (AA'BC'), by Del 12;
/. C' coincides with C^; and IC . IC' = IA. lA', &c.;
.*. I, conj. of the point at inf. on AA', is centre of the involution.
Cor. — 'Two pairs AA', BB' determine an involution.'
Ex. Prove wlien A, A' are one internal and one external to BB'.
206 DOUBLE POINTS AND RAYS. [CH. VII.
Definition 13. — The double „ „ of a ^^^S® jj^ involution are
rays j)encil
self-conjugate ^ ; i.e. each is its own conjugate.
Definition 14. — Real points in a plane are points whose dis-
tances from some given point are measured by a real number /x ;
and real lines are lines having real points continuously to infinity
in either direction.
Real lines and points can be constructed by ruler, compass, &c.
Imaginary points in a plane are points whose distances from
some given point are measured by ju,^— 1, where /x is some real
number; and imaginary lines in a plane through a given point
are lines all of whose points are imaginary except the given point.
Imaginary lines and points can be imagined, but not constructed
by the ordinary constructions for real points.
Ex. V^ is represented by a turn through a right angle. Justify this.
Theorem 26. — ' A ^gjjg^i in involution has one pair of double
pom s . ^jjggg g^j,Q harmonic conjugates of any pair, and are
real or imaginary according as the involution is continuous or
separate.'
If I is centre of involn. range or transvl. :
(i.) If the involn. is continuous, circles AAV,
points V, W on the same side of AB (1st fig.);
and I is external to A A' and to BB'.
Draw circs. VW touching AB at D, Dj ;
.'. 10^= IDi^ = lA . I A', which is positive,
= -I- jtx^j jtx a real measure.
/. ID= +/x= -IDp
Thus D, Dj are real points.
(ii.) If the involn. is separate (2nd fig.),
circs. AA'V, BB'V meet on opp. sides of AB ;
and I is internal to A A' and to BB'.
Hence if 5, 5j are double points,
182 = |g^2 ^ i;^ ^ i;^'^ which is negative,
= — ju,2, where /a is a real measure ;
/. |8 = ^ ^ - 1 = - I8j ; and 8, 8j are imaginary.
Since I A . IA'= ID^ or 18^ ; D, Dj or 8, 8^ are harm, coiijs. of A, A'.
A/
v'v,
BBV
meet in
y
Kw
D y
IB A
A' B'
/
1
I
/^ ^
N
\
\
J
B'
\
\
v.^
a\;b
/
CH. VII.]
IMAGINARY POINTS SELF-POLAR TRIANGLES.
207
Imaginary points enable us to make our theorems continuous.
Thus we say :
*A straight line meets a circle or conic in two real, two
coincident, or two imaginary points ; '
so that a straight line always meets a circle or conic in two points.
Construction 7. — ' Determine the imaginary points in which a
line outside a given circle meets it.'
If PN cuts a circ, centre O, in 8, 6j,
if ON J_ PN, and r is measure of radius ;
/. N82 = r2-ON2 = -ftS say, /x real;
thus N8= J-l jOV\^-r^= - NSj.
Ex. If Q is pole of PN, then Q5, QS^ are tangents from Q. (Cale. QS^.)
Theorem 27. — 'The pencil of sides of self-polar triangles of
a circle at a common vertex is in involution.'
'A line in the plane of a circle is cut in involution by the
sides of self-polar triangles of its pole.*
If P is the vertex of self-polar trs. PAA',
PBB', &c., of a circle, centre O, and N the foot
of its polar AA',
draw A'PM perp. to OA, (polar of A) ;
then NA:ON = MP:MO = NP:NA';
.*. NA. NA' = ON . NP (in niagn. and sign)
= simly. NB.NB', &c.;
i.e. P(AA', BB'...) and (AA', BB'...) are in
involn.
Note. If P is a point outside the circle, from
which tangents PT, PT' can be drawn, T, T' are
the points in which AA', polar of P, meets the
circle ; also, if A moves to coincidence with T, A' coincides with T,
and PT and T each is its own conjugate.
If P is inside circle and 5, 5^ double points of (AA', BB'...),
N52=ON.NP =-ON(ON-OP) = ?-2_oN2. Hence,
Cor. — * The double ^°^^ ^ of the self-polar involution of a
rays
B A
208 INVOLUTION PENCIL — RIGHT-ANGLED PAIR. [CH. VII.
These properties of self-polar triangles of a circle remain true in
projection; that is, they are true of any perspective of a circle,
except that the centre N of the involution on AA', though on the
central ray through P, the pole of AA', is not generally the foot of
the perpendicular from the centre or the point P on the line AA'.
On account of their frequent occurrence in conies, we shall
use the following abbreviations :
s.p. tr. for self-polar triangle ;
s.p. involn. of a point for the involution of sides of self-polar
triangles with this point as vertex ;
s.p. involn. of a line for the range on it of the s.p. involn. of
its pole.
Theorem 28.— 'Every pencil in involution has one right-
angled pair, and is entirely right-angled when two pairs are
right-angled.'
If AA' is a transvl. of the pencil in involn. _
V(AA', BB'...),
draw circs. VAA', VBB' meeting at W ;
determine I, centre of involn. (AA', BB'...).
Draw circ. VWR, centre on A A', cutting AA'
in R, R' ;
.*. RVR' is a rt. ang., •/ RR' is a diam.,
and IR.IR' = IV.IW=IA.IA', &c.
.-. VR, VR' is a rt.-angled pair of V(AA', BB'...).
(ii.) If a second pair VA, VA' is rt.-angled,
AA' is diam. of circ. VAA' ;
.-. AB, line of centres of VAA', VRR', is rt. bisr. of VW;
.'. centre of circ. VBB' is in AB, and BB' its diam. ;
.*. VB, VB', and simly. any other pair, is right-angled.
Note. As any two angles at a point can be projected into right
angles, any invohition pencil can be projected into a rt. pencil. But the
projection is real when only the circles which determine the vertex of
projection (Constr. 6, ii.) meet in real points— i.e. when the involution
is separate and not continuous.
Ex. Show that every right pencil is in involution.
Can the double rays of a right pencil be represented ?
CH. VII.] CIRCLE — CENTRAL INVOLUTION — CIRCULAR POINTS. 209
Theorem 29. — 'The self-polar involution of the centre of a
circle is right-angled.'
If I is the point at inf. on a central ray Ol
of a circle, then OJ, the polar of I, traverses O,
and XOI.
Sinily., if J is at infinity on OJ,
Ol is the polar of J, and lOJ a s.p. tr. ;
.*. each pair of the s.p. invohi. of O is rt.-angled.
Theorem 30. — * Two circles meet in four points ; two at
infinity, imaginary, and two other real or imaginary points.'
If O, O' are centres of two circles, and lOJ,
ro'J' right angles of pari, sides ; then
Ol, O'r, and simly. OJ, O'J', meet at inf.;
.'. s.p. involns. of line at inf. of the circs. O and
O' are congruent ;
.'. they have the same double points, which are
common points of the two circles ; also these
common points are imaginary, because the
involn. is separate.
These are the circular points at infinity.
The other two common points of the circles are the points,
real or imaginary, in which the radical axis of the circles meets
them.
Note. As the line joining O or O' to a circular point at infinity is a
double ray of a right-angled pencil, it is its o\vn perpendicular ; hence a
remarkable property of the circular points at infinity :
'The line joining any point to a circular point at infinity is
perpendicular to itself.'
The construction of rays or points in involution reduces to that for a
radical axis of two circles.
Thus, in Th. 25, the pairs A A', BB', &c. are the intersections of
the transversal AA' with the coaxial circles AVW, BVW, &c., whose
radical axis is VW ; and the radical axis VW is itself the limiting
circle of this system, whose centre is at infinity on the right bisector
of VW.
P.O. N
210
INVOLUTION — CONCURRENT CHORDS.
[CH. vir.
Theorem 31. — 'A ^^^^^i of three pairs is in involution if
the cross ratio of any four Po^^^s ^g equal to that of their
conjugates.' ^^^^
If A, A'; B, B'; C, C, are three pairs
of points of the range or on a transvl. of
the pencil ;
and if, for example, (AA'BC) = (A'AB'C),
find I, Ci so that lA . lA' = IB . IB' = IC . IC, ; XK'' 'c ^ ^"v
.'. (AA', BB', CCj) is in involn., I A B c c' b' a'
and (A'AB'Ci) = (AA'BC) = (A'AB'C) ;
.*. C' coincides with Cp and (A A', BB', CC') is in involn.
Note. This is the most useful form of test for an involution, derived
from its homographic property.
Theorem 32. — * A system of concurrent chords of a circle or
conic subtends a pencil in involution at any point on the circle.*
If clids. A A', BB', CC'... of a circle meet
at Q, and BC' meets AA' in D ; A
then, if P is a point on the curve,
P{AA'BC) = C'(AA'BC) = (AA'DQ)
= B(AA'DQ) = B(AA'C'B')
= P(B'C'A'A), reversing,
= P(A'AB'C'), interch. conjs. (Th. 12.)
.'. P(AA', BB', CC'...) is in involution.
Theorem 33.— 'A transversal is cut in involution by a circle
or conic and the opposite sides of an inscribed quadrilateral.'
If transvl. AB cuts circle and sides of
quadl. PQRS in A, A', B, B', C, C; P.
(AA'BC) = P(AA'BC') = P(AA'QS)
= R(AA'QS) = (AA'CB')
= (A'AB'C). (Th. 12.)
Note. These theorems, 32 and 33, can be
proved in exactly the same way for a conic,
Ex. Show in the same way that the diagonals PR, SQ determine a
pair in the involution. What property of a quadrilateral can you
deduce? Of a system of conies through four points? What will the
double points represent in this system ?
v
CH. VII.] RECTANGLE OF PARTS OF TRANSVERSAL. 211
Construction 8. — ' Construct the rectangle of parts from any
point of a transversal to a circle or conic'
(ii.) 'Construct the rectangle of the parts from the double
points of an involution range to any point in the range.'
If V is a point on a transvl. of a circle or an involn. range,
and if I is the centre of the given invohi. y
or of the s.p. involn. of the transvl., 5 T"*"
determine V', conj. of V.
(i.) The involn. has real double points y j
D, Dj, intersections of transvl. and curve ;
.-. D, Dj are harm, conjs. of V, V. (Th. 29.)
.'. VD . VD' = VI . W. (Th. 5, ii.)
(Note that V is first letter in each term.)
(ii.) The involn. has imaginary double
points 8, 8j, intersections of transvl. and
ciirve."^
Construct two points P, O (P between
I, O) on the perp. lO, such that Ol . IP = IV . IV' (I in each term) ;
find C, Cj such that OC2 = OP . Ol = OCj^
draw circ, centre O, rad. OC ;
.'. VV is polar of P to this circle ;
also IV.IV' = OI.IP;
.*. V, V' are a pair in the s.p. involn. of VV' to circle (Th. 27.),
and VPV' is a s.p. triangle.
.'. 5, 8j are the intersections of VV and circle.
Hence, if VP cuts circle in E, Ej,
VS . VSj = VE . VEj (rect. property of circle)
= VP. VM, by (i.) (E, E^ double points on transvl. VP),
= VI . VV, in cyclic quadl. IPMV'.
Hence the following simple principle :
Theorem 34. — 'The rectangle of the parts from a point of
a transversal to a conic or circle is equal to the rectangle of
the parts from the point to its own conjugate and the centre
of the self-polar involution of the transversal.'
This gives a simple proof of Appolonius' theorem, Ch. VIII.,
Thh. 11, 31.
* We may prove (ii.) without circle, thns : since I bisects l^i,
V5 . v^i=vi2 - 152= VI2- IV. iy'= VI2+ VI . iv'= VI . w.
212
PURE GEOMETRY : ABSOLUTE THEOREM. [CH. VII.
Our treatment of similar figures in Chh. III., V. rested on
certain theorems (Unit, Alternando, Summation) which we derived
in Ch. Y. from the properties of number. But by modifying the
fundamental axiom "^ of magnitude, we can derive all properties
of figures by pure — i.e. self-contained — geometry, without the
aid of number, as follows.
Archimedes' Principle. — ' Any magnitude, however small, can
be so repeated that the whole exceeds any given magnitude of
the same kind.'
Hence, if the difi'erence d of magnitudes X, Y cannot be so
repeated as to exceed a given magnitude Z, d must be zero
magnitude, and X = Y.
I. Repetition Theorem. — If the adjoining figure represents a
mode of repetition of a length d along a line,
parallels at successive points determine a like mode a^
of repetition of some length a whose whole repeats
into a given length Z. Hence :
'All modes of repetition of any length d are included in
those of all lengths a which repeat into a given length Z.'
II. Absolute Theorem. — * If a constant length PU, called the
absolute, is drawn from a fixed point P in a fixed line AB, and
AV parallel to BU meets PU in V, then PV is constant, and is
called the relative of PA to the absolute of PB.'
If PU' = PU, AV II BU', and py ^PV = d;
and if a is any length repeating into
PU, or PU' ;
then the equidistant parls. to BU, BU',
dividing PU, PU' into parts a, are con-
current, two and two, on PB, and have
V, V' between corresp. parls. of the two
systems.
/, cZ = PV-'PV'<a;
repeat d so that a repeats into PU ;
.-. whole of «^<PU;
and this is true for all values of a;
i.e. for all modes of repetition of d ;
.*. d is zero, and PV = PV' = const.
* Gr. u^iufAtx. or xo/v^ hvoia. (common notion).
CH. VII.] PURE GEOMETRY : RATIO BY GEOMETRY. 213
If PA'B' is any second transvl. of parls. AA', BB' (PA:t>PA'),
draw PAjBi to AV, BU, so that PBi = PB' (last fig.);
;. PA^ = PA' (as PV = PV' above) ;
.-. relative of PA' to abs. of PB' = rel. of PAj to abs. of PBj = PV.
Cor. — 'The relative of one part PA of a transversal from
a fixed point to two fixed parallels, to the absolute of the other
part PB, is constant.'
Ex. If PBi= - PB, prove PVi= - PV.
Geometrical definition of ratio. — The ratio of a line X to a
line Y is the relative of X to the absolute of Y.
Take X as PA, constr. PB = Y, PU the abs.; then PV = X : Y.
Rule of Signs. — PU is positive ; hence we see at once, drawing
PB theopp. way to PA, that X:Y= -(X:-Y)= -(-X:Y).
III. Unit Theorem.— (i.) ' If X : Y = Z : Y, then X = Z.'
If X alters to Z, Y const., PB, PU remain fixed, and A moves;
/. the ratio PV alters and Z : Y =(= X : Y ;
hence X does not alter, and Z must be equal to X.
(ii.) ' If X : Y = X : Z, then Y = Z ' (similarly).
IV. Proportional division, similar figures, and the rectangle
theorem (Eucl. YI. 2-18) derive from Absolute Theorem, Cor.,
exactly as in Ch. III., Thh. 37-53.
V. Altemando, Invertendo.
If X : Y = Z : W ; then rect. XW = YZ = ZY (Ch. III., Th. 52) ;
/. X : Z = Y : W. (Alternando.)
Also YZ = XW ; .-. Y : X = W : Z. (Invertendo.)
VI. Summation.— If X^ : Y^ = Xg : Y^ = &c.,
set off Xj, Xg, Xg..., Yj, Yo, Yg... along the sides of
an angle O, join corresp. points of divn. AP, BQ... ;
then Xj : Xg = Yj ; Yg, Altemando ;
.-. AP II BQ II CR (simly.), &c. ;
.-. (X, + X2 + X3...):X, = (Yi + Y2 + Y3...):Yi (Prop, divn.);
/. (Xi + X2 + X3...):(Yi + Y, + Y3...) = Xi:Yi = &c.
Similarly componendo, dividendo, ex aequali.
214 PURE GEOMETRY : CROSS RATIO. [CH. VII.
To make this chapter ' pure geometry ' — i.e. geometry of figure
ouly — the only alterations required in the text are to replace the
definitions of cross ratio and harmonic division by geometrical
ones, and modify the proofs of Thh. 4, 6, 7, 8, 5, 27.
Definition. — ' The cross ratio of a pencil V(ABCD)
is HB;KB, where HK||VD.'
'The cross ratio of a range (ABCD) is that of
any pencil on it V(ABCD).' "^
The fundamental theorems follow as in text.
' A pencil is harmonic when the interchange of two conju-
gates leaves its cross ratio unaltered.' (HB : KB= KB : HB, and
HB = BK.)
(A) Th. 4. EHGK is a transversal of
quadl. ACBDEF ;
.-. (FDKB)-G(FDKB) = (FCHA),onFC,
= E(FCH A) = (FBKD), on FB ;
i.e. (FDKB) is unaltered on interchang-
ing conjs. D, B.
.'. (FDKB) is harmonic.
(B) Thh. 6, 7, 8. Polar properties.
If CD is a fixed chd., AB any chd. of circ. or conic, through
fixed point G; tangents CT, TD and quadl. ACBD form inhexagon
CCADDB ; .'. intersections of opp. sides
CC, DD, i.e. T^
CA, DB, i.e. F y are collinear (Pascal, proved as in text) ;
AD, BC, i.e. E J
i.e. T, a fixed point, lies in EF, which also contains L, a fixed point
(harm. conj. of fixed point G to fixed points C, D).
.'. EF is a fixed line, cutting AGB harmy. at M ;
.'. polar of G, locus of M, is a str. line.
(C) Th. 5. Make MAB a diam. of a circle, and use polar and
tangents of M ; results follow easily.
(D) Th. 27. If GEF, GE'F' are s.p. trs. of G, the joins of vertices
EF to a fixed point C on curve determine chds. AB, A'B', &c.
through G (Constr. 2).
/. C(EF, E'F'...) and .'. G(EF, E'F'...) is in involn. (Th. 32).
* Just the inverse order of our Defs. 4, 5 above.
CH. VII.] PURE GEOMETRY, 215
The definition of ratio and the properties of proportional lines
given above are sufficient for all pure geometry — i.e. for the
geometry of figure; and the tendency of modern geometry as
treated by Townsend, Chasles, Cremona, and others has been to
make it entirely a matter of figure, independent of number. The
method, however, is only complete if and when we succeed in
defining ratio or deriving the theorems of proportion entirely by
figure, as we have just done.
jS'ow that this is done, it becomes possible for every truth of
geometry, whether originally stated and arrived at by extraneous
aid — e.g. that of number — or not, to be expressed as some property
of a figure ; and ' no truth of geometry arrived at by extraneous
aid can tell us about a figure any fact which cannot be arrived
at by geometry only.'
This is very well illustrated by our two treatments of cross
ratio.
In Def. 4 we defined the cross ratio (ABCD) as the ratio of
rectangles AB . CD : AD . CB. Then, in order to reduce this to
the ratio of two magnitudes of another kind (lines HB:KB), we
require a nexus associating ratios of different kinds of magnitude.
This nexus is number.* But we could get no property of figure
out of our definition until we had eff'ected this reduction.
Similarly with the alternative definition (product of ratios), we
can get no property of figure until we reduce to a simple ratio.
By using our second definition (the ratio HB : BK), we have just
shown that all results flow directly from the definition ; and it is
obvious, on comparing the two processes, that that of pure geometry
is the more elegant.
This does not, however, make these numerical processes useless ;
they often guide us to the right direction in which to look for a
geometrical solution of a problem. What we ought to do in any
case where we find some geometrical truth by, say, the ratio of
two areas, is to look for the geometrical solution.
Treated entirely geometrically, Euclidean geometry is a complete
self-contained science.
* Euclid's critical word in defining ratio is TriXiKorrira — * how-many-times-
ness' — and he no doubt regarded ratio as a 'quotient,' though number was not
completely understood in his day.
^l6 EXAMPLES. [CH. VII.
EXAMPLES— XLIIL
1. Having regard to sign, show that if A, B, C are points in a straight
line in any order, AB + BC = AC. Extend the theorem to n points.
2. Show that, with proper signs, the sum of projections of ?i-l
consecutive sides of an ?i-sided polygon is equal to that of the nth. side.
3. If C is a point in a straight line AB such that AC + BC = 0, and
P is any other point whatever in the line of AB, show that AP + BP
= 2CP.
4. If AB, A'B' are any two parts of a given line, M, M' their mid
• ^ xu KiiKii/ ..1 AA' + BB' AB' + BA'
points, then MM = either or — —
5. If At, V are any two real numbers, O a point in a line AB such that
fiPKO + vBO-O, and P any point whatever in the line of AB, show
that ixJKP + j'BP = (/* + »')OP. Explain the result when fx + v i& zero.
6. If the circumcircle of a triangle inverts about a vertex into a
straight line through its own centre, find the constant of inversion, and
construct a square equal to it.
7. With a vertex of a triangle and the square on a tangent from it to
the incircle as vertex and constant of inversion, the incircle inverts into
itself. Find the inverses of the opposite side and ecircle.
8. Any two circles are inverse figures about each centre of similitude.
Prove when the circles cut.
9. If two circles are inverse figures, the centre of inversion is a centre
of similitude. (Use the common tangents. )
10. Any two circles can be inverted into themselves. What is the
locus of the centre of inversion ? Extend the theorem to three circles.
11. Any two circles can be inverted into equal circles.
12. The angle of two lines or curves at a common point is equal to
that of their inverses at the inverse point.
13. Inscribe in a circle a triangle whose sides pass through three given
points.
14. Given three points A, B, C in a harmonic range, construct the
fourth by ruler and pencil.
15. If AB, CD, EF are diagonals of a complete quadrilateral, A, C, E
collinear, and AB, CD meet in P, and EP cuts the side CBF in K, then
(CKBF) = (CFBK).
16. Construct two points C, D in a line AB such that (ABCD) is
harmonic, and CD is bisected at a given point M in the line AB
produced. (MB . MA = MC2=MD2.)
CH. VII.] EXAMPLES. 217
17. If one pair of conjugates of a harmonic pencil is right-angled, it
bisects the angles of the other pair.
18. The diagonals of a parallelogram are harmonic conjugates of
parallels to the sides at the diagonal point. (Project into complete
quadl.)
19. KM, M' and N, N' are harmonic conjugates of A, C and B, D in
a parallelogram ABCD, parallels AP, AQ to MN, M'N' form a harmonic
pencil with AB, AD. (Project M'N' to inf.)
20. The internal and external bisectors of an angle form a harmonic
pencil with its sides.
21. If the line AISE contains in- and e-centres I, E of a triangle ABC,
and meets BC in S, then (AISE) is harmonic.
22. In a triangle ABC, the points S, D, X, P, P' on BC are foot of
bisector of A, foot of perpendicular, mid point, and points of contact of
in- and e-circle. Show that :
(i.) (DPSP') is harmonic, and XS. XD = XP2.
(ii. ) If ST is the second tangent from S to the incircle, and XT
meets the incircle in Y, then Y is on the nine-point circle, and
these circles touch at Y. (Invert from X, constant XP^.)
(iii.) The nine-point circle touches the ecircles.
23. Construct a tangent to a given circle from an outside point by
ruler and pencil only.
24. Construct the polar and tangents to a circle of |" radius from a
point li" from the centre.
25. Find a point 1|" from the centre in Ex. 24, forming with the given
point a side of a self-polar triangle, and complete the triangle.
26. Construct a self -polar triangle to a circle, ^" rad., one vertex 1^"
from centre, one side from this vertex If". Measure the sides.
27. Show that every self-polar triangle of a circle is obtuse. What
point is the orthocentre ?
28. Given an obtuse triangle, construct its polar circle. (To which the
triangle is self-polar. ) What point of the triangle is the centre ?
29. Construct the polar circle of a triangle. If", 1^", 1" sides. Measure
the radius.
30. The polars of a point to a system of coaxial circles are concurrent.
31. If AB, BC in order in a line are 1", V', construct D so that
(ABCD)=-f Measure CD.
32. If two sides of a variable triangle pass through fixed points, and
the vertices move on three concurrent lines, the third side passes through
a fixed point.
33. If two vertices of a variable triangle lie on fixed lines, and the
sides pass through three coUinear points, the third vertex lies ou a fixed
line.
218 EXAMPLES. [CH. VII.
34. The cross ratio P(ABCD) of the pencil of four fixed points
on a circle at a fifth point P on it, is the ratio of the rectangles of
chords AB . CD : AD . CB.
35. If O, A, B are fixed points in a fixed line, and OPQR is any
transversal of three concentric circles, centre O, show that there is one
point C on AB such that PA, QB, RC are always concurrent.
36. If E, F are points on opposite sides AB, CD of a quadrilateral, the
intersections of AF, DE and BF, CE are collinear with the diagonal
point.
37. If two triangles in a circle have fixed bases BC, EF, the line
joining the intersections of their other sides AB, DE and AC, DF
traverses a fixed point. (Pascal's theorem. )
38. If sides AB, CD of a quadrilateral in a conic meet in P ; AD,
BC in Q ; AC, BD in R ; show that PQR is a self -polar triangle.
39. Mark any four points A, B, C, D. Find a point P on any line
through A such that P(ABCD) = a given value /a. (Constr. 4.)
40. The vertex of perspective of two figures is its own perspective.
41. Project any two triangles in perspective into similarly situated
triangles. (Project axis to infinity.)
42. Project a complete quadrilateral (i.) into a rhombus, (ii.) into a
square. (Constr. 6, ii.)
43. Project a hexagon in a circle into a hexagon with opposite sides
parallel. (Project Pascal line to infinity.)
44. Two circles are in perspective to a centre of similitude and their
radical axis.
45. Kays AP, BQ, CR from the vertices of a triangle to the opposite
sides are concurrent. If angles at P, Q project into right angles, show
that R projects into one also. (Property of orthocentre. )
46. A transversal cuts a system of coaxial circles in involution.
47. If coaxial circles meet in imaginary points, show that the involu-
tion on the line of centres is separate and the double points real. What
circles do these points represent ?
48. Represent the distances of the imaginary common points of two
non-intersecting circles from the foot of their radical axis. (Constr. 7.)
49. A straight line is cut in involution by the pairs of opposite sides
and the diagonals AC, BD of a quadrilateral ABCD. (See Th. 36.)
50. Rays from a point on a circle to the ends of diameters form a
pencil in involution.
51. Show that if the s.p. involution of a point to a circle is right-
angled, that point is the centre.
52. Construct the right-angled pair and the double rays of the s.p.
involution of a point 1" from the centre of a circle of ^" radius.
219
CHAPTER YIIL
GONIGS.
FOCAL, CONJUGATE DIAMETER, POLAR, INVOLUTION,
AND CROSS RATIO PROPERTIES.
Definition 1. — A conic is a plane curve whose radii from a
fixed point, the focus, have a constant ratio, the eccentricity, to
the corresponding distances from a fixed line, the directrix."^
The focal radius of a point in the plane of a conic is the join of
the focus F to the point.
The axis is the line through the focus perpendicular to the
directrix DX.
Cor. — 'A conic is symmetrical about its axis.'
The vertices are the two points A, A' of the conic on the axis ;
and the major or transverse axis is the length A A'.
The centre is the mid point C of the major or transverse axis.
A central conic is an ellipse or hyperbola.
Definition 2. — A conic is ellipse, parabola, or hyperbola
according as eccentricity e<l, e = l, e>\.
(The figures are given on the next page.)
There are two vertices A, A' dividing FX internally and ex-
ternally in the ratio FA : AX = FA' : XA' = FP : PD = e.
In the ellipse, FA' < XA', .*. A, A' are on same side of X.
In the parabola, A' bisects XF externally at infinity ; hence
also the centre of a parabola is at infinity.
In the hyperbola, FA' > XA', .*. A, A' are on opp. sides of X.
Definition 3. — The ordinate of a point P of a conic is the
perpendicular PN to the axis; the abscissa of P is the distance
AN of the ordinate from vertex A.
The ordinate MP of a diameter QM is the parallel to the
tangent at Q. QM is its abscissa.
Definition 4. — Chord, secant or transversal, tangent, and
polar are defined as for the circle. The definitions are general.
* Some examples for plotting are given in Ch. V.
220
FORM OF THE CURVE.
CH. VIII.
Construction 1. — 'Construct a conic of given eccentricity
from focus and directrix, by chords perpendicular to the axis ;
and determine its form.'
If NP±axis FA, draw arcs, centre F, rad. 6NX, cutting NP in
P, P'; and make PD perp. to drx.
.-. FP:PD = FP:NX = e; similarly FP: P'D' = e;
.'. P, P' are on curve, and no other point of NP is on it.
Note. P, P' move up to coincidence at A, A'; hence AH, A'H' perp.
to axis are tangents at the vertices.
Form of the ellipse ; e< 1.
A, A' are on same side of X ; FP = eXN,
.*. NP meets curve when FN^eNXoreXN;
i.e. when only FN : NX>>e>FA : AX,
or FN:XN:|>FA':XA';
i.e. when only N lies between A and A^
Also, FP<XN, and is always finite; hence:
'The ellipse is a closed curve, lying en-
tirely between the tangents at the vertices.'
Form of the parabola ; e = 1.
A' is at infinity, FP = XN,
.'. NP meets curve when FN^NX or XN ;
i.e. when only N lies between A and infinity
on the same side as F ; hence :
' The parabola is an open curve lying entirely on the focus
side of the tangent at the vertex, and extending to infinity.'
Form of the hyperbola ; e > 1 .
A, A' are on opp. sides of X ;
FP = eXN,
.*. NP meets curve when FN:j>eNX
or eXN ;
i.e. when only N lies between A and inf.
on the focus side of A, or between A'
and inf. on the opp. side of A'; hence:
'The hyperbola is a curve of two open branches, lying
entirely on opposite sides of tangents at the vertices, and
extending to infinity.*
Ex. Draw (sqd. paper) tiie curves; 6=f, e = l, e = ^.
CH. VIII.] DEFINITIONS CONSTRUCTION OF FOCAL CHORD. 221
Definition 5. — A focal chord is a chord through the focus.
The latus rectum is the focal chord perpendicular to the axis.
A diameter is a central chord ; and a diameter of a parabola is
a parallel to the axis.
The asymptotes of a hsrperbola are the tangents from the centre
to the curve.
The subtangent of a point on a conic is the axial intercept
between the tangent and ordinate of the point.
Definition 6. — A normal is a perpendicular to a tangent at its
point of contact.
The subnormal of a point on a conic is the axial intercept
between the normal and ordinate of the point.
Definition 7. — The axcircle or auxiliary circle of a central
conic is the circle on the major or transverse axis as diameter.
The focircle of a conic is the circle through the inner vertex,
with the focus as centre.
Construction 2. — 'Construct the points on any focal ray of
a conic of given eccentricity, from focus and directrix.'
Draw focircle cutting ray FP in p, q ;
make Fi eql. to AX, AH tangt. at vertex,
meeting z^^ in H, make DHP perp. to drx.
Then FD = and || zH, •/ Fz = and || DH ;
.-. FP:PD = Fi?:HD = FA:AX = e.
.'. P is on the curve.
Similarly, a second point Q on the other
side of F is on the curve, and it is easily
seen that no third point of FP is on the curve.
Note. DP meets Fr in one* second point R on the curve, so that
every parallel to the axis meets the curve in two points P, R (which may
be imaginary or coincident).
In the parabola, i is on the focircle, r coincides with i, and R
is at infinity on the axis ; so that DP meets the parabola in one
finite point only. Hence :
Cor. — * A diameter of a parabola meets the curve in one finite
point and one point at infinity.'
* DF bisects suppt. of ang. PFR, by next theorem.
222
FOCAL PROPERTIES TANGENT.
[CH. VIII.
Theorem 1. — 'The focal radius of the directrix point of a
transversal of a conic bisects the supplement of the focal angle
of its chord ; and the focal angle of a tangent from a point
on the directrix is a right angle.'
If a trans vl. PQK cuts the drx. in K,
and PD, QE_Ldrx.; then
(i.) KP:KQ = PD:QE = FP:FQ;
.'. KF bisects an ang. F of tr. FQP ;
i.e. KF bisects siippt. of focal ang. QFP.
(ii.) If Q moves to coincidence with P,
KP coincides with tangt. TP,
the focal ang. QFP becomes zero, and its
suppt. two rt. angs.;
.'. TF_LFP, and focal ang. TFP is a rt. ang.
Cor. (i.). — 'A point P on a conic has one only tangent.'
Cor. (ii.). — ' Tangents at the ends of a focal chord intersect
on the directrix.'
Cor. (ill.). — 'A transversal meets a conic in two only points.*
Note. In the hyperbola, when P, Q are on different branches, KF
bisects the focal angle itself PFQ. (Draw fig. and verify.)
Ex. Show that the tangent at the point of infinity of a parabola is the
line at infinity.
Theorem 2. — * Tangents from any point to a conic have equal
focal angles.'
If TP, TQ are tangts. from T to conic,
draw TM, TN perp. to FP, FQ,
draw PTK to drx., and TE, PD perp. to drx.;
then KF_LFP || TM ;
.-. FM : FP = KT : KP = TE : PD ;
i.e. FM :TE = FP:PD = e;
_\ FM = e . TE = similarly FN.
.-. rt. tr. FTM = FTN ;
.-. ang. TFP = TFQ.
Cor. — Since any second tangt. TQ from T makes ang. TFQ = TFP,
' a point has two only tangents to a conic'
CU. VIII.] POINTS OX A LINE TANGENTS FROM A POINT.
223
Construction 3. — 'Construct the points in which a trans-
versal meets a conic of given eccentricity, from focus and
directrix.'
and tangt.
If transvl. KH meets drx,
at vertex in K, H,
draw Hpq to the focircle, pari, to KF ;
join Fp, ¥q to meet KH in P, Q,
and make HE, PD perp. to drx.
.-. FP:F^ = KP:KH = PD:HE;
i.e. FP:PD = F29:HE = FA:AX = e.
.'. P, and similarly Q, is on the curve.
Note. If Hj9 pari, to KF meets the circle in imaginary points, KH
meets the conic in imaginary points, so that a straiglit line meets
a conic in two real, two coincident (as a tangent), or two imaginary
points.
Construction 4. — 'Construct the tangents from a point to a
conic of given eccentricity, from focus and directrix.'
Dj. -.p
If T is the point, draw circ, diam. FT,
draw TE perp. to drx., make chds. FM, FN
each eql. to e . TE, make FK perp. to FM,
draw KTP to P on FM.
.*. FP:FM = KP:KT = PD:TE;
.-. FP:PD = FM:TE = e.
.*. P is on the curve, and KP its tangent.
If FN meets curve in Q, ang. TFQ = TFP ;
.*. TQ is the other tangent from T.
Note. The chords FM, FN can or cannot be drawn in the circle
according as FT ^ e . TE ;
i.e. according as T is or is not between F and the cnrve.
The part of the plane on the same side of the curve as F is
the inside, from points of which tangents cannot be drawn;
and the other part the outside of the curve.
The second branch of the hyperbola has a second focus beyond
A', so that the outside of the hyperbola is the space between the
two branches in which directrices and centre lie. Hence tangents
can be drawn from the centre of a hyperbola to the curve.
224 EXAMPLES. [CH. VIII.
EXAMPLES— XLIV.
CONICS.
1. Plot 6 ordinates, about ^" apart, and draw the conies, given
FX = 1", e = 7:3, e = l, e = 3:5.
2. With the data of Ex. 1, find the points of the conies on focal rays
making angles of 30°, 60°, 90°, 120° with the axis, and draw the curves.
3. Given FX = 1.6", e = 8:7, e = \, e = 7:9, make XD on directrix 1",
and angles XDP, XDQ, 90°, 108°. Construct the near point of DP, DQ
on each curve.
4. One end of a string is fixed on a drawing-board, and the other to
the long arm of a T-square, whose short arm moves along one edge of
the board. Find the locus of a point which keeps the string tight
against the long arm. Construct a parabola in this way.
5. On a diagonal FD from a fixed point F to a fixed straight line a
rhombus FPDQ is described, FQ having a fixed direction. Find the
locus of P. Can it be an ellipse ? Why ?
6. The vertex A of a triangle is fixed, B describes a fixed line, BC has
a given direction. Find the locus of C when AC : BC is constant.
7. Plot 4 points on the locus of Ex. 6, dist. from A to line 1 cm.,
ang. of BC to line 60°, AC:BC = 3:2. (Draw CD perp. to line,
calculate AC : CD.)
8. A circle touches a given circle and a given line. Find the locus of
its centre. Find also the locus of the centre of a circle touching the arc
and chord of a given segment of a circle.
9. A circle, centre F, cuts a conic in R, R'. A chord PQ cuts RR' in
H and the directrix in K ; Hqp parallel to KF cuts FQ, FP in q, p.
Show that q, p are on the circle. (Draw HE, PD, RM perp. to drx.)
10. P is any point, Q, R fixed points, on a conic. The chords PQ, PR
meet the directrix in H, K. Show that the focal angle HFK is constant.
11. Q. Ri S, T are four fixed points, P any point, on a conic. The
pencil P(QRST) cuts the directrix in HKLM ; show that the pencil
F(HKLM) is of given form — i.e. has given angles in order.
12. Two fixed tangents AT, BT determine an intercept AB on a
movable tangent AA'. Show that the focal angle AFB is constant.
13. A movable tangent cuts four fixed tangents in A, B, C, D. Show
that the pencil F(ABCD) has constant form.
14. A focal chord of a conic is harmonically divided by focus and
directrix.
15. The semi-latus-rectum is the harmonic mean of the parts of a
focal chord. (If PFP' cuts drx. in K, and PD, P'D'±drx., FX is H.M.
of PD, P'D' by Ex. 16.)
16. Show that the directrix is the polar of the focus. Also that the
polar of any point on the directrix is a focal ray.
17. Deduce the cross-ratio properties of a conic from Exx. 11, 13.
OH. VIII.]
THE PARABOLA : FOCAL PROPERTIES.
225
Theorem 3. — 'Of any point on a parabola,'
(i.) * The tangent bisects the angle of the focal radius and
diameter ; '
(ii.) 'The focal radius is equal to the axial intercept of
focal radius and tangent ; '
(iii.) ' The subtangent is equal to twice the abscissa ; '
(iv.) *The foot of the focal perpendicular on a tangent lies
on the tangent at the vertex.'
If tangt. at P meets axis in T, and
drx. in K, and if PD± DX, FH ± FT ; then
(i.) KFP = rt. ang. = KDP (Th. 1),
and PR = PD, '.• e = 1 ;
.-. rt. tr. FPK = DPK.
.'. tangt. TP bisects ang. FPD of focal
rad. FP and diani. DP.
(ii.) Ang. FTP = alt. ang. TPD = FPT ;
.*. focal rad. FP = axial intercept FT.
(iii.) TF = FP = DP = XN,
.*. TX = FN, and TA = AN;
i.e. subtangt. TN = 2 . AN.
(iv.) KP is rt. bisr. of FD,
.'. H is mid point of FD ;
.-. HA||DXi.FX;
i.e. H is on the tangt. at vertex A.
Cor. — * Two tangents to a parabola cannot be parallel.'
P/
D
>
>
T X
A
If n
Theorem 4. — 'Tangents to a parabola at the ends of a focal
chord intersect at right angles on the directrix.'
If PFP' is a focal chord, and FK±PP', then
the tangents at P, P' meet FK on drx. DD'
in K, say.
Also, PK, P'K bisect angs. FKD, FKD' ;
.-. PKJ_P'K.
Ex. 1. Show that the circumcircle of FHP, in Th. 3, touches AH.
Ex. 2. Perpendicular tangents to a parabola intersect on the directrix.
(Converse of Th. 4. )
Ex. 3. Show that K bisects DD' in Th. 4.
p. G. O
226
THE PARABOLA : CIRCUMSCRIBING TRIANGLE. [CH. YIIL
Theorem 5. — 'The focal triangles of two tangents from a
point to a parabola are similar.'
If TP, TQ are tangts. to a parabola, meeting
tangt. of vertex at H, K ;
then angs. A, FKQ, FHT are rt. angs.,
and ang. KFQ = KFA, at focus ;
.-. tr. FQK III FKA.
.-. ang. FQK = FKA = suppt. of FKH
= FTH, in cyclic quadl. FKHT ;
.'. ang. FQT = suppt. of FQK = FTP.
Also, ang. QFT = TFP, in trs. QFT, TFP ;
.-. tr. FQT III FTP.
Ex. Show that the triangle FTP can he derived from FQT by
rotation and multiplication. What is the multiplier ?
Show also that a FTP : a FQT = FP : FQ.
Note. This Ex. helps us to remember which angles are equal.
Theorem 6. — * The circumcircle of a circumscribing triangle of
a parabola passes through the focus.'
If tr. STV circumscribes a parabola,
and P, Q, R are points of contact ;
then tr. FQT ||| FTP ;
and FRS|||FSP; s^^.
.'. ang. FTV = FPS
= FSV ;
.*. F lies on circle STV.
Theorem 7. — * The orthocentre of a circumscribing triangle of
a parabola lies on the directrix.' (Steiner.)
The tangent at the vertex, containing feet of perps. from F on
the tangts., is the Simson line of F in triangle STV (last fig.).
.'. tangt. at vertex bisects the join FH to the orthocentre. (Ex.
iv. p. 112, Ch. ly.)
.'. H is on the directrix.
This may also be derived from Brianchon's theorem.
Ex. Construct a circum triangle, given in position the orthocentre and
two sides tangent to the parabola.
CH. VIII.] THE parabola: normal SQUARE OP ORDINATE.
227
Theorem 8. — * In a parabola,'
(i.) ' The axial intercept of the focal radius and normal of a
point is equal to the focal radius ; ' (FG = FP) ;
(ii.) 'The subnormal is equal to the semi-latus-rectum ; *
(NG = 2AF);
(ill.) * If PN is the ordinate of P, then PN^ = 4AF . AN.'
If PT, PG, DPE are tangt., normal, and diam. of P ; then
(i.) TP bisects ang. FPD,
.*. PG, perp. to TP, bisects suppt. FPE.
. ang.
FPG
= EPG = F
^GP;
'. FG
= FP.
(ii.)
FG =
FP = DP =
= XN;
NG =
XF = FL =
= 2AF
.... , PN NG , . , , 2AF
("^•)tn = pn(^^^-*^^->=pn'
/. pn2 = 2af.tn = 4af.an.
Ex, Show that the curve PN2 = 4AF.AN (A and AF fixed and
PN J. AF), is a parabola.
Theorem 9. — (i.) *A diameter of a parabola bisects chords
parallel to its tangent ; and the curve bisects the diametral
distance of a point from the chord of contact of its tangents.'
(ii.) 'If PM is an ordinate of a diameter QM, then
MP2 = 4FQ.QM.'
If TP, TP' are tangts. of cbd. PP',
T'QTi tangt. of Q, and TQM, T'M', DP,
and D'P' diams. ; then,
(i.) tr. DPT = FPT,
.*. TD = TF = simly. TD';
.', TM is rt. bisr. of DD' and bisects PP';
simly. T'M' bisects PQ, PT; Tj bisects TP';
.*. T'QTi II PP', and bisects TM ;
i.e. diam. QM bisects clid. of P pari, to QT',
and Q bisects TM.
(ii.) Ang. FQT' = TQT'; and tr. FQT' | |j FT'P;
.-. ang. FT'Q = FPT' = DPT = rTQ;
.-. tr. FQT'lllT'QT;
.-. p^ = P^^; i.e. Qr2 = FQ.TQ = FQ.QM;
.*. Mp2 = 4Qr2 = 4FQ.QM.
228 THE PARABOLA : APOLLONIUS AREA OP SEGMENT. [CH. VIII.
Theorem 10. — ' The rectangle of the parts of any transversal
from a point to a parabola is four times the rectangle of the
focal radius of its conjugate diameter and its diametral distance
from the curve.' (VP . VP' = 4FQ . VR.)
If RV, PVP' are the diam. dist. and transvl. of V
and QM the diam. conj. to PP' ;
make RN pari, to PP'.
/. VP.VP' = MV2-MP2=NR2-MP2
= 4FQ.QN-4. FQ.QM
= 4FQ.MN = 4FQ. VR. . -p
Theorem 11. — ApoUonius' theorem : * The ratio of the rect-
angles of parts of two transversals of given directions from any
point to a parabola is constant, and is equal to that of the focal
radii of their conjugate diameters.*
If PVP', QVQ' are transvls. of V in fixed
directions, their conj. diams. HM, KN, and
points H, K are fixed.
Hence, if VR is diam. dist. of V,
VP.VP' 4.FH.VR
VQ.VQ'
4.FK
FH
FK
VR
constant.
Theorem 12. — 'The area of a segment of a parabola is two-
thirds of that of the triangle of its chord and tangents.'
If TP, TP' are tangts. of chd. PP', and
TQM the diam. of T, draw tangt. T'QTj, and
diams. T'Q', TjQ^ ; and repeat the process
indefinitely.
Then A PQM = A PTQ = 2 A TQT' ;
similarly, A P'QM = 2 A TQTj,
and A PQP' = 2 A TT'Tj.
Similarly, area of inpoln. PQ'QQiP' = 2 area of
corresp, circumpoln.; and so on.
And if the process is infinitely continued, the inner perimeters
of circum- and in-polygons coincide with the curve ;
/. area of segment PQP' = 2( A TPP' - segt. PQP') - J ^ P^P'.
CH. VIII.] EXAMPLES. 229
EXAMPLES— XLV.
The Parabola.
1. Find the locus of the centre of a circle which passes through a
given point and touches a given line.
2. Focal radii of a parabola increase with their angle from FA.
3. The shortest focal chord of a parabola is the latus rectum.
4. A circle on a focal chord as diameter touches the directrix.
5. The radius of the circle of Ex. 4 through the point of contact of
the directrix meets the curve in Q. Show that the radius is parallel to
the axis, and equal to 2 . FQ.
6. Find the envelope of one side of a right angle whose vertex moves
on a line, and whose other side passes through a iixed point. (See
Ch. v., 'Plotting Loci.') Generalise this.
7. Construct the tangent and normal of a point P on a parabola by
the circle, centre F, radius FP.
8. Draw a parabola of given focus to touch a given line at a given
point.
9. Given focus, directrix, and the line of a tangent to a parabola,
construct the point of contact.
10. If PG, PD are normal and perpendicular to directrix of a point P
on a parabola, FP bisects DG.
11. If PT is the tangent of P in Ex. 10, PT and FD right-bisect each
other. What figure is FPDT ? FGPD ?
12. The tangent from A to a circle on a focal radius FP as diameter
is half the ordinate of P.
13. The envelope of the fold of a rectangle of which one vertex is
made to travel along an opposite side, is a parabola.
14. If Q is the point on a parabola of the conjugate diameter of a
focal chord PP', show that PP' = 4FQ.
15. If TP, TQ are tangents to a parabola, FT^^ FQ . FP.
16. The intersections of the joins of the ends, two and two, of two
parallel chords of a parabola lie on the conjugate diameter.
17. If R, S are the points of intersection of Ex. 16, show that the
curve bisects RS. (Use polar.)
18. The locus of mid points of focal chords of a pai-abola is a parabola
of half its dimensions.
19. The rectangle FP . FP' of a focal chord is equal to FA . PP'.
20. A focal chord PP' and its rectangle FP . FP' increase with the
angle made by the larger part with the latus rectum.
21. If a circle cuts a parabola in P, P', Q, Q', show that PP' and QQ'
make equal angles with the axis.
22. If VPP', \/QQ', two transversals from a point V to a parabola,
meet the axis in R, S, then VP . VP' : VQ . VQ - VR^ : VS^.
230
CENTRAL CONICS : SQUARE OF ORDINATE — FOCI. [cH. VIII.
Theorem 13.
Np2
If NP is an ordinate of an
the
ratio
7 is constant and ec[ual to
V' : a2
+ 62
NA.NA
the axes A A', BB'.'
Draw PAK, PA'K' to drx., join KF, K'F ;
then KF_LK'F (bisrs. of AFP', AFP) ;
.'. XF is mean propl. of KX, XK'.
NP KX NP XK'
Also,
ellipse
hyperbola'
where 2a, 2h are
NA
NP2
AX' NA'
KX . XK'
XA '
XF2
•• NA.NA' AX.XA' AX . XA' ^^^^*-
(i.) In the ellipse, the minor axis is
the diam. BCB', perp. to AA'.
.'. NP2: NA.NA' = CB2:CA.CA'= -W-\a^.
(li.) In the hyperbola, construct h
so that 62 . ^2 _ Np2 . NA . NA', which is
positive ; and make B'C = CB = 6 ;
then B'CB is the conjugate axis.
Also, if ^, p are the imaginary points
in which CB meets the curve ; then
C^ : a2= - C/32 : CA . CA'= - W : a^.
:, Cp = Cl3' = b J-i OT lb. I
14.— 'An
ellipse
hyperbola
axis and has two foci ; and the centre bisects all
is symmetrical about its
£rh.
Theorem
minor
conjugate
diameters.'
Make CM =CN, draw ordinate MQ ;
/. MA= -NA'; MA'= - NA ;
.-. MA . MA' = NA . NA', and MQ^ = MP^ ;
.*. CB is rt. bisr. of PQ, and the curve is
symmetrical about its second axis ;
,'. there is a second focus F' (CF' = FC),
and a second drx. D'X' (CX' = XC) ;
also, if QQ' is the did. bisected at M,
PCQ' is the diam. of P bisected atC.
Cor. — ' The centre of a conic is the pole of the line at infinity
For the harm. conj. of C to P, Q' is at inf.
CH. VIII.] CENTRAL CONICS : FOCAL PROPERTIES.
231
Theorem 15. — 'In an ellipse,'*
(L) 'CF:CA = e = CA:CX;'
(ii.) 'The sum of focal radii is equal to the major axis;'
(FP+F'P = A'A = 2«);
(iii.) 'FB = a; CF2 = a2_j2. fA.A'F = &2; e^ = {a^ -h'^)/aV
FA A'F A'F+FA
(i.) e =
AX AX A'X + AX
_2CA 2CF
~ 2CX °^ 2CA '
i.e. CA:CX-e = CF:CA.
(ii.) FP+F'P = eNX + eX'N=eX'X
= A'A = 2a.
(iii.) FB = F'B = i(FB + F'B) = «.
CF2 = FB2-CB2 = a2-62.
FA.A'F = CA2-CF2 = ?;2
e2 = CF2:CA2 = (a2-&2):a2.
B P
X'
/:^A.
A' F' C N F A
Theorem 16.—' In a hyperbola/
(i.) 'CF:CA = e = CA:CX;'
(ii.) 'The difference of focal radii of a point is equal to the
transverse axis ; ' (F'P - FP = A'A = 2a) ;
(iii.) ' FL = (e' - l)a ; f e^ = (aP + b'^)/a^ ; CF2 = a'' + b^ = AB' ;
FA. FA' = 61'
r\ _ AF A'F A'F + AF
^^■^ ^~XA~A^~A'X + XA
_2CF 2CA
~ 2CA °^ 2CX '
F' A' X
X A F N
i.e. CF : CA = e = CA : CX.
(ii.) F'P-FP = e(X'N-XN) = A'A = 2a.
(iii.) FL2 = e2xF2 = e2(CF - CX)2 = e^fea - -Y = {e^ - l^a^ ;
FA . FA' = CF2 - CA2 = (e2 - l)a2 ;
«2 FA . FA ^ ^ a^
:. CF2 = eV_a2+62 = AB2; FA . FA' = CF^'-CA2 = 62.
Ex. 1. The curves FP + F'P = const, are conies.
£x. 2. The tangent from F to axcircle of hyperbola ==6.
* F, A, X are on right of C for convenience of sigiu
t FL is semi-latus-rectum.
232
CENTRAL CONICS : SQUARE OF ORDINATE. [CH. VIII.
Theorem 17.— 'In an ellipse, if PM is an ordinate to BB', the
ratio MP2:^-^-CM2
hK'
NA . NA' CA2 - CN^
NP- CM=^
CN2 . ^. . MP2
(summation) = ,;2—^-,.
A F
' In a hyperbola, MP^-.U^ + C^^ = a^ : b^'
(Prove as above.)
Theorem 18. — 'The curve NP-: NA. NA' = constant, AA' given
and N a right angle, is an ellipse or hyperbola according as the
sign of the ratio is negative or positive.'
Make a = CA = A'C, and ^^ ; ^2 ^ + n p2 ; N A . N A'
(i.) For the - sign ;
make CB = h, perp. to CA, BF = a.
Construct ellipse, foe. F, vert. A, ecc. CF:CA.
This coincides with the given curve.
Note. If fe>a, make b the major axis,
(ii.) For the + sign ; make CB = b, CF = BA = JaF+W.
Construct hyperbola, foe. F, vert. A, ecc. CF : CA.
This coincides with the given curve.
Theorem 19. — 'An ellipse can be derived from its axcircle
by multiplying the ordinates of the circle to the axis by the
ratio b'.a,'
If ^PN, gQM are joint ordinates of circle —^'^
and ellipse from major axis ;
NP2:N^/= -NP2:NA. HfK' = W\a?.
.'. NP :Np = b :a = s'milj. MQ : Mg.
P,2^ or Q,(2 are corresp. points of the curves;
also, ^PN and gQM are similarly divided ;
.*. 2P) QP» MN are concurrent on the axis. Hence :
Cor. — ' The ellipse is the perspective of the axcircle, vertex at
infinity along BB', axis of perspective AAV
The tangents at corresp. points P, 2^ meet on the axis ; and the
line at inf. in either fig. is its perspective in the other.
Ex. 1. Prove the corresponding theorem, using the circle on the minor
axis as diameter.
Ex. 2. Show that Thh. 13, 15, 17 for the ellipse can be written for the
hyperbola by using - b^ for + b\
CH. VIII.] CENTRAL CONICS : TANGENT AND NORMAL.
233
Theorem 20. — (i.) 'If PG is the normal of an ellipse or
hyperbola, FG : FP = e.'
(ii.) 'The tangent and normal of a point of an ellipse or
hyperbola are the bisectors of angle of the focal radii of the
point.'
If the tangt. of P meets drx. in K,
and PD_Ldrx., KFP is a rt. ang. ;
.*. circ. on diam. KP passes through
F, D and touches normal PG at P.
.*. ang. FPG = FDP, opp. arc ;
and ang. PFG = FPD, alt, ang.,
.-. tr. FGP IJI PFD ;
.', FG : FP = FP : DP = e (without regard to sign).
Simly. F'G : FP = e = FG : FP ;
.*. PG and its perp. KP are bisectors of ang. P of tr. FPF'.
Fig. of Hyp.
(Same proof for Ell.)
Theorem 21. — (i.) 'The foot of a focal perpendicular of a
tangent of an ellipse or hyperbola lies on the axcircle.'
(ii.) 'The rectangle of focal perpendiculars on a tangent
of an ellipse or hyperbola is equal to the square on CB.'
(FM.F'M'= ±b\)
If FM, F'M'±MP the tangt. at P, then
(i.) the centre O of circle FMP is the mid
point of FP ;
.-. coil F'P ('.• C bisects F'F).
Also, ang. OMP = OPM = FPM',
.*. OM II FP, and O is on CM.
Also, OM = lFP; CO = jF'P;
.-. CM = l(FP+F'P) = a.
.'. M, and simly. M', is on the axcircle.
(ii.) If MF meets axcircle in Mj,
M'Mj is a diam., and MiF= F'M'.
.-. FM . F'M' = MjF . FM = A'F . FA = b^.
(Thh. 15, 16 (iii.).)
Note. The point of contact P of tangent from M on the axcircle is on
the circle through F touching the axcircle at M.
We can draw tangents from any point T hy drawing a circle on diam.
TF to cut the axcircle in M, K, say; TM, TK are tangts.
Fig. of Ell.
(Same proof for Hyp.
234 CENTRAL CONICS : TANGENTS DIRECTOR CIRCLE. [CH. VIII.
Theorem 22.—* Tangents to an , ® g^^®, from a point form
equal angles with the focal rays of the point.'
If TP, TQ lire tangts. from T,
draw perps. FM, FN, F'M', F'N' from foci;
.'. FM.F'M' = &^=FN.F'N'.
.-. FM : FN = FN' : F'M'.
Turn fig. TMFN over into posn. THVK;
.*. tr. VKH III F'N'M', and is simly. situated
about T ;
.'. TVF' is a str. line. ^ig- of EU.
.-. ang. FTP = VTH = F'TQ. ^^^^ P^°°^ ^""^ ^^P'^
Theorem 23. — 'Perpendicular tangents of an , ® ^^^®, inter-
sect on a fixed circle.' (The director circle.)
If perp. tangts. TP, TQ meet the axcircle in
M, M', N, N', the quadls. FT, FT are rect-
angles.
Also, CT2 = TM . TM' + CM2 (Th. 93, Ch. V.)
= FN.F'N' + CM2
Fig. of Ell.
(Same proof for Hyp.)
KA V X
.*. CT is constant, and locus of T is a circle.
In the hyperbola, CT^ = a^ — W.
Ex. Prove conversely that tangents from a point on the director circle
are perpendicular.
Note. The perp. tangts. from an axial point V of
the director circle, as VHR, make angs. of 45° with
the axis, by symmetry.
Hence, if FH±VR, and HK±CA,
KV = KH = FK. (Isosc. rt. trs.)
Thus, if C moves to infinity, F and A remaining fixed, HK coincides
with tangt. at A (limiting axcircle), the conic becomes a parabola,
KV coincides with AX, and the limiting director circle passes through X.
Thus the director circle of the parabola is the directrix. (Th. 4.)
Ex. Interpret Th. 18 for the parabola in the same way.
Definition 8. — Conjugate diameters are pairs of diameters each
of which bisects chords parallel to the other.
Any line through a point is conjugate to the diameter bisecting
a chord parallel to the line.
CH. VII I.]
THE ELLIPSE : CONJUGATE LINES.
235
Theorem 24. — * In an ellipse,'
(i.) * A diameter bisects chords parallel to its tangents ; '
(ii.) ' Conjugate diameters correspond to perpendicular diam-
eters of the axcircle, and their tangents form a parallelogram.'
pU
— ---..^
/^ p!i^— ^
"^^S?^^^^\""^-^
1/ 1
^,/^M^**'*»*^^^^V'*^*"*'"*^ ^ ^
// ■
l/ ' ^^^^^^vV"^ ^^*''*'^<k^ ^ ^
I i
/ 1 i M ^"^^^''^^^^^^^"^''^^^^w
1 1 k
1 1 1 1 '^'^^-^ ^^^^^-
A' N'CKH NA S T
(i.) If PP' is a chd. pari, to tangt. QT, 'pp\ qJ corresp. chd. and
tangt. of axcircle,* M, in mid points of chds.,
and PN, P'N', QH, MK perps. to AA'; then
KM II PN and P'N', .*. KM bisects jp^ in m,
and K??i : KM = N^? : NP = Hg' : HQ ;
.-. fig. SKMm III THQ</ ( •/ TQ || SM) ;
.'. 'pj^' II '\ql.Gm and Oq.
.'. C, ??z, g, and .*. also C, M, Q, are coUinear ;
i.e. diam. CQ bisects chds. PP' pari, to its tangt. QT.
Cor. — * Parallels of axcircle correspond to parallels of ellipse.*
(ii.) If CQ bisects chds. pari, to CR,
then CR || tangt. HQ of diam. CQ ;
and if Cq, Cr are corresp. diams. of
axcircle, fig. CrRN ||| H^QM.
.•. Cr II Hq _L Cq, and tangt. Kr ± Cr || C^ ;
.-. fig. KrRN III CqQM, and KR || CQ ;
.'. CR bisects chds. pari, to CQ ;
i.e. CQ, CR are conj. diams,, and the corresp. diams. of axcircle
C^', Cr are perpendicular.
Also, tangts. at Q, Q', R, R' form a parallelogram.
Ex. Supplemental chordsf are parallel to conjugate diameters.
Definition 9. — The eccentric angle of a point on an ellipse is
the angle of the axis and the diameter of the corresponding point
of the axcircle.
* Large and small letters are used for corresponding points,
t Chds. from a point on curve to the ends of
236
THE ELLIPSE : CONJUGATE DIAMETERS AREA. [CH. Mil.
Theorem 25. — 'In an ellipse,'
(i.) * The ordinate and central abscissa of a point of eccentric
angle <^ are b sin <^, a cos </> ; '
(ii.) 'The sum of squares of conjugate semi-diameters has
constant area a^ + b-;' (CR^ + CQ^ = a^ + 1)^) ;
(iii.) ' The rectangle of focal radii of a point is equal to the
square of its conjugate semi-diameter ; ' (FQ . F'Q = CR^) ;
(iv.) 'The parallelogram of tangents of conjugate diameters
has constant area 4a/>.'
If CQ, CR are conj., <^, </>' ecc. angs. of
Q, R, and q, r corresp. points of axcircle,
and TH, TR tangts. ; then
(i.) CM = Cq cos <^ = a cos <^ ;
MQ = -. Mq = b sin </>.
(ii.) c},' = rCH = 90° +qCH = 90° + cf, ;
:. CQ2 + CR2 = CM2 + MQ2 + NC2 + NR^
= ft2 cos^cf> + b^ sin^^ + a^ sin^^ + b^ cos^^
= a^ + b\
Cor.— 'N7=CM; HC = Mq.'
(iii.) (2a)2 = (FQ + F'QY = FQ^ + F'Q^ + 2FQ . F'Q
= 2CF2 + 2CQ2 + 2FQ.F'Q; r
.-. 2^2 = (a^ - 52) + (^2 ^ 52 _ CR2) + FQ . F'Q
.-. FQ.rQ = CR2. ^,
(iv.) zz7Tr = 4. oTC = 8. Arcq = 8. Arch
= 4 . RN . CH = 4 . 6 cos (^ . CH = 4& . Cg^
= iab.
Ex. The least sum of conjugate diameters is 2{a + b).
S^pi C F
Theorem 26. — * The area of an ellipse is Trab.'
Divide the semi-ellipse and senii-axcircle by
ordinates into narrow strips PN, ^j.N ;
then if rRL bisects trapeziums PN, ^jN,
area PN:2?N = MN. RL.MN .rL = b :a.
Hence, if the lines of division PM, QN...
move to coincidence, the number of strips becoming infinite,
area of semi-ell. : semicirc. ==b : a.
.'. area of ellipse = - x area of circ. =7rab.
MLN
CH. VIII.
THE HYPERBOLA : ASYMPTOTES.
237
Theorem 27. — 'The asymptotes of a hyperbola touch the
curve at infinity.'
Construction 5. — ' Construct the asymptotes of a hyperbola ;
and construct the conjugate axis.'
If X is foot of drx., AE tangt. at A,
and FD tangt. from F to axcircle ;
then tr. CDF^CAE.
.-. CD:CE = CA:CF = CX:CA;
.'. DX il AE_LAA', and D is on drx.
Also, since D on the axcircle is foot
of perp. from F on CD,
CD, and simly. CD', is a tangt. to the
hyperbola ;
.*. the circle through F, touching the
axcircle at D, meets CD in the point
of contact I. (JS^ote, p. 233.)
But since FD is tangent to the axcircle, this circle must be the
limiting circle, centre at inf. along CD, represented by FD ; and
it therefore cuts CD at inf.
Hence the asymptotes touch the hyperbola at infinity.
To construct the asymptotes, join C to the points D, D' in which
the axcircle cuts the directrix.
Also, b^=FA. FA' = FD^, *.• FD touches axcircle ;
.'. b = FD.
To construct the conjugate axis, join F to a point of inter-
section D of axcircle and directrix, then b = FD.
If 2a is the angle of the asymptotes.
tan a
AE
CA
DF
CD
or a = tan ^ -•
Note. Only those diameters in the angle of the asymptotes containing
A meet the curve in real points. All diameters in the angle containing
B meet the curve in imaginary points.
Definition 10. — The conjugate hyperbola is the hyperbola
whose transverse axis is BB', and conj. axis A A'.
It has the same asymptotes ; and we show later that an imagi-
nary diameter of the hyperbola in direction CQ is J - IQQ', where
QQ' is the real diam. of the conj. hyp. in this direction.
238
THE HYPERBOLA : CONJUGATE LINES.
[CH. YITL
Theorem 28. — ' In a hyperbola,*
(i.) 'If a chord PP' meets the asymptotes in D, D', then
PD. PD'= -CR-, where CR is the parallel semi-diameter of the
conjugate hyperbola ; '
(ii.) 'DP = P'D', and the asymptotal intercept of a tangent
is bisected at the point of contact ; '
(iii.) * A diameter bisects chords parallel to its tangents ;'
(iv.) 'Tangents of conjugate diameters form a parallelogram
with the asymptotes as diagonals.'
If KPK^ RHH' II CB;
then (i.) tr. PKD ||| RHC ;
and PK'D' ||| RHC.
PD.PD' CR.CR ,
*• PK.PK'~RH.RH''
and PK.PK' = NP2-NK2
= {(CN2 - CA2) - CN2} tan^a
= -CA2 tan2a= - &2.
Simly. RH. RH'= +h\
in conj. hyp. (using Th. 17) ;
.-. PD . PD' = - CR2 = simly. P'D . P'D'.
(ii.) MD2-MP2= -PD.PD'= -P'D.P'D'
= MD'2 - MP'^, if M bisects DD' ;
.-. MP = MP', DP = P'D', M bisects clid. PP'.
Also, if PP' moves pari, to CR to coincide with tangt. TQT',
then P, P' coincide with Q ; and D, D' with T, T' ;
.-. TQ = QT', and QT^ = - QT . QT' = CR2 ;
.*. QT = and || CR, hence TR = and 1| CQ.
(iii.) C, Q, M are coUinear, '.* TQT', DM D' are similarly divided;
.*. CQ bisects clid. PP' pari, to its tangt.
(iv.) Draw TR to T^ ; then CQ bisects T^T',
.•. CR bisects TTj, hence R bisects chd. STS', by (ii.),
and TRTj is tangt. at R to conj. hyperbola ;
i.e. CR and CQ bisect chds. pari, to the other in either hyperbola,
and are conj. diams. of each.
Also tangts. at Q, Q', R, R' meet on asms, and form a parm.
Note. If CR meets the first hyp. in p, p' ; then pp' meets asms, in
C, C. _
.-. Cp2=:pC.pC = PD.PD'=-CR2; and Cp^sJ-l.CR.
We may use without confusion Op or CR as the conj. semi-diam. of CQ.
CH. VIII.] THE HYPERBOLA : CONJUGATE DIAMETERS.
239
Theorem 29. — 'In a hyperbola,'
(i.) 'The triangle of a tangent and the asymptotes has con-
stant area ah; and the parallelogram of tangents of conjugate
diameters has constant area ^ah ; '
(ii.) 'The difference of sauares of two conjugate semi-
diameters has constant area a- -b^ ;' (CQ- - CR- = a^ _ ^2^ .
(iii.) 'The rectangle of focal rays of a point is equal to the
square of its conjugate semi-diameter.' (FQ . F'Q = CR^.)
(i.) If CQ, CR are conj. diams.,
and TQT', TR tangts. to hyp. and conj.,
QE II CT, 2a ang. of asms. CT, CT' ;
then RQ, CT are bisected at D,
and O RQ = A TCr = 4CD . CE sin 2a.
If KQK' II CB, then QK . QK' = - b\
and trs. QDK, K'EQ are of const, form ;
.*. QD . QE : QK . QK' = const. , /^ say ;
i.e. CD . CE = QD . QE = - fxh^ = const. ;
.*. A TCr = O RQ = 4CD . CE sin 2a = const.
= O BA = ab ('.* a, b are conj.).*
(ii.) If circ. on diam. TT' cuts CT in H,
CQ2 - CR2 = CQ2 - QT2 = CH . CT = CT'. CT cos 2a
= const. = a^ - b'^ ( '.' a, b are conj.).
(iii.) (2a)2 = (FQ-F'Q)2
= FQ2 + F'Q2-2FQ.F'Q
= 2 . CF2 + 2CQ2 - 2FQ . F'Q ;
.*. 2a2 = (a2 + z,2) + (CR2 + a2 _ ^2) _ pQ . f'Q. »^'"
.'. FQ.F'Q = CR2.
Definition 11.
equal axes.
A rectangular or equilateral hyperbola has
Theorem 30. — 'In a rectangular hyperbola,'
(i.) 'The hyperbola ^ its conjugate, from which it can be
derived by rotation through a right angle.'
(ii.) 'The asymptotes are perpendicular.'
(iii.) 'Perpendicular diameters are equal.'
(iv.) 'Conjugate diameters are equal.' ( •/ a2 - 52 = 0.)
* The movable tangt. TT' forms homographic ranges on fixed tangts. CT, CT',
C in one corresp. to I at inf. in the other. Superposing these, we have an
in vein., centre C ; .*. CT . CT'= const. (Pure geom.)
240
CENTRAL CONICS : APOLLONIUS NORMAL. [cH. VIII.
Theorem 31. — Apollonius' theorem: 'The ratio of the rect-
angles of parts of two transversals of given directions from
any point to an ellipse or hyperbola is constant, and equal to
that of the squares of semi-diameters of those directions.'*
If semi-diams. CR, CS || trans vis.
PVP', QVQ' of given dims., make HK
polar of V cutting CV, RC at T, K,
and make CM conj. to and bisecting
PP', and VL || CM ;
then polar of K on CR || conj. CM,
and traverses V, i.e. VL is polar of K ;
.*. K, L are a pair of s.p. involn. of
CR.
.'. CR2= -CK.CLt^CK. VM.
Also, VP. VP'-VH. VM
(•.• HPVP' is harm.),
VP.VP' VHVTVQ.VQ'
CK~
Fig. of Hyp.
(Same proof for EIL)
(simly.)
•• CR2 CK CT CS2
i.e. VP . VP' : VQ . VQ' = CR2 : CS^ = const.
Note. The case where QQ' is the diameter conjugate to PP' should be
specially studied.
Theorem 32. — 'If the ordinates, tangent, and normal of a
point P of an ellipse or hyperbola meet the axes in N, N', T, T',
G, G', and PG meets CR, conjugate to CP, in H ; then
(i.) CN.CT = «2^ CN'.cr= ±h^;
(ii.) PH.PG=±&^ PH.PG' = a2'
(i.) PNP', chd. of contact of tangts. from T,
is the polar of T (see Thh. 34, 36) ;
.'. CN . CT = CA2 = a^. And simly.
CN'.Cr^CB2 (ell.) = C^2 (hyp.)= +Z>2.
(ii.) Draw CLMK perp. to TT', pari, to PG;
.'. T'N'LM, TNMK are cyclic quadls.
.*. PH.PG = CM.CL = CN'.CT'= ±&^
PH . PG' = CM . CK = CN . CT = a\
Students may leain Th. 36 here, corresp. to Thh. 6, 7, 8, Ch. VII.
* This statement should be learnt ; the demonstration may be postponed,
f If R is a real point on the curve, the sign is positive.
CH. VIII.] EXAMPLES. 241
EXAMPLES— XLVI.
Central Conics.
1. Find the locus of the centre of a circle which touches two given
circles. Discuss the different cases.
2. Construct an ellipse by pencil, loop of thread, and two pins.
3. Construct part of a hyperbola by pencil, thread, a rod movable
about a fixed point in it, and a pin.
4. Focal radii of an ellipse or hyperbola increase with their angle
from FA. (Use branch of hyp. in which F lies.)
5. Focal chords of a central conic increase with the angle from tlie
latus rectum of their greater part.
6. The foci divide harmonically the axial intercept of tangent and
normal of a point on a central conic.
7. If the normal and tangent of a point P of a central conic meet the
axis in G, T, then CG . CT = CF2. (XJse Ex. 6.)
8. Find the locus of a point whose distance from a given circle is
proportional to its distance from a given straight line.
9. If FM, FN are perpendicular to the tangent and normal of a point
on a conic, MN passes through the centre.
10. If the normal PG of a point to a conic is equal to FG, then FP is
equal to the latus rectum.
11. Given the foci F, F' in position, and the axis 2a of a conic,
construct the points P, P' of any focal ray.
12. A focal radius FP of a conic is produced to Q ; show that F'Q is
divided by the tangent of P in the ratio PF' : PQ.
13. Any point on the central perpendicular to a tangent of a conic is
equidistant from the feet of the focal perpendiculars.
14. Find the greatest intercept of a tangent to a conic by its axcircle.
15. Tangents at the ends of the two latera recta of a conic intersect
on the axcircle or at infinity.
16. If the normal and tangent of a point on a conic meet CB in G', T',
thenCG'.CT'=-CF2.
17. A circle on a focal radius of a conic as diameter touches the
axcircle.
18. The two circles through the point of contact of a tangent, one
focus, and the foot of its perpendicular on the tangent, meet on the axis
and on the ordinate of the point of contact.
19. Given the foci F, F' and the line of a tangent to a conic, construct
the point of contact of the conic. What different cases are there ?
20. Find the envelope of one side of a right angle whose vertex moves
on a fixed circle and whose other side traverses a fixed point. (Ch. V.)
p. a. p
242 EXAMPLES. [CH. VIII.
21. An arc of constant angle is described on a focal radius of a conic ;
show that it touches a fixed circle. (See Ex. 20 and Ch. V.)
22. The intersections of joins of ends of two parallel chords of a
conic lie on the conjugate diameter.
23. The rectangle QP . QP' of the intercept by the axcircle on the
tangent at Q of a conic is equal to CR^-i^ where CR is conjugate
toCQ.
24. If FM, F'M' are perpendiculars from the foci to the tangent of a
point P of a conic, and PN is an ordinate, then NM : NM'=PM : PM'.
25. Supplemental chords of a hyperbola are parallel to conjugate
diameters.
26. A parabola is described through two foci of a hyperbola, with its
focus on the curve. Show that its axis is parallel to an asymptote.
27. The asymptotes and a pair of conjugate diameters of a hyperbola
form a harmonic pencil.
28. If PM is an ordinate to a diameter QQ' of an ellipse or hyperbola,
RR' the conjugate diameter, MP^ : MQ . MQ'= +CR2 : CQ\
29. If a diameter CQ meets an ordinate PM of a point P on a conic
in M, and the tangent at P in T, show that CM . CT = CQ2. (Polar.)
30. A rod has tAvo fixed studs on it, which move in two grooves forming
a right angle. Show that a pencil fixed at a point in the rod traces an
ellipse. (Trammels. ) What is the curve if the grooves form any angle ?
31. An ellipse and hyperbola have the same foci. Show that they cut
at right angles.
32. If a parallelogram is inscribed in an ellipse, the centre is its
diagonal point.
33. If a circle cuts a central conic in four points, any two common
chords make equal angles with the axis.
34. Given in position two conjugate diameters of an ellipse, construct
the curve. (If QQ', RR' are diams., constr. tangt. QT to meet director
circ. in T ; TQ = b, whence a. )
35. Given in position two conjugate diameters of a hyperbola, con-
struct the curve. (Constr. asymptotes and axes.)
36. A chord of a conic which subtends a right angle at each focus is
parallel to the axis.
37. A rectangular hyperbola circumscribing a triangle passes through
its orthocentre.
38. The hyperbola approaches its asymptotes as it extends farther and
farther from its vertex, and reaches them at infinity.
39. If two cars run, one along a hyperbola (sufficiently large), and the
other along the near asymptote, starting from tangent of vertex and keep-
ing their heads always on the same ordinate, show that they must collide.
CH. VIII.]
PERSPECTIVE : CONIC AND FOCIRCLE.
243
Theorem 33. — ' A conic is the perspective of its focircle, with
focus and tangent of vertex as centre and axis of perspective.'
If a did. PQ of a conic cuts drx. and
tangt. of A in K, H, and PD, HE _L DX ;
draw Hqp pari, to FK to meet FQ, FP.
.-. Fjp : FP = KH : KP = HE : PD = AX : PD,
,-. Fp:AX = FP:PD = ^, i.e. Fp = FA;
.', 2^f ^rid simly. q, is on focircle. Hence :
Corresp. points P, q are collr. with F,
It lines PQ, pq u cone, on AH ;
i.e. circle and conic are in persp., with F, AH as centre and axis.
Note. Small and large letters p, P, &c. on circ. and conic correspond.
Line at Infinity — Centre — Asymptotes.
Make F/ = XA, .'. H^ || FE ; .'. ray EH is persp. of Hi;
,'. persp. of I is on EH, Ft, and is pt. at inf. I on XF ; and the persp.
of ij pari, to axis of persp. is IJ at inf. (Ch. VII., Th. 20.)
Also, if c is the pole of ij\ its persp. C is the pole of I J, and is
the centre of the conic.
There are three cases, according to position of c and i.
(i.) z outside, c inside circle ; FA<2F<AX; e<l. (Above fig.)
The conic is an ellipse ; ij docs not meet the circle,
.'. I J does not meet the ellipse ; i.e. —
*An ellipse has no real points at infinity, and no real
asymptotes.'
(ii.) /, c coincide on circle ; FA = 2F = AX; e=l.
The conic is a parabola ; ij touches the circle at c.
.*. IJ touches the parabola at inf. at its centre ; i.e. —
*A parabola touches the line at infinity, which coincides
with its asymptotes.'
(iii.) i inside, c outside circle; FA>iF>AX; e>l.
The conic is a hyperbola ; ij meets the circ. in real points j, /,
from which real tangents cj, cj' are drawn from c ;
,*. I J meets hyp. in real points J, J'; CJ, CJ' are real tangts. ; i.e. —
' A hyperbola meets the line at infinity in two real points,
and has two real asymptotes.*
Ex. Construct, by perspective, the asymptotes of a hyperbola, FX = 1",
e = 3, Compare with those of Constr, 5,
244
POLAR OP CONIC : SELF-POLAR INVOLUTION. [CH. VIII.
Theorem 34. — 'In any conic or perspective of a circle,'
(i.) 'The polar of a point P in its plane is a straight line
through the points of contact of tangents from P.'
(ii.) ' If Q is on the polar of P, P is on the polar of Q.'
(iii.) * The pencil of sides of self-polar triangles at a common
vertex is in involution.*
' A line in its plane is cut in involution by the sides of self-
polar triangles of its vertex.'
(iv.) 'The double ^°^f*^ of the self-polar involution of a
rays
line „_ ., ^ points of the ray on ., ^ „„,.„« »
. . are the . . n x-l-xj. the curve,
point tangents from the point to
(Ch. YIL, Thh. 8, 9, 10, 27.)
If PRS is a transvl. of P, Q the harm,
conj. of P to R, S ; then the polar of P is
the locus of Q.
Project into a circle, denoting corresp.
points by corresp. small letters ; then,
(i.) (prqs) = {PRQS) ^ -\,
.'. locus of q is a str. line ah, the polar of ^
to circle, through points of contact of tangts.
.*. locus of Q, i.e. the polar of P, is a str.
line AB through points of contact of tangts.
PT, PT'.
(ii.) If Q is on the polar of P,
q is on polar of J9, ;p on polar of q,
.*. P is on polar of Q.
(iii.) The self-polar pencil of P has
P(AA', BB'...)=i;(aa', hh'...);
,'. P(AA', BB'...) and range (AA', BB'...) are in involn.
(iv.) ab meets circ. in double points of {aa\ hb'...),
and tangts. pt, pt' are double rays oi p{aa\ bb' ...);
.'. AB meets conic in double points of (AA', BB'...),
and tangts. PT, PT' are double rays of P(AA', BB'...).
Caution. Double points of an involn. project into double points, but
the centre does not project in general into the centre. Explain this.
Ex.* Prove in the same way the cross-ratio properties of a conic.
(Ch. VII., 21-24.)
* Important.
CH. VIII.] PERSPECTIVE OF A CIRCLE : POLAR INVOLUTION. 245
Theorem 35. — ' In any conic or perspective of a circle,'
(i.) 'Pairs of conjugate diameters are pairs in the self-polar
involution of the centre.'
(ii.) 'The polar of a point is conjugate to the central ray
of that point, and parallel to the tangents of that ray.'
(iii.) 'The foot of the polar is the centre of the self-polar
involution of that line.'
(A) Central curves. Centre at finite
distance.
(i.) If CI, CJ (I, J at inf.) are a pair
in the s.p. in vein, of the centre,
CIJ is a s.p. triangle.
.*. transvl. IPMP', pari, to CI, is divided
harmy. ;
.*. M bisects PP', and CJ, CI are conj. diams.
(ii.) If PP', polar of V, 1| CI,
then V is on polar of I, *.' I is on polar of V;
.*. V is on C J conjugate to CI or PP'.
(iii) Draw VI pari, to PP',
/. IMV is a s.p. triangle ;
.*. M, conj. of I in s.p. involn. of PP', is centre of the involn.
and also the foot of the polar of V.*
(B) Parabolic curves. Centre C at infinity.
C, on its own polar (line at inf.), is point of
contact of tangt. polar; hence corresp. point c of
circle is point of contact of corresp. tangt. ci.
(Cp. Th. 33, ii.)
And every line through c meets the circle in one
other real point q ; hence :
' Each diameter of a parabolic curve meets the curve in one
finite point ; all diameters are parallel.'
Also, if ipmp' is polar of v, cv is polar of % and iq a tangent j
also (cmqv)j {ipmp') are harmonic.
.'. PP', polar of V, II tangt. Ql (top figure, C, I at oo ) ;
M bisects PP', and is centre of its s.p. involn. ; Q bisects MV.
Hence (ii.), (iii.) above are true of parabolic curves; also :
' The diametral distance of a point from the foot of its polar
to a parabolic curve is bisected by the curve.'
* The foot of polar to a conic is on central ray.
246 PERSPECTIVE OP A CIRCLE : POLAR — INVOLUTION. [cH. VIII.
Theorem 36. — 'If the central ray of a point V to a conic
or perspective of a circle meets the curve in Q, Q\ and its polar
or chord of contact of its tangents in M, then CM . CV = CQ^.'
If TT' is polar or chd. of contact of tangts.
of a point V to the curve, then v,
(VQMQ') is a harm, range,
and C is the mid point of QQ' ;
.*. CIVI.CV = CQ2.
If VC meets the curve in imaginary points 8, Sj,
then 8, 8^ are double points and C the centre of
s.p. involn. of VC ;
.-. CM . CV = C82. (Ch. VII., Thh. 26, 27.)
The corresponding property of a parabolic
curve is VQ = QM.
Ex.* 'The focus is the pole of the directrix, and its s.p.
involution is right-angled.'
Theorem 37. — * One pair of conjugate diameters of a conic or
perspective of a circle is right-angled, and each of these is a line
of symmetry.*
The conj. diams. form the s.p. involn. of the
centre, which has one rt.-angled pair AA', BB'.
A
(Ch. YII., Th. 30.)
.*. chd. PP', pari, to AA', is bisected by its a' c
conj. CB ;
.*. curve is symmetrical about BB', and simly. ^
about AA'.
Cor. — 'If the central involution of a conic is right-angled,
the conic is a circle.'
For the curve is then symmetrical about every diameter.
Hence, if a diameter BB' bisects the angle PCP' of any two
central radii, CP = CP'.
Note. If the s.p. invohi. of a point is right-angled, it is easy to show
that the point is a focus, and its polar a directrix. (Ex. 7, below.)
These points must come on the principal axes (the right-angled pair
of the centre), and two of them are always imaginary (on the minor or
transverse axis). Tangents from a focus pass through the circular
points at infinity (Ch. VII., Th. 30).
* Important.
CH. VIII.] EVERY PERSPECTIVE OF A CIRCLE A CONIC.
247
NJA S'
Theorem 38. — ' Every perspective of a circle or conic, centre
at finite distance, is an ellipse or hyperbola.'
One at least of the right-angled
pair of conj. diams. meets the curve
in real points A, A'.* ^'
If CI (I at inf.) is conj. to AA', ^^^
and chds. PP', RR' || CI ; ^
complete quadl. PP'R'R, to I, S' ;
draw diags. PSR', P'SR.
.*. ISS' is a s.p. triangle (Ch. VII.,
Constr. 2) ; .'. S, S' are on AA' and are harm, conjs. ut A, A'.
Take A, X, Y collr. on diags. SS', PP', RR' of quadl. SRS'R';
.*. their harm, conjs. A', X', Y' are collinear (Ch. VII., Th. 22)
also MP2=MX. MX', NR2 = NY. NY' (Ch. VII., Th. 5);
MP2 MX. MX' NY.NY' . , , NR2
-. (sim. trs.)
■7 = const. ;
"MA. MA' MA. MA' NA . NA' ' "~'' NA . NA'
/. locus of P,persp. of circle, is an ellipse or hyperbola (Th. 18).
Theorem 39. — 'Every perspective of a circle or conic, centre
at infinity, is a parabola.'
If /z, tangt. at i to circ, projects to
inf. (V centre of persp.), and yj A_\/i,
and ja is the other tangt. of j ; then
aj, ai proj. into AJ, Al pari, to V;*, Vi*;
.'. JAI is a rt. ang. (Ch. VII., Constr. 6, ii.).
If chds. RR', PP' II A J ; complete quadl.
PRR'P' and diags. to J, S', S ;
.*. SS' along Al is divided harmy. at A, I.
Make MX, NY eql. to AM, AN ;
.*. A, X, Y are collr. on diags. SS', PP^
RR' of quadl. SRS'R' ;
/. their harm, conjs. I, X', Y are collinear;
/, X'Y' II Al, and MX'= NY' = const. = Z say;
/. MP2=MX.MX' = Z.MX-Z.AM.
,*. locus of P, persp. of circle, is a parabola whose vertex is A, and
whose focal distance AF is //4.
* Because two of three sides of a s.p. tr. of a circle meet it in real points.
248
CONIC THROUGH FIVE POINTS.
[CH. VIII.
Theorem 40. — 'Any five points can be projected on to a
circle;* and one only conic f can be drawn through any five
points (no three points being coUinear).'
If A, B, C, D, E are five points,
determine the line AH through A so
that A(HBCD) = E(ABCD) (Ch. VII.,
Constr. 4) ; v^ _^r-:^^^
then AH cuts externally one at least
side, BC say, of tr. BBC.
Draw any circle BCa, Ha a tangent;
join EA to K ; Ka to e ; Aa, Ee to V ;
.'. HBCea is persp. of HBCEA to vertex V,
axis BC.
Join AD to L, La to meet circ. in d.
,'. e{aBCd) = a{aBCd) = {HBCL), (aa is tangt. at a),
= A(HBCD) = E(ABCD);
and three pairs of corresponding rays meet in K, B, C on BC ;
.'. the fourth pair ed, ED must meet on BC ;
.'. triangles ead, EAD are in persp. (Ch. VII., Th. 17, ii.) ;
.*. Dd, Ee, Aa are concurrent in V ;
.*, aBCde is perspective of ABCDE.
Hence a conic, the persp. of circle aed, passes through the five
points A, B, C, D, E. M
Also if a line AP meets this conic in P, and ""^^^
any conic whatever through the five points in P'; g,, '\ •£
join PA to M ; PD, P'D to N, N' on BC. ^^C\^
;. PXABCD) = E(ABCD) = P(ABCD); ^'^ "^^P"^'
.-. (MBCN) = (MBCN'); ^^^d
.*. N coincides with N', and P' with P. '^ n'
Hence any conic through the five points coincides with this
conic — i.e. there is one only conic.
Note. The conic is the locus of points P whose cross ratio
P(ABCD)=:E(ABCD). Hence:
Cor. — ' The locus of points whose pencil to four fixed points
is constant, is a conic'
(See Ch. VII., Constr. 4, Th. 16. AH (top fig.) is tangt. at A.)
* The construction given is always real.
t A conic is assumed to have the cross-ratio properties of a circle. These
follow exactly as in Th. 34.
CH. VIII.]
RECIPROCATION.
249
.A-^
If all points and lines of a figure are replaced respectively by
their polars and poles to a fixed circle or conic, the resulting figure
is the polar reciprocal, and the process is called reciprocation.
The polar reciprocal of a curve AB is the en-
velope of polars a, h of points A, B on it. Their in-
tersection X is the pole of the chord AB ov x; and
as B moves to coincidence with A, h moves to coin-
cidence with a ; x becomes the tangent at A, and X
the intersection of coincident tangents — i.e. the point of contact
of a — on the envelope (Ch. V., Envelopes).
Theorem 41. — ' The polar reciprocal of any range is a homo-
graphic pencil.'
If A BCD is the range, P its pole to a circle,
centre O, P(a^y8) the polar reciprocal of (ABCD) ;
then P(a^78)_LO(ABCD), in the circle,
.-. P(a^y8) ^ O(ABCD) = (ABCD). /^B^X_D
And by projecting a conic into a circle, the / \\ /^^
theorem is seen to be true for any conic. ^''^
Theorem 42. — (i.) ' The polar reciprocal of a conic is a conic'
(ii.) 'One only conic can be drawn to touch five straight
lines ' (no three lines being concurrent).
If a, b, c, d, e are the lines; A', B', C,
D', E' their poles to a given conic;
draw tangts. a', b', c, d\ e to the conic
A'B'C'D'E' at these points.
The polar recpl. of this conic is a curve
touching a, 5, c, d, e at A, B, C, D, E, say ;
.'. if P is any point on this curve, its
polar recpl. p touches conic A'B'...E'.
.-. P(ABCD) =p'{a'b'c'd') = e'{a'b'cd')
(Ch. VII., Th. 25)
= E(ABCD).
.*. locus of P is a conic touching a, &, c, d, e at A, B, C, D, E.
Hence the polar reciprocal of a conic A'B'C'D'E' is a conic, and
any conic touching a, &, c, d, e must reciprocate into A'B'C'D'E^
.*. there is only one conic touching these lines.
Cor. — 'The envelope of a line forming a range of constant
cross ratio on four given lines is a conic'
•a'
250 CONICS THROUGH FOUR POINTS — PARABOLA. [CH. VIII.
Construction 6. — * Construct a conic through five points.'
We have already given two solutions
of this problem. In Ch. VII., Constr. 4,
the point P on a transversal AP of the
conic is determined by the cross -ratio
property P(ABCD) = ju,,
(M being here equal to E(ABCD).
And in Th. 40 we show how the conic may be derived from
a circle by perspective. But Pascal's theorem gives a simple
solution by drawing straight lines only, without any calculation.
Thus, given five points A, B, C, D, E ; then if P is the point of
the conic on a ray AP, ABCDEP is an in-hexagon.
Join AB, DE to L; AP, DC to N ; BC, LN to M ; EM to P.
CONICS THROUGH FoUR PoiNTS.
Since a conic can be drawn through five points, any number
can be drawn through four points. But a parabola has already
one condition given — viz. the tangent line at infinity ; hence only
a limited number of parabolas can be drawn through four points.
Construction 7. — (i.) 'Construct a conic through four points
to touch a given line ;
(ii.) Construct the parabolas through four points.*
The given tangent TT^ cuts the conic
and opp. sides AB, CD and BD, AC of
the quadl. of the points in the iiivoln. (XX',
vr...). (Ch. VII., Th. 33.)
Since TT^ touches the required conic,
the point of contact is one of the double points T, T^ of this
involution. Thus five points are known.
There are clearly two solutions.
For the parabola, construct the double rays BT, BTj of the
pencil B(XX', YY'...) on any transversal; then BT or BTj passes
through the point of contact of the parabola at infinity. Thus a
fifth point on the parabola is known — viz. the point at inf. on a
double ray — and two only parabolas can be drawn through four
points. Similarly, one parabola can be drawn to touch four lines.
A
_Ii
Y X X' Y' 1
CH. VIII.] EXAMPLES. 251
EXAMPLES— XLVIL
1.* The curve MP^ : MQ . MQ'=:/a, a constant, MP of given direction,
M on a fixed length QQ', is a conic. Interpret for parabola.
2. If a quadrilateral ABCD is inscribed in a conic, the intersections of
AB, CD ; AD, BC ; and AC, BD form a self-polar triangle.
3. Project a circle or conic into another conic so that any given point
in its plane becomes the centre of the new curve.
If the given point is on the curve, what is the conic ?
4. A conic P is projected into another P' so that a given line projects
to infinity. What do the sides of s.p. triangles of the pole of this line
become ? What is the special property of these lines ?
5. Any two angles can be projected into right angles, and at the same
time any straight line to infinity (Ch. VII., Constr. 6). Apply this to
Ex. 4 so that P' may be a circle. Is the construction always real ? (See
Th. 37, Cor.)
6. If any point is inside a conic (so that real tangents cannot be drawn
from it), every straight line through it cuts the curve.
7. If the s.p. involution of a point F to a conic is right-angled, and
DX is polar of F, if P is any point on the curve, and if FX perpendicular
to DX meets the curve in A ; show that
(i.) Every line through F meets the conic.
(ii.) A tangent PD from D on the directrix subtends a right
angle at F.
(iii.) If PA meets DX in K, then K is a bisector of AFP (polar of
K±KF and divides KAP harmonically).
(iv. ) F is a focus (show FP : PM =: FA : AX, if PM ± DX).
8. If the distances PF, QF, RF of three points on a conic from F are
proportional to their distances PL, QM, RN from the polar of F, show
that F is a focus.
9. Assuming that the s.p. involution of a focus of a conic is right-
angled, show that the focus must lie on one of the right-angled pair of
conjugate diameters.
10. Show that real tangents can always be constructed from two only
of the vertices of a s.p. triangle of a conic.
11. Real points exist on two only of the sides of a s.p. triangle. How
do these sides correspond to the vertices of Ex. 10 ? Why ?
12. Given the centre and a s.p. triangle of a conic, construct the real
points on the sides of the triangle. Simplify this when the triangle is
obtuse and the point is the orthocentre.
13. Show that there is one only conic having a given centre and a
given s.p. triangle. If the triangle is acute and the point its ortho-
centre, what is the conic ? (Show that five points are fixed.)
* Important.
252 EXAMPLES. [CH. VIII.
14. Right-angled pairs of chords of a conic from a point on it determine
a system of chords concurrent on the normal. (Ch. VII., Th. 32.)
15. The circumcircles of s. p. triangles of a point to a conic are coaxial.
16. The intercept of a tangent of a conic by the directrices is divided
harmonically by the point of contact and the pole of the normal.
Interpret this for the parabola.
17. Find the locus of the pole of the side AD to all conies circum-
scribing the quadrilateral ABCD.
18. Show that a conic can be circumscribed about a quadrilateral
ABCD so as to have any side, or the diagonal AC or BD, a diameter.
19. If three conies have a common chord, the other common chords of
each two of them are concurrent.
20. ABCD is a quadrilateral, AB is fixed, and CD of given length
moves on a straight line. Find the locus of intersections of AD and BC.
21. The locus of the point P on a transversal QXYZ of a fixed
triangle ABC, turning about a fixed point Q, such that (XYZP)=/a, a
constant, is a conic through A, B, C, Q.
22. If three sides of a variable triangle pass through fixed points, and
two vertices move on fixed lines, the locus of the third vertex is a conic.
(MacLaurin.)
23. If three vertices of a triangle move on fixed lines, and two sides
pass through fixed points, the third side envelops a conic. (Reciprocate
the last example. )
24. Two tangents from a point to a conic form an involution with the
joins of the point to the vertices of a circumscribing quadrilateral of a
conic. Reciprocate this.
25. Reciprocate Pascal's theorem and Brianchon's theorem.
26. Reciprocate with regard to a circle the theorem, 'The locus of a
point at which two fixed points subtend a right angle is a circle.'
And the following : ' The envelope of one side of a right angle whose
vertex describes a fixed line, and whose second side traverses a fixed
point, is a conic'
27. If the polars to a conic of three vertices of a triangle are concur-
rent, the polars of all points are concurrent. What is the conic ?
28. Two angles of constant magnitude move about fixed points P, Q,
and the intersection of one side of each describes a straight line ; show
that the intersection of the other sides describes a conic. (Newton.)
What is the curve if the first intersection describes a conic?
29. Find the locus of the centres of all conies passing through four
given points. Where does it meet the chords of these points ?
30. Four points A, B, C, D of a variable conic are fixed. Two iixed
lines AP, CQ meet the conic in P, Q. Show that PQ passes through a
fixed point.
253
CHAPTER IX.
SOLID GEOMETRY.
PLANES, POLYHEDRA, CONE, CYLINDER, SPHERE.
Definition 1. — The meet of two planes which have a common
point is the straight line in which they cut each other.
Definition 2. — A dihedral angle is the figure of two planes
which terminate in a common line (e.g. angle of two pages of a
partly opened book).
Definition 3. — A plane section of a figure is a figure formed
by the first figure on a plane cutting it.
Definition 4. — A perpendicular to a plane is a straight line
perpendicular to all lines in the plane which meet it.
Definition 5. — A straight line parallel to a plane is a parallel
to some line in the plane.
Definition 6. — Parallel planes are perpendicular to tlie same
straight line.
Note. It is sliown below that one only perp. to a plane exists at each
point in the plane, and that any two such perps. are pail. The direc-
tion of a perp. therefore serves to define the direction of the plane. Thus
two parallel planes as just defined have the same direction, which is
consistent with our definition of parallel straight lines. It is shoAvn
also that pari, planes either coincide or do not meet, and that non-parl.
planes meet.
Conditions which determine a Plane.
Either part of a plane may be turned about the line joining two
fixed points A, B in it ; if the two parts are turned at a suitable
rate, they may be kept always in one plane, and in making a
complete or half turn they sweep out the whole of what is called
space — i.e. in some position or other they contain any third point
C whatever ; thus a plane can be drawn through any three non-
collinear points A, B, C by moving to pass through A, B and
turning to pass through C ; and any second plane through A, B, C
coincides with the first (Th. 5, Ch. I.). Hence :
254
PLANE AND PERPENDICULAR.
[CH. IX.
Theorem 1. — 'One only plane can be drawn through any
three non-coUinear points; through any straight line and a
point not on the line ; through any two intersecting or parallel
straight lines.'
Theorem 2. — *A straight line perpendicular to two straight
lines in a plane is perpendicular to all straight lines in the
plane through their common point ; i.e. it is perpendicular to
the plane.'
If NP±NA, NB in plane ANB,
and NC is any other line in the j^lane,
produce PN to Q, make NQ eql. to NP.
Then in plane APQ, NA is rt. bisector
of PQ,
/. AP = AQ ; similarly BP = BQ ;
and AB is common to trs. APB, AQB,
/. ang. PAB = QAB,
And in trs. APC, AQC,
AQ = AP, AC is common, ang. PAC = QAC ;
.*. CP = CQ, and C is on rt. bisector of PQ in plane PCQ ;
.-. CNJ_PQ.
Hence PN J_any other line NC in plane ANB.
Cor. (i.). — 'The locus of perpendiculars to a line at a given
point is the plane perpendicular to the line at that point.'
Cor. (ii.). — 'The locus of points in space equidistant from two
fixed points is the plane bisecting their join at right angles.'
Construction 1. — 'Construct a plane perpendicular to a
given line at a given point.'
Draw a perp. to the line at the point, and rotate the figure
about the line ; the perp. sweeps out the plane.
Construction 2. — ' Construct a perpendicular to a plane at
a point in it.'
Construct any plane P by (i.) perp. to a line X ; superpose P on
the given plane so that X passes through the given point ; the
new position of X is the required perpendicular. Clearly there is
one only perp. at each point of a plane.
CH. IX.]
MEET — ANGLE OF TWO PLANES.
255
either
Theorem 3. — 'Two planes having a common point
coincide or meet in a straight line.'
If planes A, B meet at P, draw the perps.
PM, PN to A, B ; then
either (i.) PM coincides with PN, M
.'. plane A coincides \vith B ;
or (ii.) PM does not coincide with PN ;
draw PQ perp. to plane NPM,
.*. PQ±PM and PN ; i.e. PQ is in planes
A, B;
.*. planes A, B meet in a straight line PQ.
Theorem 4. — 'The angle formed hy two planes on a third
plane perpendicular to their meet iis constant, and called the
angle of the planes.'
If AOB, CQD are angles formed by
planes AQ, BQ on planes AB, CD perp.
to their meet OQ :
In plane AOB make a scale, unit
two right angles, from OA, and divide
decimally the unit, tenth, &c. contain-
ing OB.
Planes through OQ and the points of
division determine a similar scale, same
unit, from QC in plane CQD ;
and OB, QD come between the same divisions of the two scales ;
.'. ang. AOB = CQD = constant.
Note. Perpendiculars to OA, OB in plane OABj. planes AQ, BQ,
and their angle = BOA. Hence :
Cor. (L). — ' The angle of two planes is equal to that of their
perpendiculars at any point on their meet.'
Cor. (ii). — *A plane containing a perpendicular to a given
plane is perpendicular to the plane.' *
Cor. (iii.). — 'The meet of two planes each perpendicular to
a third, is perpendicular to the third plane.'
Ex. Find the shortest distance from a given point to a given
plane.
* '.' the angle of the planes is right.
256 PLANES — PARALLEL PLANES. [CH. IX.
Theorem 5.—* Two perpendiculars to a plane are parallel.'
If MP, NQ_L plane AB,
make MA, NB perp. to MN, and
make NR pari, to PM in plane PNM ;
.-. NRXMN, and MNi. plane RNB.
.'. ang. RNB = ang. of planes PNM, AB
= PMA = rt. ang.
/. NR±NB and to NM, i.e. ± plane AB ;
.*. NQ coincides with NR and || MP.
Cor. — * Perpendiculars to parallel planes are parallel.'
Theorem 6. — 'A plane perpendicular to any line is perpen-
dicular to any parallel line.'
If plane AB (last %.)i.MP, and NR || MP,
then NR is in plane PNM ;
make NQ perp. to plane AB.
.*. NQ II MP II NR ; i.e. NQ coincides with NR ;
.*. plane AB±NR.
Cor. — ' Two parallels to a third line (which need not be in
their plane) are parallel.'
Construction 3. — 'Construct a perpendicular to a given
plane AB from an outside point Q.'
Make MP±AB, QN||MP; .'. QNJLAB.
Theorem 7.— 'The meets of parallel planes with any third
plane are parallel.'
If pari, planes X, Y meet a plane AD in
CD, AB ;
make plane NAC perp. to AB, containing
the parls. CN, AM perp. to X, Y ;
and make CE pari, to AB in plane AD.
.'. CE_L plane NAC, ±NC; "V
.*. CE is in plane X, and also in plane AD ; '
.*. CE is the meet of planes X, AD ;
,*, CD coincides with CE and |j AB,
CH. IX.]
MEETING OF PLANES.
257
Theorem 8.^' Two parallel planes either coincide or do not
meet.'
Parallel planes X, Y
either (i.) have a common point N and a common perp. NP, and
therefore coincide ;
or (ii.) do not meet. (We say that they meet in the str. line at
infinity.)
Theorem 9. — ' Two non-parallel planes meet.*
If plane X not || Y, and from a point A in X,
AN is perp. to Y ;
choose some line AB in X not perp. to AN.
.'. in plane BAN, cutting Y in NC,
AB not II NC, and meets it ;
.'. the planes X, Y meet.
Theorem 10. — * Three planes, no two of which are parallel,
meet in a point, which may be at infinity.'
If a plane Z (BAN in last fig.) meets two planes X, Y in AB,
NC; then
either (i.) AB, NC in plane Z cross at a point,
or (ii.) AB, NC n Z are parallel and meet at infinity.
Theorem 11. — 'Two lines which meet form the same angle
as any two parallels to them which meet ; and the planes of
the two pairs of lines are parallel.'
If OA II QC, and OB || QD,
make ON perp. to plane AOB,
meeting plane CQD in N ; "^
draw NE, NF pari, to CQ, QD ;
/. NE II OAJ_ON ; and NF || OB±ON ;
.'. plane ENF, i.e. CQDJLON
II plane AOB.
Also, ang. AOB = ang. of planes AON, BON = ENF
= CQD.
* If QM ± plane CQD, the pari. ON meets plane CQD on the meet of th§
planes OMQ, CQD,
p.G. q
258
PARALLEL PLANES — SECTIONS.
[CH. IX.
Theorem 12. — ' The meets of planes parallel to a given line
are parallel to this line.'
For a pari, to the line, through any point on the meet of two
planes parallel to it, is in both planes and coincides with their
meet.
Theorem 13. — ' The sections by parallel planes of a system
of planes parallel to a given line are congruent.'
If a system of planes AG, BH, &c. pari, to a
given line meet in AF, BG, &c., and cut two
pari, planes in polygons ABC..., FGH...;
then AF || given line 1| BG, &c. ;
AB II FG; BO || GH, &c.;
ang. ABC
FGH, &c.; and figs. AG, BH, &c.
are parms. ;
.*. AB = FG, BC = GH, &c.;
.'. polygon ABC . ^ poln. FGH . . .
Theorem 14. — *A system of parallel planes cuts off propor-
tional parts from any two straight lines.'
If pari, planes cut two lines in A, B, C, D and
P, Q, R, S, and AS in A, E, F, S ;
then BE || CF || DS ;
AP||EQ||FR;
.-. AB : CD = AE : FS = PQ : RS.
Theorem 15. — 'The sections by parallel planes of a system
of concurrent planes are similar polygons whose areas are
proportional to the squares of their distances from the common
vertex.'
If planes AG, BH, &c. through O meet in
OA, OB, &c., and cut two pari, planes in
polns. ABC..., FGH...; andOMN±ABC;
then tr. CAB ||| OFG, and OMB ||| ONG ;
.'. AB : FG = OB : OG = BC : GH, &c. ;
ang. ABC = FGH, &c.;
.-. poln. ABC... Ill FGH....
Also, area AC : area FH = AB2 : FG^ = OM2 : ON2.
CH. IX.] PROJECTION AND PERSPECTIVE IN SPACR 259
If all the points of a figure are joined to a fixed point, forming
a pencil or cone of rays at the point, any plane section of the
pencil is a representation of the figure, and is called its perspective
or projection on the plane ; the fixed point being the origin or
vertex of projection. Thus the representation of a scene on a
painter's canvas is the projection of the scene on the plane of
the canvas from the eye as vertex. We need here only consider
the projection of plane figures.
If V is the vertex of projection, P, Q points on a given plane A ;
P', Q' their projections on another plane A',
and RS the meet of A, A';
then the plane VPQ contains P'Q', and the
meets of the planes A, A', VPQ have a
common point ;
i.e. PQ, P'Q' intersect on the meet RS of the
planes A, A'.
Hence the important principle in the pro-
jections of plane figures :
Theorem 16. — ' The joins of corresponding points of two plane
figures in perspective pass through the vertex, and the inter-
sections of corresponding straight lines are on the axis of
projection.'
It is clear, therefore, that the relations of two such figures are
precisely the same as in plane perspective.
Thus the projection of a circle is a conic, and the projection of
a conic is a conic ; also, any plane section of a cone on a circle or
conic as base is a conic.
All points at infinity in one plane correspond to a straight line
on the plane of projection, parallel to the meet of planes ; so that,
for example, a parallelogram projects into a complete quadrilateral.
Conies were originally studied as plane sections of the right cone,
whence the name ; and the modern geometry of position, developed
by Cremona and others, is based upon projection in space, of which
plane projection is a limiting case ; the conic being defined as the
projection of a circle. Students who have mastered Chh. VII.
and VIII. will have no difficulty in following the more general
theory of perspective in space.
260 ORTHOGONAL OR RIGHT PROJECTION. [CH. IX.
If perpendiculars from all points of a iigure are drawn to a
given plane, the plane figure formed is the right projection of
the given figure on the plane. By Th. 13, projections on parallel
planes are congruent.
Theorem 17. — 'The right projections of parallel lines on a
plane are parallel, and proportional to the lines.'
If A'B', C'D' are rt. projs. of AB, CD on
a plane, ^ ^^
and AM, CNi.BB', DD'; '^f'^^M c^/
then BB' || DD', and AB I| CD,
.-. plane AB' |I plane CD', /. A'B' || C'D',
and ang. B = D, .*. rt. triangle ABM ||| CDN ; c^ 6'
/. AB : CD = AM : CN = A'B' : C'D'.
Theorem 18. — *The area of the right projection of a plane
area on a plane is the product of the given area and the
cosine of the angle of the planes.'
If plane AB'D || plane of projection, meeting
plane of tr. ABC in AD,
and plane BNB'j. meet AD ;
ang. BNB' = ang. of planes = a say. ^
.'. area AB'D = JAD . B'N = JAD . BN cos a
= ABD cos a.
Similarly, AC'D = ACD cos a,
.*. sum or diffce. AB'C' = ABC cos a.
Similarly, if any area X is broken up into triangles,
proj. X' = X cos a.
Plan and Elevation.
Drawings of solid figures for machine and building construction
are made by means of right projections on horizontal and vertical
planes, called plan and elevation respectively. The methods of
construction are given in text-books of Practical Solid Geometry.
The projection of a circle, radius r, is an ellipse, axes 2r, 2?" cos a.
Also, the projection of a parallogram is a parallelogram.
CH. IX.] SOLID figures: definitions. 261
Definition 7. — A solid figure is one whose points, lines, surfaces
are not all in one plane.
Definition 8. — A conical surface is one generated by a straight
line moving so as always to pass through a fixed point.
The moving line in any particular position is a generator, and
the fixed point the vertex of the surface.
Definition 9. — A cylindrical surface is one generated by a
straight line parallel to a fixed line, and moving along a given
curve.
The fixed line is the guide, and the moving line in any par-
ticular position a generator, of the surface.
Definition 10. — A solid angle is a figure formed by three or
more planes, each meeting two others, at a point (polyhedral
angle), or by a conical surface which returns into itself.
Polyhedral angles are trihedral, tetrahedral, &c. according to
the number 3, 4, &c. of planes at the point.
Definition 11. — A polyhedron is a figure enclosed by planes.
A polygon formed with one of the planes by its neighbours is a
face, a side of a face an edge.
A regular polyhedron has all its faces congruent regular
polygons.
Polyhedra are tetrahedron, hexahedron, &c. according to the
number 4, 6, &c. of faces.
A paralhedron "^ is a hexahedron whose faces are parallel, two
and two.
Its faces are parallelograms and its edges are equal, four and
four.
A cuboid, or rectangular paralhedron, is a paralhedron whose
non-parallel faces are perpendicular.
Its faces are rectangles and its edges are equal, four and four.
A cube is an equal-faced cuboid.
Its faces are squares, and all its edges are equal.
* This name is more suggestive than parallelepiped.
262 SOLID FIGURES : DEFINITIONS. [CH. IX.
Definition 12. — A pyramid is a figure formed by a polyhedral
angle and a plane cutting its faces.
The point of the angle is vertex, the plane section base, and
the distance vertex to base the altitude, of the pyramid.
The faces meeting at the vertex are triangles.
Definition 13. — A cone is a figure formed by a conical angle
and a plane cutting it.
The point of the angle is vertex, the plane section base, and the
distance vertex to base the altitude, of the cone. A straight line
along the surface from vertex to base is a vertical edge.
A right circular cone, or cone of revolution, has its base a
circle, and its vertex on the axis or central perpendicular of the
base. The length of a vertical edge is in this case the slant
height of the cone.
Definition 14. — A prism is a polyhedron having two parallel
faces, and the others parallel to a given line called its guide.
The guide faces are parallelograms, and the guide
edges all equal; the parallel bases are congruent
polygons, and their perpendicular distance is the
altitude of the prism.
Prisms are triangular, hexagonal, &c. according to
the number of sides 3, 6, &c. of the base.
A wedge is a triangular prism.
Definition 15. — A cylinder is a figure enclosed by a cylindrical
surface and two parallel planes.
The plane sections are bases, their perp. distances the altitude,
of the cylinder, and straight lines along the surface from base to
base, pari, to the guide, are guide edges.
A right cylinder has its bases perpendicular to the guide.
A right circular cylinder is a right cylinder with circular bases.
The central perpendicular of its bases is its axis.
Note. Theorems 13 and 15 may he extended to cylinders and cones
by considering these as limiting forms of pj'ramid or prism whose vertical
or guide edges move up to coincidence. Thus :
CH. IX.] PLANE SECTIONS — SPHERE. 263
Theorem 19. — (i.) 'Parallel plane sections of a cylinder are
congruent ; (ii.) parallel plane sections of a cone are similar,
and their areas are proportional to the squares of their distances
from the vertex.'
Definition 16. — A sphere is a closed surface all points of which
are equally distant from a fixed point, called its centre.
A straight line from centre to surface is a radius, and from
surface to surface through the centre a diameter.
Note. A right circular cone, a right circular cylinder, and a sphere
can be generated by revolving respectively a right triangle, a rectangle,
and a semicircle about one side.
Theorem 20. — *A plane section of a sphere is a circle, and is
perpendicular to the diameter through the centre of the circle.'
If A B is a plane section of a sphere, centre O,
and ON J_ plane AB ;
then in rt. trs. OAN, OBN,
OA = OB, ON is common,
.-. NB = NA.
.'. B is on the circle, centre N, rad. NA ;
and the plane ± the diameter NO.
Definition 17. — A central plane section of a sphere is a great
circle ; any other plane section a small circle.
Definition 18. — A lune of a sphere is a portion of the surface
cut out by a dihedral angle through a diameter (as ACBD).
Definition 19. — A zone is a portion of a sphere a
cut out by two parallel planes.
Definition 20. — A tangent plane to a sphere
at a point is the plane containing all tangents to
great circles through the point ; and it is perpen-
dicular to the radius at that point.
By joining any number of points to form an
inpolyhedron, and drawing tangents at these points to form a
circumpolyhedron, the sphere can be seen to be the limit of an
inpolyhedron whose vertices move up to coincidence, their number
becoming infinite.
^64
POLYHEDRAL ANGLES.
[CH. IX.
Theorem 21.—' The sum of any two face angles of a trihedral
angle is greater than the third.'
If the face ang. AVC is not less than either
ang. AVB, BVC of trihedral ang. V ;
make ang. AVD eql. to AVB, and
make VB = VD.
.-. tr. AVB = AVD, and AB = AD.
But AB + BO AC, in tr. ABC,
>AD + DC;
.*. BO DC.
And in trs. BVC, DVC,
B V = D V, C V is common, BC > DC ;
.•. ang. BVODVC.
.•. angs. AVB + BVOAVD + DVOAVC.
Theorem 22. — ' The sum of face angles of a polyhedral angle
is less than four right angles.' *
If ABC... is a plane section of a polyhedral
angle V, and P a point within the polygon
ABCD... ;
then sum of angs. of all trs. at V
= sum of angs. of all trs. at P.
But sum of base angs. A, B, C... of the V trs.
> sum of base angs. A, B, C... of the P trs. .
sum of face angs. at V < sum of angs. at P
< four rt. angs.
Eegular Polyhedra.
No regular polyhedron can have face polygons of more than
five sides, since three angles of a regular hexagon make 4 rt. angs.,
so that three or more angles of a regular hexagon, heptagon, &c.
cannot form a polyhedral angle ; thus all the possibles are Tetra-
hedron, Octahedron, Icosahedron (3, 4, 5 triangles at a point) ;
Cube (3 squares at a point) ; Dodecahedron (3 pentagons at a
point). These may be studied from models.
* If there are re-entrant angles, the theorem may not be true.
CH. IX.]
AREA OF CONE.
265
Theorem 23. — 'The area of the curved surface of a right
circular cone is half the product of slant
height and circumference of base.' v
If VAB is a cone, VA, VC, VD... vertical
edges forming the pyramid VACD..., and VP
perp. to AC ; then
the surface of the cone is the limit of the sum
of tr. faces VAC, VCD..., when the edges of
the pyramid move up to coincidence ; and its
base is the limit of the polygon ACD....
1. .. ^/VP.AC VP'. CD
limit of ( ^ 1 ^ —
.*. area of cone VAB
.,.)
= VA X half length of circle AB
= half (slant height x circumf. of base).
Theorem 24. — * The curved surface of a right cylinder is the
product of altitude and circumference of base.*
Prove as the last, using guide edges.
Theorem 25. — ' The area of the curved surface of a frustum *
of a right circular cone is the product of its slant height and
circumference of median circle.'
If CDAB is a frustum of a rt circ. cone,
vertex V, slant height hj base radii r\ 7\ ; and if
h', \ are slant heights of the cones VCD, VAB ;
then diam. of median circle
= I (diam. CD + diam. AB)
= r' + i\',
.*. circumf. of median circle = 7r(/ + r-^.
Also, h! \r' = h^ \r^ = ii say ;
, area of frustum = area of VAB - area of VCD
= \{h'. 27rr' - \ . 27r?'i) = 7r/i(/2 _ r^^)
= 7r/x(r' - ri)(/ + r^) = irQi' - h^)(r' + r^
= hx circumf. of median circle.
* A part contained between the base and a pari, plane.
%6
THE SPHERE — AREAS.
[CH. IX.
Theorem 26. — (i.) * The area of the curved surface of a zoner
of a sphere is the product of its altitude and a great circle ;
(ii.) the area of a sphere is the product of a diameter and a
great circle.' (Area = 47rr2.)
If CDEF is a zone of a sphere, centre
O, and diam. AB_Lparl. planes EF, CD;
cut the zone by pari, sections into a
number of small zones as EFPQ, and
draw perps. CL, PN, EM to AB in the
plane ACB.
Then, if section PQ moves to coin-
cidence with EF, the surface of EFPQ
becomes a frustum of a rt. circ. cone
whose area is Stt EM . EP.
And EP becomes tangt. at E and_LOE ;
also EM_LMN ;
.-. ME:OE = cos MEO = MN:EP;
.-. rect. EM. EP = MN.OE;
.*. area of zone EFPQ = 27r EM . EP = 27r MN . OE
= M N X circumf . of great circle ;
.*. area of CDEF = sum of small zones
= ML X circumf. of great circle
= alt. X great circle.
And area of sphere = sum of areas of zones from A to B
= diam. x great circle.
Ex. Show that the curved surface of a zone of a sphere is equal to
that of a right cylinder of equal height, base radius that of the sphere.
Theorem 27. — ' The area of a lune of a sphere is the product
of a diameter and the median arc*
If ACBD is a lune of a sphere,
ECD the great circle perp. to diam. AB ;
then CD is the median arc.
Hence, if CD = ju. . circumf. of ECD,
area of lune = ju . area of sphere
= /x . circumf. of ECD x diam.
= CD X diam.
= diam. x median arc.
CH. IX.]
VOLUME OF PARALHEDROX — CUBOID.
!67
Theorem 28. — 'Two cuboids are congruent which have three
concurrent edges of one equal, each to each, to three concurrent
edges of the other.'
This needs no demonstration.
D F
Theorem 29. — 'The volume of a cuboid is the product of
measures of its edges.'
If A, fx, V are measures of the edges
AB, AC, AD of cuboid AE ;
from A along AB make a decimal scale
of units, and through the divisions draw
planes pari, to ACD ;
these form a similar scale of cuboids, unit
FC ; and the point B and plane BE come
between the same divisions of the two scales ;
.*. vol. AE = A.FC.
Similarly, making scales of units along AC, AD,
vol. FC = ftFG = /Ltv . GH = ftv unit cubes ;
i.e. vol. AE = A/Av.
Cor. — * Volume of a cuboid = base x altitude.'
Theorem 30. — ' The volume of a paralhedron is the product
of base and altitude.'
If AB, CD are opp. faces, AEFC a
base of paralhedron AD ;
through A, C draw pari, planes AKG,
CLH perp. to AC, making parn. AH.
Then base AKLC = base AEFC,
and fig. HCLFD = GAKEB;
.*. parns. AH, AD have equal vol.,
base, and alt.
Similarly, drawing planes through AC, KL, perp. to base AL, a
cuboid AM is formed of same vol., base, and alt. as AH.
/. vol. AD = vol. AM =base AL x alt.
= base AEFC X alt.
K E
Cor. — < Paralhedra of equal bases and altitudes are equal'
268
VOLUME OF PRISM.
[CH. IX.
Theorem 31. — * The volume of a triangular prism or wedge is
the product of base and altitude.'-
L D
v^_ VI v_ _i/
M "g N H
If ABC, DEF are tr. faces of the
wedge or prism AF,
complete paralhedron AGDH by planes
through BE, CF pari, to AF, AE ;
and through B, E draw pari, planes
BPK, EQL perp. to guide edge BE,
forming the paralhedron KMLN.
Then fig. KBPCA = LEQFD,
.*. wedge AF = wedge KQ in vol. ;
but rt. wedge KQ^rt. wedge MQ = wedge GF, in voL J
.*. wedge AF = wedge GF in vol.
= I parn. AGDH
= base ABC X alt.
We have proved incidentally :
Cor. — 'A plane diagonal of a paralhedron bisects it; i.e.
divides it into wedges of equal volume.'
Note. When the faces of a paralhedron bisected by the diagonal
plane ± the other faces, the wedges are evidently congruent, as we
have assumed in the above proof.
Theorem 32. — * The volume of a prism is the product of base
and altitude.'
If ABC..., FGH... are pari, faces of a prism,
and planes are drawn through the guide edges
and a pari, line MN ; the prism is divided into
a number of wedges MFG, MGH, &c.
.'. vol. of prism
= sum of wedges MFG, MGH, &c.
= alt. X sum of bases MAB, MBC, &C.
= alt. X area of base ABCDE.
Ex. 1. Show that the square of diagonal of a cuboid is the sum of
squares of its edges.
Ex. 2. Calculate the volume of a prism, alt. 20 ft., base a regular
octagon of 1 ft. side.
CH. IX.] VOLUME OF PYRAMID. 269
Theorem 33. — * The volume of a pyramid is one-third of the
product of base and altitude.'
If VABC is a pyramid, area of base ABC = a,
and alt. VNPyi=7i;
divide the total altitude VM into any number
n of equal parts y, so that ny = }i\
construct on each section an upper right prism,
and on all but the lowest a lower right prism,
of alt. y.
Then if S' is the volume of all the upper prisms,
Sj II II M lower II
and S It II of the pyramid ;
S>Si but<S'.
But S' - Sj = lowest prism = y . ABC, which can be made
smaller than any given cube, however small, by making ?/ small
enough ;
.'.if the points of division move to coincidence, S' - Sj, and
therefore S' - S, becomes zero ;
.*. S = limit of S' when y becomes zero.
!N"ow, the area of any section DEF is proportional to the square
of its height VN = /xVN2, say ;
.'. a = area ABC = /x^^ = irn^y'^,
and the volumes of successive prisms from V are
.-. s-=/xy3(i2 + 22 + 3^... + ^^)==/x^3^(^+y+i)^ •
= ^ + ^%'f-^multiplyingout)
_ha ya fxhy^
3 2 6 *
.*. S = -^ (limit of S' when y is zero).
Cor. — 'Pyramids of equal bases and altitudes are equal.'
Ex. Calculate the volume of a pyramid, alt. 12 cm., base a regular
hexagon of 2 cm. side ; and the volume of a frustum of this pyramid
of height 8 cm. (See p. 271.)
* By a well-known algebraic formula.
270 VOLUME OF CYLINDER — CONE — SPHERE. [CH. IX.
Theorem 34.— 'The volume of a cylinder is the product of
base and altitude.'
If ABCD is a cylinder, and AC, BD,
EF, &c. are guide edges ;
then vol. of prism AEG... CFH = base x alt.
Hence, if the guide edges AC, EF... move
to coincidence along the surface,
vol. of cylinder ABCD = base x alt.
Theorem 35. — 'The volume of a cone is one-third of the
product of base and altitude.'
If VAB is a cone, and VA, VC... vertical
V
then vol. of pyramid VAC... =^ base x alt.
Hence, if the vertical edges VA, VC... move to
coincidence along the surface,
vol. of cone VAB = J base x alt.
Cor. — ' A cone is one-third of the cylinder of the same base
and altitude.'
Note. If r is the radius of base, h the altitude of a right circular cone
or cylinder, the volume is TrrVijS or Trr^A.
Theorem 36. — ' The volume of a sphere is that of a pyramid
whose base is the surface of the sphere, and altitude the
radius.*
If the whole surface of a sphere, centre O, is
divided into small triangles ABC,
the volume of the pyramid OABC on the plane baso
ABC is J base x alt.
Hence, if the points of the bases move up to
coincidence along the sphere, the sum of all bases
becomes the surface, and of all pyramids the volume, of the
sphere, and the altitude becomes the radius ;
.'. vol. of sphere = J surface of sphere x radius.
Cor. — ' The volume of a portion of a sphere cut out by a cone
whose vertex is the centre is -J surface of portion x radius.'
CH. IX.] VOLUME OF FRUSTUM OF PYRAMID — CAP OF SPHERE. 271
Theorem 37. — ' The volume of a frustum of a pyramid or cone
is that of a pyramid whose altitude is that of the frustum, and
base the sum of the two bases and their mean proportional.'
If ABC...DEF... is a frustum of pyramid or
cone, vertex V, and the
areas of ABC, DEF are a\ a^, the
alts. It II M II Ti', \, and
alt. of the frustum h ;
then if a = fxh'^, a^ = }ih^,
mean prop, of a', a^ = J^ = fjih'h^ ;
.*. vol. of frustum AF = ^fxh'^ - \iih-^
E ^^>C
Theorem 38. — 'The volume of a segmental cap,"*^ altitude h,
of a sphere of radius r, is tttJi^ - — .'
If ACS is a segmental cap of a sphere, centre
O ; then OAB is a cone.
If Jc is alt. of cone, / the radius of plane
section AB — i.e. of base of cone — then
I^=^r^-k^ = h{2r-h).
Also, vol. of solid sector OABC
= J area ACB x rad. of sphere
And vol. of cone OAB = ^— = — ^ ^ ^
o o
.'. vol. of cap = 'nr/i2
Cor. — ' The volume of a zone can be found as the difference of
two segmental caps.'
Ex. Calculate the volume of a zone, alt. 2 cm., dist. of greater face
from centre of sphere 1 cm., lad. of sphere 4 cm.
* A cap is a portion cut off by a plane surface.
272 EXAMPLES. [CH. IX.
EXAMPLES— XLVIII.
Planes and Solid Figures.
1. The angle of two planes is that of two perpendiculars to the planes.
2. A plane is symmetrical in space about any perpendicular.
3. Find the locus of points in space equidistant from two fixed points.
4. If AP is perpendicular to the plane of an isosceles triangle ABC
(AB = AC), BC is perpendicular to the plane tlirough AP bisecting BC.
5. Find the locus of points equidistant from all points of a circle.
6. A right triangle revolves about a side of the right angle. Show
that the opposite vertex describes a circle.
7. If the distances PA, PB of a point P from fixed points A, B are
given in magnitude, P lies on a fixed circle.
8. Find the locus of points at a given distance from a given plane.
9. The angle formed by two planes on a plane perpendicular to one of
them is greatest when the third plane is perpendicular to both.
10. The locus of points in a plane at a given distance from a given
outside point is a circle.
11. A plane perpendicular to any line is perpendicular to any plane
through the line.
12. Through a given line in a plane draw a perpendicular plane.
13. Two planes which meet form equal angles on any two parallel
planes cutting them.
14. Through a point draw a line parallel to a given line in a plane.
15. Through a point draw a plane parallel to a given plane.
16. The locus of points at a given distance a from a given point, and
at a given distance h from a given plane, is a circle.
17. Find the locus of points dividing in a given ratio (i.) all lines from
a given point to a given plane, (ii.) all lines bounded by two planes.
18. Construct the common perpendicular of two non-intersecting non-
parallel straight lines.
19. The common perpendicular of two non- intersecting lines is the
least line that can be drawn from one to the other.
20. AH points dividing in a given ratio a line bounded by two straight
lines lie in a certain plane.
21. No two straight lines joining points in two non -intersecting
non-parallel lines are in the same plane.
22. If two faces of a trihedral angle slide along fixed planes, the third
face is always parallel to a fixed plane.
23. The dihedral angle of two planes is proportional to their angle.
24. Two planes which meet cross one another and form four angles.
CH. IX.] EXAMPLES. 273
■25. If two planes cross, opposite dihedral angles are equal.
26. Any three faces of a tetrahedron are together greater in area than
the fourtli.
27. Find the locus of a point whose distances from two fixed planes
(i. ) are equal, (ii. ) have a fixed ratio.
28. If the joins of a point to the contour of a given figure are multi-
plied by a given ratio, their ends form the contour of a similar and
similarly situated figure.
29. Find the locus of a point dividing in a given ratio the joins of a
given point to points on a sphere.
30. If a sector of a circle is revolved about its bisector of angle, a
cube can be inscribed in the resulting solid sector.
31. The points of contact of tangents from an outside point to a sphere
are in a plane and on a circle. (Rotate circle about diam. )
32. Tlie locus of points from which equal tangents can be drawn to
two spheres is a plane. (Radical plane.)
33. The radical planes of three spheres meet in a straight line.
34. The centroids of parallel plane sections of a trihedral angle lie on a
line through its vertex.
35. The joins of the vertices of a tetrahedron to the centroids of
opposite faces are concurrent, and divide each other in the ratio 1 : 3.
36. Find the locus of a point equidistant from three given points.
37. Circumscribe a sphere to a tetrahedron.
38. Find the locus of a point equidistant from the faces of a trihedral
angle.
39. Inscribe a sphere to a tetrahedron. How many e-spheres are
there?
40. Insciibe a sphere in a right circular cone.
41. The vertex of a right conical surface divides the line of centres of
two inscribed spheres in the ratio of the radii.
42. The lengths of all common tangents of the spheres of Ex. 41 whose
directions pass through the vertex are equal.
43. Circles of a sphere equally distant from its centre are equal.
44. Calculate the following surfaces : (i.) cube, edge 1-6" ; (ii.) regular
tetrahedron, edge 1" ; (iii.) right circular cone, height 6 cm., base radius
2-5 cm. ; (iv.) right circular cylinder, height 3-2", base radius 1-8".
45. A zone is cut out of a sphere, radius 5 cm., by two planes distant
3-2 cm., 4-8 cm. from the centre ; calculate its area.
46. Calculate the ratio to the whole surface of the earth of the part
contained between two meridians of longitude 23° 30' and 57° 50' W.
47. What is the area of the cap of the Arctic Circle of the earth?
(Earth's radius, 3960 miles ; arc from pole to edge of circle, 23J°.)
p. a. R
274 EXAMPLES. [CH. IX.
48. A heap of stones has the form of a frustum of a pyramid of square
base ; side of lower base 30 ft. , of upper base 27 ft. , height 2^ ft. ;
find the height of the vertex, and the volume of the heap.
49. If a cubic foot of the heap in Ex. 48 weighs 168 lb., what is the
total weight in tons ?
50. If the volume in Ex. 48 is calculated as a cuboid of the same
heiglit, with the square half-way up as base, what is the error?
51. A wedge can be divided into three pyramids of equal volume.
52. A peat-stack runs from a rectangular base 10 ft. by 30 ft. to a
horizontal line at the top 10 ft. high. If the ends slope at the same
angle as the sides, calculate its volume. (Divide it into a pyramid
and a wedge.)
53. A pipe, average diameter 2 ft. , brings water from a lake 70 milep
distant ; how much water is there in the pipe at a given moment ?
54. A reservoir is ^ mile long, ^ mile broad ; if 10,000,000 gallons are
drawn off in a day, how long will it take to lower the level 15 ft. ?
(Take 1 gal = .16cub. ft.)
55. The locus of a point whose distances from two points 4" apart are
in the ratio 3: 2 is a sphere. (Find locus in a plane and rotate.)
56. Find the volume of the sphere in Ex. 55.
57. A funnel has an upper diameter 6", lower diameter 1", depth of
conical part 4", length of cylindrical part 4". What surface has it ?
58. A bucket has the shape of a frustum of a cone ; upper diameter
1 ft., lower 10", depth 1 ft. Calculate its volume.
59. A stone of irregular shape is put into a cylinder, 10" diameter,
containing water, and causes the water to rise 6". Find the volume of
the stone.
60. The base of a cylindrical can is an ellipse, axes 12" and 6", and
the height 10". How many gallons does it hold? (Area of ellipse
= 7r X product of semi-axes; 1 cub. ft. =6-23 gals.)
61. Calculate the area of the right projection of a circle of radius 5 cm.
upon a plane at an angle of 50° to its plane.
62. An oak pillar 12 ft. high has as base a regular hexagon of 1 ft.
side. Calculate its surface and volume.
63. If the pillar of Ex. 62 was cut out of a trunk of 3 ft. diameter and
the same length, how much was cut away ?
64. A tree stands 100 ft. high on a base of 17 ft. diameter. Taking it
as a cone, how many tons of wood does its trunk yield ? (Take 1 cub. ft.
to weigh 50 lb. )
65. Two caps 3" and 4" deep are cut off a sphere of 12" diameter.
Calculate the volume and surface of the remainder.
CH. IX.] FUNDAMENTAL PROPERTY OP THE STRAIGHT LINE. 275
Kow that some familiarity with the figures of two planes has
been obtained, we can show that it is possible to give formal
proof (i.e. directly from the definitions of plane and straight line)
of Thh. 2, 3, and 6 of Chapter I., which we there derived
experimentally.
Theorem 2, Ch. I. — 'Two straight lines coincide entirely
when two points of each coincide.'
The folds of two planes may be placed so as to have two
points A, B common ;
one side of each plane may be kept fixed, y\ y
and their other sides X, Y moved about their
respective folds AB.
As these folds are the boundaries of the
non-moving part of the planes, neither fold
moves when X, Y move ;
if then Y is moved until some point P of the fold of X is on Y,
and the two parts X, Y are then rigidly fastened so as to move
together,
P, being on the fold of X, does not move, and therefore it is on the
fold of Y, since all points of Y move which are not on the fold —
i.e. every point P of the fold of X is on the fold of Y.
It should be noticed, however, that we here make an assumption
not generally necessary for plane geometry, but necessary for the
geometry of space, and which has already been given on the first
page of this chapter — viz. that if a plane turns completely round
the line joining two points on it, it sweeps out the whole of
space. It was because of the difficulty of this notion that this
formal proof had to be deferred.
It is impossible, of course, to prove this experimentally or
formally ; its justification is that it is consistent with all those
results of geometry which we can test, and that it enables us to
treat consistently and successfully such problems as movements of
stars which are some billions of miles away. It is assumed at
some stage or other in all systems of space geometry.
We can now also establish formally Theorems 3 and 5 of
Ch. I.
276
REVERSIBILITY OF STRAIGHT LINE AND ANGLE. [CH. IX.
Theorems 3, 5 of Ch. I. — 'Every straight line has a mid
point, and every angle a bisector, about which it can be
reversed.'
If BAG is any angle, we may suppose two lines AP, AQ to
turn in the plane from the sides AB, AC
at the same rate — i.e. so that at any
instant CAQ can reverse on to BAP.
These moving lines meet in some line
AD in the angle ;
.*. the angle DAC reverses on to DAB,
similarly DAB on to DAC ; and the
whole angle BAC reverses about the bisec-
tor AD.
Similarly, if points P, Q move from
A, B along the line AB in opposite direc-
tions at the same rate — i.e. so that at any A * ■ ■ b
instant QB can reverse on to AP —
these moving points meet in a point M, the mid point of AB,
about which the line can be reversed.
The above principle of reversibility of an angle and of a straight
line is very important, and has been entirely overlooked in our
text-books. Without it, for example, it is impossible to establish
the congruence of two triangles which have two sides and their
angle of each equal, or two angles and corresponding side equal,
when the triangles have contrary aspects.
The difficulty will be readily understood by substituting two
spherical triangles (formed by arcs of great circles on a sphere) of
contrary aspect, and trying to establish their congruence.
As the fundamental property of an isosceles triangle is either
proved by direct turning over, or by the congruence of two
triangles of contrary aspect — e.g. those formed by bisector of
vertical angle — it is evident that this property cannot be proved
without the assumption of this principle of reversibility, which
therefore ought to be explicitly given.
Note. By turning over DAC on to DAB, and then reversing it on
itself in this position, it is readily seen that CAD can be rotated into
coincidence with DAB.
CII. IX.] NOTE ON DIRECTION. 277
We can now also justify our restriction of the lines to one
plane, in the definition of lines in the same direction.
* Two non-coplanar lines cannot have the same direction ; ' * or,
' Two lines in the same direction are coplanar, and therefore
parallel.'
If AB, CD are non-coplanar lines, MN their common perp. ;
CED a plane through CD pari, to AB,
and cutting plane AMN in NE ;
then DNE = ang. of planes = a say.
Draw any transversal PMQ of AB, CD,
make QL pari, to AB and NE; plane
MNL perp. to QL.
Then if angs. PMB, MQN made by
transversal PQ with AB, CD are ^, 7 ;
QL QL QN
QM~QN*QM
= cos NQL . cos NQM = cos DNE . cos 7
= cos a . cos 7.
.*. unless cos P and cos 7 are each zero — i.e. unless ^ = 7 = 90° —
(in which case PQ is MN and not any line), we can only have
cos P = cos 7 when cos a = 1 ;
i.e. P = y when only a = — i.e. when only the planes AMN, CMN
coincide, and CD, ifVB are coplanar.
Thus two non-coplanar lines make equal angles towards tlie
same parts with their common perpendicular only ; and hence two
lines having the same direction must be coplanar — i.e. parallel.
* That is, cannot make equal angles towards the same parts with any third
line Avhatever which meets them.
cos 13 = COS MQL =
278
ANSWERS TO EXAMPLES.
I.
1. 4" ; 2i cm. + . 2. 60°. 4. 60°. 5. 2^".
6. AC, AD. 7. 45°, 135°, 90°. 9. 40°, 140° 14. 3 cm.
II.
2- 1 SaIdIIls^'ov ir'- . ^- ^^°' ^^^°- 5- ^'°' ^''°-
6. 2". 7. 129°, 51°, 129°.
III.
3. 90°. 4. 60°. 5. Altitude from base, 3-1 cm. 6. QR = l-9", or -9".
IV.
1. Alt. from base, 2J cm. 3. 70°; base, 1". 4. 60°. 5. 41°.
6. 4-8 cm. 7. 3 cm. 8. 14°.
1. 2-8 cm., 5-3 cm. 3. 33° ; long. diag. 2.1". 4. 360° ; long. diag. 5-6 cm.
6. -87", or I". 7. 4 cm. 8. R = -77", or|". 9. R =2-5 cm.
VI.
1. 61°, 30°, 22°, 14^°. 2. BP=3.5cm. 5. AP = 1.31". 6. 2-8 cm.
VII.
2. Fourth side, 1-82 mi. 3. 1/50,000. 4. 7/288.
5. Unit, 1". 6. Unit, -72". 7. 1/190,080 ; Ig".
8. 1/52,800. 9. 22i mi. 11. 3/100,000.
XI.
2. 1-45", or ItV". 3. 2-39", or 21". 4. Alt. = 1-52", or l^'\
5. 1"; 60^ 6. 1-08", or M". 7. 1-72", or If".
XIL
1. n". 2. -755", or |". 3. 1-9", or Ig". 4. 1-77", or If".
ANSWERS. 279
XV.
1. 1"; .77",orr. 2. r-
XVI.
2. 3sq. in. 3. -974 sq. in. 4. l-75sq. in., 3-5sq. in.; ly^or l-56sq. in.
XVII.
1 36.64 SQ cm 2 i lengths : 8.167", 16-96 cm., 34-56 yd.
• ^' * (Areas: 53-09 sq. in., 22-91 sq. era., 95-06 sq. yd.
o j Arcs : 1-885, 2-513, 3-77 cm. - ^o
{ Areas : 2-262, 3-016, 4-524 sq. cm.
XVIII.
1. 80° 24'. 2. 1-46". 3. 1.23". 4. 48 yd.
5. 3 ft. 6 in. 6. 35-75 ft. 10. 5145 yd. 11. 104 ft
12. 18 mi. ; parm. 13. 2 mi. E. and W. 14. 47° 10'. 15. 4-76 cm.
16. Alt. 1-15 cm. 17. Short diag. 1-69". 18. Ang. 70° 43'. 22. 3-53 cm.
23. 1-98". 24. 4-95 cm. 27. 17-32 mi. 31. Perp. 2-44 and 335 cm.
32. CD = 3-6 cm. 33. 3i, 6^ cm. 34. 3-96 cm. 35. 3-35, 4-3 cm.
36. U". 37. 1-758". 39. 60 = 1-6". 40. EF=U".
41. Other diag. 2.34". 42. Long diag. 7-27 cm. 44. 9 : 49.
46. 3-91 sq. in., 10428 sq. yd., 472-5 sq. ft. 47. 2992 sq.yd. ; diag. 83-14 yd.
4& 4-74 cm. ; 4-37"; 5 mi. ; 346 yd.
49. \/3", or 1-732"; -866"; 2-598 cm. ; 4-33 ft. ; 6-062 yd.
50. 400 yd. 51. 116-6 ft. 52. 9. 53. 9:25.
54. 18-85 ft. ; 87-96", or 7 ft. 4"; 11 ft. 55. 280, 720, 480.
56. 2827 ft., 616 sq. in., 9-625 sq. ft. 57. 1257 sq. ft.
58. -262, -393, -524, -628, 1-05 sq. in. 59. 2, LU sq. in. 60. 32 sq. cm.
61. 36 sq. cm. 62. 7-73 sq. cm. 63. 1-732", 1-5", 1-3 sq. in.
XXI.
1. 10, 12, 16 rt. angs. ; 4 rt. angs. 2. 108°, 120°, 128f°. 3. c= -804".
4. c = -88". 5. 60°; c=2-77cm. 6. AD = .72". 7. If".
11. Alt. 3.2 cm. 13. 120°, 30°; &=l-5". 14. 2-12 cm.
15. c= 4-23 cm. 1^'. a=l-92". 18. & = 2.59cm. 19. c = 2-16".
20. Side, 4-41 cm. 21. Diag. 4-67 cm. 26. Sides, 1-84", -79 ".
27. Sides, 3-21 cm., 2-09 cm. 28. Sides, -74", 114", 1-13".
29. Other diag. 2-39". 40. 7-43 mi. 42. Chd. 1-54".
43. Sides, 4-9, 3-37 cm. 44. 8-09, 5-27 mi. 45. 4-73 mi.
47. Side, 2". 48. Side, 3-97 cm. 49. Side, 2-35". 51. Diag. 3-3".
52. 1", 1-73". 54. 6 = 5-55 cm. 56. BC, 3 cm. ; BD, 4-36 cm.
57. 72°, 36°. 58. 5-5 cm. 59. CD, AD, 1"; 108°, 36°. 60. 108°.
280 ANSWERS.
XXIV.
6. 26° 34'. 8.78.2 ft. 9.50 ft. 11. AC, 543 cm. 12. 3.14", 2-66".
14. 5.05". 19. Line divides angle into 35° 24' and 27° 36'.
20. Short diag. 2.28 cm. 21. Side, 2-35 cm. 25. Middle side, -94".
26. Side, 2.4 cm. 27. 2.65. 28. 1". 29. DC = -28".
30. 1-8", -8", 1.2". 32. CD, 9 cm. 33. = 1-5 cm.
34.1:9. 35. Greatest length, 5.6". 36. -618". 37. ^^^\ or 1.618.
38.4:1. 39. A B, 162". 41. Rad. 1.44 cm. 42. Had. 2.16 cm.
43. 6=1-91 cm. 46. 4.32 sq. cm., 3-12 sq. cm., 214.63 sq. cm.
47. 2.08 cm., 1-77 cm., 14.64 cm. 48. 16-24 sq. cm.
50. Side, 3.08 cm. and 1.73 cm. 51. 164.5 sq. mi.
52. -24 sq. mi., or 153.6 .acres. 53. 4.43 sq. cm., 8.36 sq. cm., -5 sq. in.
54. Side, 3-46 cm. 55. Side, 2-2". 56. Side, 3".
57. Parts, 1-661", -339". 58. -865". 59. 1-94".
60. Side, 2.44 cm. 61. Side, 1-32". 63.3.45". 64. Third side, 1-87".
65. Alt. from diag: to rt. ang. -6". 66. Side, 5.4 cm. 67. PD = -87".
68. AC, 2-5 cm.- AD, 10 cm.; AP, 6.25cm. 7L 4". 72. IM cm.
XXVII.
5. 1-62". 7. Central dist. of chd. 1-51". 9. Rad. 1-04", 14. 3.96".
15. 2 cm. 18. Dist. 5-13 cm. 20. §", or -67".
Page 119.
1. 8100 c. in., 6000 c. ft. 2. 343 c. in., 2197 c. ft., 13-824 c. c.
Page 120.
1. 11-5 c. c. 2: 4-524 sq. cm., 27-14 c. C.
Page 121.
1. 12-15 c. in. 2. 47-12 sq. cm., 37-7 c. c.
Page 122.
1. 50-29 sq. cm., 32-17 sq. in., 314-2 sq. ft. ;
33.5 c. c, 17.15 c. in., 524 c. ft.
2. 24,883-2 mi. 3. 24,881-4 mi., 197,060,800 sq. mi.
XXVIII.
1.5184 c. ft. 2. 9504 c. in. 3. 6375 c. in. 4. 84,480 c. yd.
5. 36 c. in. 6. 216 c. in. 7. 1357. 8. 4-9.
9. 330 sq. in. 10. 3,440,853 c. yd. 11. 25-13 c. in.
12. 268-1 c. yd., 207-25 sq. yd. 13. 4-189 c. ft. 14. 2-094 c. ft.
15. 3". 16. 265. 18. 25,943 sq. yd.
ANSWERS.
281
Pages 130, 131.
A , 9-8 sq. cm. ; r, 1-23 ; r„, 2-45 ; n, 3-27 ; r^ 9-8 ; R, 3-57 ; Ol, 2 cm.
Page 132.
Lengths: 18-22, 1-619, 3-645, 11-39, 15-39.
Areas : 26-42, 2-348, 5-284, 16-51, 2231.
XXX.
1. Mean part, 2". 4. -^, 1-3 sq. in., 12 sq. cm. 5. 2-12 sq. in.
6. A, 94-1 sq. ft.; R, 9-18; r, 4-09; ra, 8-56; n, 13-44; r„ 18-82 ft.
9. 2-1 cm. 10. Dist. from cent. 2-1, 2-9 cm. U. 1§", IJ" from ends.
13. 1 cm., 5 cm. from ends. 15. 3-873, 38-73 mi. 16. 1-94".
18. Ne\v side, 1-414". 19. Side, -62". 20. Line cuts CB 1-5" from C.
22. Side, 5-265 cm. 26. Line cuts AC -409" from A.
27. 1-57 cm. from A; a. 28. 2-598, 10-392 sq. cm.
29. 8 :3V3, or 1-539. 30. Chds. 1-27 rad., 1-55 rad.
31. 3-61 sq. cm. 38. Side of sq. 7-65 cm.
XXXIV.
5. Diag. 5-23 cm. 8. 34°, 97°, 34°.
9. Short diag. 2-98 cm. 16. A, 135°; D, 60°; E, 105°; EC, 3".
17. 80°, 39°; b, 5-72". 19. b, 2-61 cm. 22. Alt. from BC, 1-53".
23. «, 802 cm. 28. Side, 1-5". 3L 2-13".
34. Diag. 3-62". 35. Longest side, 2-55 cm. 49. 6, 4-11 cm.
Page 163.
4. 2-789, 1-395. 5. -968.
Page 164.
3. 4-24 cm. 4. 86-6 ft.
Page 166.
2. -578, 1-41; -922, 2-38; -8035, -595.
XXXV.
2. (i.) 3^ ; (ii.) 2V2, or 2-828 ; (iii.) 1.
3. (i-) ^2"^, or 3-098 ; (ii.) 1 + V2, or 2-414 ; (iii.) ^i|^, ov 1-943.
6. (i-)^\or.966;(iL)l;(iii.)^.
282 ANSWERS.
7. (i.) 45°; (ii.) 60°; (iii.) 30°; (iv.) 0, or 60°; (v.) 60°.
8. (i.) -"- , t; 11.) , -; (ill.) — ^ ^ , ^ ^
11 1 _J_ _ /Q __L 1 _A
^^- 2' V2' ^ ' V2' 2' V3'
12. {i.)M + ^^{l-m{l-l^); (ii.) -Vl-/'^', -Vl--^^ -^l-fjl, -sj\-k\
13. 70-02 ft. 14. 7464 ft. 15. 46-5 ft. 16. 3° 49', 176 ft.
17.662 ft. 18. 16.66 yd., or 49-96 yd. 19. 4-276 mi. 20. 8° 28'.
Page 170.
4. (i.) C, 55°; h, 668-2 ; c, 566-7 ; R, 345-9 yd.
(ii.) B, 45°; a, 1-38 ; c, 1-027 ; R, 6405 cm.
Page 171.
1. (i.) A, 41° 48'; B, 78° 34'. (ii.) A, 44° 25'; B, 57** 7.
(iii.) A, 35° 49'; B, 48° 28'.
Page 172.
2. (i.) A, 89-3; r, 3-97; n, 11-9; R, 9-07.
(ii.) A, 228-6 ; r, 6-351 ; n, 20-78 ; R, 14-43.
XXXVI.
1. 81° 51', or 98° 9'. 2. 57° 58', or 122° 2'. 3. 46° 50'.
4. B, 44° 25'; C, 34° 3'. 5. A, 22° 20'; C, 108° 13'. 6. 3-16".
7. 11-14 ft. 8. 185-4 yd. 9. 168-8 yd. 10. 565.
11. 7335. 12. c, 550-8; «, 314-9. 13. 34° 38'. 14. 41° 48'.
15. C, 85° 40'; A, 52° 37'. 16. 76° 44'. 17. 73° 24'. 18. 46° 59' ; 4095.
19. 54° 4', or 125° 56'. 20. 42° 24', or 137° 36'.
21. A, 56° 19'; B, 77° 58'; b, 1126 : or A, 123° 41'; B, 10° 36'; 6, 211-8.
22. 95° 44'. 23. 281-5, 125-9. 24. r, 39-8, 7-215; Va, 84-41, 32-22.
25. 72,100,57,350. 26. x^J2|2. 27. 2 : V7, or -7703. 28. (2-l± V-41)/2.
29. 1-376", -7237 ; 103° 42', 31° 8'. 30. Side, 8-09 cm. ; angs. 72°, 72°, 36°.
XXXVII.
1. 487 or 413 yd. 2. 40-14 ft. 3. 4897 ft. 4. 12,000 ft.
XXXVIII.
V3-1 1-V3 1-V3 1 + V3 ,
1- ■2V2'' '2V2"' "^2'' "2^72"' ^^ ^^~^' " "^^ ^
XXXIX.
1. 2 sin 66° cos 6°; ^(cos 12° - cos 132°). 2. 96°.
ANSWERS. 283
Page 180.
7 24 1
1 — — • — 48. QQ
^' 25' 25' 49'^^^^-
XL.
1. 120°. 2. 30°. 3. 0, 180°, 8° 25'. 4. 66° 8', or 91° 41'.
5. 63° 26'. 6. -90°, or 270°; 61° 56'.
Page 182.
1760 ft. 847,066 mL
XLI.
i. J» fs. 1^. ^- 2. 120°, 45°, 144°, 54°. 3. 34-56 ml
z 12 O 4
XLII.
1. c, b. 3. 6. 8. -8506". 9. 2-598 cm. IQ. a/2 sin -, a/2 tan --
n ' n
11. ^^^^, or .309. 12. 1-464 ft. 13. 2r sin ^, r cos ^.
16. 46° 53', 54° 33', 58° 19'. 19. -9616, -2747. 20. -7454, 1-118.
26. 3°, 48°, 64°. 27. - -9397, - -766, 5671. 28. 25° 55', 48° 1', 16° 15'.
29. 30° ; ±90°, or 60° ; ±60°, or ±30°. 39. 60° 57'. 40. 10-8 ft, 30° 58'.
41. 98-57 ft. 42. 1980 mi., 12440 mi. 43. Side, 141-4 ft. ; diag. 200 ft.
44. 41° 49'. 45. 38° 11'. 50. -1749", -6527", -2404 sq. in.
51. AD, 1-368"; CD, 1-009". 54. B, 75°. 55. a, 207-6; 6, 260-4 yd.
56. 3302 m., 1,756,000 sq. m. 57. 67° 18', or 112° 42'; 236-6 or 72-2 yd.
58. A, 62°; B, 57° 42'; C, 104° 5'; D, 136° 13'.
59. 91° 2'; AD, 6014; BC, 751-4; DC, 332-8 yd.
60. 1093 m. 66. Cot 2^; -tan(a-^). 69. 2 tan 2a, 70. 0.
73. 11,041 ft., 58°. 74. 3-903 or 3-223 mi.
75. (i.) 0°, or 180°; (ii.) 30°, or 150°; (iii.) 90°; (iv.) 135°, 0°, 90°.
76. -j^, or -9931 ; ^+^-, or 1 (nearly). 83. - 63/65. 84. ^^•
87. (i.) 0, 22^°, 20°. (ii.) 30°, 60°. (iii.) 0, 90°, 180°. (iv.) 45°, 135°.
(v.) a, 90° -a. (vi.) d, 0, or tt ; 0, 0, or x; 6, 34° 3'; <t>, 101° 32'.
(vii.) d, 64° 39'; 0, 37° 3'.
92. 5-477 mi., 17-32 mi., 54-77 mi. 93. 69-12 mi. 94. 1-02".
95. 95° 33'. 96. ^, l^^.^l 72°, 70°, 30°, 40°. 99. 57-29 cm.
101. (1.) j2' l2"' l2"-' 12' "12"' l2' ^''-^ '^ 30,^1.180 +/1 30'.
.... . 9r _ IT
(HI.) g, 2w7r + ^-
284 ANSWERS.
XLIII.
24. Length of tangt., 1-3". 25. Side, 1.65". 26. M7", 1-06".
29. -61 o". 31. i".
Page 268.
2. 96-57 c. ft.
Page 269.
6-93 c. c, 667 c. c.
Page 271.
7-33 c. c.
XL VIII.
44. 15-36, 1-732, 70-68, 56-55 sq. in.
45. Curved surf., 50-266, 251-33 sq. cm.; total area, 104-3, 305-4 sq. cm.
46. 103 : 540 (or 1080). 47. 8,172,000 sq. mi. 48. 25 ft, 2032-5 c. ft.
49. 152-4 tons. 50. 1-875 c. ft. (<-l per cent.). 52. 1333 J c. ft.
53. 171,600 c. yd. 54. 32-67 days. 56. 24-43 c. in. 57. 64-44 sq. in.
58. -662 c. ft., or 1144 c. in. 59. 471-2 c. in. 60. 2-04.
61. 50-48 sq. cm. 62. 72 or 77-2 sq. ft. ; 31-18 c. ft. 63. 53-65 c. ft.
64. 169. 65. 528-8 c. in.; curved surf., 188-5 sq. in. ; total, 373-85 sq. in.
THE END.
Edinburgh :
Printed by W. & R. Chambers, Limited.
UNIVERSITY OF CALIFORNIA LIBRARY
BERKELEY
Return to desk from which borrowed.
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