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& A SCHOOL GEOMETRY.
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A SCHOOL GEOMETRY
PARTS I., II. AND III.

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SCHOOL GEOMETRY
PARTS I., II., AND III.
Part I. Limes and Angles. Rectilineal Figwres
Part II. Areas of Rectilineal Figures
Part III. Cºrcles
(Containing the substance of Euclid Book I., Book III. 1-34,
and part of Book IV.)
BY
H. S. HALL, M.A.
AND
F. H. STEVENS, M.A.
SECOND EDITION REVISED
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COPYRIGHT.
First Edition July 1903.
Reprinted October 1903 (twice),
Second Edition, Revised, March 1904.
Reprinted 1905 (twice), 1906, 1907 (twice), 1908, 1909, 1910,
1911, 1912, 1913, 1914, 1915, 1916, 1917, 1918 (twice),
1919 (twice), 1920, 1921.
GLASGow ; PRINTED AT THE UnivERSITY PREss
By RoberT MACLEhose AND co, LTD,”
PREFACE.
THE present work provides a course of Elementary Geometry
based on the recommendations of the Mathematical Association
and on the schedule recently proposed and adopted at Cambridge.
The principles which governed these proposals have been
confirmed by the issue of revised schedules for all the more
important Examinations, and they are now so generally accepted
by teachers that they need no discussion here. It is enough to
note the following points:
(i) We agree that a pupil should gain his first geometrical ideas
from a short preliminary course of a practical and experimental
character. A suitable introduction to the present book would
consist of Easy Exercises in Drawing to illustrate the subject
matter of the Definitions; Measurements of Lines and Angles;
Use of Compasses and Protractor; Problems on Bisection, Per-
pendiculars, and Parallels; Use of Set Squares; The Construction
of Triangles and Quadrilaterals. These problems should be
accompanied by informal explanation, and the results verified
by measurement. Concurrently, there should be a series of ex-
ercises in Drawing and Measurement designed to lead inductively
to the more important Theorems of Part I. [Euc. I. 1-34]* While
strongly advocating some such introductory lessons, we may point
out that our book, as far as it goes, is complete in itself, and from
the first is illustrated by numerical and graphical examples of the
easiest types. Thus, throughout the whole work, a graphical
and experimental course is provided side by side with the usual
deductive exercises. . .
(ii) Theorems and Problems are arranged in separate but parallel.
courses, intended to be studied pari passw. This arrangement is
made possible by the use, now generally sanctioned, of Hypothetical
Constructions. These, before being employed in the text, are care-
fully specified, and referred to the Axioms on which they depend.
* Such an introductory course is now furnished by our Lessons in Experi.
mental and Practical Geometry.
Vi - PREFACE.
(iii) The subject is placed on the basis of Commensurable Mag.
nitudes. By this means, certain difficulties which are wholly
beyond the grasp of a young learner are postponed, and a wide
field of graphical and numerical illustration is opened. Moreover
the fundamental Theorems on Areas (hardly less than those on
Proportion) may thus be reduced in number, greatly simplified,
and brought into line with practical applications.
(iv) An attempt has been made to curtail the excessive body of
text which the demands of Examinations have hitherto forced as
“bookwork” on a beginner's memory. Even of the Theorems
here given a certain number (which we have distinguished with
an asterisk) might be omitted or postponed at the discretion of the
teacher. And the formal propositions for which—as such—teacher
and pupil are held responsible, might perhaps be still further
limited to those which make the landmarks of Elementary Geo-
metry. Time so gained should be used in getting the pupil to
apply his knowledge; and the working of examples should be
made as important a part of a lesson in Geometry as it is so
considered in Arithmetic and Algebra.
Though we have not always followed Euclid's order of Proposi-
tions, we think it desirable for the present, in regard to the
subject-matter of Euclid Book I. to preserve the essentials of his
logical sequence. Our departure from Euclid's treatment of Areas
has already been mentioned; the only other important divergence
in this section of the work is the position of I. 26 (Theorem 17),
which we place after I. 32 (Theorem 16), thus getting rid of the
tedious and uninstructive Second Case. In subsequent Parts a freer
treatment in respect of logical order has been followed.
As regards the presentment of the propositions, we have con-
stantly kept in mind the needs of that large class of students, who,
without special aptitude for mathematical study, and under no
necessity for acquiring technical knowledge, may and do derive
real intellectual advantage from lessons in pure deductive reasoning.
Nothing has as yet been devised as effective for this purpose as the
Euclidean form of proof; and in our opinion no excuse is needed
for treating the earlier proofs with that fulness which we have
always found necessary in our experience as teachers.
• *
PREFACE. * VT1
The examples are numerous and for the most párt easy. They
have been very carefully arranged, and are distributed throughout
the text in immediate connection with the propositions on which
they depend. A special feature is the large number of examples
involving graphical or numerical work. The answers to these
have been printed on perforated pages, so that they may easily be
removed if it is found that access to numerical results is a source
of temptation in examples involving measurement.
We are indebted to several friends for advice and suggestions.
In particular we wish to express our thanks to Mr. H. C. Playne
and Mr. H. C. Beaven of Clifton College for the valuable assist-
ance they have rendered in reading the proof sheets and checking
the answers to some of the numerical exercises.
H. S. HALL.
F. H. STEVENS.
November, 1903.
PREFATORY NOTE TO THE SECOND EDITION.
IN the present edition some further steps have been taken towards
the curtailment of bookwork by reducing certain less important
propositions (e.g. Euclid I. 22, 43, 44) to the rank of exercises.
Room has thus been found for more numerical and graphical
exercises, and experimental work such as that leading to the
Theorem of Pythagoras.
Theorem 22 (page 62), in the shape recommended in the Cam-
bridge Schedule, replaces the equivalent proposition given as
Additional Theorem A (page 60) in previous editions.
In the case of a few problems (e.g. Problems 23, 28, 29) it has
been thought more instructive to justify the construction by a pre-
liminary analysis than by the usual formal proof.
|H. S. HALL.
F. H. STEVENS.
March, 1904.
CONTENTS,
PART I.
Axioms. Definitions. Postulates. - as tº º tº
HYPOTHETICAL CONSTRUCTIONS tºº *º º º sº sº
INTRODUCTORY - sº- º ſº sº º wº º & *
SYMBOLS AND ABBREVIATIONS - º - * * & * *
Lines and Angles.
THEOREM 1. [Euc. I. 13.] The adjacent angles which one
straight line makes with another straight line on one side
of it are together equal to two right angles.
CoR. l. If two straight lines cut one another, the four
angles so formed are together equal to four right angles.
CoR. 2. When any number of straight lines meet at a
point, the sum of the consecutive angles so formed is equal
to four right angles.
CoR. 3. (i) sº of the same angle are equal.
(ii) Complements of the same angle are equal.
THEOREM 2. [Euc. I. 14.] If, at a point in a straight line,
two other straight lines, on opposite sides of it, make the
adjacent angles together equal to two right angles, then
these two straight lines are in one and the same straight line.
THEOREM 3. [Euc. I. 15.] If two straight lines cut one
another, the vertically opposite angles are equal.
Triangles.
DEFINITIONS gº - - *- º -> - sº * 4-
THE COMPARISON OF Two TRIANGLES sº wº tº- - tº-
THEOREM 4. [Euc. I, 4.] If two triangles have two sides
of the one equal to two sides of the other, each to each, and
the angles included by those sides equal, then the triangles
are equal in all respects.
PAGB
10
11
11
ll
14
16
17
18
K CONTENTS.
THEOREM 5. [Euc. I. 5.] The angles at the base of an isosceles
triangle are equal. *
CoR. 1. If the equal sides of an isosceles triangle are pro-
duced, the exterior angles at the base are equal.
CoR. 2. If a triangle is equilateral, it is also equiangular.
THEOREM 6. [Euc. I. 6..] If two angles of a triangle are equal
to one another, then the sides which are opposite to the equal
angles are equal to one another.
THEOREM 7. [Euc. I. 8.] If two triangles have the three sides
of the one equal to the three sides of the other, each to each,
they are equal in all respects.
THEOREM 8. [Fuc. I. 16.1 If one side of a triangle is pro-
duced, then the exterior angle is greater than either of the
interior opposite angles.
CoR. l. Any two angles of a triangle are together less
than two right angles. .
CoR. 2. Every triangle must have at least two acute
angles.
CoR. 3. Only one perpendicular can be drawn to a
straight line from a given point outside it.
THEOREM 9. [Euc. I. 18.] If one side of a triangle is greater
than another, then the angle opposite to the greater side is
greater than the angle opposite to the less.
THEOREM 10. [Euc. I. 19.] If one angle of a triangle is
greater than another, then the side opposite to the greater
angle is greater than the side opposite to the less.
THEOREM ll. [Euc. I. 20.] Any two sides of a triangle are
together greater than the third side.
THEOREM 12. Of all straight lines from a given point to a
given straight line the perpendicular is the least.
CoR. l. If OC is the shortest straight line from O to the
straight line AB, then OC is perpendicular to AB.
CoR. 2. Two obliques OP, OQ, which cut AB at equal
distances from C the foot of the perpendicular, are equal.
CoR. 3. Of two obliques OQ, OR, if OR cuts AB at the
reater distance from C the foot of the perpendicular, then
R is greater than OQ. * -
Parallels.
PLAYFAIR's Axiom - tºº - * wº t- gº rº- gº
THEOREM 13. . [Euc. I. 27 and 28.] If a straight line cuts two
other straight lines so as to make (i) the alternate angles
PAGE
20
21
21
28
29
29
30
31
32
33
33
33
33
35
CONTENTS.
equal, or (ii) an exterior angle equal to the interior opposite
angle on the same side of the cutting line, or (iii) the interior
angles on the same side equal to two right angles; then in
each case the two straight lines are parallel.
THEOREM 14. [Euc. I. 29.] If a straight line cuts two parallel
lines, it makes (i) the alternate angles equal to one another;
(ii) the exterior angle equal to the interior opposite angle on
the same side of the cutting line; (iii) the two interior angles
on the same side together equal to two right angles.
PARALLELS ILLUSTRATED BY ROTATION. HYPOTHETICAL CON-
STRUCTION tºº * ºs tº ſº sº * sº º e
THEOREM 15. [Euc. I. 30.] Straight lines which are parallel
to the same straight line are parallel to one another.
Triangles continued.
THEOREM 16. [Euc. I. 32.] . The three angles of a triangle
are together equal to two right angles.
CoR. 1. All the interior angles of any rectilineal figure,
together with four right angles, are equal to twice as many
right angles as the figure has sides.
CoR. 2. If the sides of a rectilineal figure, which has no
re-entrant angle, are produced in order, then all the exterior
angles so formed are together equal to four right angles.
THEOREM 17. [Euc. I. 26.] If two triangles have two angles
of one equal to two angles of the other, each to each, and any
side of the first equal to the corresponding side of the other,
the triangles are equal in all respects.
ON THE IDENTICAL EQUALITY OF TRIANGLES * º 4- *
THEOREM 18. Two right-angled triangles which have their
hypotenuses equal, and one side of one equal to one side of
the other, are equal in all respects.
THEOREM 19. [Euc. I. 24.] If two triangles have two sides of
the one equal to two sides of the other, each to each, but the
angle included by the two sides of one greater than the angle
included by the corresponding sides of the other; then the
base of that which has the greater angle is greater than the
base of the other.
CoNVERSE OF THEOREM 19 - * * sº * * *
Parallelograms.
DEFINITIONS º * * tº sº * - tº sº * *
THEOREM 20. [Euc. I. 33.] The straight lines which join the
extremities of two equal and parallel straight lines towards
the same parts are themselves equal and parallel.
PAGº
36
38
39
46
50
5]
52
53
57
xii CONTENTS.
PAGA
THEOREM 21. [Euc, I. 34.] The opposite sides and angles of a
tºº are equal to one another, and each diagonal
isects the parallelogram. i 58
CoR. l. If one angle of a parallelogram is a right angle, all
its angles are right angles. 59
CoR. 2. All the sides of a square are equal; and all its
angles are right angles. 59
CoR. 3. The diagonals of a parallelogram bisect one
another. 59
THEOREM 22. *If there are three or more parallel straight lines,
and the intercepts made by them on any transversal are equal,
then the corresponding intercepts on any other transversal
are also equal. 62.
CoR. In a triangle ABC, if a set of lines Po, Qq, Rr, ...,
drawn parallel to the base, divide one side AB into equal parts,
they also divide the other side AC into equal parts. - 63
DIAGONAL SCALES - * & e * tº ſº tº tº - 66
Practical Geometry. Problems.
INTRODUCTION. NECESSARY INSTRUMENTS sº * • - 69
PROBLEMs on LINES AND ANGLES.
PROBLEM 1. To bisect a given angle. 70
PROBLEM 2. To bisect a given straight line. 7]
PROBLEM 3. To draw a straight line perpendicular to a given
straight line at a given point in it. 72
PROBLEM 4. To draw a straight line perpendicular to a given
straight line from a given external point. 74
PROBLEM 5. At a given point in a given straight line to make
an angle equal to a given angle. 76
PROBLEM 6. Through a given point to draw a straight line
parallel to a given straight line. 77
PROBLEM 7. To divide a given straight line into any number
of equal parts. 78
THE CONSTRUCTION OF TRIANGLES.
PROBLEM 8. To draw a triangle, having given the lengths of
the three sides. 80
PROBLEM 9. To construct a triangle having given two sides
and an angle opposite to one of them. 82
PROBLEM 10. To construct a right-angled triangle having given
the hypotenuse and one side. 83
THE Construction of QUADRILATERALS.
PROBLEM 11. To construct a jºinisterºl, given the lengths
of the four sides, and one angle 86
CONTENTS.
xiii
PROBLEM 12. To construct a parallelogram having given two
adjacent sides and the included angle.
PROBLEM 13. To construct a square on a given side.
Loci.
PROBLEM 14. To find the locus of a point P which moves so
that its distances from two fixed points A and B are always
equal to one another.
PROBLEM 15. To find the locus of a point P which moves so
that its perpendicular distances from two given straight lines
AB, CD are equal to one another.
INTERSECTION OF LOCI - º - - - - - *
THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.
I. The perpendiculars drawn to the sides of a triangle from
their middle points are concurrent.
II. The bisectors of the angles of a triangle are concurrent.
III. The medians of a triangle are concurrent.
CoR. The three medians of a triangle cut one another at a
point of trisection, the greater segment in each being towards
the angular point.
PART II.
Areas.
DEFINITIONS te - - - * - - º - -
THEOREM 23. AREA OF A RECTANGLE.
THEOREM 24. [Euc. I. 35.] Parallelograms on the same base
and between the same parallels are equal in area.
AREA OF A PARALLELOGRAM - * - º - - --
THEOREM 25. AREA OF A TRIANGLE.
THEOREM 26. [Euc. I. 37.] Triangles on the same base and
between the same parallels (hence, of the same altitude) are
equal in area.
THEOREM 27. [Euc. I. 39.] If two triangles are equal in area,
and stand on the same base and on the same side of it, they
are between the same parallels.
THEOREM 28. AREA OF (i) A TRAPEZIUM.
(ii) ANY QUADRILATERAL.
AREA of ANY RECTILINEAL FIGURE - - - - - tº-
THEOREM 29. [Euc., I, 47. PYTHAGORAS's THEOREM.] In a
right-angled triangle the square described on the hypotenuse
is equal to the sum of the squares described on the other two
sides.
PAGE
87
88
91
92
93
96
96
97
97
99
100
104
105
106
108
108
ll 2
112
114
118
xiv. CONTENTS.
ExPERIMENTAL PROOFS OF PYTHAGORAS's THEOREM * . sº
THEOREM 30. [Euc. I. 48.] If the square described on one side
of a triangle is equal to the sum of the squares described on
the other two sides, then the angle contained by these two
sides is a right angle.
PROBLEM 16. To draw squares whose areas shall be respectively
twice, three-times, four-times, ... , that of a given square.
Problems on Areas.
PROBLEM 17. To describe a parallelogram equal to a given
triangle, and having one of its angles equal to a given angle.
PROBLEM 18. To draw a triangle equal in area to a given
quadrilateral.
PROBLEM 19. To draw a parallelogram equal in area to a given
rectilineal figure, and having an angle equal to a given angle.
Axes of Reference. Coordinates.
ExERCISES FOR SQUARED PAPER gº º, gº wº sº º
PART III.
The Circle. Definitions and First Principles. - - -
SYMMETRY. SYMMETRICAL PROPERTIES OF CIRCLES º º
Chords.
THEOREM 31. [Euo. III. 3..] If a straight line drawn from
the centre of a circle bisects a chord which does not pass
through the centre, it cuts the chord at right angles.
Conversely, if it cuts the chord at right angles, it bisects it.
CoR. 1. The straight line which bisects a chord at right
angles passes through the centre. h
CoR. 2. A straight line cannot meet a circle at more than
two points.
CoR. 3. A chord of a circle lies wholly within it,
THEOREM 32. One circle, and only one, can pass through any
three points not in the same straight line. *
CoR. 1. The size and position of a circle are fully deter-
mined if it is known to pass through three given points.
CoR. 2. Two circles cannot cut one another in more than
two points without coinciding entirely.
HYPOTHETICAL CONSTRUCTION - * se º gº fº tº
PAGE
120
122
124
126
128
129
132
139
141
144
145
145
145
146
147
147
147
CONTENTS.
XV
THEOREM 33. [Euc. III. 9..] If from a point within a circle
more than two equal straight lines can be drawn to the
circumference, that point is the centre of the circle.
THEOREM 34. [Euc. III. 14.] Equal chords of a circle are equi-
distant from the centre.
Conversely, chords which are equidistant from the centre
are equal.
THEOREM 35. [Euc. III. 15.] Of any two chords of a circle,
that which is nearer to the centre is greater than one more
remote. -
Conversely, the greater of two chords is nearer to the
centre than the less.
CoR. The greatest chord in a circle is a diameter.
THEOREM 36. [Euc. III. 7..] If from any internal point, not
, the centre, straight lines are drawn to the circumference of a
circle, then the greatest is that which passes through the .
centre, and the least is the remaining part of that diameter.
And of any other two such lines the greater is that which
subtends the greater angle at the centre.
THEOREM 37. [Euo. III. 8.] If from any external point
straight lines are drawn to the circumference of a circle, the
greatest is that which passes through the centre, and the
least is that which when produced passes through the centre.
And of any other two such lines, the greater is that which
subtends the greater angle at the centre.
Angles in a Circle.
THEOREM 38. [Euc. III. 20.] The angle at the centre of a
circle is double of an angle at the circumference standing on
the same arc.
THEOREM 39. [Euc. III. 21.] Angles in the same segment of a
circle are equal. -
CoNVERSE or THEOREM 39. Equal angles standing on the
same base, and on the same side of it, have their vertices on
an arc of a circle, of which the given base is the chord.
THEOREM 40. [Euo. III. 22.] The opposite angles of any
quadrilateral inscribed in a circle are together equal to two
right angles.
CoNVERSE OF THEOREM 40. If a pair of opposite angles
of a quadrilateral are supplementary, its vertices are con-
cyclic. -
THEOREM 41. [Euc. III. 31.] The angle in a semi-circle is a
right angle.
CoR. The angle in a segment greater than a semi-circle is
acute; and the angle in a segment less than a semi-circle is
obtuse.
PAGE:
148
150
152
153
154
156
158
160
16]
163
164
165
xvi CONTENTS.
THEOREM 42. [Euc. III. 26.] In equal circles, arcs which sub-
tend Équal angles, either at the centres or at the circum-
ferences, are equal.
CoR. In equal circles sectors which have equal angles are
equal. º -
THEOREM 43. [Euc. III. 27.] In equal circles angles, either at
the centres or at the circumferences, which stand on equal
arcs are equal. -
THEOREM 44. [Euc. III. 28.] In equal circles, arcs which are
cut off by equal chords are equal, the major arc equal to the
major are, and the minor to the minor.
THEOREM 45. [Euc. III. 29.] In equal circles chords which
cut off equal arcs are equal. -
Tangency.
DEFINITIONS AND FIRST PRINCIPLES - sº es sº &= sº
THEOREM 46. The tangent at any point of a circle is perpendi-
cular to the radius drawn to the point of contact.
CoR. 1. One and only one tangent can be drawn to a
circle at a given point on the circumference.
CoR. 2. The perpendicular to a tangent at its point of
contact passes through the centre.
CoR. 3. The fadius drawn perpendicular to the tangent
passes through the point of contact. º
THEOREM 47. Two tangents can be drawn to a circle from an
external point. as:
CoR. The two tangents to a circle from an external point
are equal, and subtend equal angles at the centre.
THEOREM 48. If two circles touch one another, the centres and
the point of contact are in one straight line,
CoR. l. If two circles touch externally the distance be-
tween their centres is equal to the sum of their radii.
CoR. 2. If two circles touch internally, the distance be-
tween their centres is equal to the difference of their radii.
THEOREM 49. [Euc. III. 32.] The angles made by a tangent
to a circle with a chord drawn from the point of contact are
ºly equal to the angles in the alternate segments of
the circle.
g
Problems. -
GEOMETRICAL ANALYSIS - * wº- sº e- tº-3 e *
PROBLEM 20. Given a circle, or an aré of a circle, to find its
centre. .
PBOBLEM 21. To bisect a given arc.
PAGF;
166
166
167
168
169
172
174
174
174
174
176
176
178
178
178
180
182
183
183
CONTENTS. xvii
PAGE
PROBLEM 22. To draw a tangent to a circle from a given ex
ternal point. - , 184
PROBLEM 23. To draw a common tangent to two circles. 185
THE CONSTRUCTION OF CIRCLES - - - - - - - 188
PROBLEM 24. On a given straight line to describe a segment of
a circle which shall contain an angle equal to a given angle. 190
CoR. To cut off from a given circle a segment containing
a given angle, it is enough to draw a tangent to the circle,
and from the point of contact to draw a chord making with
the tangent an angle equal to the given angle. 191
Circles in Relation to Rectilineal Figures.
DEFINITIONS - - * - - - - * - - 192
PROBLEM 25. To circumscribe a circle about a given triangle. 193
PROBLEM 26. To inscribe a circle in a given triangle. 194
PROBLEM 27. To draw an escribed circle of a given triangle. 195
PROBLEM 28. In a given circle to inscribe a triangle equi-
angular to a given triangle. 196
BROBLEM 29. About a given circle to circumscribe a triangle
equiangular to a given triangle. 197
PROBLEM 30. To draw a regular polygon (i) in (ii) about a
given circle. - 200
PROBLEM 31. To draw a circle (i) in (ii) about a regular polygon. 201
Circumference and Area of a Circle - - *- - - 202
Theorems and Examples on Circles and Triangles.
THE ORTHoCENTRE OF A TRIANGLE - tº- - º tº- - 207
LOCI - * - as - • - tº- º - • - - 210
SIMSON's LINE - - & wº tº- º t <- - - 212
THE TRIANGLE AND ITs CIRCLES - * - tº º - 213
THE NINE-PoſNTs CIRCLE - - *- -> - * cº - 216
Appendix. On the Form of Some Solid Figures.
Answers to Numerical Exercises,
GEOMETRY.
PART I.
AXIOMS,
ALL mathematical reasoning is founded on certain simple
principles, the truth of which is so evident that they are
accepted without proof. These self-evident truths are called
Axioms. --~~ ~ ...º
e * * **** ~£ wº-
For instance: Å, rºº # = ~ . . . . . .
, , ) *: & à
Things which are equal to the same thing are equal to one
another. R →, sº sº, e º
The following axioms, corresponding to the first four Rules
of Arithmetic, are among those most commonly used in
geometrical reasoning.
Addition. If equals are added to equals, the sums are equal.
&) ºf . , ,
Subtraction. If equals are taken from equals, the remainders
are equal. ~ 5 ºn --
Multiplication. Things which are the same multiples of equals
are equal to one another.
For instance: Doubles of equals are equal to one another.
Division. Things which are the same parts of equals are equal
to one another.
For instance: Halves of equals are equal to one another.
The above Axioms are given as instances, and not as a
complete list, of those which will be used. They are said to
be general, because they apply equally to magnitudes of all
kinds. Certain special axioms relating to geometrical magni-
tudes only will be stated from time to time as they are
required.
H. S. G. A Qº-
2 GEQMETRy.
º
*~-
an o-oo ov, , , º 0.5°
DEFINITIONS AND FIRST PRINCIPLES.
Every beginner knows in a general way what is meant by a
point, a line, and a surface. But in geometry these terms are
used in a strict sense which needs some explanation.
^\GY, -
1. A point has position, but is said to have no magnitude.
This means that we are to attach to a point no idea of size either as
to length or breadth, but to think only where it is situated. A dot
made with a sharp pencil may be taken as roughly representing a
i. but small as such a dot may be, it still has some length and
readth, and is therefore not actually a geometrical point. The smaller
the dot however, the more nearly it represents a point.
^^
( sº
2. A line has length, but is said to have no breadth.
A line is traced out by a moving point. If the point of a pencil is
moved over a sheet of paper, the trace left represents a line. But such
a trace, however finely drawn, has some degree of breadth, and is
therefore not itself a true geometrical line. The finer the trace left by
the moving pencil-point, the more nearly will it represent a line.
3. Proceeding in a similar manner from the idea of a line
to the idea of g surface, we say that
A surface has length and breadth, but no thickness.
And finally,
§ {}^^-).
A solid has length, breadth, and thickness.
Solids, surfaces, lines and points are thus related to one another:
r fö 3- Y & -
(i) A solid is bounded by surfaces. -
(ii) A surface is bounded by lines; and surfaces meet in lines.
(iii) A line is bounded (or terminated) by points; and lines meet in
points.
V
- ** 3 \ \\
4. A line may be straight or curved.
A straight line has the same direction from point to point
throughout its whole length. $
A curved line changes its direction continually from point
to point. - - . . . . . . . ;
DEFINITIONS. 3
AxIOM. There can be only one straight line joining two given -
points: that is, -
Two straight lines cannot enclose a Space.
Sºn v $
ſº %
5. A plane is a flat surface, the test of flatness being that
if any two points are taken in the surface, the straight line
between them lies wholly in that surface.
6. When two straight lines meet at a
point, they are said to form an angle. B
2 ºv-,
The straight lines are called thé arms of the
angle; the point at which they meet is its vertex.
f\ \ 6.0
The magnitude of the angle may be thus Ó A
explained:
Suppose that the arm OA is fixed, and that OB turns about
the point O (as shewn by the arrow). Suppose also that OB
began its turning from the position OA. Then the size of the
angle AOB is measured by the amount of turning required to
bring the revolving arm from its first position OA into its
subsequent position OB.
Observe that the size of an angle does not in any way depend on the
length of its arms.
Angles which lie on either side of C - B
a common arm are said to be ad-
jacent. ** J: Sea. -
For example, the angles AOB, BOC,
which have the common arm OB, are O A
adjacent.
When two straight lines such as AB, CD
cross one another at O, the angles COA, BOD
are said to be vertically opposite. The 5 O
angles AOD, COB are also vertically opposite
to one another.
C
A

4 GEOMETRY.
7. When one straight line stands on an-
other so as to make the adjacent angles equal
to one another, each of the angles' is called a
right angle; and each line is said to be per-
pendicular to the other.
B O A
AXIOMS. (i) If O is a point in a straight line AB, then a line
OC, which turns about O from the position OA to the position OB,
must pass through one position, and only one, in which it is
perpendicular to AB.
(ii) All right angles are equal.
A right angle is divided into 90 equal parts called degrees (°); each
degree into 60 equal parts called minutes ('); each minute into 60 equal
parts called seconds (").
In the above figure, if OC revolves about oftom the
position OA into the position OB, it turns through two right
angles, or 180°. * *
If OC makes a complete revolution about O, starting from OA
and returning to its original position, it turns through four
right angles, or 360°.
8. An angle which is less than one right
angle is said to be acute. •
That is, an acute angle is less than 90°.
O
A
9. An angle which is greater B
than one right angle, but less than
two right angles, is said to be obtuse.
That is, an obtuse angle lies between
90° and 180°. O A
10. . If one arm OB of an angle turns -
until it makes a straight line with the ---,
other arm QA, the angle so formed is 5-4-5- A
called a straight angle. -
A straight angle=2 right angles= 180°.

DEFINITIONS. 5
11. An angle which is greater
than two right angles, but less than () , ,

four right angles, is said to be :-- A
reflex. T (ii) º
That is, a reflex angle lies between B
180° and 360°.
NOTE. When two straight lines meet, two angles are formed, one
greater, and one less than two right angles. The first arises by
supposing OB to have revolved from the position OA the longer way
round, marked (i); the other by supposing OB to have revolved the
shorter way round, marked (ii). Unless the contrary is stated, the
angle between two straight lines will be considered to be that which is
less than two right angles.
12. Any portion of a plane surface bounded by one or
more lines is called a plane figure. ..

13. A circle is a plane figure contained P
by a line traced out by a point which moves
so that its distance from a certain fixed /
point is always the same. * O
• Here the point P moves so that its distance P
from the fixed point O is always the same.
The fixed point is called the centre, and the bounding line
is called the circumference. . . . . . . . . . .
14. A radius of a circle is a straight line drawn from the
centre to the circumference. It follows that all radii of a
circle are equal.
15. A diameter of a circle is a straight line drawn through
the centre, and terminated both ways by the circumference.
16. An arc of a circle is any part of the circumference.
6 GEOMETRY.
17. A semi-circle is the figure bounded
by a diameter of a circle and the part of the
circumference cut off by the diameter. |
18. To bisect means to divide into two equal parts.
AxIOM's. (i) If a point O moves
from A to B along the straight line A Ö B
AB, it must pass through one posi- *
tion in which it divides AB into two equal parts.
That is to say:
(ii) If a line OP, revolving about O, turns
from OA to OB, it must pass through one
position in which it divides the angle AOB
into two equal parts.
That is to say:
HYPOTHETICAL CONSTRUCTIONS.
From the Axioms attached to Definitions 7 and 18, it
follows that we may suppose •
(i) A straight line to be drawn perpendicular to a given
straight line from any point in it. .
(ii) A finite straight line to be bisected at a point.
(iii) An angle to be bisected by a line.
SUPERPOSITION AND EQUALITY.
Axiom. Magnitudes which can be made to coincide with one
another are equal. -
*
This axiom implies that any line, angle, or figure, may be taken up
from its position, and without change in size or form, laid down upon a
second line, angle, or figure, for the purpose of comparison, and it
states that two such magnitudes are equal when one can be exactly
placed over the other without overlapping.
This process is called superposition, and the first magnitude is said to
be applied to the other.




POSTULATES. 7
POSTULATES.
In order to draw geometrical figures certain instruments
are required. These are, for the purposes of this book, (i) a
straight ruler, (ii) a pair of compasses. The following Postulates
(or requests) claim the use of these instruments, and assume
that with their help the processes mentioned below may be
duly performed.
Let it be granted:
1. That a straight line may be drawn from any one point to
any other point.
2. That a FINITE (or terminated) straight line may be
PRODUCED (that is, prolonged) to any length in that straight
line.
3. That a circle may be drawn with any point as centre and
with a radius of any length.

NoTES. (i) Postulate 3, as stated above, im-
plies that we may adjust the compasses to the
length of any straight line PQ, and with a radius
of this length draw a circle with any point O as
centre. That is to say, the compasses may be
used to transfer distances from one part of a
diagram to another.
(ii) Hence from AB, the greater of two -
8traight lines, we may cut off a part equal l
to PQ the less. A |X B
For if with centre A, and radius equal g
to PQ, we draw an arc of a circle cutting
§§ X, it is obvious that AX is equal P Q
O i*\cºs

8 GEOMETRY.
INTRODUCTORY.
$
1. Plane geometry deals with the properties of such lines
and figures as may be drawn on a plane surface.
2. The subject is divided into a number of separate dis-
cussions, called propositions.
Propositions are of two kinds, Theorems and Problems.
A Theorem proposes to prove the truth of some geometrical
statement. - -
A Problem proposes to perform some geometrical construc-
tion, such as to draw some particular line, or to construct
some required figure.
3. A Proposition consists of the following parts:
The General Enunciation, the Particular Enunciation, the
Construction, and the Proof.
(i) The General Enunciation is a preliminary statement,
describing in general terms the purpose of the proposition.
(ii) The Particular Enunciation repeats in special terms
the statement already made, and refers it to a diagram, which
enables the reader to follow the reasoning more easily.
(iii) The Construction then directs the drawing of such
straight lines and circles as may be required to effect the
purpose of a problem, or to prove the truth of a theorem.
(iv) The Proof shews that the object proposed in a problem
has been accomplished, or that the property stated in a theorem
is true.
4. The letters Q.E.D. are appended to a theorem, and stand
for Quod erat Demonstrandum, which was to be proved.
INTRODUCTORY. 9
5. A Corollary is a statement the truth of which follows
readily from an established proposition; it is therefore
appended to the proposition as an inference or deduction,
which usually requires no further proof. •
6. The following symbols and abbreviations are used in
the text of this book. -- -
In Part I.
... for therefore, a for angle,
= , is, or are, equal to, A , triangle.
After Part I.
pt. for point, perp. for perpendicular,
st. line , straight line, par" , parallelogram,
rt. A. , right angle, rectil. , rectilineal, - . . .
par' (or ||) , parallel, , , G) , circle, - . . . .
Sq. ,, Square, gº O* ,, circumference;
and all obvious contractions of commonly occurring words,
such as opp., adj., diag, etc., for opposite, adjacent, diagonal,
etc.
[For convenience of oral work, and to prevent the rather common
abuse of contractions by beginners, the above code of signs has been
introduçed gradually, and at first somewhat sparingly.]
In numerical examples the following abbreviations will
be used. º
m. for metre, cm, for centimetre,
mm. , millimetre. km. , kilometre.
Also inches are denoted by the symbol ().
. . . . Thus 5” means 5 inches.
10 GEOMETRY.
ON LINES AND ANGLEs.
THEOREM 1. [Euclid I. 13.]
The adjacent angles which one/straight line makes with another
straight line on one side of it, are together equal to two right angles, sº, so
D;

s
O
B A
Let the straight line CO make with the straight line AB the
adjacent 4 "AOC, COB. - -
It is required to prove that the 4 'AOC, COB are together equal to
two right angles. . -
Suppose OD is at right angles to BA.
Proof. Then the 4 "AOC, COB together
= the three Z "AOC, COD, DOB.
Also the 4 "AOD, DOB together .*
- . = the three z*AOC, COD, DOB.
. ... the 4 "AOC, COB together = the 4 "AOD, DOB
= two right angles.
Q.E.D.
PROOF BY ROTATION.
Suppose a straight line revolving about O turns from the position OA
into the position OC, and thence into the position OB; that is, let the
revolving line turn in succession through the 4." AOC, COB.
Now in passing from its first position OA to its final position OB,
the revolving line turns through two right angles, for AOB is a straight
line, -
Hence the 4." AOC, COB together= two right angles,
LINES AND ANGLES. 11
surn of * * *
CoROLLARY 1. . If two straight lines 2^ D
cut one another, the four angles so formed ' ". . . * , º,
are together equgl to four right angles. A O B
2, 3} {} >. X = > r2 -73, 3 º' -, *
For example, C
4. BOD + 4. DOA+ 4 AOC + 4-COB=4 right angles.
* &
* - - ~}^2,224
CoRoºſ/ARY 2. When any number of
straight lines meet at a point, the sum of
, the consecittà dhgles so formed is equal
to four right angles.
º, J - ºr, .
For a straight line §§§ about O, and turning in succession
through the Z." AOB, BOC, COD, DOE, EOA, will have made one
complete revolution, and therefore turned through four right angles.
* * e- ºn tº
Or, ºr/ACN §N, } º v
definitions. - T * .”.
g >\* J. v v. 5 of ºn 3 > * : J-, *, * > . . . . . .
(i) Two angles whose sum is two right angles, are said to
be supplementary; and each is called the supplement of the
other. ** > . . Y. & Q wº d's * * *** **, * * * *
Thus in the Fig. of Theor. 1 the angles AOC, COB are supplementary.
. Again the angle 123° is the supplement of the angle 57°. *.
• * ** tº o * } s oº ev tº , Tºa tº v tº •
(ii) Two angles whose sum is one right angle are said to
be complementary; and each is called the complement of the
other. S". . . . . . . . . sº º * : * ~ *** * * > . . . . . . . . , , , , ,
Thus in the Fig. of Theor. 1, the angle DOC is the complement of 9 ****
the angle AOC. Again angles of 34* and 56° are complementary, a a ºn
. . . . no o 2 St A, x^^ y o 5 ° tº * : * , ~'s
CoRoLLARY 3. (i) Supplements of the same angle are equal.
(ii) Complements of the same angle are equal.

12 - GEOMETRY.
THEOREM 2. [Euclid I. 14.] *
If, at a point in a straight line, two other straight lines, on
opposite sides of it, make the adjacent angles together equal to two
ºright angles, then these two straight lines are in one and the same
straight lime.
...Rºzz woº “ºver "w C
. ~q → ~ n g
Q
X
B O A

At O in the straight line CO let the two straight lines OA,
OB, on opposite sides of CO, make the adjacent 4 " AOC, COB
together equal to two right angles: (that is, let the adjacent
** AOC, COB be supplementary).
It is required to prove that OB and OA are in the same straight
lime. - •.
Produce AO beyond O to any point X; it will be shewn that
OX and OB are the same line. ’, -
Proof. Since by construction AOX is a straight line,
... the 4: COX is the supplement of the 4 COA. Theor. 1.
But, by hypothesis, -
s the A. COB is the supplement of the A. COA,
... the 4 COX = the A. COB;
.*. OX and OB are the same line.
But, by construction, Ox is in the same straight line
with OA; -
hence OB is also in the same straight line with OA.
- Q.E.D.
LINES AND ANGLES. I3
EXERCISES.
1. Write down the supplements of one-half of a right angle, four-
thirds of a fight angle; also of 46°, 149°, 83°, 101° 15'.
2. Write down the complement of two-fifths of a right angle ;
also of 27°, 38° 16', and 41°29' 30".
3. If two straight lines intersect forming four angles of which one
is known to be a right angle, prove that the other three are also right
angles. * , of
Qº’ roº, isſon, uzov j A.
4,311n the triangle ABC the angles ABC, ACB are given equal. If
the side BC is prod; both ways, shew that the exºriº; angles so
23% are equal. 2. º t vºteo
5. In the triangle ABC the angles ABC, ACB are given equal. If
AB and AC are produced beyond the base, shew that the exterior angles
so formed are equal.
Dºrismos. The lines which bisect an angle and the
adjacent angle made by producing one of its arms are called
the internal and external bisectors of the given angle.
Thus in the diagram, OX and OY are the
ºnal and external bisectors of the angle
6. Prove that the bisectors of the adjacent angles which one
straight line makes with another contain a right angle. That is to
say, the internal and extermal bisectors of an angle are at right angles
to one another.
7, Shew that the angles AOX and COY in the above diagram are
complementary.
8. Shew that the angles BOX and COX are supplementary; and
also that the angles AOY and BOY are supplementary.
9. If the angle AOB is 35°, find the angle COY.

14 GEOMETRY.
THEOREM 3. [Euclid I. 15.]
If two straight lines cut one another, the vertically opposite angles
1 are equal, 2 * 6, Ó 2. * >J
f
* -}
A D
let the straight lines AB, CD cut one another at the point O.
It is required to prove that
(i) the 4 AOC = the 4, DOB;
(ii) the 4 COB = the Z.AOD.
Proof. Because AO meets the straight line CD,
‘.. the adjacent 4 " AOC, AOD together = two right angles;
that is, the 4- AOC is the supplement of the 4 AOD.
Again, because DO meets the straight line AB,
... the adjacent 4 "DOB, AOD together = two right angles;
that is, the 4, DOB is the supplement of the 4-AOD.
Thus each of the 4 " AOC, DOB is the supplement of the 4-AOD,
... the 4 AOC = the 4 DOB.
Similarly, the 4-COB = the Z.AOD.
* Q.E.D.
PROOF BY ROTATION.
Suppose the line COD to revolve about O until OC turns into the
osition OA. Then at the same moment OD must reach the position
B (for AOB and COD are straight).
Thus the same amount of turning is required to close the Z.AOC as to
close the 4-DOB.
... the Z.AOC= the A. DOB.

LINES AND ANGLES. 15
EXERCISES ON ANGLES.
. (Numerical.)
1. Through what angles does the minute-hand of a clock turn in
(i) 5 minutes, (ii) 2.1 minutes, (iii) 43% minutes, (iv) 14 min. 10 sec. 2
And how long will it take to turn through (v) 66°, (vi) 222°?
2. A clock is started at noon : through what angles will the hour-
hand have turned by (i) 3.45, (ii) 10 minutes past 52 And what will
be the time when it has turned through 1723° 2
3. The earth makes a complete revolution about its axis in 24 hours.
Through what angle will it turn in 3 hrs. 20 min., and how long will it
take to turn through 130°2
4. In the diagram of Theorem 3
(i) If the 4-AOC=35°, write down (without measurement) the value
of each of the A_* COB, BOD, DOA. -
(ii) If the Z *COB, AOD together make up 250°, find each of the
A s COA, BOD.
(iii) If the 4–5 AOC, COB, BOD together make up 274°, find each of
the four angles at O. -
(Theoretical.)
5. If from O a point in AB two straight lines OC, OD are drawn on
opposite sides of AB so as to make the angle COB equal to the angle
AOD; shew that OC and OD are in the same straight line.
6. Two straight lines AB, CD cross at O. If OX is the bisector of
the angle BOD, prove that XO produced bisects the angle AOC. ~.
7. Two straight lines AB, CD cross at O. If the angle BOD is
bisected by OX, and AOC by OY, prove that OX, OY are in the same
straight line. - * : *
8. If OX bisects an angle AOB, shew that, by folding the diagram
about the bisector, OA may be made to coincide with OB.
How would OA fall with regard to OB, if -
(i) the 4- AOX were greater than the 4 XOB;
(ii) the 4- AOX were less than the A-XOB?
9. AB and CD are straight lines intersecting at right angles at O;
shew by folding the figure about AB, that OC may be made to fall
along OD. & - :
10. A straight line AOB is drawn on paper, which is then folded
about O, so as to make OA fall along OB; shew that the crease left in
the paper is perpendicular to AB. -
16 GEOMETRY,
ON TRIANGLES.
1. Any portion of a plane surface bounded by one or more
lines is called a plane figure.
The sum of the bounding lines is called the perimeter of the figure.
The amount of surface enclosed by the perimeter is called the area.
2. Rectilineal figures are those which are bounded by
straight lines.
3. A triangle is s plane figure bounded by three straight
lines.
4. A quadrilateral is a plane figure bounded by four straight
lines.

5. A polygon is a plane figure bounded by
more than four straight lines.
6. A rectilineal figure is said to be
equilateral, when all its sides are equal;
equiangular, when all its angles are equal ;
regular, when it is both equilateral and equiangular.
7. Triangles are thus classified with regard to their sides:
A triangle is said to be
• 2 equilateral, when all its sides are equal;
,,,,, º isosceles, when two of its sides are equal;
scalene, when its sides are all unequal.



Equilateral Triangle. Isosceles Triangle, Scalene Triangle.

In a triangle ABC, the letters A, B, C often denote
the magnitude of the several angles (as measured in 6. &
degrees); and the letters a, b, c the lengths of the
opposite sides (as measured in inches, centimetres, or
some other unit of length). B & C
TRIANGLES. 17
Any one of the angular points of a triangle may be regarded as its
vertex; and the opposite side is then called the base.
In an isosceles triangle the term vertea, is usually applied to the point
at which the equal sides intersect; and the vertical angle is the angle
included by them.
8. Triangles are thus classified with regard to their angles:
A triangle is said to be
***!?ight-angled, when one of its angles is a right angle;
42.2×obtuse-angled, when one of its angles is obtuse;
3.30%acute-angled, when all three of its angles are acute.
3.
[It will be seen hereafter (Theorem 8. Cor. 1) that every triangle mw8t
have at least two acute angles.]

~1 ~/ 2\
Right-angled Triangle. Obtuse-angled Triangle. Acute-angled Triangle.
In a right-angled triangle the side opposite to the right angle is
called the hypotenuse.
9. In any triangle the straight line joining a vertex to the
middle point of the opposite side is called a median.
THE COMPARISON OF TWO TRIANGLES.
(i) The three sides and three angles of a triangle are called
its six parts. A triangle may also be considered with regard
to its area.
(ii) Two triangles are said to be equal in all respects,
when one may be so placed upon the other as to exactly
coincide with it; in which case each part of the first triangle
is equal to the corresponding part (namely that with which
it coincides) of the other; and the triangles are equal in area.
In two such triangles corresponding sides are opposite to equal
angles, and corresponding angles are opposite to equal sides.
Triangles which may thus be made to coincide by super-
position are said to be identically equal or congruent.
H. S. G. E
18 GEOMETRY.
THEOREM 4... [Euclid I. 4. 2’
** ‘.3 × 13%, sº .# J a 2-4
If two triggles have two sides of the one equal to two sides of the
other, éâch'fö’each, and the angles included by those sides equal,
then the triangles are equal in all respects.
i gº equ 9. tºgº.
3, 2, , , . . . . .3, ºr y r*. º' Tºº. . .
> A D
! . C ! s F
E
Let ABC, DEF be two triangles in which
AB = DE,
AC = DF,
and the included angle BAC = the included angle EDF.
It is required to prove that the AABC = the A DEF in all
Tespects.
6) G),
º'º ano


Proof. Apply the AABC to the A DEF,
so that the point. A falls on the point D,
and the side AB along the side DE.
Then because AB = DE,
... the point B must coincide with the point E.
And because AB falls along DE,
and the A. BAC = the 4. EDF,
... AC must fall along DF.
And because AC = DF,
... the point C must coincide with the point F.
Then since B coincides with E, and C with F,
... the side BC must coincide with the side EF.
Hence the A ABC coincides with the A DEF,
and is therefore equal to it in all respects.
Q.E.D.
CONGRUENT TRIANGLES. 19
Obs. In this Theorem we must carefully observe what is
given and what is proved.
AB = DE,
Given that AC = DF,
and the Z. BAC = the 4. EDF.
From these data we prove that the triangles coincide on
superposition.
BC = EF,
Hence we conclude that the 4 ABC = the 2. DEF,
and the 4. ACB = the 4 DFE ;
also that the triangles are equal in area.
Notice that the angles which are proved equal in the two triangles
are opposite to sides which were given equal.
A D
NoTE. The adjoining diagram shews
that in order to make two congruent
triangles coincide, it may be necessary
to reverse, that is, turn over one of them C F
before superposition. º
B E


*... l. XXERCISES.
Jºus;*
(i) } Shew that the bisector %the º; angle of an isosceles triangle
i) bisects th9, base, (ii) is perpendicular to the e
#!". *ś %
2. Let O be the hiddle pbint of a straight line AB, and let OC be
perpendicular to it. Then if P is any point in OC, prove that PA=PB.
f Y ~ *
3. Assuming that the four sides of a square are equal, and that its
angles are all right angles, prove that in the square' ABCD, the
diagonals AC, BD are equal.
te
4. ABCD is a square, and L, M, and N are the middle points of AB,
BC, and CD : prove that
(i) LM = MN. (ii) AM = DM.
(iii) AN = AM. (iv) BN =DM.
[Draw a separate figure in each case.]
5. ABC is an isosceles triangle: from the equal sides AB, AC two
§" gº AX, AY are cut off, and BY and CX are joined. Prove that
20 GEOMETRY.
THEOREM 5. [Euclid I. 5.]
f ºr
The angles at the base of an isosceles triangle are gual. # rºo o-f
*4 wº %24 ºjºs, 24 ºf 54 at , J &
2 * Lº
A
5–5–3

Let ABC be an isosceles triangle, in which the side AB = the
side AC.
It is required to prove that the 4. ABC = the 4 ACB.
Suppose that AD is the line which bisects the 4. BAC, and
let it meet BC in D.
1st Proof. Then in the A" BAD, CAD,
BA = CA,
because AD is common to both triangles,
3. and the included Z. BAD = the included Z. CAD ;
... the triangles are equal in all respects; Théor. 4.
so that the Z. ABD = the 4- ACD. -
Q.E.D.
2nd Proof. Suppose the AABC to be folded about AD.
Then since the A-BAD = the 4. CAD,
... AB must fall along AC.
And since AB = AC,
‘. B must fall on C, and consequently DB on DC.
... the 4. ABD will coincide with the 4. ACD, and is therefore
equal to it.
Q.E.D.
ISOSCELES TRIANGLES. 21

COROLLARY 1. If the equal sides AB, AC of
an isosceles triangle are produced, the exterior
angles EBC, FCB are equal ; for they are the
supplements of the equal angles at the base. B C
E F
CoROLLARY 2. If a triangle is equilateral, it is also equi-
angular.
DEFINITION. A figure is said to be symmetrical about a
line when, on being folded about that line, the parts of the
figure on each side of it can be brought into coincidence.
The straight line is called an axis of symmetry.
That this may be possible, it is clear that the two parts of the figure
must have the same size and shape, and must be similarly placed with
regard to the axis.
Theorem 5 proves that an isosceles triangle is symmetrical about
the bisector of its VERTICAL angle.
An equilateral triangle is symmetrical about the bisector of ANY
ONE of its angles.
EXERCISES.
1. ABCD is a four-sided figure whose sides are all equal, and the
diagonal BD is drawn; shew that
(i) the angle ABD = the angle ADB ;
(ii) the angle CBD = the angle CDB ;
(iii) the angle ABC = the angle ADC.
2. ABC, DBC are two isosceles triangles drawn on the same base
BC, but on opposite sides of it: prove (by means of Theorem 5) that
the angle ABD = the angle ACD.
3. ABC, DBC are two isosceles triangles drawn on the same base
BC and on the same side of it; employ Theorem 5 to prove that
the angle ABD = the angle ACD.
4. AB, AC are the equal sides of an isosceles triangle ABC ; and
L., M, N are the middle points of AB, BC, and CA respectively: prove
that
(i) LM = NM. (ii) BN = CL.
(iii) the angle ALM = the angle ANM.
22 GEOMETRY.
THEOREM 6. [Euclid I. 6.]
If two angles of a triangle are equal to one another, then the
sides which are opposite to the equal angles are equal to one another.
f • ,
... sſe 9 º' a Y 9v 3 yºurº 22/ * *
{ **** *- p §, * *
y fºr 3~ * * * **** > syave a A
# es.” * * * -->
25 , , ,-\, g º $3 * > D
B C |

Let ABC be a triangle in which
the 4 ABC = the Z. ACB.
It is required to prove that the side AC=the side AB.
If AC and AB are not equal, suppose that AB is the greater.
From BA cut off BD equal to AC.
Join DC.
Proof, Then in the A" DBC, ACB,
ſ DB = AC,
because BC is common to both,
land the included / DBC = the included 4. ACB;
y
... the A DBC = the AACB in area, Theor. 4.
the part equal to the whole; which is absurd.
..'. AB is not unequal to AC ;
that is, AB = AC.
Q.E.D.
COROLLARY. Hence if a triangle is equiangular it is also
equilateral.
A THEOREM AND ITS CONVERSE. 23
NOTE ON THEOREMS 5 AND 6.

A
Theorems 5 and 6 may be verified ex- * \
perimentally by cutting out the given * ,
AABC, and, after turning it over, fitting , ,
it thus reversed into the vacant space left / \
in the paper. f *
C’ C B
Suppose A'B'C' to be the original position of the AABC, and let
ACB represent the triangle when reversed.
In Theorem 5, it will be found on applying A to A' that C may be
made to fall on B', and B on C".
In Theorem 6, on applying C to B' and B to C' we find that A will
fall on A'.
In either case the given triangle reversed will coincide with its own
“trace,” so that the side and angle on the left are respectively equal to
the side and angle on the right.
NOTE ON A THEOREM AND ITS converse.
The enunciation of a theorem consists of two clauses. The first
clause tells us what we are to assume, and is called the hypothesis; the
second tells us what it is required to prove, and is called the conclusion.
For example, the enunciation of Theorem 5 assumes that in a certain
triangle ABC the side AB = the side AC : this is the hypothesis. From
this it is required to prove that the angle ABC = the angle ACB : this is
the conclusion.
If we interchange the hypothesis and conclusion of a theorem, we
enunciate a new theorem which is called the converse of the first.
For example, in Theorem 5
it is assumed that “ . S : AB = AC ;
it is required to prove that the angle ABC= the angle Ace.}
Now in Theorem 6
it is assumed that the angle ABC = the angle ACB;
it is required to prove that AB = AC.
S Thus we see that Theorem 6 is the converse of Theorem 5; for the
hypothesis of each is the conclusion of the other.
In Theorem 6 we employ an indirect method of proof frequently
used in geometry. It consists in shewing that the theorem cannot be
intrue; since, if it were, we should be led to some impossible conclusion.
This form of proof is known as Reductio ad Absurdum, and is most
commonly used in demonstrating the converse of some foregoing theorem.
It must not however be supposed that if a theorem is true, its con-
verse is necessarily true. [See p. 25.]
24 GEOMETRY.
2 THEOREM 7. [Euclid I. 8.]
re-ſki
z; 2 ºf £72.
. If two triangles have the three sides of the one equal to the three
sides of the other, each to each, they are equal in all respects.


Let ABC, DEF be two triangles in which
AB = DE,
AC = DF,
BC = EF.
It is required to prove that the triangles are equal in all respects.
Proof. , Apply the AABC to the A DEF,
so that B falls on E, and BC along EF, and
so that A is on the side of EF opposite to D.
Then because BC = EF, C must fall on F.
Let GEF be the new position of the AABC.
Join DG.
Because ED = EG,
... the Z. EDG = the A. EGD. Theor. 5.
Again, because FD = FG,
... the /. FDG = the 4 FGD.
Honce the whole / EDF = the whole / EGF,
that is, the A. EDF = the A. BAC.
Then in the A" BAC, EDF;
BA = ED,
because AC = DF,
|- the included A. BAC = the included 4. EDF;
... the triangles are equal in all respects. Theor. 4.
Q.E.D.
CONGRUENT TRIANGLES. 25
Obs. In this Theorem
it is given that AB = DE, BC = EF, CA= FD;
and we prove that 4 C = 4-F, 4-A = 4. D, A-B = A. E. i i
Also the triangles are equal in area.
Notice that the angles which are proved equal in the two triangles
are opposite to sides which were given equal.
NotE 1. We have taken the case in which DG falls within the
As EDF, EGF.
Two other cases might arise:
(i) DG might fall outside the 4.5 EDF, EGF [as in Fig. 1].
(ii) DG might coincide with DF, FG [as in Fig. 2).

A D A D
2 21 F
B C E B C ~!
Fig.1. G Fig.2. G


These cases will arise only when the given triangles are obtuse-
angled or right-angled; and (as will be seen hereafter) not even then,
if we begin by choosing for superposition the greatest side of the AABC,
as in the diagram of page 24.
NOTE 2. Two triangles are said to be equiangular to one another
when the angles of one are respectively equal to the angles of the other.
Hence if two triangles have the three sides of one severally equal to the
three sides of the other, the triangles are equiangular to one another.
. The student should state the converse theorem, and shew by a
diagram that the converse is not necessarily true.
*** At this stage Problems 1-5 and 8 [see page 70] may
conveniently be taken, the proofs affording good illustrations of the
Identical Equality of Two Triangles.

26 GEOMETRY.
EXERCISES.
ON THE IDENTICAL EQUALITY OF Two TRIANGLES.
THEOREMs 4 AND 7.
(Theoretical.)
1. Shew that the straight line which joins the vertex of an isosceles
triangle to the middle point of the base,
(i) bisè& the vertical angle :
(ii) is ...; to the base.
\?

Zºnſ, a 2 `. 3. Aſ ºf 9 tº
2. If ABCD is a rhombus, º is, an eqüilateral foursided figure;
shew, by drawing the diagonal A that
(i) the jść "...º.
(ii) AC bisèëts'éâch of the angles BAD, BCD.
3. If in a quadrilateral ABCD the opposite sides are equal, namely
AB = CD and AD = CB; prove that the angle ADC = the angle ABC.
4. If ABC and DBC are two isosceles triangles drawn on the same
base BC, prove (by means of Theorem 7) that the angle ABD = the
angle ACD, taking (i) the case where the triangles are on the same side
of BC, (ii) the case where they are on opposite sides of BC.
5. If ABC, DBC are two isosceles triangles drawn on opposite
sides of the same base BG, and if AD be joined, prove that each of the
angles BAC, BDC will be § into two equal parts.
Il Y J () () r
6. Shew that the straight lines which join the extremities of the
base of an isosceles triangle to the middle points of the opposite sides,
are equal to one another.
7. Two given points in the base of an isosceles triangle are equi-
distant from the extremities of the base : shew that they are also
equidistant from the vertex.
8. Shew that the triangle formed by joining the middle points
of the sides of an equilateral triangle is also equilateral.
9. ABC is an isosceles triangle having AB equal to AC; and the
angles at B and C are bisected by BO and CO: shew that
(i) BO = CO ;
(ii) AO bisects the angle BAC, º
10. Shew that the diagonals of a rhombus [see Ex. 2) bisect one
another at right angles.
11. The equal sides BA, CA of an isosceles triangle BAC are pro-
duced beyond the vertex A to the points E and F, so that AE is equal
to AF; and FB, EC are joined: shew that FB is equal to EC.
TRIANGLES. 27
EXERCISES ON TRIANGLES.
(Numerical and Graphical.)
1. Draw a triangle ABC, having given a =2.0", b=2'1", c=l'3".
Measure the angles, and find their sum.
2. In the triangle ABC, a = 7-5 cm., b=7-0 cm., and c=6.5 cm.
Draw and measure the perpendicular from B on CA.
3. Draw a triangle ABC, in which a =7 cm., b=6 cm., C–65°.
How would you prove theoretically that any two triangles having
these parts are alike in size and shape? Invent some experimental
illustration.
4. Draw a triangle from the following data: b-2", c=2'5", A=57°;
and measure a, B, and C.
Draw a second triangle, using as data the values just found for a,
B, and C ; and measure b, c, and A. What conclusion do you draw Ż
5. A ladder, whose foot is placed 12 feet from the base of a house,
reaches to a window 35 feet above the ground. Draw a plan in which
l" represents 10 ft.; and find by measurement the length of the ladder.
6. I go due North 99 metres, then due East 20 metres. Plot my
course (scale l cm. to 10 metres), and find by measurement as nearly as
you can how far I am from my starting point.
7. When the sun is 42° above the horizon, a vertical pole casts a
shadow 30 ft. long. Represent this on a diagram (scale 1" to 10 ft.);
and find by measurement the approximate height of the pole.
8. From a point A a surveyor goes 150 yards due East to B ; then
300 yards due North to C ; finally 450 yards due West to D. Plot his
course (scale 1" to 100 yards); and find roughly how far D is from A.
Measure the angle DAB, and say in what direction D bears from A.
9. B and C are two points, known to be 260 yards apart, on a
straight shore. A is a vessel at anchor. The angles CBA, BCA are
observed to be 33° and 81* respectively. Find graphically the approxi.
mate distance of the vessel from the points B and C, and from the
nearest point on shore.
10. In ºgº º: it is required to find the distance between
two points A and B; but as a lake intervenes, a direct measurement
cannot be made. The surveyor therefore takes a third point C, from
Which both A and B are accessible, and he finds CA=245 yards,
CB=320 yards, and the angle ACB =42°. Ascertain from a plan the
approximate distance between A and B.

28 GEOMETRY.
THEOREM 8. [Euclid I. 16.]

If one side of a triangle is produced, then the exterior angle is
jºr than either of the interior opposite angles.
f / \Z^
& B a. *
Let ABC be a triangle, and let BC be produced to D.
It is required to prove that the exterior 4. ACD is greater than
either of the interior opposite 4 "ABC, BAC.
Suppose E to be the middle point of AC.
Join BE; and produce it to F, making EF equal to BE.
Join FC.
Proof. Then in the A"AEB, CEF,
N AE = CE,
because EB = EF,
and the A. AEB = the vertically opposite / CEF;
... the triangles are equal in all respects; Theor, 4.
so that the Z. BAE = the 4. ECF.
But the 4, ECD is greater than the 4. ECF;
... the 4. ECD is greater than the 4. BAE;
that is, the 4. ACD is greater than the A. BAC.
In the same way, if AC is produced to G, by supposing A to
be joined to the middle point of BC, it may be proved that
the 4 BCG is greater than the 4 ABC.
But the 4 BCG = the vertically opposite / ACD.
N. 4. ACD is greater than the 4. ABC.
Q.E.D.

*
\
EXTERIOR AND INTERIOR TRIANGLES. 29
COROLLARY 1. Any two angles of a triangle are together less
than two right angles.

* A
For the A. ABC is less than the A-ACD : Proved.
to each add the 4-ACB.
Then the 4-5 ABC, ACB are less than the 4–5 ACD, ACB,
therefore, less than two right angles.
B C D
COROLLARY 2. Every triangle must have at least two acute
angles.
For if one angle is obtuse or a right angle, then by Cor. 1.each of the
other angles must be less than a right angle.
COROLLARY 3. Only one perpendicular can be drawn to a
straight line from a given point outside it. g

P
If two perpendiculars could be drawn to AB from
P, we should have a triangle PQR in which each of
the Z. PQR, PRQ would be a right angle, which is
impossible.
A Q R B
EXERCISES.
1. Prove Corollary 1 by joining the vertex A to any point in the
base BC.
2. ABC is a triangle and D any point within it. If BD and CD are
joined, the angle BDC is greater than the angle BAC. Prove this
(i) by producing BD to meet AC. .
(ii) by joining AD, and producing it towards the base.
3. If any side of a triangle is produced both ways, the exterior
angles so formed are together greater than two right angles.
4... To a given straight line there cannot be drawn from a point out-
side it more than two straight lines of the same given length.
5, . If the equal sides of an isosceles triangle are produced, the
exterior angles must be obtuse.
30 g GEOMETRY.
THEOREM 9. [Euclid I. 18.]
If one side of a triangle is greater than another, then the angle
opposite to the greater side is greater than the angle opposite to the
less.
A

B C
Let ABC be a triangle, in which the side AC is greater than
the side AB.
It is required to prove that the 4. ABC is greater than the LACB.
From AC cut off AD equal to AB.
Join BD.
Proof. Because AB = AD,
... the 4. ABD = the 4. ADB. Theor. 5.
But the exterior 4. ADB of the A BDC is greater than the
interior opposite / DCB, that is, greater than the 4- ACB.
... the 4. ABD is greater than the 4. ACB.
Still more then is the Z.ABC greater than the 4. ACB.
$ & Q.E.D.
Obs. The mode of demonstration used in the following Theorem
is known as the Proof by Exhaustion. It is applicable to cases in which
one of certain suppositions must necessarily be true; and it consists in
shewing that each of these suppositions is false with one exception:
hence the truth of the remaining supposition is inferred.
INEQUALITIES. zºº 31
THEOREM 10. [Euclid I. 19.]
If one angle of a triangle is greater than another, then the side
opposite to the greater angle is greater than the side opposite to the
less.

A
B’ C
Let ABC be a triangle, in which the 4. ABC is greater than
the 4. ACB.
It is required to prove that the side AC is greater than the
side AB.
Proof. If AC is not greater than AB,
it must be either equal to, or less than AB.
Now if AC were equal to AB,
then the 4. ABC would be equal to the 4 ACB; Theor. 5.
but, by hypothesis, it is not.
Again, if AC were less than AB,
then the Z.ABC would be less than the 4 ACB ; Theor. 9.
but, by hypothesis, it is not.
That is, AC is neither equal to, nor less than AB.
... AC is greater than AB. Q.E.D.
[For Exercises on Theorems 9 and 10 see page 34.]
32 GEOME1RY.
THEOREM 11. [Euclid I. 20.]
Any two sides of a triangle are together greater than the third
side.

B C
Let ABC be a triangle.
It is required to prove that any two of its sides are together
greater than the third side.
It is enough to shew that if BC is the greatest side, then
BA, AC are together greater than BC.
Produce BA to D, making AD equal to AC.
Join DC.
Proof. Because AD = AC,
... the 4- ACD = the 4. ADC. Theor. 5.
But the A. BCD is greater than the 4. ACD;
... the A. BCD is greater than the A-ADC,
that is, than the 4. BDC.
Hence from the ABDC, *
BD is greater than BC. Theor. 10.
But BD = BA and AC together;
..". BA and AC are together greater than BC.
Q.E.D.
NoTE. This proof may serve as an exercise, but the truth of the
Theorem is really self-evident. For to go from B to C along the
straight line BC is clearly shorter than to go from B to A and then
from A to C. In other words
The shortest distance between two points is the straight line which joine
them.
INEQUALITIES. 33
THEOREM 12.
Of all straight lines drawn from a given point to a given straight
line the perpendicular is the least.

A R Q C P B s
Let OC be the perpendicular, and OP any oblique, drawn from
the given point O to the given straight line AB.
It is required to prove that OC is less than OP.
Proof. In the AOCP, since the 4. OCP is a right angle,
... the A. OPC is less than a right angle ; Theor. 8. Cor.
that is, the A. OPC is less than the z OCP.
... OC is less than OP. Theor. 10.
Q.E.D.
COROLLARY 1. Hence conversely, since there can be only
one perpendicular and one shortest line from O to AB,
If OC is the shortest straight lime from O to AB, then OC is
perpendicular to AB.
COROLLARY 2. Two obliques OP, OQ, which cut AB at equal
distances from C the foot of the perpendicular, are equal.
The A• OCP, OCQ may be shewn to be congruent by Theorem 4
- hence OP=OQ.
COROLLARY 3. Of two obliques OQ, OR, if OR cuts AB at the
greater distance from C the foot of the perpendicular, then OR is
greater than OQ. *
The Z.OQC is acute, ... the Z.OQR is obtuse;
‘.. the 4-OQR is greater than the Z.ORQ ;
" ... OR is greater than OQ.
H. S. G. C
ſe
y
34 GEOMETRY.
EXERCISES ON INEQUALITIES IN A TRIANGLE.
1. The hypotenuse is the greatest side of a right-angled triangle.
2. The greatest side of any triangle makes acute angles with each of
the other sides.
3. If from the ends of a side of a triangle, two straight lines are
drawn to a point within the triangle, then these straight lines are together
less than the other two sides of the triangle.
4. BC, the base of an isosceles triangle ABC, is produced to any
point D ; shew that AD is greater than either of the equal sides.
in sºiâû the • -
5. If in a àuadrīſātehā the greatest and least sides are opposite
to one another, then each of the angles adjac to the least side is
greater than its opposite angle. - #" sºo
6. In a triangle ABC, if AC is not greater than AB, shew that
any straight line drawn through the vertex A and terminated by the
base BC, is less than AB. 214 ſtrº) Gl fi/03]?()))
7. ABC is a triangle, in which OB, OC bisect the angles ABC,
ACB respectively: shew that, if AB is greater than AC, then OB is
greater than OC.
8. The difference of any two sides of a triangle is loss than the
third side. -
*
9. The sum of the distances of any point from the three angular
points of a triangle is greater than half its perimeter.
10. The perimeter of a quadrilateral is greater than the sum of
its diagonals.
3/3/0/06)
11. ABC is a triangle, and the vettical angle BAC is bisected by
a line which meets BC in X; shew that BA is greater than BX, and
CA greater than CX. Hence obtain a proof of Theorem 11.
; ºn of Yay.'íſ
... 12. The sum of the distances of any point within a triangle from
its angular points is less than the perimeter of the triangle.
13. . The sum of the diagonals of a quadrilateral is less than the
sum of the four straight lines drawn from the angular points to any
given point. Prove this, and point out the exceptional case.
14. In a triangle any two sides are together greater than twice the
*
median which bisects the remaining side.
[Produce the median, and complete the construction after the
manner of Theorem 8.] -
15. In any triangle the 8wm of the medians is less than the perimeter.
PARALLELS. 35
PARALLELS.
• 9 º' {{{}A &YYYi \
flºº 3. - twu. . .
EFINITION. Parallel straight lines are such as, being in \
the same plane, do not meet however far they are produced Wyn Gio
.###" § its n aph Y
w gº

vºy")
OTE. §araj § must be in the 8ame plane. For instance, two
straight lines, one of which is drawn on a table and the other on the
floor, would never meet if produced; but they are not for that reason
necessarily parallel. e., - -
Óºb s
A®”Tºo inº, straight lines cannot both be parallel
to a third straight line. -
In other words: -
Through a given point there can be only one straight line parallel
to a given Straight line.
This aś ion is known as Playfair's Aaciom.
DEFINITION. When two straight lines AB, CD are met by
a third straight line EF, eight angles are formed, to which for
the sake of distinction particular names are given.

Thus in the adjoining figure }{ E
1, 2, 7, 8 are called ; Øſ. A 1/2
3, 4, 5, 6 are called interior angles, - 4/3 B
4 and 6 are said to be alternate angles;
so also the angles 3 and 5 are alternate
* 5/6
to one another. * C 8/7 D
Of the angles 2 and 6, 2 is referred
to as the exterior angle, and 6 as the F
interior opposite angle on the same side x.
of EF. Such angles are also known as corresponding angles.
Similarly 7 and 3, 8 and 4, 1 and 5 are pairs of corresponding
angles. g
36 GEOMETRY.
THEOREM 13. [Euclid I. 27 and 28.] ; :
If a straight line cuts two other straight lines so as to make
(i) the alternate angles equal,
or (ii) an eaterior angle equal to the interior opposite angle on the
Same side of the cutting line,
or (iii) the interior angles on the same side equal to two right
angles; r
them in each case the two straight lines are parallel.

(i) Let the straight line EGHF cut the two straight lines
AB, CD at G and H so as to make the alternate 4 "AGH, GHD
equal to one another. :
It is required to prove that AB and CD are parallel.
Proof. If AB and CD are not parallel, they will meet, if
produced, either towards B and D, or towards A and C. -
If possible, let AB and CD, when produced, meet towards B
and D, at the point K. - - - .
Then KGH is a triangle, of which one side KG is produced to A;
... the exterior 4. AGH is greater than the interior opposite
4-GHK; but, by hypothesis, it is not greater. -
...; AB and CD cannot meet when produced towards B and D. .
Similarly it may be shewn that they cannot meet towards
A and C:
..". AB and CD are parallel.
PARALLELS. 37

(ii) Let the exterior / EGB = the interior opposite 4 GHD.
It is required to prove that AB and CD are parallel.
Proof. Because the A. EGB = the 4 GHD,
and the Z.EGB = the vertically opposite LAGH;
... the 4- AGH = the 4 GHD :
and these are alternate angles;
... AB and CD are parallel.
(iii) Let the two interior Z" BGH, GHD be together equai to
two right angles.
It is required to prove that AB and CD are parallel.
Proof. Because the Zºº BGH, GHD together = two right
angles;
and because the adjacent / "BGH, AGH together = two right
angles; -
... the 4 "BGH, AGH together = the 4 "BGH, GHD.
From these equals take the Z. BGH ;
then the remaining / AGH = the remaining 4 GHD :
and these are alternate angles;
..". AB and CD are parallel.
Q.E.D.
* & , '
DEFINITION. A straight line drawn across a set of given
lines is called a transversal.
For instance, in the above diagram the line EGHF, which crosses
the given lines AB, CD is a transversal.
38 GEOMETRY.
THEOREM 14. [Euclid I. 29.]
If a straight line cuts two parallel lines, it makes
(i) the alternate angles equal to one another; -
(ii) the exterior angle equal to the interior opposite angle on the
same side of the cutting line ;
(iii) the two interior angles on the same side together equal to two
Tight angles.

Let the straight lines AB, CD be parallel, and let the
straight line EGHF cut them.
It is required to prove that
(i) the 4 AGH = the alternate 4 GHD;
(ii) the eaterior A. EGB = the interior opposite 4 GHD;
(iii) the two interior A * BGH, GHD together = two right angles.
Proof (i) If the 4. AGH is not equal to the 4 GHD,
suppose the LPGH equal to the 4 GHD, and alternate to it;
then PG and CD are parallel. Theor. 13.
But, by hypothesis, AB and CD are parallel;
... the two intersecting straight lines AG, PG are both parallel
to CD: which is impossible. Playfair's Aa.iom.
... the 4-AGH is not unequal to the 4 GHD;
that is, the alternate 4 "AGH, GHD are equal.
(ii) Again, because the A. EGB = the vertically opposite
4 AGH ;
and the 4. AGH = the alternate 2 GHD ; Proved.
... the exterior 4. EGB = the interior opposite 4 GHD.
PARALLELS. 39
(iii) Lastly, the 4. EGB = the 4 GHD; Proved.
add to each the Z. BGH ;
then the 4 "EGB, BGH together = the angles BGH, GHD.
But the adjacent / " EGB, BGH together = two right angles;
... the two interior a "BGH, GHD together = two right angles.
Q.E.D.
PARALLELS ILLUSTRATED BY ROTATION.
The direction of a straight line is determined by the angle which
it makes with some given line of reference.
Thus the direction of AB, relatively to the given line YX, is given by
the angle APX.
Now suppose that AB and CD in *
the adjoining diagram are parallel; C
then we have learned that
the ext. Z-APX = the int. opp. / CQX;
that is, AB and CD make equal angles &
with the line of reference YX. Y Q * P X
This brings us to the leading idea
connected with parallels: -
J’arallel straight lines have the 8ame D B
DIRECTION, but differ in Position.
h The same idea may be illustrated
thus :
Suppose AB to rotate about P through the 4-APX, so as to take the
position XY. Thence let it rotate about Q the opposite way through
the equal / XQC : it will now take the position CD. Thus AB may be
brought into the position of CD by two rotations which, being equal
and opposite, involve no final change of direction.
A
HYPOTHETICAL CONSTRUCTION. In the above diagram let
AB be a fixed straight line, Q a fixed point, CD a straight
line turning about Q, and YGPX, any transversal through Q.
Then as CD rotates, there must be one position in which the
4. CQX == the fixed z. APX.
Hence through any given point we may assume a line to pass
parallel to any given straight line.
Obs. If AB is a straight line, movements from A towards
B, and from B towards A are said to be in opposite senses
of the line AB,

40 geometry.
THEOREM 15. [Euclid I. 30.]
Straight lines which are parallel to the same straight line are
parallel to one another. - -

E
A a/
C H/
P K
Z
Let the straight lines AB, CD be each parallel to the straight
line PQ. .
It is required to prove that AB and CD are parallel to one
another. .
Draw a straight line EF cutting AB, CD, and PQ in the
points G, H, and K. -
B
D
Q
Proof. Then because AB and PQ are parallel, and EF meets
them, :
... the 4. AGK = the alternate Z. GKQ.
And because CD and PQ are parallel, and EF meets them,
... the exterior 4 GHD = the interior opposite Z. GKQ.
... the 4 AGH = the 4 GHD ;
and these are alternate angles;
..'. AB and CD are parallel. -
- Q.E.D.
NoTE. If PQ lies between AB and CD, the Proposition needs no
proof; for it is inconceivable that two straight lines, which do not
meet an intermediate straight line, should meet one another.
The truth of this Proposition may be readily deduced from Playfair's
"Axiom, of which it is the converse. - - -
For if AB and CD were not parallel, they would meet when produced.
Then there would be two intersecting straight lines both parallel to a
third straight line: which is impossible. - -
Therefore AB and CD never meet; that is, they are parallel.
PARALLELS. 41
EXERCISES ON PARALLELS.
- cºw Aniº")
1. In the diºm of the previous page, if the angle EGB is 55°,
4.63express in degrees each of the angles GH C, HKQ, QKF. -
2. Straight lines which are perpendicula; to the same straight line are
parallel to one another.
3. If a straight line meet two or more parallel straight lines, and is
perpendicular to one of them, it is also perpendicular to all the others.
4. Angles of which the arms are parallel, each to each, are either
equal or supplementary.
N. 5. Two straight lines AB, CD bisect one another at O. Shew that
the straight lines joining AC and BD are parallel.
N. V.O3-( .
– 6. Any straight line drawn parallel to the base of an isosceles tri-
angle makes equal angles with the sides. -
7. If from any point in the bisector of an angle a straight line is
drawn parallel to either arm of the angle, the triangle thus formed is
isosceles.
8. From X, a point in the base BC of an isosceles triangle ABC, a
straight line is drawn at right angles to the base, cutting AB in Y, and
CA produced in Z: show the triangle AYZ is isosceles.
9. If the straight line which bisects an exterior angle of a triangle
is parallel to the opposite side, shew that the triangle is isosceles.
10. The straight lines drawn from any point in the bisector of an
angle parallel to the arms of the angle, and terminated by them, are
equal; and the resulting figure is a rhombus. ' "
11. AB and CD are two straight lines intersecting at D, and the
adjacent angles so formed are bisected : if through any point X in DC a
straight line YXZ is drawn parallel to AB and meeting the bisectors in
Y and Z, shew that XY is equal to XZ.
12. Two straight rods PA, QB revolve about pivots at Pand Q, PA
making 12 complete revolutions a minute, and QB making 10. If they
start parallel and pointing the same way, how long will it be before they
are again parallel, (i) pointing opposite ways, (ii) pointing the same way?
42 GEOMETRY.
THEOREM 16. [Euclid I. 32.]
The three angles of a triangle are together equal to two right
angles. *
Let ABC be a triangle.
It is required to prove that the three 4 "ABC, BCA, CAB together
= two right angles.
Produce BC to any point D; and suppose CE to be the line
through C parallel to BA.
Proof. Because BA and CE are parallel and AC meets them,
... the 4. ACE = the alternate A. CAB.
Again, because BA and CE are parallel, and BD meets them,
... the exterior 4. ECD = the interior opposite 4. ABC,
... the whole exterior 4. ACD = the sum of the two interior opposite
Z_* CAB, ABC.
To each of these equals add the A. BCA;
then the 4 "BCA, ACD together = the three / "BCA, CAB, ABC,
But the adjacent / "BCA, ACD together = two right angles.
... the 4 "BCA, CAB, ABC together = two right angles.
Q.E.D.
Obs. In the course of this proof the following most im-
portant property has been established.
If a side of a triangle is produced the 60terior angle is equal to
| the two interi Osite amales.
Namely, the ext. A. ACD = the ACAB+ the 4. ABC.

THE ANGLES OF A TRIANGLE. 43
INFERENCES FROM THEOREM 16.
1. If A, B, and C denote the number of degrees in the angles of
a triangle,
then A+ B+ C = 180°.
2. If two triangles have two angles of the one respectively cqual
to two angles of the other, then the third angle of the one is equal to
the third angle of the other.
3. In any right-angled triangle the two acute angles are comple-
mentary.
4. If one angle of a triangle is equal to the sum of the other
two, the triangle is right-angled.
5. The Sum of the angles of any quadrilateral figure is equal to
four right angles.
EXERCISES ON THEOREMI 16.
1. Each angle of an equilateral triangle is two-thirds of a right
angle, or 60°.
2. In a right-angled isosceles triangle each of the equal angles
is 45°.
3. Two angles of a triangle are 36° and 123° respectively: deduce
the third angle; and verify your result by measurement.
4. In a triangle ABC, the A. B=111°, the 4-C=42°; deduce the 4-A,
and verify by measurement.
5. One side BC of a triangle ABC is produced to D. If the exterior
angle ACD is 134°; and the angle BAC is 42°; find each of the remaining
interior angles.
6. In the figure of Theorem 16, if the Z.ACD=118°, and the
4. B=51°, find the A-3A and C ; and check your results by measurement.
7. Prove that the three angles of a triangle are together equal to two
night angles by supposing a line drawn through the vertea parallel to the
base.
8. If two straight lines are perpendicular to two other straight lines,
each to each, the acute angle between the first pair is equal to the acute
angle between the 8econd pair.
44 GEOMETRY.

cººl jihnºlºgy
COROLLARY 1. All theft, angles of any whº,
together with four right angles, are equal to twice as many right
angles as the figure has sides.
D

C
A B
Let ABCDE be a rectilineal figure of n sides.
It is required to prove that all the interior angles + 4 rt. A "
= 2m rt. A ".
Take any point O within the figure, and join O to each of
its vertices.
Then the figure is divided into n triangles.
And the three 4 " of each A together = 2 rt. A ".
Hence all the 4 " of all the A* together = 2n rt. A ".
But all the 4 " of all the A" make up all the interior angles
of the figure together with the angles at O, which = 4 rt. A ".
... all the int. A * of the figure +4 rt. A "= 2n rt. A ".
Q.E.D.
DEFINITION. A regular polygon is one which has all its
sides equal and all its angles equal.
Thus if D denotes the number of degrees in each angle of
a regular polygon of n sides, the abové result may be stated
thus:
mD + 360° = m. 180°.
EXAMPLE.
Find the number of degrees in each angle of
s’ (i) a regular hexagon (6 sides);
(ii) a regular octagon (8 sides);
*(iii) a regular decagon (10 sides).
THE ANGLES OF RECTILINEAL FIGURES. 45
EXERCISES ON THEOREM 16.
(Numerical and Graphical.)
1. ABC is a triangle in which the angles at B and C are re-
spectively double and treble of the angle at A: find the number of
degrees in each of these angles.
2. Express in degrees the angles of an isosceles triangle in which
(i) Each base angle is double of the vertical angle;
(ii) Each base angle is four times the vertical angle.
X 3. The base of a triangle is produced both ways, and the exterior
angles are found to be 94° and 126°; deduce the vertical angle. Con-
struct such a triangle, and check your result by measurement.
4. The sum of the angles at the base of a triangle is 162°, and their
difference is 60°: find all the angles.
5. The angles at the base of a triangle are 84° and 62°; deduce
(i) the vertical angle, (ii) the angle between the bisectors of the base
angles. Check your results by construction and measurement. ,
6. . In a triangle ABC, the angles at B and C are 74° and 62°: if AB
and AC are produced, deduce the angle between the bisectors of the
exterior angles. Check your result graphically.
7. Three angles of a quadrilateral are respectively 114%.”, 50°, and
75%’; find the fourth angle.
8. In a quadrilateral ABCD, the angles at B, C, and D are re-
spectively equal to 2A, 3A, and 4A ; find all the angles.
9. Four angles of an irregular pentagon (5 sides) are 40°, 78°, 122°,
and 135°; find the fifth angle.
10. In any regular polygon of n sides, each angle contains 2(n-2)
right angles. 7?,
(i) Deduce this result from the Enumciation of Corollary 1.
(ii) Prove it independently by joining one vertex A to each of the
others (except the two immediately adjacent to A), thus dividing the
polygon into n-2 triangles.
ll. How many sides have the regular polygons each of whose
angles is (i) 108°, (ii) 156°2
12. Shew that the only regular figures which may be fitted together
So as to form a plane surface are (i) equilateral triangles, (ii) squares,
(iii) regular heavagons.
46 GEOMETRY.
COROLLARY 2. If the sides of a rectilineal figure, which has
To re-entrant angle, are produced in order, them all the eaterior
angles so ſº age together equal to four|right angles. J|}|JPlötſ
T 14 "rh $) i > 6) 25 tº jejlſ 1
1st Proof. Suppose, as before, that the figure has n sides;
and consequently n vertices.
Now at each vertex
the interior 4-4 the exterior Z = 2 rt. A ";
and there are n vertices,
... the sum of the int. A '+ the sum of the ext. A "- 2n rt. A ".
Dut by Corollary 1,
the sum of the int. Z * + 4 rº. Z." = 2n rt. A ";
... the sum of the ext. A * = 4 rt. A ".
Q.I.D.
2nd Proof.
C
©
Take any point O, and suppose Oa, Ob, Oc, Od, and Oe, are
lines parallel to the sides marked, A, B, C, D, E (and drawn
from O in the sense in which those sides were produced).
Then the exterior / between the sides A and B = the Zaob.
And the other exterior A * = the 4 "bCC, cod, dOe, eOa,
respectively.
... the sum of the ext. A "a the sum of the 4 " at O
= 4 rt. A ".

THE ANGLES OF RECTILINEAL FIGURES. 47
EXERCISES.
1. If one side of a regular hexagon is produced, shew that the
exterior angle is cqual to the interior angle of an equilateral triangle.
2. Express in degrees the magnitude of each eacterior angle of
(i) a regular octagon, (ii) a regular decagon.
3. How many sides has a regular polygon if each exterior angle is
(i) 30°, (ii) 24°3
4. If a straight line meets two parallel straight lines, and the two
interior angles on the same side are bisectod, shew that the bisectors
meet at right angles.
5. If the base of any triangle is produced both ways, shew that the
sum of the two exterior angles minus the vertical angle is equal to two
right angles.
6. In the triangle ABC the base angles at B and C are bisected by
BO and CO respectively. Shew that the angle Boc-904.
7. In the triangle ABC, the sides AB, AC are produced, a
exterior angles are bisected by BO and CO. Shew that
A
– 90° – ' '.
BOC – 90 2
8. The angle contained by the bisectors of two adjacent angles of
a quadrilateral is equal to half the sum of the remaining angles.
9. A is the vertex of an isosceles triangle ABC, and BA is produced
to D, so that AD is equal to BA; if DC is drawn, shew that BCD is a
right angle.
10. The straight line joining the middle point of the hypotenuse
ºf a right-angled triangle to the right angle is equal to half the
hypotenuse.
ExpºRIMENTAL proof THEOREM 16. (A + B -C=180°.]
In the A ABC, AD is p BC the
greatest side. AD is bi t right
º” by ZY; and YP, rp", on
If now the A is folded
dotted lines, the zºº A, B,
cide with the 4 ºz.D.Y., ZDQ,
, their sum is 180°.
three
ll coin-







48 - GEOMETRY.
THEOREM 17. [Euclid I. 26.] . -
If two triangles have two angles of one equal to two angles of the
other, each to each, and any side of the first equal to the corresponding
side of the other, the triangles are equal in all respects.Göyüşo 3'ſ a
*::. . . & ºf !.
p me wºº Fº,
*
#
TX; \ -
B . C E F
Let ABC, DEF be two triangles in which
the 4-A = the 4. D,
the A. B = the 4 E,
also let the side BC = the corresponding side EF.
***A&
It is required to prove that the A' ABC, DEF are equal in all
Tespects.
Proof. The sum of the 4 "A, B, and C ºr
= 2 rt. A ". Theor. 16.
= the sum of the 4 "D, E, and F;
and the 4 "A and B = the 4 "D and E respectively,
... the 4 C = the 4. F. -
Apply the AABC to the A DEF, so that B falls, on E, and
BC along EF. - º
Then because BC = EF,
... C must coincide with F.
And because the A. B = the Z.E,
... BA must fall along ED.
And because the 4.6–the A.F,
... CA must fall along FD.
... the point A, which falls both on ED and on FD, must coin-
cide with D, the point in which these lines intersect.
... the AABC coincides with the A DEF,
and is therefore equal to it in all respects.
So that AB = DE, and AC = DF; .
and the AABC = the A DEF in area. Q.E.D.


CONGRUENT TRIANGLES. 49
EXERCISES.
ON THE IDENTICAL Equality OE TRIANGLES.
1. Shew that the perpendiculars drawn from the extremities of the
base of an isosceles triangle to the opposite sides are equal.
2. Any point on the bisector of an angle is equidistant from the arms
of the angle.
3. Through O, the middle point of a straight line AB, any straight
line is drawn, and perpendiculars AX and BY are dropped upon it from
A and B : shew that AX is equal to BY.
4. If the bisector of the vertical angle of a triangle is at right
angles to the base, the triangle is isosceles.
5. If in a triangle the perpendicular from the vertex on the base
bisects the base, then the triangle is isosceles.
6. If the bisector of the vertical angle of a triangle also bisects the
base, the triangle is isosceles.
[Produce the bisector, and complete the construction after the
manner of Theorem 8.]
7. The middle point of any straight line which meets two parallel
straight lines, and is terminated by them, is equidistant from the
parallels.
8. A straight line drawn between two parallels and terminated by
them, is bisected; shew that any other straight line passing through
the middle point and terminated by the parallels, is also bisected at
that point.
9. If through a point equidistant from two parallel straight lines,
two straight lines are drawn cutting the parallels, the portions of the
latter thus intercepted are equal.
10. In a quadrilateral, ABCD, if AB = AD, and BC = DC : shew that
the diagonal AC bisects each of the angles which it joins; and that AC
is perpendicular to BD. §
11. A surveyor wishes to ascertain the breadth of a river which he
cannot cross. Standing at a point A near the bank, he notes an object B
immediately opposite on the other bank. He lays down a line AC of any
length at right angles to AB, facing a mark at O the middle point of AC.
From C he walks along a line perpendicular to AC until he reaches a point
D from which O and B are seen in the same direction. He now measures
CD: prove that the result gives him the width of the river.
H.S. G. D
60 GEOMETRY.
ON THE IDENTICAL EQUALITY OF TRIANGLES.
Three cases of the congruence of triangles have been dealt
with in Theorems 4, 7, 17, the results of which may be
summarised as follows:
Two triangles are equal in all respects when the following
three parts in each are severally equal:
1. Two sides, and the included angle. Theorem 4.
2. The three sides. Theorem 7.
3. Two angles and one side, the side given in one triangle
CORRESPONDING to that given in the other. Theorem 17.
Two triangles are not, however, necessarily equal in all
respects when any three parts of one are equal to the corre-
sponding parts of the other.
For example:
(i) When the three angles of one are
equal to the three angles of the other,
each to each, the adjoining diagram
shews that the triangles need not be
equal in all respects.
(ii) When two sides and one angle in one are equal to two
sides and one angle of the other, the given angles being opposite
to equal sides, the diagram below shews that the triangles
need not be equal in all respects.
B C E É. F
For if AB = DE, and AC = DF, and the 4 ABC = the A DEF, it
will be seen that the shorter of the given sides in the
triangle DEF may lie in either of the positions DF or DF'.
NOTE. From these data it may be shewn that the angles opposite
to the equal sides AB, DE are either equal (as for instance the Alº ACB,
DF'E) or 8wpplementary (as the Z." ACB, DFE); and that in the former
case the triangles are equal in all respects. . This is called the
ambiguous case in the congruence of triangles. [See Problem 9, p. 82.]
If the given angles at B and E are right angles, the ambiguity dis-
appears. This exception is proved in the following Theolem.

CONGRUENT TRIANGLES. 51
THEOREM 18.
Two right-angled triangles which have their hypotenuses equal,
and one side of one equal to one side of the other, are equal in all
Tespects.

A D
B C C’ É F
Let ABC, DEF be two right-angled triangles, in which
the 4 "ABC, DEF are right angles,

the hypotenuse AC = the hypotenuse DF,
and AB = DE,
It is required to prove that the A"ABC, DEF are equal in all
Tespects.
Proof. Apply the AABC to the A DEF, so that AB falls
on the cqual line DE, and C on the side of DE opposite to F.
Let C' be the point on which C falls.
Then DEC' represents the AABC in its new position.
Since each of the 4 "DEF, DEC' is a right angle,
... EF and EC' are in one straight line.
And in the AC'DF, because DF = DC' (i.e. AC),
... the Z DFC' = the A DC'F. Theol. 5,
Hence in the A*DEF, DEC',
the 4: DEF = the 4. DEC', being right angles;
because the Z. DFE = the 4 DC'E, Proved.
and the side DE is common.
... the A*DEF, DEC' are equal in all respects; Theor. 17.
that is, the A*DEF, ABC are equal in all respects.
& Q.E.D.
52 GEOMETRY,
*THEOREM 19. [Euclid I. 24.]
If two triangles have two sides of the one equal to two sides of the
other, each to each, but the angle included by the two sides of ome
greater than the angle included by the corresponding sides of
the other; them the base of that which has the greater angle is
greater than the base of the other.

* \,
*

B C E K G
F
Let ABC, DEF be two triangles, in which
BA = ED,
and AC = DF,
but the A. BAC is greater than the A. EDF.
It is required to prove that the base BC is greater than the
base EF.
Proof. Apply the AABC to the A DEF, so that A falls on D,
and AB along DE.
Then because AB = DE, B must coincide with E.
Let DG, GE represent AC, CB in their new position.
Then if EG passes through F, EG is greater than EF;
that is, BC is greater than EF.
But if EG does not pass through F, suppose that DK bisects
the 4 FDG, and meets EG in K. Join FK.
Then in the A* FDK, GDK, \
FD = GD,
because DK is common to both,
|- the included / FDK = the included A. GDK;
*. FK = GK. Theor, 4.
Now the two sides EK, KF are greater than EF;
that is, EK, KG are greater than EF.
‘. EG (or BC) is greater than EF. Q.E.D.
CONVERSE OF THEOREM. 19. 53
*s
“ſhan * *
conversely, ãf two triangles have two sides of the one equal to two
sides of the other, each to each, but the base of one greater than the
base of the other; then the angle contained by the sides of that which
has the greater base, is greater than the angle contained by the
corresponding sides of the other.

A D
• *.
B C E
F
Let ABC, DEF be two triangles in which
BA = ED,
and AC = DF,

but the base BC is greater than the base EF.
It is required to prove that the 4. BAC is greater than the 4. EDF,
Proof. If the A, BAC is not greater than the 4. EDF,
it must be either equal to, or less than the A. EDF.
Now if the Z. BAC were equal to the A EDF,
then the base BC would be equal to the base EF; Theor. 4,
but, by hypothesis, it is not.
Again, if the A. BAC were less than the 4. EDF,
then the base BC would be less than the base EF; Theor. 19.
but, by hypothesis, it is not.
Thät is, the 4. BAC is neither equal to, nor less than the A. EDF;
... the A. BAC is greater than the A. EDF.
Q.E.D.
.* Theorems marked with an asterisk may be omitted or postponed at the
discretion of the teacher.
* 54 GEOMETRY.
REVISION LESSON ON TRIANGLES,
1. State the properties of a triangle relating to
(i) the sum of its interior angles;
(ii) the sum of its exterior angles.
What property corresponds to (i) in a polygon of n sides? With
what other figures does a triangle share the property (ii) 2 A
*
TU a " . . * * * *
2. Classify triangles with regard to their angles. Enumciate any
Theorem or Corollary assumed in the classification.
~, 3 * .# , , -", " . . e
3. Enunciate two Theorems in which from data relating to the sides
a conclusion is drawn relating to the angles. #
Q In the triangle ABC, if a = 3-6 cm., b=2.8 cm., c=3.6 cm., arrange
the angles in order of their sizes (before measurement); and prove that
the triangle is acute-angled.
u / 4. Enunciate two Theorems in which from data relating to the
angles a conclusion is drawn relating to the sides.
| In the triangle ABC, if
(i) A=48° and B-51°, find the third angle, and name the greatest
side. *
(ii) A=B-62}", find the third angle, and arrange the sides in order
of their lengths.
5. From which of the conditions given below may we conclude that
the triangles ABC, A'B'C' are identically equal? Point out where
ambiguity arises; and draw the triangle ABC in each case.
A = A' =71°. a = a' = 4:2 cm. A = A' =36°,
(i) ; B–B'-46°. (ii) { b- b'-2'4 cm. (iii) { B= B'- 121°.
a = a' = 3-7 cm. C = C' = 8.1°. C = C'—23°.
a = a' = 3.0 cm. B= B’= 53°. C = C'— 90°.
(iv) { b = b'=52 cm. (v) { b =b' = 4:3 cm. (vi) { c = c = 5 cm.
C = C =
5
* = 4.5 cm. c= c’ = 5.0 cm. a = a' = 3 cm.
º
6. Summarise the results of the last question by stating generally
under what conditions two triangles
(i) are necessarily congruent ;
& (ii) may or may not be congruent.
/ 7. If two triangles have their angles equal, each to each, the triangles
are not necessarily equal in all respects, because the three data are not
independent. Carefully explain this statement.
, EXERCISES ON TRIANGLES. 55
(Miscellaneous Eacamples.)
8. (i) The perpendicular is the shortest line that can be drawn to a
given straight line from a given point.
*
x (ii) obliques which make equal angles with the perpendicular are
equal.
X (iii) Of two obliques the less is that which makes the Smaller angle with
the perpendicular.
9. If two triangles have two sides of the one equal to two sides of the
other, each to each, and have likewise the angles opposite to one pair of
equal sides equal, then the angles opposite to the other pair of equal sides
are either equal or 8wpplementary, and in the former case the triangles are
equal in all respects. * * * * , *; 2. & 2
10. PQ is a perpendicular (4 cm. in length) to a straight line XY.
Draw through P a series of obliques making with PQ the angles 15°,
30°, 45°, 60°, 75°. Measure the lengths of these obliques, and tabulate ; c.
the results.
11. PAB is a triangle in which AB and AP have constant lengths
4 cm. and 3 cm. If AB is fixed, and AP rotatés about A, trace the
changes in PB, as the angle A increases from 0° to 180°.
Answer this question by drawing a series of figures, increasing A by
increments of 30°. Measure PB in each case, and tabulate the results.
12. From B the foot of a flagstaff AB a horizontal line is drawn
passing two points C and D which are 27 feet apart. The angles BCA
and BDA are 65° and 40° respectively. Represent this on a diagram
(scale 1 cm. to 10 ft.), and find by measurement the approximate height
of the flagstaff.
13. From P, the top of a lighthouse PQ, two boats A and B are
seen at anchor in a line due south of the lighthouse. It is known that
PQ = 126 ft., 4 PAQ =57°, 4 PBQ =33°; hence draw a plan in which
l" represents 100 ft., and find by measurement the distance between A
and B to the nearest foot.
14. From a lighthouse L two ships A and B, which are 600 yards
apart, are observed in directions S.W. and 15° East of South respec-
tively. At the same time B is observed from A in a S.E. direction.
Draw a plan (scale 1" to 200 yds.), and find by measurement the distance
of the lighthouse from each ship.
56 GEOMETRY.
PARALLELOGRAMS. zºº - ºr
& " " +
, --, * : DEFINITIONS, º
2 * ca. . . . . ' ' ' '9 * > * <2* J & J

1. A quadrilateral is a plane figure bounded
by four straight lines. -3
The straight line which joins opposite angular
points in a quadrilateral is called a diagonal. 2, encº
,& 5 ". . * {..., " .
2. A parallelogram is a quadrilateral
whose opposite sides are parallel. , sº-
[It will be proved hereafter that the opposite / 7
sides of a parallelogram are equal, and that its

opposite angles are equal.]
f
g a = r
3. A rectangle is a parallelogram which
has one of its angles a right angle. . . . .

[It will be proved hereafter that all the angles of CŞ.
a rectangle are right angles. See page 59.]
ty 3. * *
4. A square is a rectangle which has two
adjacent sides equal.

[It will be proved that all the sides of a square are
equal and all its angles right angles. See page 59.]
dry 3 +. “ , ”

5. A rhombus is a quadrilateral which
has all its sides equal, but its angles are
not right angles.
£, , -, *
6. A trapezium is a quadrilateral which has
one pair of parallel sides.
PARALLELOGRAMS. 57
THEOREM 20. [Euclid I. 33.]
& gy 3 tº 3, . "
The straight lines which join the extremities of two equal and
parallel straight lines towards the same parts are themselves equal
and parallel. Yvº, c = ... º.
A B

C D
Let AB and CD be equal and parallel straight lines; and let
them be joined towards the same parts by the straight lines
AC and BD.
It is required to prove that AC and BD are equal and parallel.
**. Join BC.
Proof. Then because AB and CD are parallel, and BC meets
th €II) 3. 2/3/* ce".
... the 4. ABC = the alternate 4. DCB.
Now in the A* ABC, DCB,
| AB = DC,
because { BC is common to both ;
land the 4 ABC = the A. DCB ; Proved.
... the triangles are equal in all respects;
so that AC = DB, ......................... (i)
and the 4 ACB = 4 DBC.
But these are alternate angles;
..'. AC and BD are parallel. ................., (ii)
That is, AC and BD are both equal and parallel.
Q.E.D.
58 GEOMETRY.
THEOREM 21. [Euclid I. 34.] }
The opposite sides and angles of a parallelogram are equal to one
another, and each diagonal bisects the parallelogram.
\!
Let ABCD be a parallelogram, of which BD is a diagonal.
It is required to prove that
(i) AB = CD, and AD = CB, *
(ii) the A. BAD = the A. DCB,
(iii) the Z, ADC = the 4 CBA,
(iv) the AABD = the A CDB in area.
Proof. Because AB and DC are parallel, and BD meets them,
... the 4. ABD = the alternate 4-CDB.
Again, because AD and BC are parallel, and BD meets them,
... the 4. ADB = the alternate / CBD.
Hence in the A"ABD, CDB,
the Z. ABD = the 4 CDB,
because {the Z. ADB = the A CBD, Proved.
and BD is common to both ;
... the triangles are equal in all respects; Theor. 17.
so that AB = CD, and AD = CB; ..................... (i)
and the 4. BAD = the 4. DCB ; ........................ (ii)
and the AABD = the A CDB in area. .............. (iv)
And because the Z. ADB = the 4. CBD, Proved.
and the Z. CDB = the 4. ABD,
... the whole / ADC = the whole / CBA........., (iii)
Q.E.D.
*****~~

3. gº* *
59
gº' PARALLELOGRAMS.
COROLLARY 1. If one angle of a parallelogram is a right
angle, all its angles are right angles.
In other words : . .
All the angles of a rectangle are right angles.
For the sum of two consecutive 4–2 rt, 4. ; (Theor. 14.)
..., if one of these is a rt. angle, the other must be a rt. angle.
A And the opposite angles of the par" are equal;
'', ... all the angles are right angles.
* CoRollARY 2. All the sides of a square are equal; and all
tts aples are right angles. ---
“,
% &Yº
COROLLARY 3. The diagonals of a parallelogram bisect one
another. D = C

Let the diagonals AC, BD of the par"
ABCD intersect at O.
2" To prove AO = OC, and BO =OD.
“… * In the A* AOB, COD, A-, B
{...}} ~ * f \ , , the Z.OAB = the alt. A OCD,
because { the 4-AOB =vert. opp. / COD,
+-
and AB = the opp. side CD ;
... OA = OC ; and OB = OD. Theor. 17,
EXERCISES.
1. If the opposite sides of a quadrilateral are equal, the figure is a
parallelogram.
2. ... If the opposite angles of a quadrilateral are equal, the figure is a
parallelogram.
3. ... If the diagonals of a quadrilateral bisect each other, the figure is a
parallelogram.
4. The diagonals of a rhombus bisect one another at right angles.
5. If the diagonals of a parallelogram are equal, all its angles are
right angles.
6. In a parallelogram which is not rectangular the diagonals are
unequal.
60 GEOMETRY.
EXERCISES ON PARALLELS AND PARALLELOGRAMS,
(Symmetry and Superposition.) ;
1. Shew that by folding a rhombus about one of its diagonals the
triangles on opposite sides of the crease may be made to coincide.
d That is to say, prove that a rhombus is symmetrical about either
iagonal.
2. Prove that the diagonals of a square are aves of symmetry. Name
two other lines about which a square is symmetrical.
3. The diagonals of a rectangle divide the figure into two congruent
triangles: is the diagonal, therefore, an axis of symmetry? About
what two lines is a rectangle symmetrical ?
4. Is there any axis about which an oblique parallelogram is sym-
metrical ? Give reasons for your answer.
5. In a quadrilateral ABCD, AB = AD and CB=CD; but the sides
are not all equal. Which of the diagonals (if either) is an axis of
symmetry :
6. Prove by the method of superposition that
(i) Two parallelograms are identically equal if two adjacent sides of
one are equal to two adjacent sides of the other, each to each, and one
angle of ome equal to one angle of the other.
(ii) Two rectangles are equal ºf two adjacent 8ides of one are equal to
two adjacent Sides of the other, each to each.
7. Two quadrilaterals ABCD, EFGH have the sides AB, BC, CD, DA
equal respectively to the sides EF, FG, GH, HE, and have also the
angle BAD equal to the angle FEH. Shew that the figures may be
made to coincide with one another.
(MIiscellaneous Theoretical Eacamples.)
X 8. Any straight line drawn through the middle point of a diagonal
of a parallelogram and terminated by a pair of opposite sides, is bisected
at that point.
* 9. In a parallelogram the perpendiculars drawn from one pair of
opposite angles to the diagonal which joins the other pair are equal.
sy 10. If ABCD is a parallelogram, and X, Y respectively the middle
points of the sides AD, BC; shew that the figure AYCX is a paral-
lelogram.
PARALLELS AND PARALLELOGRAMS. 61
11. ABC and DEF are two triangles such that AB, BC are respec-
tively equal to and parallel to DE, EF; shew that AC is equal and
parallel to DF.
12. ABCD is a quadrilateral in which AB is parallel to DC, and AD
equal but not parallel to BC; shew that
(i) the 4-A + the Z-C= 180°– the A-B-H the A. D.;
(ii) the diagonal AC= the diagonal BD ;
(iii) the quadrilateral is symmetrical about the straight line joining
the middle points of AB and DC.
13. AP, BQ are straight rods of equal length, turning at equal
rates (both clockwise) about two fixed pivots A and B respectively. If
the rods start parallel but pointing in opposite senses, shew that
(i) they will always be parallel;
(ii) the line joining PQ will always pass through a certain fixed
point.
(Miscellaneous Numerical and Graphical Examples.)
14. Calculato the angles of the triangle ABC, having given :
int. A.A– 3 of ext. 4-A ; 3B =4C.
15. A yacht sailing due East changes her course successively by 63°,
by 78°, by 119°, and by 64°, with a view to sailing round an island.
What further change must be made to set her once more on an Easterly
course ?
16. If the sum of the interior angles of a rectilineal figure is equal
to the sum of the exterior angles, how many sides has it, and why?
17. Draw, using your protractor, any five-sided figure ABCDE,
in which
A B = 110°, 4 C = 115°, 4- D =93°, 4 E = 152°.
Verify by a construction with ruler and compasses that AE is parallel
to BC, and account theoretically for this fact.
18. A and B are two fixed points, and two straight lines AP, BQ,
unlimited towards P and Q, are piyoted at A and B. AP, starting from
the direction AB, turns about A clockwise at the uniform rate of 74° a
second ; and BQ, starting simultaneously from the direction BA, turns
about B counter-clockwise at the rate of 3; a second.
(i) How many seconds will elapse before AP and BQ are parallel?
(ii) Find graphically and by calculation the angle between AP and
BQ twelve seconds from the start.
(iii) At what rate does this angle decrease?
62 GEOMETRY.
THEOREM 22. * *
sº a º
§ § d §º: , sº jº
If there are three or mºſºftgºgllel straight lines, and % intercepts
made by them on any transversäl are equal, then the corresponding
$ntercepts on any other transversal are also equal. $
§
A Pl NX B ñº
+

‘... !' 's 2 T-/
C al N D *)
| M N
{^NZ
E R N NTF
I
Let the parallels AB, CD, EF cut off equal intercepts PQ, QR
from the transversal PQR; and let XY, YZ be the corresponding
intercepts cut off from any other transversal XYZ.
It is required to prove that XY = YZ.
Through X and Y let XM and YN be drawn parallel to PR.
Proof. Since CD and EF are parallel, and XZ meets them,
... the 4 XYM = the corresponding / YZN.
And since XM, YN are parallel, each being parallel to PR, #
... the 4 MXY = the corresponding / NYZ.
Now the figures PM, QN are parallelograms,
... XM = the opp. side PQ, and YN = the opp. side QR ;
and since by hypothesis PQ = QR,
.*. XM = YN.
Then in the A*XMY, YNZ,
the Z XYM = the Z. YZN,
because { the A MXY = the 4. NYZ,
and XM = YN ;
... the triangles are identically equal; Theor. 17.
.*. XY = YZ.
Q.E.D.
PARALLELS AND PARALLELOGRAMS. 63
26,
COROLLARY. In a triangle ABC, if a set of lines Pp, Q4, Rr, ...,
drawn parallel to the base, divide one side AB into equal parts, they
also divide the other side AC into equal parts.

P
Q g
l ſ
R ! Yºr
| | | N.
| | |
B 1 2 3 C
, may thus be expressed
The lengths of the parallels Po, Qq, Rr,
in terms of the base BC.
Through p, q, and r let pl. q2, r3 be drawn parl to AB.
Then, by Theorem 22, these par” divide BC into four equal parts, of
which Pp evidently contains one, Qq two, and Rr three.
In other words, \ < * *S, CTs k → ;
Po–. BC; Qq=%. BC; Rr=}. BG
4 2 4 3. 4 º
Similarly if the given par” divide AB into n equal parts,
Re- • BC; and so on.
Qq=#. BC,
-
Pp=} BC,
*...* Problem 7, p. 78, should now be worked.
DEFINITION. *
- º * *\ wº º w
mities of a straight line AB perpendiculars
a G.
If from the extre
AX, BY are drawn to a straight line PQ of indefinite length,
then XY is said to be the orthogonal projection of AB on PQ.
B
A -
B
x 2^T
Y Q P LT Y Q


5 X
A
64 GEOMETRY.
EXERCISES ON PARALLELS AND PARALLELOGRAMS.
*
1. The straight line drawn through the middle point of a side of a
triangle, parallel to the base, bisects the remaining side.
[This is an important particular case of
Theorem 22. *
In the A ABC, if Z is the middle point of
AB, and ZY is drawn parl to BC, we have to
prove that AY=YC.
Draw YX parl to AB, and then prove the
As ZAY, XYC congruent.]
W
2. The straight line which joins the A
middle points of two sides of a triangle is
parallel to the third side.
[In the AABC, if Z, Y are the middle 2 Y V
points of AB, AC, we have to prove ZY f
parl to BC. /
Produce ZY to V, making YV equal to
ZY, and join CV. Prove the A* AYZ, B C
CYV congruent ; the rest follows at once.]
3. The straight line which joins the middle points of two sides of a
triangle is equal to half the third side.
4. Shew that the three straight lines, which join the middle points
of the sides of a triangle, divide it into four triangles which are identi-
cally equal.
{
5. Any straight line drawn from the vertex of a triangle to the base
is bisected by the straight line which joins the middle points of the other
Sides of the triangle. & *
6. ABCD is a parallelogram, and X, Y are the middle points of
§§ opposite sides AD, BC: shew that BX and DY trisect the diagonal
7. If the middle points of adjacent sides of any quadrilateral are
joined, the figure thus formed is a parallelogram.
8. Shew that the straight lines which join the middle points of
opposite sides of a quadrilateral, bisect one another.

PARALLELS AND PARALLELOGRAMS. 65
9. From two points A and B, and from O the mid-point between
them, perpendiculars AP, BQ, OX are drawn to a straight line CD.
If AP, BQ measure respectively 42 cm. and 5.8 cm., deduce the length
of OX, and verify your result by measurement.
Shew that OX= }(AP+ BQ) or # (AP- BQ), according as A and B
are on the same side, or on opposite sides of CD.
10. When three parallels cut off equal intercepts from two trans-
versals, shew that of the three parallel lengths between the two
transversals the middle one is the Arithmetic Mean of the other two.
ll. The parallel sides of a trapezium are a centimetres and b centi-
metres in length. Prove that the line joining the middle points of the
oblique sides is parallel to the parallel sides, and that its length is
#(a+b) centimetres.
12. OX and OY are two straight lines, and along OX five points
1, 2, 3, 4, 5 are marked at equal distances. Through these points
parallels are drawn in any direction to meet OY. Measure the lengths
of these parallels: take their average, and compare it with the length
of the third parallel. Prove geometrically that the 3rd parallel is the
mean of all five.
State the corresponding theorem for any odd number (2n + 1) of
parallels so drawn.
13. From the angular points of a parallelogram perpendiculars
are drawn to any straight line which is outside the parallelogram :
shew that the sum of the perpendiculars drawn from one pair of
opposite angles is equal to the sum of those drawn from the other pair.
[Draw the diagonals, and from their point of intersection suppose
a perpendicular drawn tº the given straight line.]
S 14. The sum of the perpendiculars drawn from any point in the
base of an isosceles triangle to the equal sides is equal to the perpen-
dicular drawn from either extremity of the base to the opposite side.
[It follows that the sum of the distances of any point in the base
of an isosceles triangle from the equal sides is constant, that is, the
same whatever point in the base is taken.]
How would this property be modified if the given point were taken
in the base produced 2 , * .*
15. The sum of the perpendiculars drawn from any point within
an equilateral triangle to the three sides is equal to the perpendicular
drawn from any one of the angular points to the opposite side, and is
therefore constant.
16. Equal and parallel lines have equal projections on any other
straight line.
H. S. G. E
66 GEOMETRY.
DIAGONAL SCALES.
Diagonal scales form an important application of Theorem 22.
We shall illustrate their construction and use by describing a
Decimal Diagonal Scale to shew Inches, Tenths, and Hundredths.
A straight line AB is divided (from A) into inches, and the
points of division marked 0, 1, 2, .... The primary division
0A is subdivided into tenths, these secondary divisions being
numbered (from 0) 1, 2, 3, ... 9. We may now read on AB
inches and tenths of an inch.
A 9 8 7 4 3 2 1 1 B
In order to read hundredths, ten lines are taken at any equal
intervals parallel to AB; and perpendiculars are drawn through
0, 1, 2, ... .
The primary (or inch) division corresponding to 0A on the
tenth parallel is now subdivided into ten equal parts; and
diagonal lines are drawn, as in the diagram,
joining 0 to the first point of subdivision on the 10" parallel.
,, 1 to the Second ,
,, 2 to the third ,
and so on.
33 3.9 y? y
92 35 33 ;
The scale is now complete, and its use is shewn in the
following example.
Eacample. To take from the scale a length of 2:47 inches.
(i) Place one point of the dividers at 2 in AB, and extend them till
the other point reaches 4 in the subdivided inch 0A. We have now
2'4 inches in the dividers.
(ii) To get the remaining 7 hundredths, move the right-hand point
up the perpendicular through 2 till it reaches the 7th parallel. Then
extend the dividers till the left point reaches the diagonal 4 also on the
7th parallel. We have now 2:47 inches in the dividers.

DIAGONAL SCALES. 67
REASON FOR THE ABOVE PROCESS.
The first step needs no explanation. The reason of the
second is found in the Corollary of Theorem 22.
Joining the point 4 to the corresponding point 5 4
on the tenth parallel, we have a triangle 4,4,5,
of which one side 4,4 is divided into ten equal
parts by a set of lines parallel to the base 4,5.
Therefore the lengths of the parallels between
4,4, and the diagonal 4,5 are Hº Hº, #, ... of the
base, which is 1 inch. &
Hence these lengths are respectively
'01, .02, '03, ... of 1 inch.
4 3 2 1
Similarly, by means of this scale, the length of a given
straight line may be measured to the nearest hundredth of an
inch.
Again, if one inch-division on the scale is taken to represent
10 feet, then 2:47 inches on the scale will represent 24-7 feet.
And if one inch-division on the scale represents 100 links, then
2.47 inches will represent 247 links. Thus a diagonal scale is
of service in preparing plans of enclosures, buildings, or field-
works, where it is necessary that every dimension of the actual
object must be represented by a line of proportional length on
the plan.
NOTE. §
The subdivision of a diagonal scale need not be decimal.
For instance we might construct a diagonal scale to read centimetres,
millimetres, and quarters of a millimetre; in which case we should take
four parallels to the line AB. *
[For Exercises on Linear Measuremenus see the following page.]


68 GEOMETRY.
EXERCISES ON LINEAR MEASUREMENTS ,
1. Draw straight lines whose lengths are lºff inches, 272 inches,
3:08 inches.
2. Draw a line 2.68 inches long, and measure its length in centi-
metres and the nearest millimetre. &
3. Draw a line 5-7 cm. in length, and measure it in inches (to the
nearest hundredth). Check your result by calculation, given that
1 cm. = 0-3937 inch.
4. Find by measurement the equivalent of 3-15 inches in centi-
metres and millimetres. Hence calculate (correct to two decimal
places) the value of 1 cm. in inches."
5. Draw lines 2-9 cm. and 6.2 cm. in length, and measure them in
inches. Use each equivalent to find the value of 1 inch in centimetres
and millimetres, and take the average of your results.
6. A distance of 100 miles is represented on a map by 1 inch,
Draw lines to represent distances of 336 miles and 408 miles.
7. If I inch on a map represents 1 kilometre, draw lines to represent
850 metres, 2980 metres, and 1010 metres.
8. A plan is drawn to the scale of 1 inch to 100 links. Measure in
centimetres and millimetres a line representing 417 links.
9. Find to the nearest hundredth of an inch the length of a line
which will represent 42-500 kilometres in a map drawn to the scale of
1 centimetre to 5 kilometres.
10. The distance from London to Oxford (in a direct line) is
55 miles. If this distance is represented on a map by 2-75 inches, to
what scale is the map drawn 2 That is, how many miles will be
represented by 1 inch 7 How many kilometres by 1 centimetre 2
[1 cm. = 0-3937 inch ; 1 km. =# mile, nearly.]
11. On a map of France drawn to the scale 1 inch to 35 miles, the
distance from Paris to Calais is represented by 4-2 inches. Find the
distance accurately in miles, and approximately in kilometres, and
express the scale in metric measure. [1 km. =# mile, nearly.]
12. The distance from Exeter to Plymouth is 37% miles, and
appears on a certain map to be 24"; and the distance from Lincoln to
York is 88 km., and appears on another map to be 7 cm. Compare the
scales of these maps in miles to the inch.
13, Draw a diagonal scale, 2 centimetres to represent 1 yard,
shewing yards, feet, and inches,
PRACTICAL GEOMETRY. 69
PRACTICAL GEOMETRY.
PROBLEMS.
The following problems are to be solved with ruler and
compasses only. No step requires the actual measurement
of any line or angle ; that is to say, the constructions are to be
made without using either a graduated scale of length, or a
protractor.
The problems are not merely to be studied as propositions;
but the construction in every case is to be actually performed
by the learner, great care being given to accuracy of drawing.
Each problem is followed by a theoretical proof; but the
Tesults of the work should always be verified by measurement,
as a test of correct drawing. Accurate measurement is also
required in applications of the problems.
In the diagrams of the problems lines which are inserted
only for purposes of proof are dotted, to distinguish them
from lines necessary to the construction.
For practical applications of the problems the student
should be provided with the following instruments:
1. A flat ruler, one edge being graduated in centimetres
and millimetres, and the other in inches and tenths.
2. Two set squares; one with angles of 45°, and the other
with angles of 60° and 30°.
3. A pair of pencil compasses.
4. A pair of dividers, preferably with screw adjustment.
5. A semi-circular protractor.
70 GEOMETRY.
*.
}
f
TROBLEM 1.
To bisect a given angle.
Let BAC be the given angle to be bisected.
Construction. With centre A, and any radius, draw an
arc of a circle cutting AB, AC at P and Q.
With centres P and Q, and radius PQ, draw two arcs cutting
at O. Join AO.
Then the A. BAC is bisected by AO.
Proof. Join PO, QC.
In the A*APO, AQO,
AP=AQ, being radii of a circle,
because PO =QO, , 25 equal circles,
and AO is common;
... the triangles are equal in all respepts; Theor. 7.
so that the A. PAO = the A. QAO ;
that is, the Z. BAC is bisected by AO.
NOTE. PQ has been taken as the radius of the arcs drawn from the
centres P and Q, and the intersection of these arcs determines the
point O. Any radius, however, may be used instead of PQ, provided
that it is great enough to secure the intersection of the arcs.

PROBLEMS ON LINES AND ANGLES. 71
PROBLEM 2.
To bisect a given straight line.

Sº
Ś
,' \
/ \
V.
/ \,
e *
* *
Aé–Hºb
* O *
Let AB be the line to be bisected.
Construction. With centre A, and radius AB, draw two
arcs, one on each side of AB.
With centre B, and radius BA, draw two arcs, one on each
side of AB, cutting the first arcs at P and Q.
Join PQ, cutting AB at O.
Then AB is bisected at O.
Proof. Join AP, AQ, BP, BQ.
In the A*APQ, BPQ,
| AP=BP, being radii of equal circles,
because AQ = BQ, for the same reason,
and PQ is common; *
... the A. APQ = the LBPQ. "Theor. 7.
Again in the A*APO, BPO,
AP=BP,
because PO is common,
and the A. APO = the 4 BPO ;
* ... AO = OB; Theor. 4.
that is, AB is bisected at O.
NoTES. (i) AB was taken as the radius of the arcs drawn from the
centres A and B, but any radius may be used provided that it is great
gº* secure the intersection of the arcs which determine the points
'P and Q.
(ii) From the congruence of the A* APO, BPO it follows that the
4. AOP=the 4 BOP. As these are adjacent angles, it follows that PQ
bisects AB at right angles. *
72 GEOMETRY.
PROBLEM 3.
To draw a straight line perpendicular to a given straight line at
a given point in it. #
O
><
A *
& \
/ \.
,’ \
* \
Af *
& *
f \
* \
& *
/ \\
A P X Q B

Let AB be the straight line, and X the point in it at which
a perpendicular is to be drawn.
Construction. With centre X cut off from AB any two
equal parts XP, XQ.
With centres P and Q, and radius PQ, draw two arcs cutting
at O. *
Join XO.
Then XO is perp. to AB.
Proof. Join OP, OG).
In the A* OXP, OXQ,
XP=XQ, by construction,
because
OX is common,
and PO = QO, being radii of equal circles;
... the Z.OXP = the Z.OXQ. Theor, 7.
And these being adjacent angles, each is a right angle;
that is, XO is perp. to AB.
*
'0's. If the point x is near one end of AB, one or other of
the alternative constructions on the next page should be used.
PROBLEMS ON LINES AND ANGLES. 73
PROBLEM 3. SECOND METHOD.
Construction. Take any point C
outside AB.
With centre C, and radius CX, draw
a circle cutting AB at D.
Join DC, and produce it to meet
the circumference of the circle at O.
Join XO. >
Then XO is perp. to AB. A DSG D2 x B
Proof. Join CX.
Because CO = CX; . . the Z. CXO = the A. COX;
and because CD = CX; . . the Z-CXD = the 4 CDX.
‘.. the whole / DXO = the Z_XOD + the A-XDO
* = } of 180°
= 90°.
‘. XO is perp. to AB.
TROBLEM 3. THIRD METHOD.
Construction. With centre X XK
and any radius, draw the arc CDE,
cutting AB at C.
With centre C, and with the
'same radius, draw an arc, cutting
the first arc at D.
With centre D, and with the
same radius, draw an arc, cut-
ting the first arc at E. A C X B
Bisect the 4 DXE by XO. Prob. 1.
Then XO is perp. to AB, *
Proof. Each of the 4 " CXD, DXE may be proved to be 60°;
and the 4 DXO is half of the 4 DXE;
... the Z CXO is 90°. *.
That is, XO is perp. to AB.
#

74 GEOMETRY.
PROBLEM 4. }
, ſº ºvº is . . . .
To draw a straight line perpendicular to a given straight line
from a given eaternal point.
* F ºr ; , , ; } X

Let X be the given external point from which a perpen-
dicular is to be drawn at AB.
Construction. Take any point.C. on the side of AB remote
from X.
With centre X, and radius XC, draw an arc to cut AB at P
and Q.
With centres P and Q, and radius PX, draw arcs cutting at
Y, on the side of AB opposite to X.
Join XY cutting AB at O.
Then XO is perp. to AB.
Proof. Join PX, QX, PY, QY,
In the A' PXY, QXY,
PX = QX, being radii of a circle,
because
PY = QY, for the same reason,
and XY is common ;
... the A. PXY = the A. QXY. Theor. 7.
Again, in the A' PXO, QXO,
PX = QX,
because | XO is common,
and the Z PXO = the Z QXO;
... the A-XOP = the A-XOQ. Theor. 4,
And these being adjacent angles, each is a right angle,
that is, XO is perp. to AB.
PROBLEMS ON LINES AND ANGLES. 75
Obs. When the point X is nearly opposite one end of AB,
one or other of the alternative constructions given below
should be used.
PROBLEM 4. SECOND METHOD.

Construction. Take any point D in X
AB. Join DX, and bisect it at C. C
With centre C, and radius CX, draw
a circle cutting AB at D and O.
* ..., A DNJ’o B
Then XO is perp. to AB.
For, as in Problem 3, Second Method, the LXOD is a right
angle.
PROBLEM 4. THIRD METHOD,

Construction. Take any two points te X
D and E in AB. .2%
With centre D, and radius DX, draw ...”.”
an arc of a circle, on the side of AB ... /
opposite to X. F-E-F-5 B
With centre E, and radius EX, draw `ss º
another arc cutting the former at Y. `,
Join XY, cutting AB at O. 2%
Then XO is perp. to AB.
(i) Prove the Aº XDE, YDE equal
in all respects by Theorem 7,
so that the A-XDE = the 4-YDE.
(ii) Hence prove the A'XDO, YDO equal in all respects
by Theorem 4, so that the adjacent / DOX, DOY are equal.
That is, XO is perp. to AB.
76 GEOMETRY,
PROBLEM 5.
At a given point in a given Straight line to make an angle equal
to a given angle.
A Eë Fo Q G
Let BAC be the given angle, and FG the given straight line;
and let O be the point at which an angle is to be made equal
to the Z. BAC. *
Construction. With centre A, and with any radius, draw
an arc cutting AB and AC at D and E.
With centre O, and with the same radius, draw an arc
cutting FG at Q.
With centre Q, and with radius DE, draw an are cutting tho
former arc at P.
Join OP.
Then POQ is the required angle.
*
Proof. Join ED, PQ.
In the A' POQ, EAD,
OP = AE, being radii of equal circles,
because {OQ = AD, for the same reason,
PQ = ED, by construction;
... the triangles are equal in all respects;
so that the A. POQ = the LEAD. Theor, 7.

PROBLEMS ON LINES AND ANGLES.
PROBLEM 6.
Through a given point to draw a straight line parallel to a given
straight line.
‘P O

X A |D TY
Let XY be the given straight line, and O the given point,
through which a straight line is to be drawn par' to XY.
Construction. In XY take any point A, and join OA.
Using the construction of Problem 5, at the point O in
the line AO make the 4. AOP equal to the 4 OAY and alternate
to it.
Then OP is parallel to XY.
Proof. Because AO, meeting the straight lines OP, XY,
makes the alternate 4 "POA, OAY equal ;
... OP is par' to XY.
*** The constructions of Problems 3, 4, and 6 are not usually
followed in practical applications. Parallels and perpendiculars
may be more quickly drawn by the aid of Set Squares. (See LESSONs
IN ExPERIMENTAL GEOMETRY, pp. 36, 42.)
78 GEOMETRY.
PROBLEM 7.
To divide a given straight line into any number of equal parts.
Let AB be the given straight line, and suppose it is required
to divide it into five equal parts.
Construction. From A draw AC, a straight line of unlimited
length, making any angle with AB.
From AC mark off five equal parts of any length, AP, PQ,
QR, RS, ST.
Join TB; and through P, Q, R, S draw par" to TB, meeting
AB in p, q, r, S.
Then since the par" Pp, Qq, Rr, SS, TB cut off five equal parts
from AT, they also cut off five equal parts from AB.
(Theorem 22.)
SECOND METHOD,
From A draw AC at any angle with
AB, and on it mark off four equal parts
AP, PQ, QR, RS, of any length.
From B draw BD par to AC, and on *
it mark off BS', S'R', R'Q', Q'P', each
equal to the parts marked on AC. A f \ q
Join PP, QQ', RR', SS' meeting AB
in p, q, r, s. Then AB is divided into
five equal parts at these points. w
[Prove by Theorems 20 and 22.] D



PROBLEMS. 79
EXERCISES ON LINES AND ANGLEs.
(Graphical Eacercises.)
1. Construct (with ruler and compasses only) an angle of 60°.
By repeated bisection divide this angle into four equal parts.
2. By means of Exercise 1, trisect a right angle ; that is, divide it
into three equal parts.
Bisect each part, and hence shew how to trisect an angle of 45°.
[No construction is known for exactly trisecting any angle.]
3. Draw a line 6-7 cm. long, and divide it into five equal parts.
Measure one of the parts in inches (to the nearest hundredth), and verify
your work by calculation. [1 cm. = 0-3937 inch.]
4. From a straight line 3'72" long, cut off one seventh. Measure
the part in centinetres and the nearest millimetre, and verify your
work by calculation. .*
5. At a point X in a straight line AB draw XP perpendicular to AB,
making XP 1.8" in length. From P draw an oblique PQ, 3.0" long,
to meet AB in Q. Mcasure XQ.
(Problems. State your construction, and give a theoretical proof.)
6. In a straight line XY find a point which is equidistant from two
given points A and B.
When is this impossible?
7. In a straight line XY find a point which is equidistant from two
intersecting lines AB, AC.
When is this impossible?
8. From a given point P draw a straight line PQ, making with a
given straight line AB an angle of given magnitude.
9. From two given points P and Q on the same side of a straight
line AB, draw two lincs which meet in AB and make equal angles
with it.
[Construction. From P draw PH perp. to AB, and produce PH to P',
making HFºequal to PH. Join PQ cutting AB at K. Join PK.
Prove that PK, QK are the required lines.]
10. Through a given point P draw a straight line such that the
perpendiculars drawn to it from two points A and B may be equal.
Is this always possible 7
80 GEOMETRY.
THE CONSTRUCTION OF TRIANGLES.
PROBLEM 8.
/2)^2 / CP}2
To draw a triangle having given the lengths of the three sides.

A
C Ö
©
6
B (Z !C X &
Let a, b, c be the lengths to which the sides of the required
triangle are to be equal.
Construction. Draw any straight line BX, and cut off from
it a part BC equal to a.
With centre B, and radius c, draw an arc of a circle.
With centre C, and radius b, draw a second arc cutting the
first at A.
Join AB, AC.
Then ABC is the required triangle, for by construction the
sides BC, CA, AB are equal to a, b, c respectively.
Obs. The three data a, b, c may be understood in two
ways: either as three actual lines to which the sides of the
triangle are to be equal, or as three numbers expressing the
lengths of those lines in terms of inches, centimetres, or some
other linear unit.
*
NoTES. (i) In order that the construction may be possible it is
necessary that any two of the given sides should be together greater
than the third side (Theorem ll); for otherwise the arcs drawn from
the centres B and C would not cut.
(ii) The arcs which cut at A would, if continued, cut again on the
other side of BC. Thus the construction gives two triangles on opposite
sides of a common base.
THE CONSTRUCTION OF TRIANGLES. 81
ON THE CONSTRUCTION OF TRIANGLES.
It has been seen (page 50) that to prove two triangles
identically equal, three parts of one must be given equal to
the corresponding parts of the other (though any three parts
do not necessarily serve the purpose). This amounts to saying
that to determine the shape and size of a triangle we must know
three of its parts: or, in other words,
To construct a triangle three independent data are required.
For example, we may construct a triangle
(i) When two sides (b, c) and the included angle (A) are given.
The method of construction in this case is obvious.
(ii) When two angles (A, B) and one side (a) are given.
Here, since A and B are given, we at once know C ;
for A+ B + C = 180°.

Hence we have only to draw the base equal \
^c
L
to a, and at its ends make angles equal to 26
B and C ; for we know that the remaining B \
angle must necessarily be equal to A. Q.

(iii) If the three angles A, B, C are given (and no side), the
problem is indeterminate, that is, the number of solutions
is unlimited.
For if at the ends of any base we make angles equal to
B and C, the third angle is equal to A.
This construction is indeterminate, because the three data
are not independent, the third following necessarily from the
other two.
H. S. G. - F
82 GEOMETRY.
PROBLEM 9.
To construct a triangle having given two sides and an angle
opposite to one of them.
B c-c, x
Let b, c be the given sides and B the given angle.
Construction, Take any straight line BX, and at B make
the 4 XBY equal to the given A. B.
From BY cut off BA equal to c.
With centre A, and radius b, draw an arc of a circle.
If this are cuts BX in two points C, and C, both on the
same side of B, both the A' ABC, ABC, satisfy the given con-
ditions.
This double solution is known as the Ambiguous Case, and
will occur when b is less than c but greater than the perp.
from A on BX.
EXERCISE.
Draw figures to illustrate the nature and number of solutions in the
following cases:
(i) When b is greater than c.
(ii) When b is equal to c.
(iii) When b is equal to the perpendicular from A on BX.
(iv) When b is less than this perpendicular.

THE CONSTRUCTION OF TRIANGLES. 83
PROBLEM 10,
To construct a right-angled triangle having given the hypotenuse
and one side.
A O B
Let AB be the hypotenuse and P the given side.
Construction. Bisect AB at O; and with Čentre O, and
radius OA, draw a semicircle.
With centre A, and radius P, draw an arc to cut the semi-
circle at C.
Join AC, BC.
Then ABC is the required triangle.
Proof. Join OC.
Because OA = OC;
... the 4. OCA = the Z.OAC.
And because OB = OC;
... the 4. OCB = the A. OBC.
... the whole / ACB = the z- OAC + the Z.OBC
* = } of 180° v Theor. 16.
= 90°.

84 GEOMETRY.
ON THE CONSTRUCTION OF TRIANGLES.
{Graphical Eacercises.)
1. Draw a triangle whose sides are 7.5 cm., 6.2 cm., and 53 cm.
Draw and measure the perpendiculars dropped on these sides from
the opposite vertices.
[N.B. The perpendiculars, if correctly drawn, will meet at a point,
as will be seen later. See page 207.]
2. Draw a triangle ABC, having given a-3:00", b=2'50", c=275".
Bisect the angle A by a line which meets the base at X. Measure
BX and XC (to the nearest hundredth of an inch); and hence calculate
the value of É. to two places of decimals. Compare your result with
the value of ;
3. Two sides of a triangular field are 315 yards and 260 yards, and
the included angle is known to be 39°. Draw a plan (1 inch to
100 yards) and find by measurement the length of the remaining side
of the field.
4. ABC is a triangular plot of ground, of which the base BC is
75 metres, and the angles at B and C are 47° and 68° respectively. Draw
a plan (scale 1 cm. to 10 metres). Write down without measurement
the size of the angle A ; and by measuring the plan, obtain the approxi-
mate lengths of the other sides of the field; also the perpendicular
drawn from A to BC.
Y 5. A yacht on leaving harbour steers N.E. sailing 9 knots an hour.
After 20 minutes she goes about, steering N.W. for 35 minutes and
making the same average speed as before. How far is she now from
the harbour, and what course (approximately) must she set for the
run home? Obtain your results from a chart of the whole course,
scale 2 cm. to 1 knot.
` 6. Draw a right-angled triangle, given that the hypotenuse
c=10.6 cm. and, one side a = 5-6 cm. Measure the third side b ; and
find the value of Ncº-a%. Compare the two results.
* 7. Construct a triangle, having given the following parts: B=34°,
b=5-5 cm., c=8'5 cm. Shew that there are two solutions. Measure
the two values of a, and also of C, and shew that the latter are
supplementary.
* 8. In a triangle ABC, the angle A=50°, and b-6-5 cm. Illustrate
by figures the cases which arise in constructing the triangle, when
(i) a =7 cm. (ii) a = 6 cm. (iii) a = 5 cm. (iv) a = 4 cm.
THE CONSTRUCTION OF TRIANGLES. 85
9. Two straight roads, which cross at right angles at A, are carried
over a straight canal by bridges at B and C. The distance between the
bridges is 461 yards, and the distance from the crossing A to the bridge
B is 261 yards. Draw a plan, and by measurement of it ascertain the
distance from A to C.
(Problems. State your construction, and give a theoretical proof.)
10. Draw an isosceles triangle on a base of 4 cm., and having an
altitude of 6-2 cm. Prove the two sides equal, and measure them to
the nearest millimetre.
X il. Draw an isosceles triangle having its vertical angle equal to a
given angle, and the perpendicular from the vertex on the base equal to
a given straight line.
Hence draw an equilateral triangle in which the perpendicular from
one vertex on the opposite side is 6 cm. Measure the length of a side
to the nearest millimetre.
12. Construct a triangle ABC in which the perpendicular from A on
BC is 5.0 cm., and the sides AB, AC are 5-8 cm. and 9.0 cm. respectively.
Measure BC. *
13. Construct a triangle ABC having the angles at B and C equal
to two given angles L and M, and the perpendicular from A on BC
equal to a given line P.
14." Construct a triangle ABC (without protractor) having given two
angles B and C and the side b.
15.- . On a given base construct an isosceles triangle having its
vertical angle equal to the given angle L.
16... Construct a right-angled triangle, having given the length of
the hypotenuse c, and the sum of the remaining sides a and b.
If c = 5.3 cm., and a +b=73 cm., find a and b graphically; and
calculate the value of Na”--b”.
17. . Construct a triangle having given the perimeter and the angles
at the base. For example, a +b+c+ 12 cm., B=70°, C=80°.
18. Construct a triangle ABC from the following data ;
a = 6-5 cm., b + c = 10 cm., and B = 60°.
Measure the lengths of b and c.
19. Construct a triangle ABC from the following data:
a =7 cm., c – b = 1 cm., and B = 55°.
Measure the lengths of b and c.
86 GEOMETRY.
THE CONSTRUCTION OF QUADRILATERALS.
It has been shewn that the shape and size of a triangle are
completely determined when the lengths of its three sides are
given. A quadrilateral, however, is not completely determined
by the lengths of its four sides. From what follows it will
appear that five independent data are required to construct a
quadrilateral.
PROBLEM 11.
To construct a quadrilateral, given the lengths of the four sides,
and one angle.
Let a, b, c, d be the given lengths of the sides, and A the
angle between the sides equal to a and d.
Construction. Take any straight line AX, and cut off from
it AB equal to a.
Make the Z. BAY equal to the Z.A.
From AY cut off AD equal to d.
With centre D, and radius c, draw an arc of a circle.
With centre B and radius b, draw another arc to cut the
former at C.
Join DC, BC.
Then ABCD is the required quadrilateral; for by construction
the sides are equal to a, b, c, d, and the 4. DAB is equal to the
given angle.

CONSTRUCTION OF QUADRILATERALS. 87
PROBLEM 12.
To construct a parallelogram having given two adjacent sides and
the included angle.
>JC .
Let P and Q be the two given sides, and A the given angle.
Construction 1. (With ruler and compasses.) Take a line
AB equal to P; and at A make the 4. BAD equal to the 4-A, and
make AD equal to Q.
With centre D, and radius P, draw an arc of a circle.
With centre B, and radius Q, draw another are to cut the
former at C.
Then ABCD is the required par”.
Proof. Join DB.
In the A" DCB, BAD,
DC = BA,
because CB = AD,
and DB is common :
... the 4 CDB = the Z. ABD ; Theor, 7.
and these are alternate angles,
‘. DC is par' to AB,
Also DC = AB;
... DA and BC are also equal and parallel. Theor. 20.
... ABCD is a par".
Construction 2. (With set squares.) Draw AB and AD as
before; then with set squares through D draw DC par' to AB,
and through B draw BC par to AD.
By construction, ABCD is a par” having the required parts.

88 GEOMETRY.
PROBLEM 13.
To construct a square on a given side.

:
K. C
J-
A B
Let AB be the given side.
Construction 1. (With ruler and compasses.) At A draw AX
perp. to AB, and cut off from it AD equal to AB.
With B and D as centres, and with radius AB, draw two arcs
cutting at C.
Join BC, DC.
Then ABCD is the required square.
Proof. As in Problem 12, ABCD may be shewn to be a par”.
And since the Z. BAD is a right angle, the figure is a rectangle,
Also, by construction all its sides are equal.
... ABCD is a square.
Construction 2. (With set squares.) At A draw AX perp. to
AB, and cut off from it AD equal to AB.
Through D draw DC par' to AB, and through B draw BC
par to AD meeting DC in C.
Then, by construction, ABCD is a rectangle. [Def. 3, page 56.]
Also it has the two adjacent sides AB, AD equal.
‘.. it is a square.
CONSTRUCTION OF QUADRILATERALS. 89
EXERCISES.
ON THE ConSTRUCTION OF QUADRILATERALS.
1. Draw a rhombus each of whose sides is equal to a given straight
line PQ, which is also to be one diagonal of the figure.
Ascertain (without measurement) the number of degrees in each
angle, giving a reason for your answer.
2. Draw a square on a side of 2:5 inches. Prove theoretically that
its diagonals are equal; and by measuring the diagonals to the nearest
hundredth of an inch test the correctness of your drawing.
3. Construct a square on a diagonal of 3:0", and measure the lengths
of each side. Obtain the average of your results.
4. Draw a parallelogram ABCD, having given that one side
AB=5-5 cm., and the diagonals AC, BD are 8 cm., and 6 cm. respectively.
Measure AD.
5. The diagonals of a certain quadrilateral are equal, (each 6-0 cm.),
and they bisect one another at an angle of 60°. Shew that five inde-
pendent data are here given.
Construct the quadrilateral. Name its species; and give a formal
proof of your answer. Measure the perimeter. If the angle between
the diagonals were increased to 90°, by how much per cent, would the
perimeter be increased ?
6. In a quadrilateral ABCD,
AB = 5-6 cm., BC = 2.5 cm., CD = 4.0 cm., and DA=3-3 cm.
Shew that the shape of the quadrilateral is not settled by these data.
Draw the quadrilateral when (i) A=30°, (ii) A=60°. Why does the
construction fail when A = 100° 2
Determine graphically the least value of A for which the con-
struction fails.
7. Shew how to construct a quadrilateral, having given the lengths
of the four sides and of one diagonal. What conditions must hold
among the data in order that the problem may be possible 2
Illustrate your method by constructing a quadrilateral ABCD, when"
(i) AB =30", BC =1.7", CD=2.5", DA=2'8", and the diagonal
BD=2'6". Measure AC.
(ii) AB =3.6 cm., BC=7-7 cm., CD=6-8 cm., DA=5.1 cm., and the
diagonal AC=8-5 cm. Measure the angles at B and D.
90 GEOMETRY.
LOCI.
DEFINITION. The locus of a point is the path traced out
by it when it moves in accordance with some given law.

P
Earample 1. Suppose the point P to move so
that its distance from a fixed point O is constant
(say 1 centinetre).
Then the locus of P is evidently the circum-
ference of a circle whose centre is O and radius
1 cm.
Eacample 2. Suppose the point P
moves at a constant distance (say 1 cm.)
from a fixed straight line AB.
Then the locus of P is one or other of
two straight lines parallel to AB, on
either side, and at a distance of 1 cm.
from it.
B.
Thus the locus of a point, moving under some given con-
dition, consists of the line or lines to which the point is
thereby restricted; provided that the condition is satisfied by
every point on such line or lines, and by no other.
When we find a series of points which satisfy the given
law, and through which therefore the moving point must pass,
we are said to plot the locus of the point.
LOCI. 91
PROBLEM 14.
To find the locus of a point P which moves so that its distances
from two fixed points A and B are always equal to one another.

A O B
Here the point P moves through all positions in which PA = PB;
. one position of the moving point is at O'the middle point
of AB.
Suppose P to be any other position of the moving point :
that is, let PA = PB.
Join OP.
Then in the A' POA, POB,
PO is common,
because OA = OB,
and PA = PB, by hypothesis;
... the A POA = the A POB. Theor. 7.
Hence PO is perpendicular to AB.
That is, every point P which is equidistant from A and B lies on
the straight line bisecting AB at right angles.
Likewise it may be proved that every point on the perpen-
dicular through O is equidistant from A and B.
This line is therefore the required locus.
92 GEOMETRY.
PROBLEM 15.
To find the locus of a point P which moves so that its perpen-
dicular distances from two given straight lines AB, CD are equal to
one another.

A D
N
C. º M 5
Let P be any point such that thé perp. PM = the perp. PN.
Join P to O, the intersection of AB, CD.
Then in the A' PMO, PNO,
the 4 PMO, PNO aré right angles,
because {the hypotenuse OP is common,
and one side PM = one side PN ;
... the triangles are equal in all respects; Theor. 18.
so that the Z. POM = the A PON.
Hence, if P lies within the Z. BOD, it must be on the bisector
of that angle ;
and, if P is within the Z.AOD, it must be on the bisector of
that angle.
It follows that the required locus is the pair of lines which bisect
the angles between AB and CD. r
LOCI. 93
INTERSECTION OF LOCI.
The method of Loci may be used to find the position of a
point which is subject to two conditions. For corresponding
to each condition there will be a locus on which the required
point must lie. Hence all points which are common to these
two loci, that is, all the points of intersection of the loci, will
satisfy both the given conditions.
ExAMPLE 1. To find a point equidistant from three given points
A, B, C, which are not in the same straight line.
(i) The locus of points equidistant from
A and B is the straight line PQ, which
bisects AB at right angles.
(ii) Similarly, the locus of points equi-
distant from B and C is the straight line
RS which bisects BC at right angles.
Hence the point common to PQ and RS
must satisfy both conditions: that is to
say, X the point of intersection of PQ and A | B
§§ will be equidistant from A, B, and C. Q
ExAMPLE 2. To construct a triangle, having given the base, the
altitude, and the length of the median which bisects the base.
Let AB be the given base, and P and
Q the lengths of the altitude and median
respectively. C E2 TNF D
Then the triangle is known if its verteac
is known.
(i) Draw a straight line CD parallel to
AB, and at a distance from it equal to P:
then the required vertea, must lie on CD. A O B
(ii) Again, from O the middle point of P
B as centre, with radius equal to Q, Q
describe a circle: * tº º
then the required vertea must lie on this circle.
Hence any points which are common to CD and the circle, satisfy
both the given conditions: that is to say, if CD intersect the circle in
E, F, each of the points of intersection might be the vertex of the
required triangle. This supposes the length of the median Q to be
greater than the altitude.
It may happen that the data of the problem are so related to one
another that the resulting loci do not intersect. In this case the
problem is impossible.

94 GEOMETRY.
Obs. In examples on the Intersection of Loci the student
should make a point of investigating the relations which must
exist among the data, in order that the problem may be
possible; and he must observe that if under certain relations
two solutions are possible, and under other relations no solu-
tion exists, there will always be some intermediate relation
under which the two solutions combine in a single solution.
EXAMPLES ON LOCI.
1. Find the locus of a point which moves so that its distance , ..."
(measured radially) from the circumference of a given circle is constant. 1" trº
º \\ *** * # / ,” *.
S 2. A point P moves along a straight line RQ ; find the position in
which it is equidistant from two given points A and B.
> 3. A and B are two fixed points within a circle : find points on the
circumference equidistant from A and B. How many such points are
there 2
4. A point P moves along a straight line RQ ; find the position in
which it is equidistant from two given straight lines AB and CD.
5. A and B are two fixed points 6 cm. apart. Find by the method,
of loci two points which are 4 cm. distant from A, and 5 cm. from B.
6. AB and CD are two given straight lines. Find points 3 cm.
distant from AB, and 4 cm. from CD. How many solutions are there?
7. A straight rod of given length slides between two straight
rulers placed at right angles to one another.
Plot the locus of its middle point ; and shew that this locus is the
fourth part of the circumference of a circle. [See Problem 10.]
8. On a given base as hypotenuse right-angled triangles are described.
Find the locus of their vertices.
S 9. A is a fixed point, and the point X moves on a fixed straight
line BC.
Plot the locus of P, the middle point of AX; and prove the locus to
be a straight line parallel to BC.
N 10. A is a fixed point, and the point X moves on the circumference
of a given circle.
Plot the locus of P, the middle point of AX; and prove that this
locus is a circle. [See Ex. 3, p. 64.]
EXAMPLES ON LOCI. 95
11. AB is a given straight line, and AX is the perpendicular drawn
from A to any straight line passing through B. If BX revolve about B,
find the locus of the middle point of AX.
12. Two straight lines OX, OY cut at right angles, and from P, a
point within the angle XOY, perpendiculars PM, PN are drawn to
OX, OY respectively. Plot the locus of P when
(i) PM + PN is constant (=6 cm., say):
(ii) PM – PN is constant (=3 cm., say).
And in each case give a theoretical proof of the result you arrive at
experimentally.
13. Two straight lines OX, OY intersect at right angles at O; and
from a movable point P perpendiculars PM, PN are drawn to OX, OY.
Plot (without proof) the locus of P, when
(i) PM =2 PN:
(ii) PM =3 PN.
14. Find a point which is at a given distance from a given point,
and is equidistant from two given parallel straight lines.
When does this problem admit of two solutions, when of one only,
and when is it impossible?
15. S is a fixed point 2 inches distant from a given straight line
MX. Find two points which are 2; inches distant from S, and also
2# inches distant from MX.
16. Find a series of points equidistant from a given point S and
a given straight line MX. Draw a curve freehand passing through all
the points se found.
17. On a given base construct a triangle of given altitude, having
its vertex on a given straight line.
18. Find a point equidistant from the three sides of a triangle.
19. Two straight lines OX, OY cut at right angles; and Q and R
are points in OX and OY respectively. Plot the locus of the middle
point of QR, when
(i) OQ --OR = constant.
(ii) OQ – OR= constant.
20. S and S' are two fixed points. Find a series of points P such
that
(i) SP+S'P=constant (say 3-5 inches).
(ii) SP – S'P= constant (say l'É inch).
In each case draw a curve freehand passing through all the points
So found.
96 GEOMETRY.
ON THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.
I. The perpendiculars drawn to the sides of a triangle from their
middle points are concurrent. -
Let ABC be a A, and X, Y, Z the middle
points of its sides.
From Z and Y draw perps. to AB, AC,
meeting at O. Join OX.
It is required to prove that OX is perp.
to BC.
Join OA, OB, OC. B X C
Proof. , Because YO bisects AC at right angles,
‘.. it is the locus of points tºwn from A and C ;
... OA = OC.
g
Again, because ZO bisects AB at right angles,
... it is the locus of points equidistant from A and B;
... OA = OB.
Hence OB = OC:
... O is on the locus of points equidistant from B and C ;
that is, OX is perp, to BC.
• Hence the perpendiculars from the mid-points of the sides meet at O.
Q. E. D.
II. The bisectors of the angles of a triangle are concurrent.
Let ABC be a A. Bisect the A." ABC,
BCA by straight lines which meet at O.
Join AO.
It is required to prove that AO bisects the
A. BAC.
From O draw OP, OQ, OR perp. to the
sides of the A. B P C
Proof. Because BO bisects the A. ABC,
... it is the locus of points gºint from BA and BC;
•. OP= OR
Similarly CO is the locus of points equidistant from BC and CA;
... OP=OQ.
Hence OR = OQ.
... O is on the locus of points equidistant from AB and AC:
that is, OA is the bisector of the A-BAC,
Hence the bisectors of the angles meet at O. Q.E.D.


THE CONCURRENCE OF LINES IN A TRIANGLE. 97
III. The medians of a triangle are concurrent.
Let ABC be a A.
Let BY and CZ be two of its medians, and let
them intersect at O.
Join AO,
and produce it to meet BC in X.
It is required to shew that AX is the remaining
median of the A.
Through C draw CK parallel to BY;
produce AX to meet CK at K.
Join BK.
Proof. In the A AKC,
oecause Y is the middle point of AC, and YO is parallel to CK,
... O is the middle point of AK. Theor. 22.
Again in the AABK,
since Z and O are the middle points of AB, AK,
... ZO is parallel to BK,
that is, OC is parallel to BK,
... the figure BKCO is a parin.
But the diagonals of a parin bisect one another;
... X is gº. point of BC.
That is; 'AX is a median of the A.
Hence the three medians meet at the point O. Q.E.D.
DEFINITION. The point of intersection of the medians is called the
centroid of the triangle.
** ** --~~ * ,
CoRollary. The three medians of a triangle cut one another at a
point of trisection, the greater segment in each being towards the angular
potºvt. \ ^
For in the above figure it has been proved that
A
also that OX is half of OK;
... OX is half of OA :
that is, OX is one third of AX.
Similarly OY is one third of SY,
and OZ is one third of CZ. Q.E.D.
By means of this Corollary it may be shewn that in any triangle
the shorter median bisects the greater side.
, , NoTE. It will be proved hereafter that the perpendiculars drawn
from the vertices of a triangle to the opposite sides are concurrent.
H. S. G. G

98 GEOMETRY.
MISCELLANEOUS PROBLEMS.
(A theoretical proof is to be given in each case.)
, 1. A is a given point, and BC a given straight line. From A draw
a straight line to make with BC an angle equal to a given angle X.
How many such lines can be drawn?
2. Draw the bisector of an angle AOB, without using the vertex O
in your construction.
, 3, . P is a given point within the angle AOB. Draw through P a
straight line terminated by OA and OB, and bisected at P.
2
2
Z 4, OA, OB, OC are three straight lines meeting at O. Draw a
transversal terminated by OA and OC, and bisected by OB.
5. Through a given point A draw a straight line so that the part
intercepted between two given parallels may be of given length.
When does this problem admit of two solutions? When of only one?
And when is it impossible?
*:A
--" 1 *
* ... tº . * e *
/ 6. In a triangle ABC inscribe a rhombus having one of its angles
coinciding with the angle A.
7. Use the properties of an equilateral triangle to trisect a given
straight line.
(Construction of Triangles.)
8. Construct a triangle, having given
(i) The middle points of the three sides.
(ii) The lengths of two sides and of the median which bisects the
third side.
(iii) The lengths of one side and the medians which bisect the other
two sides.
(iv) The lengths of the three medians.
AREAS. 99
PART II.
ON AREAS.
DEFINITIONS.
1. The altitude (or height) of a parallelogram with refer.’
ence to a given side as base, is the perpendicular distance
between the base and the opposite side.
2. The altitude (or height) of a triangle with reference to
a given side as base, is the perpendicular distance of the
opposite vertex from the base.
NOTE. It is clear that parallelograms or triangles which are between
the same parallels have the same altitude.
For let AP and DQ be the alti- G A D H
tudes of the A• ABC, DEF, which | |
are between the same parallels BF, . .
! |
Then the fig. APQD is evidently . !
a rectangle ; ! !
... AP = DQ. B P C Q E F

3. The area of a figure is the amount of surface contained
within its bounding lines.

4. A square inch is the area of a S
Square drawn on a side one inch in quare
length. inch
:
5. Similarly a square centimetre is the area of Sq.
a square drawn on a side one centimetre in length. CIII.
. The terms square yard, square foot, square metre are to be understood
in the same sense.
6. Thus the unit of area is the area of a square on a side
of unit length,
100 GEOMETRY.
THEOREM 23.
Area of a rectangle. If the number of units in the length of a
rectangle is multiplied by the number of units in its breadth, the
product gives the number of square units in the area.
D * C

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Let ABCD represent a rectangle whose length AB is 5 feet,
and whose breadth AD is 4 feet.
Divide AB into 5 equal parts, and BC into 4 equal parts, and
through the points of division of each line draw parallels to
the other.
The rectangle ABCD is now divided into compartments
cach of which represents one square foot.
Now there are 4 rows, each containing 5 squares,
... the rectangle contains 5 × 4 square feet.
Similarly, if the length = a linear units, and the breadth =b
linear units
the rectangle contains ab units of area.
And if each side of a square = a linear units,
the Square contains a” units of area. }
These statements may be thus abridged: . . . . .
the area of a rectangle = length x breadth ........... (i),
the area of a square = (side)* ................. ... (ii).
Q.E.D
COROLLARIES, (i) Rectangles which have equal lengths and
equal breadths have equal areas.
(ii) Itectangles which have equal areas and equal lengths have
also equal breadths.
AREA OF A TRIANGLE. 10]
*_* *
tº * : * * * *.
NOTATION.
The rectangle ABCD is said to be contained by AB, AD; for
these adjacent sides fix its size and shape.
A rectangle whose adjacent sides are AB, AD is denoted by
rect. AB, AD, or simply AB x AD.
A square drawn on the side AB is denoted by sq. on AB, or AB”.
EXERCISES.
(On Tables of Length and Area.)
1. Draw a figure to shew why
(i) 1 sq. yard=3° sq.feet.
(ii) I sq. foot = 12” sq. inches.
(iii) l sq., cm. = 10° sq. mm.
2. Draw a figure to shew that the square on a straight line is four
times the square on half the line.
3. Use squared paper to shew that the square on 1"= 10° times the
square on 0.1".
4. If 1" represents 5 miles, what does an area of 6 square inches
represent 2
EXTENSION OF THEOREM 23.
The proof of Theorem 23 here given supposes that the length and
breadth of the given rectangle are expressed by whole numbers ; but the
formula holds good when the length and breadth are fractional.
This may be illustrated thus:
Suppose the length and breadth are 32 cm. and 2.4 cm.; we shall
shew that the area is (3.2 x 2-4) sq. cm.
For length = 3-2 cm. = 32 mm.
breadth=2'4 cm. =24 mm.
32 × 24
... area = (32 × 24) sq. mm. = Sq. cm.
10?
=(3.2 x2'4) sq. cm.
102 GEOMETRY.
EXERCISES.
(On the Area of a Rectangle.)
Draw on squared paper the rectangles of which the length (a) and
breadth (b) are given below. Calculate the areas, and verify by the
actual counting of squares.
1. a =2", b = 3". 2. a = 1°5", b = 4".
3. a = 0-8", b = 3.5". 4. a =2'5", b = 1'4".
5. a =22", b=1.5". 6. a = l'6", b=2.1".
Calculate the areas of the rectangles in which
7. a = 18 metres, b = 11 metres. 8. a = 7 ft., b=72 in.
9. a = 2.5 km., b = 4 metres. 10. a =# mile, b = 1 inch.
11. The area of a rectangle is 30 sq. cm., and its length is 6 cm.
Find the breadth. Draw the rectangle on squared paper; and verify
your work by counting the squares.
12. Find the length of a rectangle whose area is 3.9 sq. in., and
breadth l'5". Draw the rectangle on squared paper; and verify your
work by counting the squares.
13. (i) When you treble the length of a rectangle without altering
its breadth, how many times do you multiply the area 7
(ii) When you treble both length and breadth, how many times do
you multiply the area 7
Draw a figure to illustrate your answers; and state a general rule.
14. In a plan of a rectangular garden the length and breadth
are 3-6" and 2'5", one inch standing for 10 yards. Find the area of the
garden.
If the area is increased by 300 sq. yds., the breadth remaining the
same, what will the new length be? And how many inches will repre-
sent it on your plan 2
15. Find the area of a rectangular enclosure of which a plan
(scale 1 cm. to 20 metres) measures 6-5 cm. by 4-5 cm.
16. The area of a rectangle is 1440 sq. yds. If in a plan the sides
of the rectangle are 32 cm. and 4.5 cm., on what scale is the plan
drawn 2
17. The area of a rectangular field is 52000 sq. ft. On a plan of
this, drawn to the scale of 1" to 100 ft., the length is 3'25". hat is
the breadth ?
EXERCISES ON RECTANGLES.
103
Calculate the areas of the enclosures of which plans are given below.
All the angles are right angles, and the dimensions are marked in feet.
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104 GEOMETRY.
THEOREM 24. [Euclid I. 35.]
Parallelograms on the same base and between the same parallels
are equal in area.
A E Dr F

B C
Let the par” ABCD, EBCF be on the same base BC, and
between the same par" BC, AF.
It is required to prove that
the par” ABCD = the par” EBCF in area.
Proof. In the A* FDC, EAB,
DC = the opp. side AB; Theor. 21.
because {the ext. Z. FDC = the int. opp. / EAB; Theor. 14.
Uthe int. A DFC = the ext. Z.AEB;
... the AFDC = the AEAB. Theor. 17.
Now, if from the whole fig. ABCF the AFDC is taken, the
remainder is the par" ABCD.
And if from the whole fig. ABCF the A EAB is taken, the
remainder is the parº EBCF.
... these remainders are equal;
that is, the parº ABCD = the parº EBCF. Q.E.D.
EXERCISE.
In the above diagram the sides AD, EF overlap. Draw diagrams in
which (i) these sides do not overlap; (ii) the ends E and D coincide.
Go through the proof with these diagrams, and ascertain if it applies
to them without change. A
AREAS. 105
THE AREA OF A PARALLELOGRAM.
Let ABCD be a parallelogram, F D E C
and ABEF the rectangle on the
same base AB and of the same
altitude BE. Then by Theorem 24,

area of par” ABCD = area of rect. ABEF A. B
= AB X BE
=base x altitude.
COROLLARY. Since the area of a parallelogram depends
only on its base and altitude, it follows that
Parallelograms on equal bases and of equal altitudes are equal
270 Qſì'60.
EXERCISES.
(Numerical and Graphical.)
1. Find the area of parallelograms in which
(i) the base=5-5 cm., and the height=4 cm.
(ii) the base=2'4", and the height=1.5".
2. Draw a parallelogram ABCD having given AB=2}", AD=1#",
and the 4-A = 65°. Draw and measure the perpendicular from D on AB,
and hence calculate the approximate area. Why approacimate 2
Again calculate the area from the length of AD and the perpendicular
on it from B. Obtain the average of the two results.
3. Two adjacent sides of a parallelogram are 30 metres and 25 metres,
and the included angle is 50°. Draw a plan, 1 cm. representing
5 metres; and by measuring each altitude, make two independent
aalculations of the area. Give the average result.
4. The area of a parallelogram ABCD is 4.2 sq. in., and the base
AB is 2-8". Find the height. If AD=2", draw the parallelogram.
5. Each side of a rhombus is 2", and its area is 3-86 sq. in. , Calculate
an altitude. Hence draw the rhombus, and measure one of its acute
angles.
106 GEOMETRY.
THEOREM 25.
The Area of a Triangle. The area of a triangle is half the area
of the rectangle on the same base and having the same altitude.
D E D
A
E


sºme
*
*
*-*.
wº
*
B
C
B
C
Fig. 1. Fig. 2.
Let ABC be a triangle, and BDEC a rectangle on the same
base BC and with the same altitude AF.
It is required to prove that the AABC is half the rectangle BDEC.
Proof. Since AF is perp. to BC, each of the figures DF, EF
is a rectangle.
Because the diagonal AB bisects the rectangle DF,
... the AABF is half the rectangle DF.
Similarly, the AAFC is half the rectangle FE.
".. adding these results in Fig. 1, and taking the difference in
Fig. 2, *
the AABC is half the rectangle BDEC.
Q.E.D.
COROLLARY. A triangle is half any parallelogram on the same
base and between the same parallels.

For the AABC is half the rect. BCED. & º A E
And the rect. BCED=any par" BCHG
on the same base and between the same
par”.
... the AABC is half the par" BCHG.
AREAS. 107
THE AREA OF A TRIANGLE,
If BC and AF respectively contain a units and p units of
length, the rectangle BDEC contains ap units of area.
... the area of the AABC =#ap units of area.
This result may be stated thus:
Area of a Triangle = }. base x altitude.
EXERCISES ON THE AREA OF A TRIANGLE.
(Numerical and Graphical.)
1. Calculate the areas of the triangles in which
(i) the base =24 ft., the height=15 ft.
(ii) the base=4'8", the height=3:5".
(iii) the base = 160 metres, the height=125 metres.
2. Draw triangles from the following data. In each case draw and
measure the altitude with reference to a given side as base: hence cal-
culate the approximate area.
(i) a = 8.4 cm., b = 6.8 cm., c=4.0 cm.
(ii) b = 5.0 cm., c = 6.8 cm., A=65°.
(iii) a =6.5 cm., B=52°, C=76°.
3. ABC is a triangle right-angled at C ; shew that its area =#BC x CA.
Given a = 6 cm., b=5 cm., calculate the area.
Draw the triangle and measure the hypotenuse c ; draw and measure
the perpendicular from C on the hypotenuse; hence calculate the
-- approximate area, -> <-
Note the érror in your approximate result, and express it as a pers
centage of the true value.
4. Repeat the whole process of the last question for a right-angled
triangle ABC, in which a =2'8" and b =4'5"; C being the right angle
as before.
5. In a triangle, given
(i) Area = 80 sq. in., base = 1 ft. 8 in, ; calculate the altitude.
(ii) Area = 10.4 sq. cm., altitude=16 cm.; calculate the base.
6. Construct a triangle ABC, º; given a = 3.0", b=2-8", c=2'6".
, Draw and measure the perpendicular from A on BC; hence calculate
the approximate area.
108 GEOMETRY.
THEOREM 26. [Euclid I. 37.]
Triangles on the same base and between the same parallels
(hence, of the same altitude) are equal in
(ºff'00. D A E G
Let the A' ABC, GBC be on the
same base BC and between the same
par" BC, AG.
It is required to prove that
the AABC = the A GBC in area. B C
Proof. If BCED is the rectangle on the base BC, and
between the same parallels as the given triangles,
the AABC is half the rect. BCED; Theor. 25.
also the A GBC is half the rect. BCED;
... the AABC = the A GBC. Q.E.D.
Similarly, triangles on equal bases and of equal altitudes are
equal in area.
*ś
* , ,
*** * ºf *.
THEOREM 27. [Euclid I. 39.]
wº
If two triangles are equal in area, and stand on the same base
and on the same side of it, they are between the same parallels.
Let the A* ABC, GBC, standing on . A G
the same base BC, be equal in area ; :
and let AF and GH be their altitudes. !
It is required to prove that AG and i :
BC are par’. B F C H
Proof. The AABC is half the rectangle contained by BC
and AF :
and the AGBC is half the rectangle contained by BC
and GH;
the rect. BC, AF = the rect. BC, GH;
.*. AF = GH. Theor. 23, Cor. 2.
Also AF and GH are par';
hence AG and FH, that is BC, are par'. Q.E.D.

AREAS. 109
EXERCISES ON THE AREA OF A TRIANGLE.
(Theoretical.)
1. ABC is a triangle and XY is drawn parallel to the base BC,
cutting the other sides at X and Y. Join BY and CX; and shew that
(i) the AXBC = the A YBC;
(ii) the A BXY = the A CXY;
(iii) the A ABY = the A ACX.
If BY and CX cut at K, shew that
(iv) the A BKX=the A CKY.
2. Shew that a median of a triangle divides it into two parts of
equal area.
How would you divide a triangle into three equal parts by straight
lines drawn from its vertex?
3. Prove that a parallelogram is divided by its diagonals into four
triangles of equal area.
4. ABC is a triangle whose base BC is bisected at X. If Y is any
point in the median AX, shew that
*3. the A ABY = the AACY in area.
5. ABCD is a parallelogram, and BP, DQ are thiºperpendiculars
from B and D on the diagonal AC. . A
Shew that BP= DQ. * > *-*.*.*.*-
Hence if X is any point in AC, or AC produced,
prove (i) the A ADX = the AABX;
| (ii) the A CDX = the A CBX.
6. Prove by means of Theorems 26 and 27 that the straight line
joining the middle points of two sides of a triangle is parallel to the third
side.
7. The straight line which joins the middle points of the oblique
sides of a trapezium is parallel to each of the parallel sides.
8. ABCD is a parallelogram, and X, Y are the middle points of the
sides AD, BC; if Z is any point in XY, or XY produced, shew that the
triangle AZB is one quarter of the parallelogram ABCD.
9. If ABCD is a parallelogram, and X, Y any points in DC and AD
respectively: shew that the triangles AXB, BYC are equal in area.
10. ABCD is a parallelogram, and P is any point within it ; shew
that the sum of the triangles PAB, PCD is equal to half the parallelo-
gram.
110 GEOMETRY.
*
EXERCISES ON THE AREA OF A TRIANGLE.
(Numerical and Graphical.)
X 1. The sides of a triangular field are 370 yds., 200 yds., and 190
yds. Draw a plan (scale 1" to 100 yards). Draw and measure an
altitude ; hence calculate the approximate area of the field in Square
yards.
Y.”
X 2. Two sides of a triangular erſäläsäre are 124 metres and 144
metres respectively, and the included angle is observed to be 45°.
Draw a plan (scale 1 cm. to 20 metres). Make any necessary measure-
ment, and calculate the approximate area.
3. In a triangle ABC, given that the area=6-6 sq. cm., and the base
BC=5-5 cm., find the altitude. Hence determine the locus of the
vertex A.
If in addition to the above data, BA=2.6 cm., construct the tri-
angle ; and measure CA.
4. In a triangle ABC, given area=3:06 sq. in., and a =3-0". Find
the altitude, and the locus of A. Given C=68°, construct the triangle;
and measure b.
* 5. ABC is a triangle in which BC, BA have constant lengths 6 cm.
and 5 cm. If BC is fixed, and BA revolves about B, trace the changes
in the area of the triangle as the angle B increases from 0° to 180°. --
Answer this question by drawing a series of triangles, increasińg
B by increments of 30°. Find the area in each case and tabulate the
results. Y &
(Theoretical.)
> 6. If two triangles have two sides of one respectively equal to two
sides of the other, and the angles contained by those sides supple-
mentary, shew that the triangles are equal in area. Can such triangles
ever be identically equal?
$ 7. Shew how to draw on the base of a given triangle an isosceles
triangle of equal area.
> 8. If the middle points of the sides of a quadrilateral are joined in
order, prove that the parallelogram so formed [see Ex. 7, p. 64], is half
the quadrilateral.
9. ABC is a triangle, and R, Q the middle points of the sides
AB, AC ; shew that if BQ and CR intersect in X, the triangle BXC
is equal to the quadrilateral AQXR.
y 10. Two triangles of equal area stand on the same base but on
opposite sides of it: shew that the straight line joining their vertices a cº-
is bisected by the base, or by the base produced.
THE AREA OF A TRIANGLE. 111
s
e. ". º J
[The method given below may be omitted from a first course. In
any case it must be postponed till Theorem 29 has been read.]
The Area of a Triangle. Given the three sides of a triangle,
to calculate the area.
ExAMPLE. Find the area of a triangle whose sides measure 21 m.,
17 m., and 10 m.
Let ABC represent the given
triangle.
Draw AD perp. to BC, and
denote AD by p.
We shall first find the length 1 O A 17
of BD.
Let BD = ac metres; then DC
=21 – a metres. B 32
From the right-angled AADB,
we have by Theorem 29

AD2=AB2 – BD2–102 — a 2.
And from the right-angled AADC,
AD?=AC” – DC”=17? – (21–2)”;
... 102 – a 2–172-(21–2)?
D 21 - 3: C
Of 100 — a 4–289 – 441 + 422 – 2:”;
whence ac - 6.
Again, AD?= AB2 – BD”;
OT p?s= 10%–6% =64;
... p = 8.
Now Area of triangle = }. base x altitude
= (# x 21 x 8) sq. m. =84 sq. m.
EXERCISES.
Find by the above method the area of the triangles, whose sides
are as follows:
1. 20 ft., 13 ft., 11 ft. 2. 15 yds., 14 yds., 13 yds.
3. 21 m., 20 m., 13 m. 4. 30 cm., 25 cm., 11 cm.
5. 37 ft., 30 ft., 13 ft. 6. 51 m., 37 m., 20 m.
7. If the given sides are a, b and c units in length, prove
* a2+ c2 – bº gº º 2 + c2 — b2Y 2
(i) 2-tº-º: (ii) *-a-(*.* ;
(iii) A=4N(a+b+c)(-a-F5-Fc)(a-b-Fic)(a+b+c).
112 GEOMETRY.
THE AREA OF QUADRILATERALS.
THEOREM 28,
To find the area of (i) a trapezium.
(ii) any quadrilateral.
(i) Let ABCD be a trapezium, having
the sides AB, CD parallel. Join BD, D C
and from C and D draw perpendiculars |
CF, DE to AB. !
Let the parallel sides AB, CD measure l
a and b units of length, and let the A E F B
height CF contain h units.
Then the area of ABCD = A ABD + A DBC
I l
=3AB. DE +jpc. CF
I
I h
That is,
the area of a trapezium =} height x (the Sum of the parallel sides).
(ii) Let ABCD be any quadrilateral.
Draw a diagonal AC ; and from B
and D draw perpendiculars BX, DY, to
* AC. *
If AC contains d units of length, and #
BX, DY p and q units respectively,
the area of the quad'ABCD = A ABC + A ADC
l 1
=5Ac. Bx+3AC. Dy
1 I I º
=;lp + dº = d(p + q).
That is to say, 2 2 2
the area of a quadrilateral= ; diagonal x (sum of offsets).

EXERCISES ON QUADRILATERALS. 113
EXERCISES.
(Numerical and Graphical.)
1. Find the area of the trapezium in which the two parallel sides
are 4.7" and 3'3", and the height l'5".
2. In a quadrilateral ABCD, the diagonal AC= 17 feet; and the
offsets from it to B and D are ll feet and 9 feet. Find the area.
3. In a plan ABCD of a quadrilateral enclosure, the diagonal AC
measures 8-2 cm., and the offsets from it to B and D are 3-4 cm. and
2.6 cm. respectively. If 1 cm. in the plan represents 5 metres, find
the area of the enclosure.
4. Draw a quadrilateral ABCD from the ad-
joining rough plan, the dimensions being given in
inches.
Draw and measure the offsets to A and C from
the diagonal BD ; and hence calculate the area of
the quadrilateral.
§ 5. Draw a quadrilateral ABCD from the
details given in the adjoining plan. The
2 y - dimensions are to be in centimetres.
Make any necessary measurements of your
figure, and calculate its area.
A 7.7 B
6. Draw a trapezium ABCD from the following data ; AB and CD
are the parallel sides. AB = 4".; AD = BC =2"; the Z.A= the A. B=60°.
Make any necessary measurements, and calculate the area.
7. Draw a trapezium ABCD in which AB and CD are the parallel
sides; and AB =9 cm., CD = 3 cm., and AD = BC = 5 cm.
Make any necessary measurement, and calculate the area.
\,
8. From the formula area of quad" =# diag. x (sum of offsets) shew
that, if the diagonals are at right angles,
area =# (product of diagonals).
9. Given the lengths of the diagonals of a quadrilateral, and the
angle between them, prove that the area is the same wherever they
intersect.
H. S. G. H


114 GEOMETRY.
THE AREA OF ANY RECTILlNEAL FIGURE.
1“METHOD. A rectilineal figure
may be divided into triangles whose
areas can be separately calculated
from suitable measurements. The
sum of these areas will be the area
of the given figure.
Eacample. The measurements re-
Äää to find the area of the figure
BCDE are AC, AD, and the offsets BX,
DY, EZ.
2"|METHOD. The area of a rectilineal figure is also found
by taking a base-line (AD in the diagram below) and offsets
from it. These divide the figure into right-angled triangles
and right-angled trapeziums, whose areas may be found after
measuring the offsets and the various sections of the base-line.
Eacample. Find the area of the enclosure ABCDEF from the plan
and measurements tabulated below.
YARDs,
AD = 56
VC = 12 || AV = 50
AZ = 40 || ZE = 18
YB = 20 AY = 18
AX = 10 | XF = 15
The measurements are made from
A along the base line to the points
from which the offsets spring.
Here A AXF =#. AX x XF
A AYB =#. AYx YB
A DZE=#. DZx ZE
A DVC= }. DVx VC
..., by addition, the fig. ABCDEF=
Sq, yas. Sq. yds.
=# x 10 x 15= 75
=# x 18 x 20= 180
=} x 16 x 18 = 144
=} x 6 x 12= 36
trap" XFEZ=}. XZx (XF-ZE)=#x30 x 33= 495
trap" YBCV =#. YV × (YB+ VC)=#x32X 32= 512
1442 sq. yds.


EXERCISES ON RECTILINEAL FIGURES. 115
EXERCISES.
1. Calculate the areas of the figures (i) and (ii) from the plans and
dimensions (in cms.) given below.
D
* D ii §
(i) E (ii) !
|
C ! C
:
!
A B A X B
AC=6 cm., AD = 5 cm. AB = BD = DA = 6 cm.
Lengths of offsets figured EY = CZ = 1 cm.
in diagram. DX=52 cm.
2. Draw full size the figures whose plans and dimensions are given
below ; and calculate the area in each case.
D
., E tº e
(i) D (ii)
|
|
12.5"
|
l
l
|
A X Y. B
The fig. is equilateral : AX=1}", XY=2#",
each side to be 2%". YB=1}".
3. Find the area of the figure ABCDEF from the following measure-
ments and draw a plan in which 1 cm. represents 20 metres.
THE PLAN.
METRES. C
to C
180 D B
80 to D 150
40 to E 120 50 to B
60 to F 50 F
From A






116 GEOMETRY.
EXERCISES ON QUADRILATERALS,
* (Theoretical.)
\
1. ABCD is a rectangle, and PQRS the figure formed by joining in
order the middle points of the sides.
Prove (i) that PQRS is a rhombus;
(ii) that the area of PQRS is half that of ABCD.
Hence shew that the area of a rhombus is half the product of its
diagonals.
Is this true of any quadrilateral whose diagonals cut at right angles 2
Illustrate your answer by a diagram.
2. Prove that a parallelogram is bisected by any straight line which
passes through the middle point of one of its diagonals.
Hence shew how a parallelogram ABCD may be bisected by a
straight line drawn
(i) through a given point P;
(ii) perpendicular to the side AB;
(iii) parallel to a given line Q.R.
\ 3. In the trapezium ABCD, AB is parallel to DC; and X is the
middle point of BC. Through X draw PQ parallel to AD to meet AB
and DC produced at P and Q. Then prove
(i) trapezium ABCD=parm APQD.
(ii) trapezium ABCD=twice the AAXD.
(Graphical.)
\ 4. The diagonals of a quadrilateral ABCD cut at right angles, and
measure 3.0" and 2'2" respectively. Find the area.
Shew by a figure that the area is the same wherever the diagonals
cut, so long as they are at right angles.
\ 5. In the parallelogram ABCD, AB = 8-0 cm., AD=32 cm., and the
perpendicular distance between AB and DC-3.0 cm. Draw the par-
allelogram. Calculate the distance between AD and BC; and check
your result by measurement.
6. One side of a parallelogram is 2'5", and its diagonals are 3'4"
and 2'4". Construct the parallclogram ; and, after making any neces-
sary measurement, calculate the area.
7. ABCD is a parallelogram on a fixed base AB and of constant
area. Find the locus of the intersection of its diagonals.
EXPERIMIENTAL, EXERCISES. 117
EXERCISES LEADING TO THEOREM 29.
In the adjoining diagram, ABC is a triangle A
right-angled at C ; and squares are drawn on the
three sides. Let us compare the area of the square T
on the hypotenuse AB with the sum of the squares C B
on the sides AC, CB which contain the right angle.
1. Draw the above diagram, making Ac=3 cm., and BC = 4 cm.;
Then the area of the square on AC=3°, or 9 sq. cm.
and ............... the square on BC = 4”, or 16 sq. cm.
‘.. the sum of the squares on AC, BC = 25 sq. cm.
Now measure AB; hence calculate the area of the square on AB, and
compare the result with the sum already obtained.
2. Repeat the process of the last exercise, making AC= 1.0", and
=2'4”.
3. If a = 15, b=8, c=17, shew arithmetically that cº-a2+b^.
Now draw on squared paper a triangle ABC, whose sides a, b, and c
are 15, 8, and 17 units of length ; and measure the angle ACB.
4. Take any triangle ABC, right-
angled at C ; and draw squares on AC,
CB, and on the hypotenuse AB.
Through the mid-point of the square A
on CB (i.e. the intersection of the dia-
gonals) draw lines parallel and perpen- 5
dicular to the hypotenuse, thus dividing
the square into four congruent quadri- C TIB
laterals. These, together with thesquare & i /
on AC, will be found exactly to fit into > < , 4.
the square on AB, in the way indicated o ,' 'S
by corresponding numbers. - , *
These experiñéâts point to the conclusion that:
In any right-angled triangle the square on the hypotenuse is equal
to the sum of the squares on the other two sides.
A formal proof of this theorem is given on the next page.

118 GEOMETRY.
THEOREM 29. [Euclid I, 47.]
53 25- ) 2
In a right-angled triangle the square described on the hypotenuse
is equal to the Sum of the squares described on the other two sides.
G
F H
C
\ -->K
g” Nº.
A B
D L E

Let ABC be a right-angled A, having the angle ACB a rt. 4.
It is required to prove that the Square on the hypotenuse AB = the
Sum of the squares on AC, CB.
On AB describe the sq. ADEB; and on AC, CB describe the
sqq. ACGF, CBKH.
Through C draw CL par' to AD or BE.
Join CD, FB.
Proof. Because each of the Z." ACB, ACG is a rt. 4-,
‘. BC and CG are in the same St. line.
Now the rt. A BAD = the rt. A FAC ;
add to each the A. CAB :
then the whole / CAD = the whole / FAB.
Then in the A* CAD, FAB,
CA = FA,
because AD = AB,
and the included 4 CAD = the included 4- FAB;
... the A CAD = the A FAB Theor. 4.
THEOREM OF PYTHAGORAS. 119
Now the rect. AL is double of the A CAD, being on the
same base AD, and between the same par” AD, CL.
And the sq. GA is double of the A FAB, being on the same
base FA, and between the same par” FA, GB.
... the rect. AL= the sq. GA.
Similarly by joining CE, AK, it can be shewn that
the rect. BL = the sq. HB.
... the whole sq. AE = the sum of the sqq. GA, HB :
that is, the square on the hypotenuse AB = the sum of the
squares on the two sides AC, CB.
Q.E.D.
Obs. This is known as the Theorem of Pythagoras. The
result established may be stated as follows:
AB2 = BC?--CA2.
That is, if a. and b denote the lengths of the sides containing
the right angle; and if c denotes the hypotenuse,
c2 = a” + l2.
}
Hence a?= c” – b”; and b2 =cº — a”.
NoTE 1. The following important results should be noticed.
If CL and AB intersect in O, it has been shewn in the course of the
proof that
the sq. GA= the rect. AL;
that is, AC”= the rect. contained by AB, AO. ............ (i)
Also the sq. HB = the rect. BL ;
that is, BC*=the rect. contained by BA, BO. ............ (ii)
NoTE 2. It can be proved by superposition that squares standing on
equal sides are equal in area.
Hence we conclude, conversely,
If two squares are equal in area they stand on equal 3ides.
120 , GEOMETRY.
EXPERIMENTAL PROOFS OF PYTHAGORAS'S THEOREM,
I. Here ABC is the given
rt.-angled A ; and ABED is
§§ square on the hypotenuse
B.
By drawing lines par! to the
sides BC, CA, it is easily seen
that the sq. BD is divided
into 4 rt.-angled A*, each
identically equal to ABC, tos
gether with a central square.
Hence
sq. on hypotenuse c-4 rt. 4°A"
+ the central square
=4. #ab + (a – b)*
=2ab-i-a” – 2db + b%
= u2+b^.
II. Here ABC is the given
rt.-angled A, and the figs.
CF, HK are the sqq. on CB,
CA placed side by side.
FE is made equal to DH
or CA; and the two sqq.
CF, HK are cut along the
lines BE, ED.
Then it will be found that
the ADHE may be placed
so as to fill up the space ACB ;
and the A BFE may be made
to fill the space AKD.
Hence the two sqq. CF,
HK may be fitted together
so as to form the single fig.
ABED, which will be found
to be a perfect square, namely
the square on the hypotenuse


THEOREM OF PYTHAGORAS. 121
EXERCISES.
(Numerical and Graphical.)
1. Draw a triangle ABC, right-angled at C, having given :
(i) a = 3 cm., b = 4 cm.;
S (ii) a = 2.5 cm., b = 60 cm.;
-> (iii) a = 1-2", b = 3'5".
In each case calculate the length of the hypotenuse c, and verify a
your result by measurement.
2. Draw a triangle ABC, right-angled at C, having given :
(i) c-3'4", a = 3-0"; [See Problem 10]
(ii) c-5-3 cm., b=4'5 cm.
In each case calculate the remaining side, and verify your result by
measurement.
(The following examples are to be solved by calculation ; but in each
case a plan should be drawn on Some suitable scale, and the calculated
result verifted by measurement.) * * * *
3. A ladder whose foot is 9 feet from the front of a house reaches
to a window-sill 40 feet above the ground. What is the length of the
ladder ?
4. A ship sails, 33 miles due South, and then 56 miles due West.
How far is it then from its starting point %
5. Two ships are observed from a signal station to bear respectively
N.E. 6-0 km. distant, and N.W. 11 km. distant. How far are they
apart 2
6. A ladder 65 feet long reaches to a point in the face of a house
63 feet above the ground. How far is the foot from the house?
7. B is due East of A, but at an unknown distance. C is due South
of B, and distant 55 metres. AC is known to be 73 metres. Find AB.
8. A man travels 27 miles due South ; then 24 miles due West;
finally 20 miles due North. How far is he from his starting point?
9. From A go West 25 metres, then North 60 metres, then East
80 metres, finally South 12 metres. How far are you then from A2
10. A ladder 50 feet long is placed so as to reach a window 48 feet
high ; and on turning the ladder over to the other side of the street, it
reaches a point 14 feet high. Find the breadth of the street.
122 GEOMETRY.
THEOREM 30. [Euclid I, 48.]
If the square described on one side of a triangle is equal to the
sum of the squares described on the other two sides, then the angle
contained by these two sides is a right angle.
A D


|B C E F
Let ABC be a triangle in which
the sq. on AB = the sum of the sqq. on, BC, CA.
It is required to prove that ACB is a right angle.
Make EF equal to BC.
Draw FD perp to EF, and make FD equal to CA.
Join ED.
Proof. Because EF = BC,
... the sq. on EF = the sq. on BC.
And because FD = CA,
... the sq. on FD = the sq. on CA. *
Hence the sum of the sqq. on EF, FD = the sum of the sqq.
on BC, CA.
But since EFD is a rt. Z-,
... the sum of the sqq. on EF, FD = the sq. on DE: Theor. 29.
And, by hypothesis, the sqq. on BC, CA = the sq. on AB,
... the sq. on DE = the sq. on AB.
‘. DE = AB.
Then in the A* ACB, DFE,
AC = DF,
because | CB = FE,
and AB = DE ;
... the Z ACB = the A.DFE. Theor. 7.
But, by construction, DFE is a right angle;
... the 4 ACB is a right angle.
Q.E.D.
THEOREM OF PYTHAGORAS AND ITS CONVERSE. 123
EXERCISES ON THEOREMS 29, 30.
(Theoretical.)
1. Shew that the square on the diagonal of a given square is double
of the given square.
$ 2. In the AABC, AD is drawn perpendicular to the base BC. If
the side c is greater than b,
shew that c2 – b%- BD2 – DC”.
3. If from any point O within a triangle ABC, perpendiculars OX,
OY, OZ are drawn to BC, CA, AB respectively: shew that
AZ24-BX2 + CY2=AY2+ CX2 + BZ2.
4. ABC is a triangle right-angled at A.; and the sides AB, AC are
wintersected by a straight line PQ, and BQ, PC are joined. Prove that
-* A *- BQ2+ PC2=BC2+ PQ2.
5. In a right-angled triangle four times the sum of the squares on
the medians drawn from the acute angles is equal to five times the
square on the hypotenuse. *
6. Describe a square equal to the sum of two given squares.
> 7. Describe a square equal to the difference between two given
squares.
8. Divide a straight line into two parts so that the square on one
part may be twice the square on the other.
9. Divide a straight line into two parts such that the sum of their
squares shall be equal to a given square.
(Numerical and Graphical.)
10. Determine which of the following triangles are right-angled :
(i) a = 14 cm., b = 48 cm., c = 50 cm.; " *.
(ii) a = 40 cm., b = 10 cm., c= 41 cm.; X
(iii) a = 20 cm., b = 99 cm., c=101 cm. vº
11. ABC is an isosceles triangle right-angled at C ; deduce from
Theorem 29 that
AB2 =2AC2.
Illustrate this result graphically by drawing both diagonals of the
square on AB, and one diagonal of the square on AC.
If AC = BC =2", find AB to the nearest hundredth of an inch, and
verify your calculation by actual construction and measurement.
* }
12. Draw a square on a diagonal of 6 cm. Calculate, and also
measure, the length of a side. Find the area.
124 GEOMETRY.
PROBLEM 16.
To draw squares whose areas shall be respectively twice, three-times,
four-times, ..., that of a given Square. *
Hence find graphically approximate values of V2, V3, V4, V5, ...,
Y

Take OX, OY at right angles P
to one another, and from them
mark off OA, OP, each one
unit of length. Join PA.
O A B C X
Then PA* = OP2 + OA2 = 1 + 1 = 2.
... PA = x/2.
From OX mark off OB equal to PA, and join PB;
then PB2 = OP2 + OB% = 1 + 2 = 3.
... PB = V3.
From OX mark off OC equal to PB, and join PC;
then PC” – OP24 OCº-1 +3= 4.
... PC = W4.
The lengths of PA, PB, PC may now be found by measurement;
and by continuing the process we may find V5, V6, V7, ... .
ExERCISES on THEOREMS 29, 30 (Continued).
13. Prove the following formula: sº
Diagonal of square=side x N2.
Hence find to the nearest centimetre the diagonal of a square on a
side of 50 metres.
Draw a plan (scale 1 cm. to 10 metres) and obtain the result as
nearly as you can by measurement.
14. ABC is an equilateral triangle of which each side=2m units,
and the perpendicular from any vertex to the opposite side=p.
Prove that p = mx/3.
Test this result graphically, when each side = 8 cm.
THEOREM OF PYTHAGORAS. 125
15. If in a triangle a=m”— n”, b = 2mm, c=m^+n”; prove algebraic-
ally that cº–a4+b”. w
Hence by giving various numerical values to m and m, find sets of
numbers representing the sides of right-angled triangles.
16. In a triangle ABC, AD is drawn perpendicular to BC. Let p
denote the length of AD.
(i) If a = 25 cm., p=12 cm., BD =9 cm.; find b and c.
(ii) If b =41", c=50", BD=30"; find p and a.
And prove that Nbº-pº-i-Vcº-º-a.
17. In the triangle ABC, AD is drawn perpendicular to BC.
Prove that
c2 – BD2–b? — CD2.
If a = 51 cm., b=20 cm., c = 37 cm.; find BD.
Thence find p, the length of AD, and the area of the triangle ABC.
18. Find by the method of the last example the areas of the triangles
whose sides are as follows:
(i) a = 17”, b = 10", c=9". (ii) a = 25 ft., b = 17 ft., c = 12 ft.
(iii) a =41 cm., b=28 cm., c= 15 cm. (iv) a = 40 yd., b=37 yd., c=13 yd.
* 19. A straight rod PQ slides between two straight rulers OX, OY
laced at right angles to one another. In one position of the rod
§º. cm., and OQ=3-3 cm. If in another position OP=4.0 cm.,
find OQ graphically; and test the accuracy of your drawing by
calculation.
20. ABC is a triangle right-angled at C, and p is the length of the
perpendicular from C on AB. By expressing the area of the triangle in
two ways, shew that
pc=ab.
1 1 ||
Hence deduce 2 =a. + 52'
126 GEOMETRY.
PROBLEMS ON AREAS.
PROBLEM 17.
To describe a parallelogram equal to a given triangle, and having
one of its angles equal to a given angle.
A F/ G
Let ABC be the given triangle, and D the given angle. '
It is required to describe a parallelogram equal to ABC, and
having one of its angles equal to D.
Construction. Bisect BC at E.
At E in CE, make the 4 CEF equal to D ;
through A draw AFG par to BC;
and through C draw CG par" to EF.
Then FECG is the required par”.
Proof. Join AE.
Now the A* ABE, AEC are on equal bases BE, EC, and of
the same altitude;
... the AABE = the AAEC.
... the AABC is double of the AAEC.
But FECG is a par" by construction;
and it is double of the AAEC,
being on the same base EC, and between the same par" EC
and AG.
... the par" FECG = the AABC;
- and one of its angles, namely CEF, - the given A.D.

PROBLEMS ON AREAS. I27
EXERCISES.
(Graphical.)
1. Draw a square on a side of 5 cm., and make a parallelogram of
equal area on the same base, and having an angle of 45°.
Find (i) by calculation, (ii) by measurement the length of an oblique
side of the parallelogram.
2. Draw any parallelogram ABCD in which AB=2#" and AD=2";
and on the base AB draw a rhombus of equal area.
DEFINITION. In a parallelogram
ABCD, if through any point K in the
diagonal AC parallels EF, HG are drawn
to the sides, then the figures EH, GF
are called parallelograms about AC, and
the figures EG, HF are said to be their
complements. -
3. In the diagram of the preceding definition shew by Theorem 21 that
the complements EG, HF are equal in area.
Hence, given a parallelogram, EG, and a straight line HK, deduce a
construction for drawing on HK as one side a parallelogram equal and
equiangular to the parallelogram E.G.
4. Construct a rectangle equal in area to a given rectangle CDEF,
and having one side equal to a given line AB.
If AB=6 cm., CD=8 cm., CF = 3 cm., find by measurement the
remaining side of the constructed rectangle.
5. Given a parallelogram ABCD, in which AB=2'4", AD=1-8", and
the 4-A=55°. Construct an equiangular parallelogram of equal area,
the greater side measuring 27". Measure the shorter side.
Repeat the process giving to A any other value; and compare your
results. What conclusion do you draw:
6. Draw a rectangle on a side of 5 cm. equal in area to an
equilateral triangle on a side of 6 cm.
Measure the remaining side of the rectangle, and calculate its
approximate area,

128 GEOMETRY.
PROBLEM 18.
To draw a triangle equal ºn area to a given quadrilateral.
\\
A E; X
D

Let ABCD be the given quadrilateral.
It is required to describe a triangle equal to ABCD in area.
Construction. Join DB.
Through C draw CX par' to DB, meeting AB produced in X.
Join DX.
Then DAX is the required triangle.
Proof. Now the A* XDB, CDB are on the same base DB and
between the same par” DB, CX;
... the AXDB = the A CDB in area.
To each of these equals add the AADB ;
then the A DAX = the fig. ABCD.
CoroLLARY. In the same way it is always possible to draw
a rectilineal figure equal to a given rectilineal figure, and
having fewer sides by one than the given figure; and thus
step by step, any rectilineal figure may be reduced to a
triangle of equal area. D

For example, in the adjoining dia-
gram the five-sided fig. EDCBA is equal C
in area to the four-sided fig. EDXA.
The fig. EDXA may now be reduced
to an equal A DXY. Y A B X
E
PROBLEMS ON AREAS. 129
PROBLEM 19.
To draw a parallelogram equal in area to a given rectilineal
figure, and having an angle equal to a given angle.
K D H

\
Let ABCD be the given rectil. fig, and E the given angle.
A G B F
It is required to draw a par” equal to ABCD and having an
angle equal to E.
Construction. Join DB.
Through C draw CF par' to DB, and meeting AB produced
in F.
Join DF.
Then the A DAF = the fig. ABCD. Prob. 18.
Draw the parº AGHK equal to the AADF, and having the
4. KAG equal to the A. E. Prob. 17.
Then the parº KG = the AADF
= the fig. ABCD ;
and it has the A. KAG equal to the A. E.
NoTE. If the given rectilineal figure has more than four sides, it
must º be reduced, step by step, until it is replaced by an equivalent
triangle.
130 GEOMETRY.
EXERCISES.
(Reduction of a Rectilineal Figure to an equivalent Triangle.)
S. 1. Draw a quadrilateral ABCD from the following data :
AB = BC = 5.5 cm.; CD = DA=4-5 cm.; the Z_A=75°.
Reduce the quadrilateral to a triangle of equal area. Measure the
base and altitude of the triangle ; and hence calculate the approximate
area of the given figure.
§ 2. Draw a quadrilateral ABCD having given :
AB=2-8", BC=3'2", CD=3'3", DA=3-6", and the diagonal BD=3-0".
Construct an equivalent triangle; and hence find the approximate
area of the quadrilateral.
& r
3. On a base AB, 4 cm. in length, describe an equilateral pentagon
(5 sides), having each of the angles at A and B 108°.
Reduce the figure to a triangle of equal area; and by measuring its
base and altitude, calculate the approximate area of the pentagon.
4. A quadrilateral field ABCD has the following measurements:
AB=450 metres, BC = 380 m., CD =330 m., AD = 390 m.,
and the diagonal AC=660 m.
Draw a plan (scale 1 cm. to 50 metres). Reduce your plan to an
equivalent triangle, and measure its base and altitude. Hence estimate
the area of the field.
(Problems. State your construction, and give a theoretical proof.)
5. Reduce a triangle ABC to a triangle of equal area having its
base BD of given length. (D lies in BC, or BC produced.)
~ 6. Construct a triangle equal in area to a given triangle, and
having a given altitude.
7. ABC is a given triangle, and X a given point. Draw a triangle
equal in area to ABC, having its vertex at X, and its base in the same
straight line as BC.
8. Construct a triangle equal in area to the quadrilateral ABCD,
having its vertex at a given point X in DC, and its base in the same
straight line as AB.
9. Shew how a triangle may be divided into n equal parts by straight
lines drawn through one of its angular points.
\
PROBLEMS ON AREAS. } 31
10. Bisect a triangle by a straight line drawn through a given point
in one of its sides.
[Let ABC be the given A, and P the A
given point in the side AB.
Bisect AB at Z; and join CZ, CP.
Through Z draw ZQ parallel to CP.
Join PQ.
Then PQ bisects the A.]
B Q C
1]... , Trisect a triangle by straight lines drawn from a given point in
one of its sides.
[Let ABC be the given A, and X the
given point in the side BC.
Trisect BC at the points P, Q. Prob. 7.
Join AX, and through P and Q draw PH and
QK parallel to AX.
Join XH, XK.
These straight lines trisect the A ; as may
be shewn by joining AP, AQ.] B P X Q C
12. Cut off from a given triangle a fourth, fifth, sixth, or any part
required by a straight line drawn from a given point in one of its sides.
13. Bisect a quadrilateral by a straight line drawn through an angular
point.
[Reduce the quadrilateral to a triangle of equal area, and join the
vertex to the middle point of the base.]
14. Cut off from a given quadrilateral a third, a fourth, a fifth, or
any part required, by a straight line drawn through a given angular
point.


132 GEOMETRY.
AXES OF REFERENCE, COORDINATES.
EXERCISES FOR SQUARED PAPER.
If we take two fixed straight lines XOX, YOY" cutting one
another at right angles at O, the position of any point P with
reference to these lines is known when we know its distances
from each of them. Such lines are called axes of reference,
XOX' being known as the axis of x, and YOY" as the axis of y.
Their point of intersection O is called the origin.

i
F S
{
The lines XOX', YOY" are usually drawn horizontally an
vertically. º
In practice the distances of P from the axes are estinated
thus:
From P, PM is drawn perpendicular to X'X; and OM and
PM are measured.
OM is called the abscissa of the point P, and is denoted by a.
PM 95 ordinate 35 35 99 3/.
The abscissa and ordinate taken together are called the
coordinates of the point P, and are denoted by (a, y).
We may thus find the position of a point if its coordinates
are known.
ExAMPLE. Plot the point whose coordinates are (5,4).
Along OX mark off OM, 5 units in length.
At M draw MP perp. to OX, making MP 4 units in length.
Then P is the point whose coordinates are (5,4).
The axes of reference divide the plane into four regions XOY,
YOx', X'OY", Y'OX, known respectively as the first, second,
third, and fourth quadrants.
AXES. COORDINATES. 133
It is clear that in each quadrant there is a point whose
distances from the axes are equal to those of P in the above
diagram, namely, 5 units and 4 units.
The coordinates of these points are distinguished by the use
of the positive and negative signs, according to the following
system : &
Abscissae measured along the a-axis to the right of the
origin are positive, those measured to the left of the origin
are negative. Ordinates which lie above the a-axis (that is, in
the first and second quadrants) are positive; those which lie
below the 2-axis (that is, in the third and fourth quadrants)
are negative.
Thus the coordinates of the points Q, R, S are
(–5, 4), (–5, – 4), and (5, – 4) respectively.
NoTE. The coordinates of the origin are (0, 0).
In practice it is convenient to use squared paper. Two
intersecting lines should be chosen as axes, and slightl
thickened to aid the eye, then one or more of the length-
divisions may be taken as the linear unit. The paper used in
the following examples is ruled to tenths of an inch.
ExAMPLE 1. The coordinates of the points A and B are (7, 8) and
(–5, 3): plot the points and find the distance between them.
After plotting the points as
in the diagram, we may find
AB approximately by direct
measurement.
Or we may proceed thus:
Draw through B a line parl
to XX' to meet the ordinate
of A at C. Then ACB is a
º A in which BC = 12,
and AC = 5.
Now AB2 = BC? --AC2
= 122 + 5°
= 144+25
= 169.
..". AB = 13.

134 GEOMETRY.
ExAMPLE 2. The coordinates of A, B, and C are (5, 7), (–8, 2), and
(3, - 5); plot these points and find the area of the triangle of which these
are the vertices.
Having plotted the points as in
the diagram, we may measure AB,
and draw and measure the perp.
from C on AB. Hence the approxi-
mate area may be calculated.
Or we may proceed thus:
Through A and B draw. AP, BQ
parl to YY".
xºrough C draw PQ parl to
Then the AABC = the trap" APQB – the two rt.-angled A* APC, BQC
= #PQ(AP+BQ)–3. AP. PC – #. BQ. QC
= } x 13 × 19 –3 x 12x2 – 4 × 7 × 11
=73 units of area.
EXERCISES.
1. Plot the following sets of points:
(i) (6, 4), (–6, 4), (–6, –4), (6, –4);
(ii) (8, 0), (0, 8), ( – 8, 0), (0, -8);
(iii) (12, 5), (5, 12), (— 12, 5), ( - 5, 12).
2. Plot the following points, and shew experimentally that each set
lie in one straight line.
(i) (9, 7), (0, 0), (–9, -7); (ii) (–9, 7), (0, 0), (9, -7).
Explain these results theoretically.
3. Plot the following pairs of points; join the points in each case,
and measure the coordinates of the mid-point of the joining line.
(i) (4, 3), (12, 7); (ii) (5,4), (15, 16).
Shew why in each case the coordinates of the mid-point are respec-
tively half the sum of the abscissae and half the sum of the ordinates of
the given points.
4. Plot the following pairs of points; and find the coordinates of
the mid-point of their joining lines.
(i) (0, 0), (8, 10); (ii) (8, 0), (0, 10);
(iii) (0, 0), (–8, - 10); (iv) (–8, 0), (0, - 10).

EXERCISES FOR SQUARED PAPER. H35
5. Find the coordinates of the points of trisection of the line joining
(0, 0) to (18, 15).
6. Plot the two following series of points:
(i) (5, 0), (5, 2), (5, 5), (5, – 1), (5, –4);
(ii) ( — 4, 8), ( – 1, 8), (0, 8), (3, 8), (6, 8).
Shew that they lie on two lines respectively parallel to the axis of y,
and the axis of 2. Find the coordinates of the point in which they
intersect.
7. Plot the following points, and calculate their distances from the
origin.
(i) (15, 8); (ii) ( – 15, -8); (iii) (2-4”, “7”); (iv) ( – ’7”, 2'4").
Check your results by measurement.
8. Plot the following pairs of points, and in each case calculate the
distance between them.
(i) (4,0), (0, 3); (ii) (9, 8), (5, 5);
(iii) (15, 0), (0, 8); (iv) (10, 4), ( – 5, 12);
(v) (20, 12), ( – 15, 0); (vi) (20, 9), ( – 15, -3).
Verify your calculation by measurement.
9. Shew that the points (–3, 2), (3, 10), (7, 2) are the angular
points of an isosceles triangle. Calculate and measure the lengths of
the equal sides.
10. Plot the eight points (0, 5), (3, 4), (5,0), (4, -3), (–5, 0),
(0, -5), (–4, 3), (–4, -3), and shew that they all lie on a circle whose
centre is the origin.
11. Explain by a diagram why the distances between the following
pairs of points are all equal.
(i) (a, 0), (0, b); (ii) (b, 0), (0, a); (iii) (0, 0), (a, b).
12. Draw the straight lines joining
(i) (a, 0) and (0, a); (ii) (0, 0) and (a, a);
and prove that these lines bisect each other at right angles.
13. Shew that (0, 4), (12, 9), (12, — 4) are the vertices of an isosceles
triangle whose base is bisected by the axis of ac.
14. Three vertices of a rectangle are (14, 0), (14, 10), and (0, 10);
find the coordinates of the fourth vertex, and of the intersection of the
diagonals.
15. Prove that the four points (0, 0), (13, 0), (18, 12), (5, 12) are the
angular points of a rhombus. Find the length of each side, and the
coordinates of the intersection of the diagonals.
16. Plot the locus of a point which moves so that its distances from
, the points (0, 0) and (4,-4) are always equal to one another. Where
does the locus cut the axes 2
I 36 GEOMETRY.
17. Shew that the following groups of points are the vertices of
rectangles. Draw the figures, and calculate their areas.
(i) (4, 3), (17, 3), (l7, 12), (4, 12);
(ii) (3, 2), (3, 15), (– 6, 15), (–6, 2);
(iii) (5, 1), ( – 8, 1), ( – 8, -8), (5, – 8).
18. Join in order the points (1", 0), (0, 1"), ( – 1", 0), (0, -1"). Of
what kind is the quadrilateral so formed ? Find its area.
If a second figure is formed by joining the middle points of the first,
find its area.
19. Plot the triangles given by the following sets of points; and find
their areas.
(i) (10, 10), (4, 0), (18, 0); (ii) (10, -10), (4, 0), (18, 0);
(iii) (— 10, 10), (–4, 0), (— 18, 0) ; (iv) (— 10, tºº 10), (–4, 0), (— 18, 0).
20. Draw the triangles given by the points
(i) (0, 0), (5, 3), (6, 0); (ii) (0, 0), (3, 0), (0, 6).
Find their areas; and measure the angles of the first triangle.
21. Plot the triangles given by the following sets of points. Shew
that in each case one side is parallel to one of the axes. ence find the
8,I'68,.
(i) (0, 0), (12, 10), (12, -6); (ii) (0, 0), (5, 8), ( – 15, 8);
(iii) (0, 0), ( – 12, 12,), ( – 12, -8); (iv) (0, 0), (-6, -8), (20, -8).
22. In the following triangles shew that two sides of each are
parallel to the axes. Find their areas.
(i) (5, 5), (15, 5), (15, 15); (ii) (8, 3), (8, 18), (0, 18);
(iii) (4, 8), ( – 16, - 4), (4, -4); (iv) (1, 15), ( – 11, 15), (1, -7).
23. Shew that (–5, 5), (7, 10), (10, 6), (–2, 1) are the angular
points of a parallelogram. Find its sides and area.
24. Shew that each of the following sets of points gives a trapezium.
Find the area of each.
(i) (3, 0), (3, 3), (9, 0), (9, 6) ; (ii) (0, 3), ( tº-º-º: 5, 3), ( ſº 2, +-tº- 3), (0, sº 3);
(iii) (8,4), (4,4), (11, - 1), (3,-1); (iv) (0, 0), (-1, 5), (-4, 5), (–8, 0).
25. Find the area of the triangles given by the following points:
(i) (5, 5), (20, 10), (12, 14); (ii) (7, 6), (— 10, 4), (–4, –3);
(iii) (0, – 6), (0, –3), (14, 5); (iv) (6, 4), (— 7, – 6), (–2, +--- 15).
26. Shew that (-5, 0), (7, 5), (19, 0), (7, - 5) are the angular
points of a rhombus. Find its sides and its area.
EXERCISES FOR SQUARED PAPER. 137
27. Join the points (0, -5), (12, 0), (4, 6), ( – 8, -3), in the order
given. Calculate the lengths of the first three sides and measure the
fourth. Find the areas of the portions of the figure lying in the first
and fourth quadrants.
28. The coordinates of four points A, B, C, D are respectively
(-4, –4), (— 10, 4), (— 10, 13), (5, 5).
Calculate the lengths of AB, BC, CD, and measure AD. Also calculate
the area of ABCD by considering it as the difference of two triangles.
29. Draw the figure whose angular points are given by
(0, –3), (8, 3), (-4, 8), (–4, 3), (0, 0).
Find the lengths of its sides, taking the points in the above order.
Also divide it into three right-angled triangles, and hence find its area.
30. A plan of a triangular field ABC is drawn on squared paper
(scale 1"= 100 yds.). On the plan the coordinates of A, B, § 8.Te
(– 1", -3"), (3", 4"), (–5", -2") respectively. Find the area of the field,
the length of the side represented by BC, and the distance from this
side of the opposite corner of the field.
31. Shew that the points (6, 0), (20, 6), (14, 20), (0, 14) are the
vertices of a square. Measure a side and hence find the approximate
area. Calculate the area exactly (i) by drawing a circumscribing
square through its vertices; (ii) by subdividing the given square as in
the first figure on page 120.
138 GEOMETRY.
*
*.
*
&
MISCELLANEOUS EXERCISES.
1. AB and AC are unequal sides of a triangle ABC; AX is the
median through A, AP bisects the angle BAC, and AD is the perpen-
dicular from Å to BC. Prove that AF is intermediate in position and
magnitude to AX and AD. ſov v.
2. In a triangle if a perpendicular is drawn from one extrºity of
the base to the bisector of the vertical angle, (i) it will make with
either of the sides containing the vertical angle an angle equal to half
the sum of the angles at the base; (ii) it will make with the base an
angle equal to half the difference of the angles at the base.
3. In any triangle the angle contained by the bisector of the
vertical angle and the perpendicular from the vertex to the base is
equal to half the difference of the angles at the base.
4. Construct a right-angled triangle having given the hypotenuse
and the difference of the other sides.
N 5. Construct a triangle, having given the base, the difference of the
angles at the base, and (i) the difference, (ii) the sum of the remaining
sides.
6. Construct an isosceles triangle, having given the base and the
sum of one of the equal sides and the perpendicular from the vertex to
the base.
7. Shew how to divide a given straight line so that the square on
one part may be double of the square on the other.
8. ABCD is a parallelogram, and O is any point without the angle
BAD or its opposite vertical angle; shew that the triangle OAC is equal
to the sum of the triangles OAD, OAB.
If O is within the angle BAD or its opposite vertical angle, shew
that the triangle OAC is equal to the difference of the triangles
OAD, OAB.
9. The area of a quadrilateral is equal to the area of a triangle
having two of its sides equal to the diagonals of the given figure, and
the included angle equal to either of the angles between the diagonals.
10. Find the locus of the intersection of the medians of triangles
described on a given base and of given area.
11. On the base of a given triangle construct a second triangle
equal in area to the first, and having its vertex in a given straight
line.
12. ABCD is a parallelogram made of rods connected by hinges. If
AB is fixed, find the locus of the middle point of CD.
PART III.
THE CIRCLE.
DEFINITIONS AND FIRST PRINCIPLES. -T
**-
*rº.
\\
1. A circle is a plane figure contained by a line traced out
by a point which moves so that its distance from a certain
fixed point is always the same.
The fixed point is called the centre, and the bounding line
is called the circumference. N
\-
NOTE. According to this definition the term circle strictly applies
to the figure contained by the circumference; it is often used however
for the circumference itself when no confusion is likely to arise.
2. A radius of a circle is a straight line drawn from the
centre to the circumference. It follows that all radii of a
circle are equal.
3. A diameter of a circle is a straight line drawn through
the centre and terminated both ways by the circumference.
4. A semi-circle is the figure bounded by a diameter of
a circle and the part of the circumference cut off by the
diameter.
It will be proved on page 142 that a diameter divides a circle into
two identically equal parts.
a 7” ( ...ºf rz 22
5. Circles that have the same centre are said to be
concentric. & Abe a 3.6%
*
140 GEOMETRY.
From these definitions we draw the following inferences:
~
*J
(i) A circle is a closed curve; so that if the circumference
is crossed by a straight line, this line if produced will cross
the circumference at a second point.
(ii) The distance of a point from the centre of a circle
is greater or less than the radius according as the point is
without or within the circumference.
(iii) A point is outside or inside a circle according as its
distance from the centre is greater or less than the radius.
(iv) Circles of equal radii are identically equal. For by
superposition of one centre on the other the circumferences
must coincide at every point.
º N
(v) Concentric circles of unequal radii cannot intersect, for
the distance from the centre of every point on the smaller
circle is less than the radius of the larger.
(vi) If the circumferences of two circles have a common
point they cannot have the same centre, unless they coincide
altogether.
6. An arc of a circle is any part of the circumference.
7. A chord of a circle is a straight line joining any two
points on the circumference. §
NoTE. From these definitions it may be seen
that a chord of a circle, which does not pass through
the centre, divides the circumference into two un-,
equal arcs; of these, the greater is called the major
arc, and the less the minor arc. Thus the major
arc is greater, and the minor arc less than the semi-
circumference.
The major and minor arcs, into which a cir- YS *
cumference is divided by a chord, are said to be `-----~~
conjugate to one another.

SYMMETRY OF A CIRCLE. 141
$.” Ža *, *
* - , SyMMETRY. . .
& Some elementary properties of circles are easily proved by
Čonsiderations of symmetry. For convenience the definition
given on page 21 is here repeated. ‘a -
DEFINITION 1. A figure is said to be symmetrical about a
line when, on being folded about that line, the parts of the
figure on each side of it can be brought into coincidence. , ,-
The straight line is called an axis of symmetry.
That this may be possible, it is clear that the two parts of the figure
must have the same size and shape, and must be similarly placed with
regard to the axis.
DEFINITION 2. Let AB be a straight line and P a point
outside it.
P

- Q
From P draw PM perp. to AB, and produce it to Q, making
MQ equal to PM.
Then if the figure is folded about AB, the point P may be
made to coincide with Q, for the 4. AMP = the 4-AMQ, and
MP = MQ.
The points P and Q are said to be symmetrically opposite
s with regard to the axis AB, and each point is said to be the
* image of the other in the axis.
, , NoTE. A point and its image are equidistant from every point on
the axis. See Prob. 14, page, 91.
142 GEOMETRY.
SOME SYMMETRICAL PROPERTIES OF CIRCLES.
I. A circle is symmetrical about any diameter.

Let APBQ be a circle of which O is the centre, and AB any
diameter. 'S -
tº
It is required to prove that the circle is symmetrical about AB.
Proof. Let OP and OQ be two radii making any equal
4." AOP, AOQ on opposite sides of OA.
Then if the figure is folded about AB, OP may be made to
fall along OQ, since the 4. AOP= the 4 AOQ. -
And thus P will coincide with Q, since OP = OQ.
Thus every point in the arc APB must coincide with some
point in the arc AQB ; that is, the two parts of the circum-
ference on each side of AB can be made to coincide.
... the circle is symmetrical about the diameter AB.
COROLLARY. If PQ is drawn cutting AB at M, then on
folding the figure about AB, since P falls on Q, MP will
coincide with MQ, - -
•. MP = MQ;
. and the 4. OMP will coincide with the 4 UMQ,
s ... these angles, being adjacent, are rt. A ';
... the points P and Q are symmetrically opposite with
‘Sregard to AB. -
Hence, conversely, if a circle passes through a given point P,
it also passes through the symmetrically opposite point with regard
to any diameter.

DEFINITION. The straight line passing through the centres
of two circles is called the line of centres. .
SYMMETRICAL PROPERTIES. 143
II. Two circles are divided symmetrically by their line of centres.
Let O, O' be the centres of two circles, and let the st. line
through O, O' cut the O* at A, B and A, B'. Then AB and
A'B' are diameters and therefore axes of symmetry of their
respective circles. That is, the line of centres divides each
circle symmetrically.
III. If two circles cut at one point, they must also cut at a
second point; and the common chord is bisected at right angles by
the line of centres.
Let the circles whose centres are O, O' cut at the point P.
Draw PR perp. to OO', and produce it to Q, so that
RQ = RP.
Then P and Q are symmetrically opposite points with
regard to the line of centres OO';
‘. since P is on the O* of both circles, it follows that Q is
also on the O* of both. [I. Cor.]
And, by construction, the common chord PQ is bisected at
right angles by OO'.


144 GEOMETRY. .
ON CHORDS.
THEOREM 31. [Euclid III. 3..]
If a straight line drawn from the centre of a circle bisects a
chord which does not pass through the centre, it cuts the chord at
Tight angles.
Conversely, if it cuts the chord at right angles, it bisects it.
Let ABC be a circle whose centre is O ; and let OD bisect
a chord AB which does not pass through the centre.
It is required to prove that OD is perp. to AB.
Join OA, OB.
Proof. Then in the A* ADO, BDO,
AD = BD, by hypothesis,
because OD is common,
and OA=OB, being radii of the circle;
... the 4ADO = the A. BDo; Theor. 7.
and these are adjacent angles,
... OD is perp. to AB. Q.E.D.
Conversely. Let OD be perp. to the chord AB.
It is required to prove that OD bisects AB.
Proof. . In the A* ODA, ODB,
. the 4: ODA, ODB are right angles,
because {the hypotenuse OA = the hypotenuse OB,
and OD is common; -
... DA = DB ; Theor. 18,
that is, OD bisects AB at D. Q.E.D.

CHORD PROPERTIES. - 145
CoROLLARY 1. The straight line which bisects a chord at
right angles passes through the centre.
COROLLARY 2. A straight line cannot meet a circle at more
than two points. .
For suppose a st. line meets a O
circle whose centre is O at the points *
A and B.
Draw OC perp. to AB.
Then AC = CB A-C-E---B

Now if the circle were to cut AB in a third point D, AC
would also be equal to CD, which is impossible.
CoroLiary 3. A chord of a circle lies wholly within it.
EXERCISES.
(Numerical and Graphical.)
1. In the figure of Theorem 31, if AB = 8 cm., and OD=3 cm., find
OB. Draw the figure, and Yºrity your result by measurement.
{K} N&as a 3
2. Calculate the length of a chord which stands at a distance 5"
from the centre of a circle whose radius is 13".
* - 3. . In a circle of 1" radius draw two chords 1.6" and 12" in length.
Calculate and measure the distance of each from the centre. .
4. Draw a circle whose diameter is 8'0 cm. and place in it a chord
6.0 cm. in length. Calculate to the nearest millimetre the distance of
the chord from the centre; and verify your result by measurement.
5. Find the distance from the centre of a chord 5 ft. 10 in. in length
in a circle whose diameter is 2 yds. 2 in. Verify the result graphically
by drawing a figure in which 1 cm. represents 10".
6. AB is a chord 2'4" long in a circle whose centre is O and whose
radius is 13"; find the area of the triangle OAB in square inches.
'? 7. Two points P and Q are 3" apart. Draw a circle with radius 1.7"
to pass through P and Q. Calculate the distance of its centre from the
chord PQ, and verify by measurement.
H.S. G. K
146 GEOMETRY.
THEOREM 32.
One circle, and only one, can pass through any three points not
in the same straight line.
TB
Let A, B, C be three points not in the same straight line.
It is required to prove that one circle, and only one, can pass
through A, B, and C. -
Join AB, BC,
Let AB and BC be bisected at right angles by the lines
DF, EG. -
Then since AB and BC are not in the same st. line, DF and
EG are not par'.
Let DF and EG meet in O.
Proof. Because DF bisects AB at right angles,
... every point on DF is equidistant from A and B.
Prob. 14.
Similarly every point on EG is equidistant from B and C.
... O, the only point common to DF and EG, is equidistant
from A, B, and C ;
and there is no other point equidistant from A, B, and C.
... a circle having its centre at O and radius OA will pass
through B and C.; and this is the only circle which will pass
through the three given points. Q.E.D.


CHORD PROPERTIES. 147
S COROLLARY 1. The size and position of a circle are fully
Šdetermined if it is known to pass through three given points; for
then the position of the centre and length of the radius can
be found.
COROLLARY 2. Two circles cannot cut one another in more
than two points without coinciding entirelift för if they cut at
three points they would have the same centre and radius.
HYPOTHETICAL CONSTRUCTION. From Theorem 32 it appears
that we may suppose a circle to be drawn through any three points
not in the same straight line.
For example, a circle can be assumed to pass through the vertices of
any triangle. -
DEFINITION. The circle which passes through the vertices
of a triangle is called its circum-circle, and is said to be
circumscribed about the triangle. The centre of the circle is
called the circum-centre of the triangle, and the radius is called
the circum-radius.
EXERCISES ON THEOREMS 31 AND 32.
(Theoretical.) , . . . . . .”
1. The parts of a straight line intercepted between the circum-
ferences of two Congºngjo girdles are equal, . . . . . . .
2. Two circles, whose centres are at A and B, intersect at C, D; and
M is the middle point of the common chord. Shew that AM and BM
are in the same straight line. *
Hence prove that the line of centres bisects the common chord at right
angles.
3. AB, AC are two equal chords of a circle; shew that the straight
line which bisects the angle BAC passes through the centre.
4. Find the locus of the centres of all circles which pass through two
gºver points. ( (. . . . . . . . . . . . . . . .xºn & Y. . . .
. 5. Describe a circle that shall pass through two given points and have
its centre in a given straight line. -
When is this impossible?
6. Describe a circle of given radius to pass through two given points. .
When is this impossible? .
148 - GEOMETRY.
*THEOREM 33. [Euclid III. 9..] .
If from a point within a circle more than two equal straight
limes can be drawn to the circumference, that point is the centre
of the circle. .*
Let ABC be a circle, and O a point within it from which
more than two equal st. lines are drawn to the O*, namely
OA, OB, OC. - - -
It is required to prove that O is the centre of the circle ABC.
Join AB, BC.
Let D and E be the middle points of AB and BC respectively.
- Join OD, OE.
Proof. In the A* ODA, ODB,
DA = DB,
because DO is common,
and OA= OB, by hypothesis;
... the 4. ODA = the 4 ODB ; Theor. 7.
... these angles, being adjacent, are rt. A ".
Hence Do bisects the chord AB at right angles, and therefore
passes through the centre. Theor. 31, Cor. 1.
Similarly it may be shewn that EO passes through the
centre. - ... "
... O, which is the only point common to DO and EO, must
be the centre. n - Q.E.D.


CHORD PROPERTIES. 149
EXERCISES ON CHORDS.
(Numerical and Graphical.)
1. AB and BC are lines at right angles, and their lengths are 1-6."
and 3.0" respectively. Draw the circle through the points A, B, and C.;
find the length of its radius, and verify your result by measurement.
2. Draw a circle in which a chord 6 cm. in length stands at a
distance of 3 cm. from the centre.
Calculate (to the nearest millimetre) the length of the radius, and
verify your result by measurement.
3. Draw a circle on a diameter of 8 cm., and place in it a chord
equal to the radius. #
Calculate (to the nearest millimetre) the distance of the chord from
the centre, and verify by measurement.
4. Two circles, whose radii are respectively 26 inches and 25
inches, intersect at two points which are 4 feet apart. Find the
distances between their centres. .
Draw the figure (scale 1 cm. to 10"), and verify your result by
measurement.
5. Two parallel chords of a circle whose diameter is 13" are re-
spectively 5" and 12" in length ; shew that the distance between them
is either 8'5" or 3-5". - -
6. Two parallel chords of a circle on the same side of the centre are
6 cm. and 8 cm. in length respectively, and the perpendicular distance
between them is 1 cm. Calculate and measure the radius.
7. Shew on squared paper that if a circle has its centre at any point
on the 2-axis and passes through the point (6, 5), it also passes through
the point (6, - 5). [See page 132.]
( Theoretical ..)
8. The line joining the middle points of two parallel chords of a
circle passes through the centre.
9. Find the locus of the middle points of parallel chords in a circle.
10. Two intersecting chords of a circle cannot bisect each other
unless each is a diameter. -
11. If a parallelogram can be inscribed in a circle, the point of inter-
section of its diagonals must be at the centre of the circle.
. 12. Shew that rectangles are the only parallelograms that can be
inscribed in a circle. - -

150 * GEOM ETRY.
THEOREM 34. [Euclid III. 14.]
Equal chords of a circle are equidistant from the centre.
Conversely, chords which are equidistant from the centre are

Let AB, CD be chords of a circle whose centre is O, and let
OF, OG be perpendiculars on them from O.
First. Let AB = CD.
It is required to prove that AB and CD are equidistant from O.
Join OA, OC.
IProof. Because OF is perp. to the chord AB,
... OF bisects AB; Theor. 31.
..'. AF is half of AB.
Similarly CG is half of CD.
But, by hypothesis, AB = CD,
.*. AF = CG.
Now in the A* OFA, OGC,
the 4 "OFA, OGC are right angles,
because {the hypotenuse OA = the hypotenuse OC,
and AF = CG ;
... the triangles are equal in all respects; Theor. 18.
so that OF = OG;
that is, AB and CD are equidistant from O.
Q.E.D.
*ść & 3.
CHORD PROPERTIES. • 151
Conversely. - Let OF = OG.
It is required to prove that AB = CD.
Proof. As before it may be shewn that AF is half of AB,
and CG half of CD.
Then in the A* OFA, OGC,
ſº 4 "OFA, OGC are right angles,
because {the hypotenuse OA = the hypotenuse OC,
l and OF = OG;
... AF = CG ; Theor. 18.
... the doubles of these are equal;
that is, AB = CD.
Q.E.D.
EXERCISES.
(Theoretical.)
1. Find the locus of the middle points of equal chords of a circle. º
2. If two chords of a circle cut one another, and make equal
angles with the straight line which joins their point of intersection to
the centre, they are equal.
3. If two equal chords of a circle intersect, shew that the segments a
of the one are equal respectively to the segments of the other.
4. In a given circle draw a chord which shall be equal to one given
straight line (not greater than the diameter) and parallel to another.
5. PQ is a fixed chord in a circle, and AB is any diameter : shew
that the sum or difference of the perpendiculars let fall from A and B
on PQ is constant, that is, the same for all positions of AB.
, i... . , Jr. º. [See Ex. 9, p. 65.]
(Graphical.)
6. In a circle of radius 4:1 cm. any number of chords are drawn
each 19 cm. in length. Shew that the middle points of these chords
all lie on a circle. Calculate and measure the length of its radius, and
draw the circle.
7. The centres of two circles are 4" apart, their common chord is
2'4" in length, and the radius of the larger circle is 3.7". Give a con-
struction for finding the points of intersection of the two circles, and
find the radius of the smaller circle.
1
5
2
GEOMETRY.
#
TIIEOREM 35. [Euclid III. 15.]
Of any two chords of a circle, that which is nearer to the centre
is greater than one more remote.º.º. 23, F-
Conversely, the greater of two chords is nearer to the centre than
the less.

A C
B
Let AB, CD be chords of a circle whose centre is O, and let
OF, OG be perpendiculars on them from O.
It is required to prove that
(i) if OF is less than OG, then AB is greater than CD;
(ii) if AB is greater than CD, then OF is less than OG.
Join OA, OC.
Proof. Because OF is perp. to the chord AB,
... OF bisects AB; \
... AF is half of AB.
Similarly CG is half of CD.
Now OA = OC,
... the sq. on OA = the sq. on OC,
But since the ZoPA is a rt. angle,
... the sq. on OA = the sqq. on OF, FA.
Similarly the sq. on OC = the sqq. on OG, GC,
... the sqq. on OF, FA = the sqq. on OG, GC,
CHORD PROPERTIES. * 153
(i) Hence if OF is given less than OG;
the sq. on OF is less than the sq. on OG.
... the sq. on FA is greater than the sq. on GC ;
... FA is greater than GC:
... AB is greater than CD.
(ii) But if AB is given greater than CD,
that is, if FA is greater than GC ;
then the sq. on FA is greater than the sq. on GC,
... the sq. on OF is less than the sq. on OG;
... OF is less than OG. Q.E.D.
CoroLLARY. The greatest chord in a circle is a diameter.
à-
EXERCISES.
(Miscellaneous.) Y, , ,
* 1. Through a given point within a circle draw the least possible
chord.
$ 2. Draw a triangle ABC in which a =3:5", b=1:2", c=3.7". Through
the ends of the side a draw a circle with its centre on the side c.
Calculate and measure the radius.
Y. 3. Draw the circum circle of a triangle whose sides are 2-6", 2-8",
and 3'0". Measure its radius.
S. 4. . AB is a fixed chord of a circle, and XY any other chord having
its middle point Z on AB; what is the greatest, and what the least
length that XY may have 2 -
" . , t
Shew that XY increases, as Z approaches the middle point of AB.
5. Shew on squared paper that a circle whose centre is at the
origin, and whose radius is 3'0", passes through the points (2-4", l'8"),
(1:8", 24").
Find (i) the length of the chord joining these points, (ii) the co-
ordinates of its middle point, (iii) its perpendicular distance from the
origin.
154 GEOMETRY.
*THEOREM 36. [Euclid III. 7.]
If from any internal point, not the centre, straight limes are
drawn to the circumference of a circle, then the greatest is that
which passes through the centre, and the least is the remaining part
of that diameter.
And of any other two such lines the greater is that which Sub-
tends the greater angle at the centre.

Let ACDB be a circle, and from P any internal point, which
is not the centre, let PA, PB, PC, PD be drawn to the O*,
so that PA passes through the centre O, and PB is the remain-
ing part of that diameter. Also let the 4. POC at the centre
subtended by PC be greater than the 4. POD subtended by PD.
It is required to prove that of these st. lines
(i) PA is the greatest,
(ii) PB is the least,
(iii) PC vs greater than PD.
Join OC, OD,
Proof. (i) In the A POC, the sides PO, OC are together
greater than PC. Theor. 11.
But OC = OA, being radii;
..". PO, OA are together greater than PC;
that is, PA is greater than PC.
Similarly PA may be shewn to be greater than any other
st line drawn from P to the C*;
. PA is the greatest of all such lines.
DISTANCE OF A POINT TO THE CIRCUMFERENCE. 155
(ii) In the A OPD, the sides OP, PD are together greater
than OD.
But OD = OB, being radii;
... OP, PD are together greater than OB.
Take away the common part OP;
then PD is greater than PB.
Similarly any other st. line drawn from P to the O* may
be shewn to be greater than PB; *
‘. PB is the least of all such lines.
(iii) In the A' POC, POD,
PO is common,
because OC = OD, being radii,
but the 4. POC is greater than the 4. POD;
... PC is greater than PD. Theor. 19.
Q.E.D.
EXERCISES.
(Miscellaneous.)
1. All circles which pass through a fixed point, and have their centres
on a given Straight line, pass also through a second fiaced point.
2. If two circles which intersect are cut by a straight line parallel
to the common chord, shew that the parts of it intercepted between
the circumferences are equal.
3. If two circles cut one another, any two parallel straight lines
drawn through the points of intersection to cut the circles are equal.
4. If two circles cut one another, any two straight lines drawn
through a point of section, making equal angles with the common
chord, and terminated by the circumferences, are equal.
\
5. Two circles of diameters 74 and 40 inches respectively have a
common chord 2 feet in length : find the distance between their centres.
Draw the figure (1 cm. to represent 10") and verify your result by
measurement. *
6. Draw two circles of radii l'0" and l'7", and with their centres
2:1" apart. Find by calculation, and by measurement, the length of
, the common chord, and its distance from the two centres.
156 GEOMETRY.
*THEOREM 37. [Euclid III. 8.]
If from any easternal point Straight lines are drawn to the
circumference of a circle, the greatest is that which passes through
the centre, and the least is that which when produced passes through
the centre.
And of any other two Such lines, the greater is that which sub-
tends the greater angle at the centre.

Let ACDB be a circle, and from any external point P let the
lines PBA, PC, PD be drawn to the O*', so that PBA passes
through the centre O, and so that the 4. POC subtended by PC
at the centre is greater than the A POD subtended by PD.
It is required to prove that of these St. lines *
(i) PA is the greatest,
(ii) PB is the least,
(iii) PC is greater than PD.
Join OC, OD.
Proof (i) In the A POC, the sides Po, OC are together
greater than PC.
But OC = OA, being radii;
... PO, OA are together greater than PC;
that is, PA is greater than PC,
Similarly PA may be shewn to be greater than any other st.
line drawn from P to the O*;
that is, PA is the greatest of all such lines.
DISTANCE OF A POINT TO THE CIRCUMFERENCE. 157
(ii) In the A POD, the sides PD, DO are together greater
than PO.
But OD = OB, Toeing radii;
... the remainder PD is greater than the remainder PB.
Similarly any other st. line drawn from P to the O* may be
shewn to be greater than PB;
that is, PB is the least of all such lines.
(iii) In the A POC, POD,
PO is common,
because OC = OD, being radii;
|but the . Focis greater than the A POD;
. PC is greater than PD. Theor. 19.
Q.E.D.
EXERCISES.
(Miscellaneous.)
1. Find the greatest and least straight lines which have one ex-
tremity on each of two given circles which do not intersect.
2. If from any point on the circumference of a circle straight lines
are drawn to the circumference, the greatest is that which passes
through the centre; and of any two such lines the greater is that
which subtends the greater angle at the centre.
3. Of all straight lines drawn through a point of intersection of two
circles, and terminated by the circumferences, the greatest is that which
is parallel to the line of centres.
4. Draw on squared paper any two circles which have their centres
on the ac-axis, and cut at the point (8, -11). Find the coordinates of
their other point of intersection.
5. Draw on squared paper two circles with centres at the points
(15, 0) and (–6, 0) respectively, and cutting at the point (0, 8). Find
the lengths of their radii, and the coordinates of their other point of
intersection. *
6. Draw an isosceles triangle OAB with an angle of 80° at its vertex
O. With centre O and radius OA draw a circle, and on its circum-
ference take any number of points P, Q, R, ... on the same side of
AB as the centre. Mgasure the angles subtended by the chord AB at
the points P, Q, R, 2.47 Repeat the same exercise with any other given
angle at O. What inference do you draw 2
º, ºn X, sº
* { 2
158 GEOMETRY.
`s.
*. *
*.
*~s. *.
r
\s *.
ON ANGLES IN SEGMENTS, AND ANGLES AT THE
CENTRES AND CIRCUMEERENCES OF CIRCLES.
THEOREM 38. [Euclid III. 20.]
The angle at the centre of a circle is double of an angle at the
circumference standing on the same arc.
A
Fig. I. Fig. 2.
Let ABC be a circle, of which O is the centre ; and let BOC
be the angle at the centre, and BAC an angle at the O*,
standing on the same arc BC.
It is required to prove that the 4. BOC is twice the A. BAC.
Join AO, and produce it to D.
Proof. In the A OAB, because OB = OA,
... the Z.OAB = the A. OBA.
... the sum of the 4 "OAB, OBA = twice the 4 OAB.
But the ext. A BOD = the sum of the 4 "OAB, OBA.;
... the A BOD = twice the 4 OAB.
Similarly the 4 DOC = twice the 4 OAC.
..., adding these results in Fig. 1, and taking the difference
in Fig. 2, it follows in each case that
the A. BOC = twice the A. BAC. Q.E.D.

ANGLE PROPERTIES. 159

A A -
/2^N. ×1N.
B º I C
D E E
Fig. 3. Fig. 4.

Obs. If the arc BEC, on which the angles stand, is a semi-
circumference, as in Fig. 3, the A. BOC at the centre is a
straight angle ; and if the arc BEC is greater than a semi-
circumference, as in Fig. 4, the A. BOC at the centre is reflea.
But the proof for Fig. 1 applies without change to both these
cases, shewing that whether the given arc is greater than,
equal to, or less than a semi-circumference,
the 4 BOC = twice the A. BAC, on the same arc BEC.
DEFINITIONS.

A segment of a circle is the figure bounded
by a chord and one of the two arcs into which
the chord divides the circumference.
NoTE. The chord of a segment is sometimes called * ...”
its base.

An angle in a segment is one formed by two
straight lines di ºwn from any point in the arc
of the segment to the extremities of its chord.
We have seen in Theorem 32 that a circle may be drawn
through any three points not in a straight line. But it is
only under certain conditions that a circle can be drawn
through more than three points. . . . . .
DEFINITION. If four or more points are so placed that a
circle may be drawn through them, they are said to be
concyclic.
160 GEOMETRY,
THEOREM 39. [Euclid III, 21.]
Angles in the same segment of a circle are equal.
Fig.1. t Fig.2.
Let BAC, BDC be angles in the same segment BADC of a
circle, whose centre is O. -
It is required to prove that the L BAC = the 4. BDC.
- Join BO, O.C.
Proof. Because the 4 BOC is at the centre, and the 4. BAC
at the O*, standing on the same arc BC, -
•'. the A-BOC = twice the 4. BAC, Theol. 38.
Similarly the A. BOC = twice the 4. BDC.
... the 4. BAC = the 4. BDC. Q.E.D.
NoTE. The given segment may be greater than a semicircle as in
Fig. 1, or less than a semicircle as in Fig. 2; in the latter case the angle
BOC will be reflex. But by virtue of the extension of Theorem 38.
#. on the preceding page, the above proof applies equally to both
gures. **...Y. A ‘. . .
Śrī ‘Y ºn tº

ANGLE PROPERTIES. 16]
CONVERSE OF THEOREM 39.
Equal angles standing on the same base, and on the 8ame side of it,
have their vertices on an arc of a circle, of which the given base is the
chord.

Let BAC, BDC be two equal angles standing on A EAD
the same base BC, and on the same side of it.
It is required to prove that A and D lie on an arc
of a circle having BC as its chord.
Let ABC be the circle which passes through
the three points A, B, C ; and suppose it cuts BD B C
or BD produced at the point E. - -
Join EC.
Proof. Then the Z. BAC= the Z BEC in the same segment.
But, by hypothesis, the Z. BAC= the 4. BDC;
... the Z BEC= the Z. BDC ; !,
which is impossible unless E coincides with D ;
‘.. the circle through B, A, C must pass through D.
CoRoDLARY. The locus of thé vertices of triangles drawn on the same
8ide of a given base, and with equal vertical angles, is an arc of a circle.
EXERCISES ON THEOREM 39.
& 1. In Fig. 1, if the angle BDC is 74°, find the number of degrees in
each of the angles BAC, BOC, OBC.
* 2. In Fig. 2, let BD and CA intersect at X. If the angle DXC=40°,
and the angle XCD=25°, find the number of degrees in the angle BAC
and in the reflex angle BOC.
* 3. In Fig. 1, if the angles CBD, BCD §§§ 43° and 82°,
find the number of degrees in the angles BAC, OBD, OCD.
4. Shew that in Fig. 2 the angle OBC is always less than the angle
BAC by a right angle.
[For further Exercises on Theorem 39 see page 170.]
H. S. G. L
162 GEOMETRY.
* THEOREM 40. [Euclid III. 22.]
The opposite angles of any quadrilateral inscribed in a circle
are together equal to two right angles.
Let ABCD be a quadrilateral inscribed in the G) ABC.
It is required to prove that -
(i) the 4 "ADC, ABC together = two ri. angles.
- (ii) the 4 "BAD, BCD together = two ri. angles.
Suppose O is the centre of the circle.
Join OA, OC.
Proof. Since the 4 ADC at the O* = half the Z AOC at the
centre, standing on the same arc ABC ;
and the 4 ABC at the O* = half the reflex 4-AOC at the centre,
standing on the same arc ADC ; º
... the 4 "ADC, ABC together = half the sum of the 4-AOC and
the reflex / AOC. -
But these angles make up four rt. angles.
... the 4 "ADC, ABC together = two rt, angles.
Similarly the 4 "BAD, BCD together = two rt. angles.
- Q.E.D.
Note. The results of Theorems 39 and 40 should be carefully
compared. -
From Theorem 39 we learn that angles in the same segment are
equal. º ... , A&so
From Theorem 40 we learn that angles in cºnjugate segments are
supplementary.
DEFINITION. A quadrilateral is called cyclic when a circle
can be drawn through its four vertices. - - -

ANGLE PROPERTIES. 163
ſº
CoNVERSE of THEOREM 40.
If a pair of opposite angles of a quadrilateral are supplementary, its
vertices are concyclic.

Let ABCD be a quadrilateral in which the A
opposite angles at B and D are supplementary. E D
It is required to prove that the points A, B, C, D
are concyclic.
Let ABC be the circle which passes through
the three points A, B, C ; and suppose it cuts AD B C
or AD produced in the point E.
Join EC.
Proof. Then since ABCE is a cyclic quadrilateral,
‘.. the A. AEC is the supplement of the 4–ABC.
But, by hypothesis, the A-ADC is the supplement of the 4- ABC;
‘.. the Z. AEC = the A-ADC ;
which is impossible unless E coincides with D.
... the circle which passes through A, B, C must pass through D :
that is, A, B, C, D are concyclic. Q. E. D.
EXERCISES ON THEOREM 40.
1. In a circle of 1-6" radius inscribe a quadrilateral ABCD, making
the angle ABC equal to 126". Measure the remaining angles, and
hence verify in this case that oppºsite angles are supplementary.
§ {}}
2. Prove Theorem 40 by the iá of Theorems 39 and 16, after first
joining the opposite vertices of the quadrilateral.
S3. If a circle can be described about a parallelogram, the parallelo-
gram must be rectangular.
4. ABC is an isosceles triangle, and XY is drawn parallel to the
base BC cutting the sides in X and Y: shew that the four points B, C,
X, Y lie on a circle.
Y. 5. If one side of a cyclic quadrilateral is produced, the exterior angle
is equal to the opposite interior angle of the quadrilateral.
164 GEOMETRY.
THEOREM 41. [Euclid III. 31.]
The angle in a semi-circle is a right angle.
wº
Let ADB be a circle of which AB is a diameter and O the
centre ; and let C be any point on the semi-circumference ACB.
It is required to prove that the 4. ACB is a ri. angle.
1st Proof. The 4 ACB at the O* is half the straight angle
AOB at the centre, standing on the same arc ADB ;
and a straight angle = two rt. angles:
... the 4. ACB is a rt. angle. Q.E.D.
2nd Proof. Join OC.
Then because OA = OC,
... the Z. OCA = the Z.OAC. Theor. 5.
And because OB = OC,
... the 4 OCB = the A. OBC.
... the whole / ACB = the 4. OAC + the A. OBC.
But the three angles of the AACB together = two rt. angles;
... the 4 ACB = one-half of two rt. angles
= one rt. angle. Q.E.D.

ANGLE PROPERTIES. 1.65

CoROLLARY. The angle in a segment greater than a semi-circle
is acute; and the angle in a Segment less than a semi-circle is obtuse.

D O .
The 4. ACB at the O* is half the 4 AOB at the centre, on the
same are ADB. *
(i) If the segment ACB is greater than a semi-circle, then
ADB is a minor are ;
... the 4. AOB is less than two rt. angles;
... the 4. ACB is less than one rt. angle.
(ii) If the segment ACB is less than a semi-circle, then ADB
is a major arc ; *
... the 4-AOB is greater than two rº, angles;
... the 4 ACB is greater than one rt. angle.
EXERCISES ON THEOREM 41.
I. A circle described on the hypotenuse of a right-angled triangle as
diameter, passes through the opposite angular point.
* 2. Two circles intersect at A and B; and through A two diameters
AP, AQ are drawn, one in each circle: shew that the points P, B, Q
are collinear.
3. A circle is described on one of the equal sides of an isosceles
triangle as diameter. Shew that it passes through the middle point of
the base. s
4. Circles described on any two sides of a triangle as diameters
intersect on the third side, or the third side produced.
5. . A straight rod of given length slides between two straight rulers.
placed at right angles to one another; find the locus of its middle point.
6. Find the locus of the middle points of chords of a circle drawn
through a fixed point. Distinguish between the cases when the given point
... is within, on, or without the circumference.
*
... • * * ~
º & DEFINITION. A sector of a circle is a figure ,’ \
* bounded by two radii and the arc intercepted !
between them. - * *
I66 GEOMETRY.
THEOREM 42. [Euclid III. 26.]
In equal circles, arcs which Subtend equal angles, either at the
centres or at the circumferences, are equal.
Let ABC, DEF be equal circles, and let the 4. BGC = the 4 EHF
at the centres; and consequently , s \sº
the ABAc=the LEDF at the O*. Theor. 38.
It is required to prove that the arc BKC = the arc ELF.
Proof. Apply the G) ABC to the G) DEF, so that the centre G
falls on the centre H, and GB falls along HE.
Then because the A BGC = the 4 EHF,
... GC will fall along HF.
And because the circles have equal radii, B will fall on E,
and C on F, and the circumferences of the circles will coincide
entirely.
... the arc BKC must coincide with the arc ELF;
... the arc BKC = the arc ELF. Q.E.D.
COROLLARY. In equal circles Sectors which have equal angles
are equal. - t
Obs. It is clear that any theorem relating to arcs, angles,
and chords in equal circles must also be true in the same circle.

ARCS AND ANGLES. 167
THEOREM 43. [Euclid III. 27.]
In equal circles angles, either at the centres or at the circum-
ferences, which stand on equal arcs are equal.
Let ABC, DEF be equal circles;
and let the arc BKC = the arc ELF.
It is required to prove that
the LBGC = the 4 EHF at the centres;
also the 4 BAC = the A. EDF at the O*.
Proof. Apply the G) ABC to the G) DEF, so that the centre G
falls on the centre H, and GB falls along HE.
Then because the circles have equal radii,
‘. B falls on E, and the two O* coincide entirely.
And, by hypothesis, the arc BKC = the arc ELF.
... C falls on F, and consequently GC on HF ;
... the A BGC = the A EHF.
And since the A. BAC at the O* = half the A. BGC at the centre;
and likewise the A. EDF = half the 4 EHF ;
. . the 2BAC-the 2EDF, Q.E.D.
- SN
f ~... .
3. •. *, *
cos -

168 GEOMETRY.
THEOREM 44. [Euclid III. 28.]
In epual circles, arcs which are cut off by equal chords are equal,
the major arc equal to the major arc, and the minor to the minor.
D
Let ABC, DEF be equal circles whose centres are G and H ;
and let the chord BC = the chord EF.
It is required to prove that *
the major arc BAC = the major arc EDF,
and the minor arc BKC = the minor arc ELF.
Join BG, GC, EH, HF.
Proof. In the A* BGC, EHF,
| BG = EH, being radii of equal circles,
because GC = HF, for the same reason,
and BC = EF, by hypothesis;
... the A BGC = the 4 EHF ; Theor. 7.
... the arc BKC = the are ELF; Theor. 42.
and these are the minor arcs.
But the whole O* ABKC = the whole O* DELF;
... the remaining arc BAC = the remaining arc EDF:
and these are the major arcs. Q.E.D.

ARCS AND CHORDS. 169
THEOREM 45. [Euclid III. 29.]
!
In equal circles chords which cut off equal arcs are equal.
Let ABC, DEF be equal circles whose centres are G and H ;
and let the arc BKC = the arc ELF.
It is required to prove that the chord BC = the chord EF.
* Join BG, EH.
Proof. Apply the G)ABC to the G) DEF, so that G falls on H
and GB along HE.
Then because the circles have equal radii,
... B falls on E, and the O* coincide entirely.
And because the arc BKC = the arc ELF,
... C falls on F.
... the chord BC coincides with the chord EF;
... the chord BC = the chord EF. Q.E.D.

170 GEOMETRY.
EXERCISES ON ANGLES IN A CIRCLE.
1. P is any point on the arc of a segment of which AB is the chord.
Shew that the sum of the angles PAB, PBA is constant.
X 2. PQ and RS are two chords of a circle intersecting at,X: prove
that the triangles PXS, RXQ are equiangular to one another.
A ****)
> 3. Two circles intersect at A and B; and through A any straight
line PAQ is drawn terminated by the circumferences: shew that PQ
subtends a constant angle at B.
4. Two circles intersect at A and B; and through A any two
straight lines PAQ, XAY are drawn terminated by the circumferences;
shew that the arcs PX, QY subtend equal angles at B.
7 5. P is any point on the arc of a segment whose chord is AB; and
the angles PAB, PBA are bisected by straight lines which intersect at O.
Find the locus of the point O.
6. If two chords intersect within a circle, they form an angle equal to
that at the centre, subtended by half the sum of the arcs they cut off.
7. If two chords intersect without a circle, they form an angle equal
to that at the centre subtended by half the difference of the arcs they cut off.
8. The sum of the arcs cut off by two chords of a circle at right
angles to one another is equal to the semi-circumference.
* 9. If AB is a faced chord of a circle and P any point on one of the
arcs cut off by it, then the bisector of the angle APB cuts the conjugate arc
in the same point for all positions of P.
10. AB, AC are any two chords of a circle; and P, Q are the middle
points of the minor arcs cut off by them ; if PQ is joined, cutting AB
in X and AC in Y, shew that AX=AY.
11. A triangle ABC is inscribed in a circle, and the bisectors of the
angles meet the circumference at X, Y, Z. Show that the angles of
the triangle XYZ are respectively
e A a B s C
90 T 5’ 90 ~ 2. 90 T2'
* 12. Two circles intersect at A and B; and through these points
lines are drawn from any point P on the circumference of one of the
circles: shew that when produced they intercept on the other circum-
ference an arc which is constant for all positions of P.
EXERCISES ON ANGLES IN A CIRCLE. 171
13. The straight lines which join the extremities of parallel chords
in a circle (i) towards the same parts, (ii) towards opposite parts, are
equal.
14. Through A, a point of intersection of two equal circles, two
straight lines PAQ, XAY are drawn; shew that the chord PX is equal
to the chord QY.
15. Through the points of intersection of two circles two parallel
straight lines are drawn terminated by the circumferences: shew that
the straight lines which join their extremities towards the same parts
re equal.
16. Two equal circles intersect at A and B; and through A any
straight line PAQ is drawn terminated by the circumferences: shew
that BP= BQ.
S 17. ABC is an isosceles triangle inscribed in a circle, and the
bisectors of the base angles meet the circumference at X and Y. Shew
that the figure BXAYC must have four of its sides equal.
What relation must subsist among the angles of the triangle ABC, in
order that the figure BXAYC may be equilateral?
18. ABCD is a cyclic quadrilateral, and the opposite sides AB, DC
are produced to meet at P, and CB, DA to meet at Q : if the circles
circumscribed about the triangles PBC, QAB intersect at R, shew that
the points P, R, Q are collinear.
19. P, Q, R are the middle points of the sides of a triangle, and X is
the foot of the perpendicular let fall from one vertear on the opposite side :
-shew that the four points P, Q, R, X are concyclic.
[See page 64, Ex. 2.; also Prob. 10, p. 83.]
-. e * *
20. Use the preceding eacercise to shew that the middle points of the
sides of a triangle and the feet of the perpendiculars let fall from the
vertices on the opposite sides, are concyclic.
21. If a series of triangles are drawn standing on a fixed base, and
having a given vertical angle, shew that the bisectors of the vertical
angles all pass through a fixed point.
22. ABC is a triangle inscribed in a circle, and E the middle point
of the are subtended by BC on the side remote from A : if through E
a diameter ED is drawn, shew that the angle DEA is half the difference
of the angles at B and C.
172 GEOMETRY.
&
TANGENCY.
DEFINITIONS AND FIRST PRINCIPLES.
1. A secant of a circle is a straight line of indefinite
length which cuts the circumference at two points.
2. If a secant moves in such a way that the two points
in which it cuts the circle continually approach one another,
then in the ultimate position when these two points become
one, the secant becomes a tangent to the circle, and is said to
touch it at the point at which the two intersections coincide.
This point is called the point of contact.
For instance :
(i) Let a secant cut the circle at the points P
P and Q, and suppose it to recede from the
centre, moving always parallel to its original
position ; then the two points P and Q will
clearly approach one another and finally coin- Q
cide.
In the ultimate position when P and Q Q
become one point, the straight line becomes a
tangent to the circle at that point.
Ö
(ii) Let a secant cut the circle at the points
P and Q, and suppose it to be turned about
the point P so that while P remains fixed, Q
moves on the circumference nearer and nearer
to P. Then the line PQ in its ultimate
position, when Q coincides with P, is a
tangent at the point P.
Since a secant can cut a circle at two points only, it is clear
that a tangent can have only one point in common with the
circumference, namely the point of contact, at which two
points of section coincide. Hence we may define a tangent as
follows:
3. A tangent to a circle is a straight line which meets
the circumference at one point only ; and though produced
indefinitely does not cut the circumference.

TANGENCY, 173


TNC T I

\\
Fig.1. Fig. 2. Fig. 3.
4. Let two circles intersect (as in Fig. 1) in the points
P and Q, and let one of the circles turn about the point P,
which remains fixed, in such a way that Q continually
approaches P. Then in the ultimate position, when Q coin-
cides with P (as in Figs. 2 and 3), the circles are said to
touch one another at P.
Since two circles cannot intersect in more than two points,
two circles, which touch one another cannot have more than
one point in common, namely the point of contact at which the
two points of section coincide. Hence circles are said to touch
one another when they meet, but do not cut one another.
NoTE. When each of the circles which meet is outside the other, as
in Fig. 2, they are said to touch ome another externally, or to have
external contact: when one of the circles is within the other, as in
Fig. 3, the first is said to touch the other internally, or to have
internal contact with it.
INFERENCE FROM DEFINITIONS 2 AND 4.
If in Fig. 1, T2P is a common chord of two circles one
of which is made to turn about P, then when Q is brought
into coincidence with P, the line TP passes through two coinci-
dent points on each circle, as in Figs. 2 and 3, and therefore
becomes a tangent to each circle. Hence
Two circles which touch one another have a common tangent at
their point of contact.
174 GEOMETRY.
THEOREM 46.
The tangent at any point of a circle is perpendicular to the
radius drawn to the point of contact. , , , , ,

P Q T
Ilet PT be a tangent at the point P to a circle whose centre
is O.
It is required to prove that PT is perpendicular to the radius OP.
Proof. Take any point Q in PT, and join OQ.
Then since PT is a tangent, every point in it except P is
outside the circle.
... OQ is greater than the radius OP.
And this is true for every point Q in PT;
... OP is the shortest distance from O to PT.
Hence OP is perp. to PT. Theor. 12, Cor. 1,
Q.E.D.
COROLLARY 1. Since there can be only one perpendicular
to OP at the point P, it follows that one and only one tangent
can be drawn to a circle at a given point on the circumference.
COROLLARY 2. Since there can be only one perpendicular
to PT at the point P, it follows that the perpendicular to a
tangent at its point of contact passes through the centre,
COROLLARY 3. Since there can be only one perpendicular
from O to the line PT, it follows that the radius drawn perpen-
dicular to the tangent passes through the point of contact.
TANGENCY, 175
THEOREM 46. [By the Method of Limits.]
The tangent at any point of a circle is perpendicular to the
Tadius drawn to the point of contact.
(Q)
Fig.1. Fig.2.
Let P be a point on a circle whose centre is O.
It is required to prove that the tangent at P is perpendicular to
the radius OP.
Let RQPT (Fig. 1) be a secant cutting the circle at Q and P.
Join OQ, OP.
Proof. Because OP = OQ,
... the Z OQP = the OPQ ;
... the supplements of these angles are equal;
that is, the 4 OQR = the 4 OPT,
and this is true however near Q is to P.
Now let the secant QP be turned about the point P so that
Q continually approaches and finally coincides with P; then
in the ultimate position,
(i) the secant RT becomes the tangent at º Fig. 2
(ii) OG) coincides with OP; 8. As
and therefore the equal / "OQR, OPT become adjacent,
... OP is perp. to RT. Q.E.D.
NotE. The method of proof employed here is known as the Method
of Limits.

176 GEOMETRY.
THEOREM 47.
Two tungents can be drawn to a circle from an external point.
Let PQR be a circle whose centre is O, and let T be an
external point.
It is required to prove that there can be two tangents drawn to
the circle from T.
Join OT, and let TSO be the circle on OT as diameter.
This circle will cut the G) PQR in two points, since T is
without, and O is within, the G) PQR. Let P and Q be these
points.
Join TP, TG); OP, OQ.
Proof. Now each of the 4 "TPO, TGO, being in a semi-
circle, is a rt. angle;
... TP, TG, are perp. to the radii OP, OQ respectively.
..". TP, TG, are tangents at P and Q. Theor. 46.
Q.E.D.
COROLLARY. The two tangentsº a circle from an eſºternal
point are equal, and subtend equal aºles at the centre.
For in the A*TPO, TGO,
the 4 "TPO, TCAO are right angles,
because {the hypotenuse TO is common,
and OP=OQ, being radii;
... TP = TQ,
and the Z, TOP = the 4. TOQ. Theor. 18.

TANGENCY, 177
EXERCISES ON THE TANGENT.
(Numerical and Graphical.)
1. Draw two concentric circles with radii 5-0 cm. and 3:Q cm.
Draw a series of chords of the former to touch the latter. Calculate
and measure their lengths, and account for their being equal.
2. In a circle of radius 1'0" draw a number of chords each l'6" in
length. Shew that they all touch a concentric circle, and find its
radius.
3. The diameters of two concentric circles are respectively 10.0 cm.
and 5.0 cm.; find to the nearest millimetre the length of any chord of
the outer circle which touches the inner, and check your work by
measurement.
4. In the figure of Theorem 47, if OP=5", TO=13", find the length:
of the tangents from T. Draw the figure (scale 2 cm. to 5"), and
measure to the nearest degree the angles subtended at O by the
tangents.
5. The tangents from T to a circle whose radius is 0.7" are each 2'4”
in length. Find the distance of T from the centre of the circle. Draw
the figure and check your result graphically.
(Theoretical.)
6. The centre of any circle which touches two intersecting straight
lines must lie on the bisector of the angle between them.
7. AB and AC are two tangents to a circle whose centre is O ; shew
that AO bisects the chord of contact BC at right angles.
8. If PQ is joined in the figure of Theorem 47, show that the angle
PTQ is double the angle OPQ.
9. Two parallel º to a circle intercept on any third tangent
a segment which subtends a right angle at the centre.
10. The diameter of a circle bisects all chords which are parallel to
the tangent at either extremity.
11. Find the locus of the centres of all circles which touch a given
8traight line at a given point.
12. Find the locus of theºcontres of all circles which touch each of
two parallel straight lines.
, 13. Find the locus of the centres of all circles which touch each of two
intersecting straight limes of unlimited length.
14. In any quadrilateral circumscribed about a circle, the sum of one
pair of opposite sides is equal to the sum of the other pair.
State and prove the converse theorem.
15. If a quadrilateral is described about a circle, the angles sub-
tended at the centre by any two opposite sides are supplementary.
II. S. G. M
178 GEOMETRY.
THEOREM 48.
If two circles touch one another, the centres and the point of
contact are in one straight line.
*
T
Let two circles whose centres are O and Q touch at the
point P.
It is required to prove that O, P, and Q are in one straight line.
Join OP, QP.
Proof. Since the given circles touch at P, they have a
common tangent at that point. Page 173.
Suppose PT to touch both circles at P.
Then since OP and QP are radii drawn to the point of
Contact,
‘. OP and QP are both perp. to PT ;
... OP and QP are in one st. line. Theor. 2.
That is, the points O, P, and Q are in one st. line. Q.E.D.
COROLLARIES. (i) If two circles touch eaternally the distance
between their centres is equal to the sum of their radii.
(ii) If two circles touch internally the distance between their
centres is equal to the difference of their radii.

THE CONTACT OF CIRCLES. 179
EXERCISES ON THE CONTACT OF CIRCLES.
(Numerical and Graphical.)
*1. From centres 2.6 apart draw two circles with radii l'7" and 0:9"
respectively. Why and where do these circles touch one another ?
If circles of the above radii are drawn from centres 0-8" apart, prove
that they touch. How and why does the contact differ from that in
the former case?
*
2. Draw a triangle ABC in which a = 8 cm., b=7 cm., and c=6 cm.
From A, B, and C as centres draw circles of radii 2.5 cm., 3-5 cm., and
4-5 cm. respectively; and shew that these circles touch in pairs.
3. In the triangle ABC, right-angled at C, a = 8 cm. and b = 6 cm.;
and from centre A with radius 7 cm. a circle is drawn. What must be
the radius of a circle drawn from centre B to touch the first circle 2
4. A and B are the centres of two fixed circles which touch in-
ternally. If P is the centre of any circle which touches the larger circle
internally and the smaller externally, prove that AP+BP is constant.
If the fixed circles have radii 5-0 cm. and 3.0 cm, respectively, verify
the general result by taking different positions for P.
5. AB is a line 4" in length, and C is its middle point. On AB,
AC, CB semicircles are described. Shew that if a circle is inscribed in
the space enclosed by the three semicircles its radius must be #".
(Theoretical.)
6 A straight line is drawn through the point of contact of two circles
whose centres are A and B, cutting the circumferences at P and Q re-
spectively; shew that the radii AP and BQ are parallel.
#
7. Two circles touch externally, and through the point of contact a
straight line is drawn terminated by the circumferences; shew that the
tangents at its extremities are parallel.
X8. Find the locus of the centres of all circles
(i) which touch a given circle at a given point ;
(ii) which are of given radius and touch a given circle.
9. From a given point as centre describe a circle to touch a given
circle. How many solutions will there be?
10. Describe a circle of radius a to touch a given circle of radius b
at a given point. How many solutions will there be?
180 GEOMETRY,
THEOREM 49. [Euclid III. 32.]
The angles made by a tangent to a circle with a chord drawn
from the point of contact are respectively equal to the angles in
the alternate Segments of the circle.
A

E -5 F
Let EF touch the G) ABC at B, and let BD be a chord drawn
from B, the point of contact.
It is required to prove that
(i) the 4- FBD = the angle in the alternate segment BAD;
(ii) the 4 EBD = the angle in the alternate segment BCD.
Let BA be the diameter through B, and C any point in the
arc of the segment which does not contain A.
Join AD, DC, CB.
Proof. Because the LADB in a semi-circle is a rt. angle,
... the 4 "DBA, BAD together = a rt. angle.
But since EBF is a tangent, and BA a diameter,
... the A. FBA is a rt. angle.
... the 4 FBA = the 4 "DBA, BAD together.
Take away the common A. DBA,
then the 4- FBD = the A. BAD, which is in the alternaté segment.
Again because ABCD is a cyclic quadrilateral,
... the A. BCD = the supplement of the A. BAD
= the supplement of the 4- FBD
= the A. EBD ;
... the 4 EBD = the A. BCD, which is in the alternate segment.
Q.E D.
ALTERNATE SUCGMENT. 181
EXERCISES ON THEOREM 49.
1. In the figure of Theorem 49, if the 4- FBD =72°, write down the
values of the 4 "BAD, BCD, EBD.
2. Use this theorem to shew that tangents to a circle from an
external point are equal.
3. Through A, the point of contact of two circles, chords APQ,
AXY are drawn; shew that PX and QY are parallel.
Prove this (i) for internal, (ii) for external contact.
4. AB is the common chord of two circles, one of which passes
through O, the centre of the other: prove that OA bisects the angle
between the common chord and the tangent to the first circle at A.
5. Two circles intersect at A and B; and through P, any point on
one of them, straight lines PAC, PBD are drawn to cut the other at C
and D : shew that CD is parallel to the tangent at P.
6. If from the point of contact of a tangent to a circle a chord is
drawn, the perpendiculars dropped on the tangent and chord from the
middle point of either arc cut off by the chord are equal.
EXERCISES ON THE METHOD OF LIMITS,
1. Prove Theorem, 49 by the Method of Limits.
[Let ACB be a segment of a circle of
which AB is the chord; and let PAT’ be any
secant through A. Join PB.
Then the A BCA= the Z-BPA;
Theor. 39.
º this is true however mear P approaches
to A.
If P moves up to coincidence with A,
then the secant PAT’ becomes the tangent
AT, and the A. BPA becomes the A- BAT.
..., ultimately, the A. BAT = the A. BCA, in
the alt. segment.]
2. From Theorem 31, prove by the
Method of Limits that
The Straight line drawn perpendicular to the diameter of a circle at its
extremity is a tangent.
3. Deduce Theorem 48 from the property that the line of centres
bisects a common chord at right angles.
4. Deduce Theorem 49 from Ex. 5, page 163.
5. Deduce Theorem 46 from Theorem 41.
$

182 GEOMETRY.
PROBLEMS.
GEOMETRICAL ANALYSIS.
Hitherto the Propositions of this text-book have been
arranged Synthetically, that is to say, by building up known
Tesults in order to obtain a new result.
But this arrangement, though convincing as an argument,
in most cases affords little clue as to the way in which the
construction or proof was discovered. We therefore draw the
student's attention to the following hints.
In attempting to solve a problem begin by assuming the
required result; then by working backwards, trace the conse-
quences of the assumption, and try to ascertain its dependence
on some condition or known theorem which suggests the
necessary construction. If this attempt is successful, the
steps of the argument may in general be re-arranged in
reverse order, and the construction and proof presented in a
synthetic form.
This unravelling of the conditions of a proposition in order
to trace it back to some earlier principle on which it depends,
is called geometrical analysis: it is the natural way of attack-
ing the harder types of exercises, and it is especially useful in
solving problems.
Although the above directions do not amount to a method,
they often furnish a very effective mode of searching for a
suggestion. The approach by analysis will be illustrated in
some of the following problems. [See Problems 23, 28, 29.]
PROBLEMS ON CIRCLES. 183
PROBLEM 20. -
Given a circle, or an arc of a circle, to find its centre.
Let ABC be an arc of a circle
whose centre is to be found.
Construction. Take two chords
AB, BC, and bisect them at right
angles by the lines DE, FG, meeting
at O. Prob. 2.
Then O is the required centre.
Proof. Every point in DE is equi.
distant from A and B. Prob. 14. >~
And every point in FG is equidistant from B and C.
... O is equidistant from A, B, and C.
... O is the centre of the circle ABC. Theor. 33.
PROBLEM 21.
To bisect a given arc.
Let ADB be the given arc to be bisected. D
e g © &
Construction. Join AB, and bisect it at
right angles by CD meeting the arc at D.
Prol. 2.,
Then the arc is bisected at D. A B
Proof. Join DA, DB. X
Then every point on CD is equidistant from A and B;
- Prob. 14.
... DA = DB ;
... the 4 DBA = the 4. DAB ; Theorem 6.
... the arcs, which subtend these angles at the O*, are equal ;
that is, the arc DA = the arc DB. -

184 GEOMETRY.
PROBLEM 22.
To draw a tangent to a circle from a given external point.
Let PQR be the given circle, with its centre at of and let T
be the point from which a tangent is to be drawn.
Construction. Join TO, and on it describe a semi-circle TPO
to cut the circle at P.
Join TP. -
Then TP is the required tangent.
Proof. Join OP.
Then since the Z TPO, being in a semi-circle, is a rt. angle,
... TP is at right angles to the radius OP.
... TP is a tangent at P. Theor. 46.
Since the semi-circle may be described on either side of TO,
a second tangent TQ can be drawn from T, as shewn in the
figure.
NoTE. Suppose the point T to approach the given circle, then the
angle PTG gradually increases. When T reaches the circumference,
the angle PTQ becomes a straight angle, and the two tangents coincide.
When T enters the circle, no tangent can be drawn. [See Obs. p. 94.]

COMMON TANGENTS. 185
PROBLEM 23.
To draw a common tangent to two circles.
Let A be the centre of the greater circle, and a its radius;
and let B be the centre of the smaller circle, and b its radius.
Analysis. Suppose DE to touch the circles at D and E.
Then the radii AD, BE are both perp. to DE, and therefore
par' to one another. º
Now if BC were drawn par' to DE, then the fig, DB would
be a rectangle, so that CD = BE =b. --
And if AD, BE are on the same side of AB,
then AC = a – b, and the 4 ACB is a rt. angle.
These hints enable us to draw BG first, and thus lead to the
following construction.
Construction. With centre A, and radius equal to the
difference of the radii of the given circles, describe a circle,
and draw BC to touch it. -
Join AC, and produce it to meet the circle (A) at D.
Through B draw the radius BE par' to AD and in the same
Sense. Join DE.
Then DE is a common tangent to the given circles.
Obs. Since two tangents, such as BC, can in general be
drawn from B to the circle of construction, this method will
, furnish two common tangents to the given circles. These are
called the direct common tangents.

186 GEOMETRY.
PROBLEM 23. (Continued.)

C
<2
Again, if the circles are external to one another two more
common tangents may be drawn.
Analysis. In this case we may suppose DE to touch the
circles at D and E so that the radii AD, BE fall on opposite sides
of AB.
Then BC, drawn par to the supposed common tangent DE,
would meet AD produced at C ; and we should now have
AC = AD + DC = a + b ; and, as before, the 2 ACB is a rt. angle.
Hence the following construction.
Construction. With centre A, and radius equal to the sum
of the radii of the given circles, describe a circle, and draw
BC to touch it.
Then proceed as in the first case, but draw BE in the sense
opposite to AD.
Obs. As before, two tangents may be drawn from B to the
circle of construction ; hence two common tangents may be
thus drawn to the given circles. These are called the
transverse common tangents.
[We leave as an exercise to the student the arrangement of the proof
in synthetic form.] *
COMMON TANGENTS. 187
EXERCISES ON COMMON TANGENTS.
(Numerical and Graphical.)
1. How many common tangents can be drawn in each of the
following cases?
(i) when the given circles intersect;
(ii) when they have external contact;
(iii) when they have internal contact.
Illustrate your answer by drawing two circles of radii l’4” and 1 0"
respectively,
(i) with l'0" between the centres;
(ii) with 2'4" between the centres;
(iii) with 0'4" between the centres;
(iv) with 3'0" between the centres.
Draw the common tangents in each case, and note where the general
construction fails, or is modified. *
2. Draw two circles with radii 2.0" and 0.8", placing their centres
2:0° apart. Draw the common tangents, and find their lengths between
the points of contact, both by calculation and by measurement.
3. Draw all the common tangents to two circles whose centres are
l'8" apart and whose radii are 0.6" and 12" respectively. Calculate and
measure the length of the direct common tangents.
4. Two circles of radii l'7" and 10" have their centres 2-1" apart.
Draw their common tangents and find their lengths. Also find the
length of the common chord. Produce the common chord and shew
by measurement that it bisects the common tangents.
5. Draw two circles with radii l'6" and 0.8" and with their centres
3'0" apart, Draw all their common tangents.
6. Draw the direct common tangents to two equal circles.
( Theoretical. )
7. If the two direct, or the two transverse, common tangents are
drawn to two circles, the parts of the tangents intercepted between the
points of contact are equal.
8. If four common tangents are drawn to two circles external to
one another, shew that the two direct, and also the two transverse,
tangents intersect on the line of centres.
9. Two given circles have external contact at A, and a direct common
, tangent is drawn to touch them at P and Q : shew that PQ subtends a
right angle at the point A.
ºf
188 GEOMETRY.
ON THE CONSTRUCTION OF CIRCLES.
In order to draw a circle we must know (i) the position of
the centre, (ii) the length of the radius.
(i) To find the position of the centre, two conditions are
needed, each giving a locus on which the centre must lie ; so
that the one or more points in which the two loci intersect
are possible positions of the required centre, as explained on
page 93.
(ii) The position of the centre being thus fixed, the radius
is determined if we know (or can find) any point on the
circumference.
Hence in order to draw a circle three independent data are
required.
For example, we may draw a circle if we are given
(i) three points on the circumference;
or (ii) three tangent lines;
or (iii) one point on the circumference, one tangent, and its point
of contact.
It will however often happen that more than one circle can be drawn
satisfying three given conditions. *
Before attempting the constructions of the next Exercise the
student should make himself familiar with the following loci.
(i) The locus of the centres of circles which pass through two
given points. *
(ii) The locus of the centres of circles which touch a given straight
line at a given point.
(iii) The locus of the centres of circles which touch a given circle
at a given point.
(iv) The locus of the centres of circles which touch a given straight
line, and have a given radius. #
(v) The locus of the centres of circles which touch a given circle,
and have a given radius.
(vi) The locus of the centres of circles which touch two given
straight lines.
THE CONSTRUCTION OF CIRCLES. 189
EXERCISES.
1. Draw a circle to pass through three given points.
2. If a circle touches a given line PQ at a point A, on what line
must its centre lie Ż
If a circle passes through two given points A and B, on what line
must its centre lie Ż
Hence draw a circle to touch a straight line PQ at the point A, and
to pass through another given point B.
3. If a circle touches a given circle whose centre is C at the point A,
on what line must its centre lie Ż
Draw a circle to touch the given circle (C) at the point A, and to pass
through a given point B.
4. A point P is 4.5 cm. distant from a straight line AB. Draw two
circles of radius 3:2 cm. to pass through P and to touch AB.
5. Given two circles of radius 3.0 cm. and 2.0 cm. respectively,
their centres being 6-0 cm. apart ; draw a circle of radius 3-5 cm. to
touch each of the given circles externally.
How many solutions will there be? What is the radius of the
smallest circle that touches each of the given circles externally 2
6. If a circle touches two straight lines OA, OB, on what line must
its centre lie 2
Draw OA, OB making an angle of 76°, and describe a circle of radius
l'2" to touch both lines.
7. Given a circle of radius 3-5 cm., with its centre 5.0 cm. from a
given straight line AB; draw two circles of radius 2.5 cm. to touch the
given circle and the line AB.
8. Devise a construction for drawing a circle to touch each of two
parallel straight lines and a transversal.
Shew that two such circles can be drawn, and that they are equal.
9. Describe a circle to touch a given circle, and also to touch a
given straight line at a given point. [See page 311.]
10. Describe a circle to touch a given straight line, and to touch a
given circle at a given point.
11. , Shew how to draw a circle to touch each of three given straight
lines of which no two are parallel.
How many such circles can be drawn 2
[Further Examples on the Construction of Circles will be found on
pp. 246, 311.]
190 GEOMETRY.
PROBLEM 24.
On a given straight line to describe a segment of a circle which
shall contain an angle equal to a given angle.
Let AB be the given st. line, and C the given angle.
It is required to describe on AB a segment of a circle containing
an angle equal to C.
Construction. At A in BA, make the 4. BAD equal to the Z.O.
From A draw AG perp. to AD.
Bisect AB at rt. angles by FG, meeting AG in G. Prob. 2.
Proof. Join GB.
Now every point in FG is equidistant from A and B;
Prob. 14.
.*. GA = GB.
With centre G, and radius GA, draw a circle, which must
pass through B, and touch AD at A. Theor. 46.
Then the segment AHB, alternate to the 4. BAD, contains an
angle equal to C. Theor. 49.
NOTE. In the particular case when the given angle is a rt. angle, the
segment required will be the semi-circle on AB as diameter. [Theorem.41.]

PROBLEMS. 191
COROLLARY. To cut off from a given circle a segment containing
a given angle, it is enough to draw a tangent to the circle, and from
the point of contact to draw a chord making with the tangent an
angle equal to the given angle.
It was proved on page 161 that
The locus of the vertices of triangles which stand on the same base
and have a given vertical angle, is the arc of the segment standing
on this base, and containing an angle equal to the given angle.
The following Problems are derived from this result by the
Method of Intersection of Loci [page 93].
EXERCISES.
1. Describe a triangle on a given base having a given vertical angle
and having its verteac on a given Straight line.
2. Construct a triangle having given the base, the vertical angle, and
(i) one other side.
(ii) the altitude.
(iii) the length of the median which bisects the base.
(iv) the foot of the perpendicular from the verteac to the base.
3. Construct a triangle having given the base, the vertical angle, and
the point at which the base is cut by the bisector of the vertical angle.
[Let AB be the base, X the given point in it, and K the given angle.
On AB describe a segment of a circle containing an angle equal to K;
gomplete the Oce by drawing the are APB. Disect the are APB at P:
join PX, and produce it to meet the O* at C. Then ABC is the
required triangle.]
4. Construct a triangle having given the base, the vertical angle, and
the sum of the remaining sides.
[Let AB be the given base, K the given angle, and H a line equal to
the sum of the sides. On AB describe a segment containing an angle
equal to K, also another segment containing an angle equal to half the
A. K. With centre A, and radius H, describe a circle cutting the are of
the latter segment at X and Y. Join AX (or AY) cutting the arc of the
first segment at C. Then ABC is the required triangle.]
5. Construct a triangle having given the base, the vertical angle, and
the difference of the remaining sides.
. 192 GEOMETRY.
CIRCLES IN RELATION TO RECTILINEAL FIGURES.
DEFINITIONS.
1. A Polygon is a rectilineal figure bounded by more than
four sides.
A Polygon of five sides is called a Pentagon,
25 sia, sides 92 Hexagon,
; } Seven sides 25 Heptagon,
33 eight sides 9) Octagon,
53 ten sides 75 Decagon,
22 twelve sides 33 Dodecagon,
25 fifteen sides » Quindecagon.
2. A Polygon is Regular when all its sides are equal, and
all its angles are equal.
3. A rectilineal figure is said to be in-
scribed in a circle, when all its angular points
are on the circumference of the circle ; and a
circle is said to be circumscribed about a recti-
lineal figure, when the circumference of the
circle passes through all the angular points of
the figure.

4. A circle is said to be inscribed in a
rectilineal figure, when the circumference of
the circle is touched by each side of the figure;
and a rectilineal figure is said to be circum-
scribed about a circle, when each side of the
figure is a tangent to the circle. sº-º

PROBLEMS ON TRIANGLES AND CIRCLES. 193
PROBLEM 25.
To circumscribe a circle about a given triangle.
Let ABC be the triangle, about which a circle is to be
drawn.
Construction. Bisect AB and AC at rt. angles by DS and
ES, meeting at S. Prob. 2.
Then S is the centre of the required circle.
Proof. Now every point in DS is equidistant from A
and B; Prob. 14.
and every point in ES is equidistant from A and C ;
... S is equidistant from A, B, and C.
With centre S, and radius SA describe a circle ; this will
pass through B and C, and is, therefore, the required circum-
circle.
Obs. It will be found that if the given triangle is acute-
angled, the centre of the circum-circle falls within it; if it
is a right-angled triangle, the centre falls on the hypotenuse:
if it is an obtuse-angled triangle, the centre falls without the
triangle.
NoTE. From page 94 it is seen that if S is joined to the middle
point of BC, then the joining line is perpendicular to BC.
Hence the perpendiculars drawn to the sides of a triangle from their
middle points are concurrent, the point of intersection being the centre
of the circle circumscribed about the triangle.
H. S. G. N

194 GEOMETRY.
PROBLEM 26.
To inscribe a circle in a given triangle.
Let ABC be the triangle, in which a circle is to be inscribed.
Construction, Bisect the 4 " ABC, ACB by the st. lines Bl,
Cl, which intersect at 1. Prob. 1.
Then I is the centre of the required circle.
Proof. From 1 draw ID, IE, IF perp. to BC, CA, AB.
Then every point in BI is equidistant from BC, BA; Prob. 15.
.*. ID = |F.
And every point in Cl is equidistant from CB, CA;
" . ID = |E.
‘. ID, IE, IF are all equal.
With centre I and radius ID draw a circle ;
this will pass through the points E and F.
Also the circle will touch the sides BC, CA, AB,
because the angles at D, E, F are right angles.
... the G) DEF is inscribed in the AABC.
NOTE. From II., p. 96 it is seen that if Al is joined, then Al bisects
the angle BAC : hence it follows that
The bisectors of the angles of a triangle are concurrent, the point of
intersection being the centre of the inscribed circle.
DEFINITION.
A circle which touches one side of a triangle and the other
two sides produced is called an escribed circle of the triangle.

PROBLEMS ON TRIANGLES AND CIRCLES. 195
PROBLEM 27.
To draw an escribed circle of a given triangle.
i
!
A B F 'TD
Let ABC be the given triangle of which the sides AB, AC are
produced to D and E.
It is required to describe a circle touching BC, and AB, AC
produced. *
Construction. Bisect the AECBD, BCE by the st. lines Bl,
Cli which intersect at 11.
Then li is the centre of the required circle.
Proof. From 11 draw I, F, IG, I, H perp. to AD, BC, AE.
Then every point in Bl, is equidistant from BD, BC; Prob. 15,
... liF = lig.
Similarly 1,6=l, H.
. 11F, I, G, 11B are all equal.
With centre 1, and radius 11F describe a circle;
this will pass through the points G and H.
Also the circle will touch AD, BC, and AE,
because the angles at F, G, H are rt. angles.
... the G) FGH is an escribed circle of the AABC.
NoTE 1. It is clear that every triangle has three escribed circles.
Their centres are known as the Ex-centres.
NoTE 2. It may be shewn, as in II., page 96, that if Ali is joined,
then Ali bisects the angle BAC; hence it follows that
.The bisectors of two eacterior angles of a triangle and the bisector of the
third angle are concurrent, the point of intersection being the centre of an
escribed circle.

196 GEOMETRY.
PROBLEM 28.
In a given circle to inscribe a triangle equiangular to a given
triangle.
G
Let ABC be the given circle, and DEF the given triangle.
Analysis. A A ABC, equiangular to the A DEF, is inscribed
in the circle, if from any point A on the O* two chords AB, AC
can be so placed that, on joining BC, the A-B = the 4-E, and
the 4 C = the A. F.; for then the Z A = the A. D. Theor. 16.
Now the A- B, in the segment ABC, suggests the equal angle
between the chord AC and the tangent at its extremity
(Theor. 49.); so that, if at A we draw the tangent GAH,
then the A. HAC = the A. E.;
and similarly, the 4. GAB = the A. F.
Reversing these steps, we have the following construction.
Construction. At any point A on the O* of the G) ABC
draw the tangent GAH. Prob. 22.
At A make the A. GAB equal to the 4- F,
and make the A. HAC equal to the A. E.
Join BC.
Then ABC is the required triangle.
NOTE. In drawing the figure on a larger scale the student should
shew the construction lines for the tangent GAH and for the angles
GAB, HAC. A similar remark applies to the next Problem.

PROBLEMS ON CIRCLES AND TRIANGLES. 197
PROBLEM 29.
About a given circle to circumscribe a triangle equiangular to
a given triangle.
L


C
A #
M B N
Let ABC be the given circle, and DEF the given triangle.
Analysis. Suppose LMN to be a circumscribed triangle in
which the A. M = the 4-E, the A N = the 4. F, and consequently,
the A. L = the A. D.
Let us consider the radii KA, KB, KC, drawn to the points of
contact of the sides ; for the tangents LM, MN, NL could be
drawn if we knew the relative positions of KA, KB, KC, that is,
if we knew the z* BKA, BKC.
Now from the quad' BKAM, since the 4 × B and A are rt. A ',
the A BKA = 180° – M = 180° – E ;
similarly the 4 BKC = 180° – N = 180° – F.
Hence we have the following construction.
Construction. Produce EF both ways to G and H.
Find K the centre of the G) ABC,
and draw any radius KB.
At K make the Z. BKA equal to the A DEG ;
and make the Z. BKC equal to the Z. DFH.
Through A, B, C draw LM, MN, NL perp. to KA, KB, KC.
Then LMN is the required triangle.
[The student should now arrange the proof synthetically. ]
198 GEOMETRY.
EXERCISES.
ON CIRCLES AND TRIANGLEs.
(Inscriptions and Circumscriptions.)
1. In a circle of radius 5 cm. inscribe an equilateral triangle; and
about the same circle circumscribe a second equilateral triangle. In
each case state and justify your construction.
2. Draw an equilateral triangle on a side of 8 cm., and find by
calculation and measurement (to the nearest millimetre) the radii of
the inscribed, circumscribed, and escribed circles.
Explain why the second and third radii are respectively double and
treble of the first.
3. Draw triangles from the following data :
(i) a =2'5", B=66°, C=50°;
(ii) a =2'5", B=72°, C=44°;
(iii) a =2'5", B=41°, C=23°.
Circumscribe a circle about each triangle, and measure the radii
to the nearest hundredth of an inch. Account for the three results
being the same, by comparing the vertical angles.
4. In a circle of radius 4 cm. inscribe an equilateral triangle.
Calculate the length of its side to the nearest millimetre; and verify
by measurement.
Find the area of the inscribed equilateral triangle, and shew that it
is one quarter of the circumscribed equilateral triangle.
5. In the triangle ABC, if I is the centre, and r the length of the
radius of the in-circle, shew that
A 15C =#ar; A ICA=$br; A IAB =#cr.
Hence prove that A ABC =# (a +b+c) r.
Verify this formula by measurements for a triangle whose sides are
9 cm., 8 cm., and 7 cm.
6. If ri is the radius of the ex-circle opposite to A, prove that
AABC=# (b+ c – a)ri.
If a = 5 cm., b=4 cm., c=3 cm., verify this result by measurement.
7. Find by measurement the circum-radius of the triangle ABC in
which a = 6.3 cm., b = 3.0 cm., and c=5.1 cm.
Draw and measure the perpendiculars from A, B, C to the opposite
sides. If their lengths are represented by p, p, ps, verify the following
statement :
bc ca ab
circum-radius =::-- * = *-.
2p, 2p2 2ps
PROBLEMS ON CIRCLES AND SQUARES. 199
EXERCISES. e
ON CIRCLES AND SQUAREs.
(Inscriptions and Circumscriptions.)
1. Draw a circle of radius l'5", and find a construction for inscribing
a square in it.
Calculate the length of the side to the nearest hundredth of an inch,
and verify by measurement.
Find the area of the inscribed square.
2. Circumscribe a square about a circle of radius l'5", shewing all
lines of construction.
Prove that the area of the square circumscribed about a circle is
double that of the inscribed square.
3. Draw a square on a side of 7.5 cm., and state a construction for
inscribing a circle in it.
Justify your construction by considerations of symmetry.
4. Circumscribe a circle about a square whose side is 6 cm.
Measure the diameter to the nearest millimetre, and test your
drawing by calculation.
5. In a circle of radius l'8" inscribe a rectangle of which one side
measures 3 0". Find the approximate length of the other side.
Of all rectangles inscribed in the circle shew that the square has the
greatest area.
6. A square and an equilateral triangle are inscribed in a circle.
If a and b i. the lengths of their sides, shew that
3a*=2b°.
7. ABCD is a square inscribed in a circle, and P is any point on the
are AD; shew that the side AD subtends at P an angle three times as
great as that subtended at P by any one of the other sides.
(Problems. State your construction, and give a theoretical proof.)
8. Circumscribe a rhombus about a given circle.
9. Inscribe a square in a given square ABCD, so that one of its
angular points shall be at a given point X in AB.
10. In a given square inscribe the square of minimum area.
ll. Describe (i) a circle, (ii) a square about a given rectangle.
12. Inscribe (i) a circle, (ii) a square in a given quadrant.
*
*
200 GEOMETRY.
ON CIRCLES AND REGULAR POLY GONS.
PROBLEM 30.
To draw a regular polygon (i) in (ii) about a given circle.
Let AB, BC, CD, ... be consecutive
sides of a regular polygon inscribed in
a circle whose centre is O.
*
Then AOB, BOC, COD, ... are con-
gruent isosceles triangles. And if
the polygon has n sides, each of the
A "AOB, BOC, COD, ... = sº º
(i) Thus to inscribe a polygon of n sides in a given circle,"
360°
draw an angle AOB at the centre equal to This gives
the length of a side AB; and chords equal to AB may now be
set off round the circumference. The resulting figure will
clearly be equilateral and equiangular.
(ii) To circumscribe a polygon of n sides about the circle,
the points A, B, C, D, ... must be determined as before, and
tangents drawn to the circle at these points. The resulting
figure may readily be proved equilateral and equiangular.
NOTE. This method gives a strict geometrical construction only when
the angle sº can be drawn with ruler and compasses.
EXERCISES.
1. Give strict constructions for inscribing in a circle (radius 4 cm.)
(i) a regular hexagon; (ii) a regular octagon; (iii) a regular dodecagon.
2. About a circle of radius 1:5" circumscribe
(i) a regular hexagon; (ii) a regular octagon,
Test the constructions by measurement, and justify them by proof.
3. An equilateral triangle and a regular hexagon are inscribed in a
given circle, and a and b denote the lengths of their sides: prove that
(i) area of triangle=# (area of hexagon); (ii) a”=3b*.
4. By means of your protractor inscribe a regular heptagon in a
circle of radius 2". . Calculate and measure one of its angles; and
measure the length of a side.

PROBLEMS, ON CIRCLES AND POLYGONS. 201
*
PROBLEM 31.
To draw a circle (i) in (ii) about a regular polygon.
Let AB, BC, CD, DE, ... be con-
secutive sides of a regular polygon
of n sides.
E
Bisect the 4 "ABC, BCD by BO, ſ D
CO meeting at O.
Then O is the centre both of the
inscribed and circumscribed circle.
B C
Outline of Proof. : Join OD; and from the congruent
A*OCB, OCD, shew that OD bisects the Z. CDE. Hence we
conclude that
All the bisectors of the angles of the polygon meet at O.
(i) Prove that OB = OC = OD = ...; from Theorem 6.
Hence O is the circum-centre.
(ii) Draw OP, OQ, OR, ... perp. to AB, BC, CD, ... .
Prove that OP= OQ = OR = ...; from the congruent A. OBP,
OBQ, ... . Hence O is the in-centre.
EXERCISES.
1. Draw a regular hexagon on a side of 2:0". Draw the inscribed
and circumscribed circles. Calculate and measure their diameters to
the nearest hundredth of an inch.
2. Shew that the area of a regular hexagon inscribed in a circle is
three-fourths of that of the circumscribed hexagon.
Find the area of a hexagon inscribed in a circle of radius 10 cm. to
the nearest tenth of a sq. cm.
3. If ABC is an isosceles triangle inscribed in a circle, having each
of the angles B and C double of the angle A ; shew that BC is a side of
a regular pentagon inscribed in the circle.
4. On a side of 4 cm. construct (without protractor)
(i) a regular hexagon; (ii) a regular octagon.
In each case find the approximate area of the figure.

202 GEOMETRY.
THE CIRCUMFERENCE OF A CIRCLE.
By experiment and measurement it is found that the length
of the circumference of a circle is roughly 3} times the length of
its diameter: that is to Say
circumference
diameter
and it can be proved that this is the same for all circles.
A more correct value of this ratio is found by theory to be
3.1416; while correct to 7 places of decimals it is 3.1415926.
Thus the value 3} (or 3:1428) is too great, and correct to 2
places only.
The ratio which the circumference of any circle bears to its
diameter is denoted by the Greek letter-T ; so that
circumference = diameter × T.
Or, if r denotes the radius of the circle,
circumference = 2r x t = 2irr;
where to Twe are to give one of the values 3}, 3.1416, or
3.1415926, according to the degree of accuracy required in
the final result.
— 2 l *
= 3} nearly ;
NoTE. The theoretical methods by which tris evaluated to any
required degree of accuracy cannot be explained at this stage, but its
value may be easily verified by experiment to two decimal places.
For example: round a cylinder wrap a strip of paper so that the
ends overlap. At any point in the overlapping area prick a pin
through both folds. Unwrap and straighten the strip, then measure
the distance between the pin holes: this gives the length of the circum-
ference. Measure the diameter, and divide the first result by the second.

Ex. 1. From these data CIRCUMFERENCE, DIAMETER. VALUE OF tr.
find and record the value
of m. 16'0 cm. 5°l cm.
Find the mean of the 8.8% 2-8”
three results. 13.5" 4'3"
Ex. 2. A fine thread is wound evenly round a cylinder, and it is
found that the length required for 20 complete turns is 75'4". The
diameter of the cylinder is 1:2": find roughly the value of tr.
Ex. 3. A bicycle wheel, 28" in diameter, makes 400 revolutions
in travelling over 977 yards. From this result estimate the value of tr.
CIRCUMFERENCE AND AREA OF A CIRCLE. 203
THE AREA OF A CIRCLE.
Let AB be a side of a polygon of
n sides circumscribed about a circle
whose centre is O and radius r. Then
we have
Area of polygon
= m, . /\ AOB
= n. #AB x OD
= }. mAB x r
= }(perimeter of polygon) x r);
and this is true however many sides the polygon may have.
Now if the number of sides is increased without limit, the
perimeter and area of the polygon may be made to differ from
the circumference and area of the circle by quantities smaller
than any that can be named ; hence ultimately
Area of circle = }. circumference x r
= , , 27trix r
= r^2.
ALTERNATIVE METHOD,
W
Suppose the circle divided into any even number of sectors having
equal central angles: denote the number of sectors by m.
Let the sectors be placed side by side as represented in the diagram ;
then the area of the circle= the area of the fig. ABCD ;
and this is true however great n may be.
Now as the number of sectors is increased, each arc is decreased;
, so that (i) the outlines AB, CD tend to become straight, and
(ii) the angles at D and B tend to become rt. angles.

204 GEOMETRY.
Thus when n is increased without limit, the fig, ABCD ultimately
becomes a 2.ectangle, whose length is the semi-circumference of the circle,
and whose breadth is its radius.
... Area of circle=#. circumference x radius
=#.2irrx ratrº. .
THE AREA OF A SECTOR.

C
If two radii of a circle make an angle of 1°, they cut off
(i) an arc whose length = sāg of the circumference;
and (ii) a sector whose area = sān of the circle;
‘.. if the angle AOB contains D degrees, then
(i) the arc AB -: of the circumference;
(ii) the sector AOB -; of the area of the circle
== * of (; circumference x radius)
= }. arc AB x radius.
THE AREA OF A SEGMENT.
The area of a minor segment is found by subtracting from
the corresponding sector the area of the triangle formed by
the chord and the radii. Thus
Area of Segment ABC = Sector OACB – triangle AOB.
The area of a major segment is most simply found by
subtracting the area of the corresponding minor segment from
the area of the circle.
CIRCUMIFERENCE AND AREA OF A CIRCLE. 205
EXERCISES.
[In each case choose the value of ºr so as to give a result of the assigned
degree of accuracy.]
1. Find to the nearest millimetre the circumferences of the circles
whose radii are (i) 4-5 cm. (ii) 100 cm.
2. Find to the nearest hundredth of a square inch the areas of the
circles whose radii are (i) 2.3". (ii) 10'6".
3. Find to two places of decimals the circumference and area of a
circle inscribed in a square whose side is 3:6 cm.
4. In a circle of radius 7-0 cm. a square is described: find to the
nearest square centimetre the difference between the areas of the circle
and the square.
5. Find to the nearest hundredth of a square inch the area of the
circular ring formed by two concentric circles whose radii are 5’7” and
6. Shew that the area of a ring lying between the circumferences of
two concentric circles is equal to the area of a circle whose radius is
the length of a tangent to the inner circle from any point on the outer.
7. A rectangle whose sides are 80 cm. and 6.0 cm. is inscribed in a
circle. Calculate to the nearest tenth of a square centimetre the total
area of the four segments outside the rectangle.
8. Find to the nearest tenth of an inch the side of a square whose
area is equal to that of a circle of radius 5".
9. A circular ring is formed by the circumference of two concentric
circles. The area of the ring is 22 square inches, and its width is 10";
taking r as *.*, find approximately the radii of the two circles.
10. Find to the nearest hundredth of a square inch the difference
between the areas of the circumscribed and inscribed circles of an
equilateral triangle each of whose sides is 4".
11. Draw on squared paper two circles whose centres are at the
points (l'5", 0) and (0, '8"), and whose radii are respectively '7" and
1-0". Prove that the circles touch one another, and find approximately
their circumferences and areas.
12. Draw a circle of radius l'0" having the point (1.6", l.2") as
centre. Also draw two circles with the origin as centre and of radii
1'0" and 3.0" respectively. Shew that each of the last two circles
touches the first.
*
206 GEOMETRY.
EXERCISES.
ON THE INSCRIBED, CIRCUMSCRIBED, AND ESCRIBED CIRCLES OF A
TRIANGLE.
(Theoretical.)
1. Describe a circle to touch two parallel straight lines and a third
straight line which meets them. Shew that two such circles can be
drawn, and that they are equal.
2. Triangles which have equal bases and equal vertical angles have
equal circumscribed circles.
3. ABC is a triangle, and I, S are the centres of the inscribed and
circumscribed circles; if A, ), S are collinear, shew that AB = AC.
4. The sum of the diameters of the inscribed and circumscribed
circles of a right-angled triangle is equal to the sum of the sides con-
taining the right angle.
5. If the circle inscribed in the triangle ABC touches the sides
at D, E, F ; shew that the angles of the triangle DEF are respectively
A B C
90-3, 90-5, 90-5.
6. If I is the centre of the circle inscribed in the triangle ABC,
and I, the centre of the escribed circle which touches BC; shew that
l, B, 11, C are concyclic.
7. In any triangle the difference of two sides is equal to the
difference of the segments into which the third side is divided at the
point of contact of the inscribed circle.
8. In the triangle ABC, I and S are the centres of the inscribed and
circumscribed circles: shew that IS subtends at A. an angle cqual to
half the difference of the angles at the base of the triangle.
Hence shew that if AD is drawn perpendicular to BC, then Al is the
bisector of the angle DAS.
9. The diagonals of a quadrilateral ABCD intersect at O; shew
that the centres of the circles circumscribed about the four triangles
AOB, BOC, COD, DOA are at the angular points of a parallelogram.
10. In any triangle ABC, if I is the centre of the inscribed circle,
and if Al is produced to meet the circumscribed circle at O; shew that
O is the centre of the circle circumscribed about the triangle BIC.
11. Given the base, altitude, and the radius of the circumscribed
circle; construct the triangle.
12. Three circles whose centres are A, B, C touch one another
externally two by two at D, E, F : shew that the inscribed circle of
the triangle ABC is the circumscribed circle of the triangle DEF.
THE ORTHOCENTRE OF A TRIANGLE. 207
THEOREMS AND EXAMPLES ON CIRCLES AND
TRIANGLES.
THE ORTHOCENTRE OF A TRIANGLE.
I. The perpendiculars drawn from the vertices of a triangle to the
opposite sides are concurrent.
In the AABC, let AD, BE be the
perpº drawn from A and B to the opposite
sides; and let them intersect at O. A

Join CO; and produce it to meet AB at F
F. E
It is required to shew that CF is perp. O
to AB. &
Join DE. B D C
Then, because the Z_* OEC, ODC are rt.
angles,
... the points O, E, C, D are concyclic :
‘.. the A. DEC= the 4-DOC, in the same segment;
= the vert. opp. / FOA.
Again, because the Z." AEB, ADB are rt. angles,
... the points A, E, D, B are concyclic :
... the A. DEB=the Z-DAB, in the same segment.
... the sum of the Z_* FOA, FAO = the sum of the Z_* DEC, DEB
=a rt. angle :
‘.. the remaining Z-AFO = a rt. angle : Theor. 16.
that is, CF is perp. to AB.
g Hence the three perpº AD, BE, CF meet at the point O.
Q. E. D.
DEFINITIONS.
(i) The intersection of the perpendiculars drawn from the
vertices of a triangle to the opposite sides is called its
orthocentre.
(ii) The triangle formed by joining the feet of the perpen-
diculars is called the pedal or orthocentric triangle.
208 GEOMETRY.
II. In an acute-angled triangle the perpendiculars drawn from the
vertices to the opposite sides bisect the angles of the pedal triangle through
which they pass.
In the acute-angled AABC, let AD, BE, A
CF be the perpº drawn from the vertices to
the opposite sides, meeting at the Ortho-
It is required to prove that N
AD, BE, CF bisect respectively
the Z = F DE, DEF, EFD. e
It may be shewn, as in the last theorem,
that the points O, D, C, E are concyclic ; B
‘.. the 4-ODE = the Z.OCE, in the same segment.

E
(\
C
%
centre O ; and let DEF be the pedal triangle. E
D
Similarly the points O, D, B, F are concyclic;
... the 4-ODF = the A. OBF, in the same segment.
But the 4-OCE = the A. OBF, each being the compt of the A. BAC.
‘.. the A. ODE = the A. ODF.
Similarly it may be shewn that the Z." DEF, EFD are bisected by
BE and CF. Q. E. D.
CoRoDLARY. (i) Every two sides of the pedal triangle are equally
inclined to that side of the original triangle in which they meet.
For the 4-EDC= the compt of the 4. ODE
= the compt of the Z.OCE
= the A. BAC.
Similarly it may be shewn that the Z-FDB = the A. BAC,
... the 4- EDC = the A. FDB = the Z.A.
In like manner it may be proved that
the A. DEC= the A. FEA = the A-B,
and the A. DFB = the A- EFA = the A.C.
CoRoll ARY. (ii) The triangles DEC, AEF, DBF are equiangular to
one another and to the triangle ABC.
NoTE. If the angle BAC is obtuse, then the perpendiculars BE, CF
bisect eacternally the corresponding angles of the pedal triangle.
THE ORTHOCENTRE OF A TRIANGLE. 209
EXERCISES.
l. If O is the orthocentre of the triangle ABC and if the perpendicular
AD is produced to meet the circum-circle in G, prove that OD=DG.
2. In an acute-angled triangle the three sides are the eacternal bisectors
of the angles of the pedal triangle: and in an obtuse-angled triangle the
sides containing the obtuse angle are the internal bisectors of the correspond-
ing angles of the pedal triangle.
3. If O is the orthocentre of the triangle ABC, shew that the angles
BOC, BAC are 8wpplementary.
4. If O is the orthocentre of the triangle ABC, them any one of the
four points O, A, B, C is the orthocentre of the triangle whose vertices are
the other three.
5. The three circles, which pass through two vertices of a triangle and
its orthocentre are each equal to the circum-circle of the triangle.
6. D, E are taken on the circumference of a semi-circle described
on a given straight line AB : the chords AD, BE and AE, BD intersect
(produced if necessary) at F and G : shew that FG is perpendicular
to AB.
7. ABC is a triangle, O is its orthocentre, and AK a diameter of the
circum-circle : shew that BOCK is a parallelogram.
8. The orthocentre of a triangle is joined to the middle point of the
base, and the joining line is produced to meet the circum-circle : prove
that it will meet it at the same point as the diameter which passes
through the vertex.
9. The perpendicular from the vertex of a triangle on the base, and
the straight line joining the Orthocentre to the middle_point of the
base, are produced to meet the circum-circle at P and Q : shew that
PQ is parallel to the base.
10. The distance of each vertea of a triangle from the orthocentre is
double of the perpendicular drawn from the centre of the circum-circle to
the opposite side.
ll. Three circles are described each passing through the Orthocentre
of a triangle and two of its vertices: shew that the triangle formed by
joining their centres is equal in all respects to the original triangle.
12. Construct a triangle, having given a vertex, the orthocentre,
and the centre of the circum-circle.
H. S. G.I. -III, O
210 GEOMETRY
LOCI.
III. Given the base and vertical angle of a triangle, find the locus of
its orthocentre.
Let BC be the given base, and X the

given angle; and let BAC be any triangle X A
en the base BC, having its vertical Z. A
equal to the Z-X. F.
Draw the perp BE, CF, intersecting at E
the orthocentre O.
It is required to find the locus of O. O
Proof. Since the A_* OFA, OEA are rt.
angles, B C
‘.. the points O, F, A, E are concyclic ;
... the A. FOE is the supplement of the 4-A :
... the vert. opp. 4- BOC is the supplement of the 4-A.
But the Z.A is constant, being always equal to the A-X;
... its supplement is constant ;
that is, the A BOC has a fixed base, and constant vertical angle;
hence the locus of its vertex O is the arc of a segment of which BC is
the chord.
IV. Given the base and vertical angle of a triangle, find the locus of
the in-centre.
Let BAC be any triangle on the given
base BC, having its vertical angle equal to A
the given 4-X; and let Al, Bl, Cl be the X
bisectors of its angles. Then l is the in- /\
Centre.

It is required to find the locus of l.
Proof. Denote the angles of the AABC
by A, B, C ; and let the 4. BIC be denoted
by I. B C
Then from the A BIC,
(i) | + #B + $C = two rt, angles; Theor. 16.
and from the AABC,
A+B+C = two rt. angles;
(ii) so that AA++B+ #C= one rt. angle.
..., taking the differences of the equals in (i) and (ii),
| – #A = one rt. angle :
Or, | = one rt. angle + A.
But A is constant, being always equal to the A-X;
i is constant :
‘.. the locus of l is the arc of a segment on the fixed chord BC
EXERCISES ON LOCI. 211
EXERCISES ON LOCI.
1. Given the base BC and the vertical angle A of a triangle; find
the locus of the ex-centre opposite A.
2. Through the extremities of a given straight line AB any two
parallel straight lines AP, BQ are drawn; find the locus of the inter
section of the bisectors of the angles PAB, QBA.
3. Find the locus of the middle points of chords of a circle drawn
through a fixed point.
Distinguish between the cases when the given point is within, on,
or without the circumference.
4. Find the locus of the points of contact of tangents drawn from
a fixed point to a system of concentric circles.
5. Find the locus of the intersection of straight lines which pass
through two fixed points on a circle and intercept on its circumference
an arc of constant length.
6. A and B are two fixed points on the circumference of a circle,
: §§ is any diameter: find the locus of the intersection of PA
and QB.
7. BAC is any triangle described on the fixed base BC and having
a constant vertical angle ; and BA is produced to P, so that BP is
equal to the sum of the sides containing the vertical angle: find the
locus of P.
8. AB is a fixed chord of a circle, and AC is a moveable chord
passing through A; if the parallelogram CB is completed, find the
locus of the intersection of its diagonals.
9. A straight rod PQ slides between two rulers placed at right
angles to one another, and from its extremities PX, QX are drawn
perpendicular to the rulers: find the locus of X.
10. Two circles intersect at A and B, and through P, any point on
the circumference of one of them, two straight lines PA, PB are
drawn, and produced if necessary, to cut the other circle at X and Y:
find the locus of the intersection of AY and BX.
11. Two circles intersect at A and B; HAK is a fixed straight line
drawn through A and terminated by the circumferences, and PAQ is
any other straight line similarly drawn : find the locus of the inter
Section of HP and QK.
212 GEOMETRY,
iº SIMSON'S LINE.
V. The feet of the perpendiculars drawn to the three sides of g
triangle from any point on its circum-circle are collinear.

Let P be any point on the circum-circle of
the A ABC ; and let PD, PE, PF be the perps. F
drawn from P to the sides. P
It is required to prove that the points D, E, F E
are collinear.
Join FE and ED :
then FE and ED will be shewn to be in the
same straight line.
Join PA, PC. *N 978
Proof. Because the 4–5 PEA, PFA are rt. angles,
... the points P, E, A, F are concyclic :
... the A. PEF=the 4. PAF, in the same segment
= the suppº of the 4. PAB
= the A. PCD,
since the points A, P, C, B are concyclic.
Again because the 4A PEC, PDC are rt. angles,
... the points P, E, D, C are concyclic.
... the A. PED = the suppº of the 4. PCD
= the suppº of the 4. PEF.
‘. FE and ED are in one st. line.
Obs. The line FED is known as the Pedal or Simson's Line of the
triangle ABC for the point P.
EXERCISES.
1. From any º P on the circum-circle of the triangle ABC,
perpendiculars PD, PF are drawn to BC and AB; if FD, or FD
produced, cuts AC at E, shew that PE is perpendicular to AC.
2. Find the locus of a point which moves so that if perpendiculars
are drawn from it to the sides of a given triangle, their feet are
collinear.
3. ABC and AB'C' are two triangles with a common angle, and
their circum-circles meet again at P; shew that the feet of perpen-
diculars drawn from P to the lines AB, AC, BC, B'C' are collinear.
4. A triangle is inscribed in a circle, and any point P on the circum-
ference is joined to the orthocentre of the triangle: shew that this
joining line is bisected by the pedal of the point P.
THE TRIANGLE AND ITS CIRCLES, 213
THE TRIANGLE AND ITS CIRCLES.
VI. D, E, F are the points of contact of the inscribed circle of the
triangle ABC, and D1, E1, F1 the points of contact of the escribed circle,
which touches BC and the other sºdes produced: a, b, c denote the length
of the sides BC, CA, AB; s the semi-perimeter of the triangle, and r, r,
the radii of the inscribed and escribed circles.
Prove the following equalities:
(i) AE = AF = 8 –a,
BD = BF = 8 – b,
CD = CE = 8 – c.
(ii) AE, =AF, F S.
(iii) CD1 – CE1 =s—b,
(iv) CD = BD, and BD=CDr.
(v) EE =FF1=a.
(vi) The area of the A ABC=rs
=ri (8-a).
(vii). Draw the above figure in the case when C is a right angle, and
prove that r= 8 – c.; r1=8–b.

214 GEOMETRY.
VII. In the triangle ABC, l is the centre of the inscribed circle, and
1, la, is the centres of the escribed circles touching respectively the sides
36,85, Ağ mºmeº.
laws, *
\\ | * > .
\ ^
Prove the following properties:
(i) The points A, 1, 11 are collinear : 80 are B, I, 12; and C, l, le.
(ii) The points lz, A, is are collinear ; 80 are la, B, [1; and 11, C, lo-
(iii) The triangle3 Bl, C, Cl,A, AlsB are equiangular to one another.
(iv) The triangle lila's is equiangular to the triangle formed by
joining the points of contact of the inscribed circle.
(v) Of the four points 1, 11, 12, is, each is the orthocentre of the
triangle whose vertices are the other three.
(vi) The four circles, each of which passes through three of the
points l, ll, la, is, are all equal.



THE TRIANGLE AND ITS CIRCLES, 215
EXERCISES.
1. With the figure given on page 214 shew that if the circles whose
centres are 1, 11, 12, is touch BC at D, D1, D2, D3, then
(i) DD2= DiD3–b. (ii) DD3- DiD2–c.
(iii) D2Ds=b+c. (iv) DD1=b - c.
2. Shew that the orthocentre and vertices of a triangle are the centres
of the inscribed and escribed circles of the pedal triangle.
3. Given the base and vertical angle of a triangle, find the locus of the
centre of the escribed circle which touches the base.
4. Given the base and vertical angle of a triangle, shew that the centre
of the circum-circle is faced.
5. Given the base BC, and the vertical angle A of the triangle, find
the locus of the centre of the escribed circle which touches AC.
6. Given the base, the vertical angle, and the point of contact with
the base of the in-circle; construct the triangle.
7. Given the base, the vertical angle, and the point of contact with
the base, or base produced, of an escribed circle ; construct the triangle,
8. is the centre of the circle ºnscribed in a triangle, and 11, ſ2, Is the
centres of the escribed circles ; shew that Ill, lla, Ils are bisected by the
circumference of the circum-circle.
9. ABC is a triangle, and le, la the centres of the escribed circles
which touch AC, and ÅB respectively: shew that the points B, C, 12, is
lie upon a circle whose centre is on the circumference of the circum-
circle of the triangle ABC.
10. With three given points as centres describe three circles touch-
ing one another two by two. How many solutions will there be?
11. Given the centres of the three escribed circles; construct the
triangle.
12. Given the centre of the inscribed circle, and the centres of two
escribed circles; construct the triangle.
13. Given the vertical angle, perimeter, and radius of the inscribed
circle; construct the triangle.
14. Given the vertical angle, the radius of the inscribed circle, and
the length of the perpendicular from the vertex to the base; construct
the triangle.
15. In a triangle ABC, l is the centre of the inscribed circle; shew
that the centres of the circles circumscribed about the triangles BIC,
Cl/A, AlB lie on the circumference of the circle circumscribed about the
given triangle.
216 GEOMETRY
THE NINE-POINTS CIRCLE.
VIII. In any triangle the middle points of the sides, the feet of the
perpendiculars from the vertices to the opposite sides, and the middle
points of the lines joining the orthocentre to the vertices are concyclic.
In the AABC, let X, Y, Z be the
middle points of the sides BC, CA,
AB; let D, E, F be the feet of the
erp" to these sides from A, B, C ; let
§ be the orthocentre, and a, 8, Y the
middle points of OA, OB, OC.
It is required to prove that
the nine points X, Y, Z, D, E, F, a, 8, y
Wre concyclic.
Join XY, XZ, Xa, Ya, Za.
Now from the A ABO,
since AZ= ZB, and Aa = a C, B X D C
... Za is parl to BO. Ex. 2, p. 64. º
And from the AABC, since BZ=ZA, and BX=XC,
... ZX is parl to AC.
but BO produced makes a rt. angle with AC ;
... the A-XZa is a rt, angle.
Similarly, the A-XYa is a rb, angle.
‘.. the points X, Z, a, Y are concyclic :
that is, a lies on the O* of the circle which passes through X, Y, Z;
and Xa is a diameter of this circle.
Similarly it may be shewn that 8 and y lie on the O* of this circle.
- Again, since a DX is a rt. angle,
... the circle on Xa as diameter passes through D.
Similarly it may be shëwn that E and F lie on the O* of this circle;
... the points X, Y, Z, D, E, F, a, 8, Y are concyclic. Q.E.D.
Obs. From this property the circle which passes through the middle
points of the sides of a triangle is called the Nine-Points Circle; many
of its properties may be derived from the fact of its being the circum-
circle of the pedal triangle. .

THE NINE-POINTS CIRCLE, 217
To prove that
. . .(i) the centre of the nine-points circle is the middle point of the
straight line which joins the orthocentre to the circum-centre.
†- (ii) the radius of the nine-points circle is half the radius of the
circum-circle. -
(iii) the centroid is collinear with the circum-centre, the nine-points
centre, and the orthocentre.
In the AABC, let X, Y, Z be the
middle points of the sides; D, E, F
the feet of the perp"; O the ortho-
centre; S and, N the centres of the
circumscribed and nine-points circles
respectively.
(i) To prove that N is the middle
point of SO. -
It may be shewn that the perp. to
XD from its middle point bisects SO;
Theor. 22.
Similarly the perp. to EY at its
middle point bisects SO :
that is, these perp" intersect at the middle point of SO:
And since XD and EY are chords of the nine-points circle,
... the intersection of the lines which bisect XD and EY at rt. angles is
its centre : Theor. 31, Cor. 1,
... the centre N is the middle point of SO. Q. E. D.
B
w" (ii) To prove that the radius of the nine-points circle is half the
rodeus of the circum-circle.
By the last Proposition, Xa is a diameter of the nine-points circle.
‘. . the middle point of Xa is its centre :
but the middle point of SO is also the centre of the nine-points circle.
(Proved.)
Hence Xa and SO bisect one another at N.
Then from the A• SNX, ONa,
SN=ON,
because and NX = Na,
and the Z-SNX= the A-ONa :
... SX = Oa.
=Aa.
And SX is also parl to Aa,
..". =Xa.
But SA is a radius of the circum-circle ;
and Xa is a diameter of the nine-points circle;
j, the radius of the nine-points circle is half the radius of the circum-
circle. [See also p. 267, Examples 2 and 3.] Q. E. D.,

218 GEOMETRY.
(iii) To prove that the centroid is collinear with points S, N, O.
Join AX and draw ag parl to SO.
Let AX meet SO at G.
Then from the AAGO, since Aa = a C,
and ag is parl to OG,
... Ag-gG. Ex. 1, p. 64.
And from the AXag, since a N = NX,
and NG is parl to ag,
* = GX
..". g\i = \ix.
... AG=# of AX;
... G is the centroid of the triangle ABC.
Theor. III., Cor., p. 97. B X D C
That is, the centroid is collinear with
the points S, N, O. Q.E.D.
EXERCISES.
1. Given the base and vertical angte of a triangle, find the locus of the
centre of the nine-points circle.
2. The nine-points circle of any triangle ABC, whose orthocentre is
O, is also the nine-points circle of each of the triangles AOB, BOC,
COA. . :
3. If I, 11, 12, la are the centres of the inscribed and escribed circles
of a triangle ABC, then the circle circumscribed about ABC is the
nine-points circle of each of the four triangles formed by joining three
of the points 1, 11, 12, is: z
4. All triangles which have the same orthocentre and the same
circumscribed circle, have also the same nine-points circle.
5. Given the base and vertical angle of a triangle, shew that one
angle and one side of the pedal triangle are constant. -
6. Given the base and vertical angle of a triangle, find the locus of
the centre of the circle which passes through the three escribed
centres.
NOTE. For some other important properties of the Nine-points
Circle see Ex. 54, page 310.

APPENDIX.
APPENDIX.
ON THE FORM OF SOME SOLID FIGURES.
(Rectangular Blocks.)
A B
The solid whose shape you are probably most familiar with
is that represented by a brick or slab of hewn stone. This
solid is called a rectangular block or cuboid. Let us examine
its form more closely.
How many faces has it? How many edges? How many
corners, or vertices?
• The faces are quadrilaterals: of what shape?
Compare two opposite faces. Are they equal? Are they
parallel?
We may now sum up our observations thus:
A cuboid has six faces; opposite faces being equal rectangles
in parallel planes. It has twelve edges, which fall into three
groups, corresponding to the length, the breadth, and the height
of the block. The four edges in each group are equal and
parallel, and perpendicular to the two faces which they cut.
The length, breadth, and height of a rectangular block are
called its three dimensions.
Ex. 1. If two dimensions of a rectangular block are equal, say,
the breadth AC and the height AD, two faces take a particular
shape. Which faces? What shape?
Ex. 2. If the length, breadth, and height of a rectangular block
are all equal, what shapes do the faces take 3

SOLID FIGURES, CUBES, 3
E
G
C
F
A B
A rectangular block whose length, breadth, and height are
all equal is called a cube. Its surface consists of six equal
Squares.
We will now see how models of these solids may be
constructed, beginning with the cube, as being the simpler
figure..
Suppose the surface of the cube to be cut along the upright
edges, and also along the edge HG; and suppose the faces to
be unfolded and flattened out on the plane of the base. The
surface would then be represented by a figure consisting of six
squares arranged as below. -
G E C F G
This figure is called the net of the cube: it is here drawn
on half the scale of the cube shewn in outline above.
To make a model of a cube, draw its net on cardboard.
Cut out the net along the outside lines, and cut partly through
along the dotted lines. Fold the faces over till the edges come
together; then fix the edges in position by strips of gummed
paper.
Ex. 3. Make a model of a cube each of whose edges is 60 cm.

4 * APPENDIX ON SOLID FIGURES.
Ex. 4. . Make a model of a rectangular block, whose length is
4", breadth 3", height 2". - •
First draw the net which will consist of six rectangles arranged
as below, and having the dimensions marked in the diagram.
*
Now cut the net out, fold the faces along the dotted lines,
and secure the edges with gummed paper, as already explained.
(Prisms.)
Let us now consider a solid whose side-faces (as in a rect-
angular block) are rectangles, but whose ends (i.e. base and
top), though equal and parallel, are not necessarily rectangles
Such a solid is called a prism.
The ends of a prism may be any congruent figures:
these may be triangles, quadrilaterals, or polygons of any
number of sides. The diagram represents two prisms, one on
a triangular base, the other on a pentagonal base.
Ex. 5. Draw the net of a triangular prism, whose ends are
equilateral triangles on sides of 5 cm., and whose side-edges measure
7 cm.

SOLID FIGURES. PYRAMIDS, 5
w
(Pyramids.)
S
A B
The solid represented in this diagram is called a pyramid.
The base of a pyramid (as of a prism) may have any number
of sides, but the side-faces must be triangles whose vertices are
at the same point.
The particular pyramid shewn in the Figure stands on a
square base ABCD, and its side-edges SA, SB, SC, SD are all
equal. In this case the side faces are equal isosceles triangles;
and the pyramid is said to be right, for if the base is placed on
a level table, then the vertex lies in an upright line through
the mid-point of the base.
Ex. 6. Make a model of a right pyramid standing on a square
base. Each edge of the base is to measure 3", and each side-edge
of the pyramid is to be 4".
To make the necessary net, draw a
square on a side of 3". This will form
the base of the pyramid. Then on the
sides of this square draw isosceles
triangles making the equal sides in each
triangle 4" long.
Explain why the process of folding
about the dotted lines brings the four
vertices together.


6 APPENDIX ON SOLID FIGURES.
Another important form of pyramid has as base an equi.
lateral triangle, and all the side edges are equal to the edges
of the base.
FIG. 2.
How many faces will such a pyramid have? How many
edges? What sort of triangles will the side-faces be? Fig. 3
shews the net on a reduced scale.
A pyramid of this kind is caked a regular tetrahedron
(from Greek words meaning four-faced).
Ex. 7. Construct a model of a regular tetrahedron, each edge
of which is 3" long.
Ex. 8. What is the smallest number of plane faces that will
enclose a space 7 What is the smallest number of curved surfaces
that will enclose a space #
(Cylinders.)
C B
D
A
s
9
FIG. I. - FIG. 2.
The solid figure here represented is called a cylinder.


SOLID FIGURES. CYLINDERS. 7
On examining the model of which the last diagram is a
drawing, you will notice that the two ends are plane, circular,
equal, and parallel.
The side-surface is curved, but not curved in every
direction; for it is evidently possible in one direction to rule
straight lines on the surface: in what direction!
Let us take a rectangle ABCD (see Fig. 2), and suppose it
to rotate about one side AB as a fixed axis.
What will BC and AD trace out, as they revolve about AB!
Observe that CD will move so as always to be parallel to
the axis AB, and to pass round the curve traced out by D. As
CD moves, it will generate (that is to say, trace out) a surface.
What sort of surface?
We now see why in one direction, namely parallel to the
axis AB, it is possible to rule Straight lines on the curved
surface of a cylinder.
It is easy to find a plane surface to represent the curved
surface of a cylinder.
R Q
S P
Cut a rectangular strip of paper, making the width PQ
equal to the height of the cylinder. Wrap the paper round
the cylinder, and carefully mark off the length PS that will
make the paper go exactly once round. Cut off all that
overlaps; and then unwrap the covering strip. You have now
a rectangle representing the curved surface of the cylinder,
and having the same area.
IH. S. G. I.-III. P

8 APPENDIX on solid FIGUREs.
(Comes.)
FIG. T.
We have now to examine the model of a cone, of which a
drawing is given above.
Its surface consists of two parts; first a plane circular base,
then a curved surface which tapers from the circumference of
the base to a point above it called the vertex. Thus the form
of a cone suggests a pyramid standing on a circular instead of
a rectilineal base.
Let us take a triangle ABC right-angled at B (Fig. 2), and
suppose it to rotate about one side AB as a fixed axis. . What
will BC trace out as the triangle revolves? Notice that AC
will always pass through the fived point A, and move round
the curve traced out by C. As AC moves, it will generate a
surface. What sort of surface? -
We now see that the kind of cone represented in the -
diagram is a solid generated by the revolution of a right-
angled triangle about one side containing the right angle.
Ex. 9. Why must the AABC, rotating about AB, be right-
angled at B, in order to generate a cone 7 $
What would be generated by the revolution of an obtuse-angled
triangle about one side forming the obtuse angle 7
Ex. 10. What would be generated by an oblique parallelogram
revolving about one side 3


SOLID FIGURES. CONES. SPHERES. 9
The curved surface of a cone may be represented by a plane
figure thus:
A
Taking the slant-height AC of the cone as radius, draw a
circle. Cut it out from your paper; call its centre A; and cut
it along any radius AC. If you now place the centre of
the circular paper at the vertex of the cone, you will find that
you can wrap the paper round the cone without fold or crease.
Mark off from the circumference of your paper the length CD
that will go exactly once round the base of the come ; then cut
through the radius AD. We have now a plane figure ACD
(called a sector of a circle) which represents the curved surface
of the cone, and has the same area.
(Spheres.)
The last solid we have to consider is the sphere, whose
shape is that of a globe or billiard ball.
B
*
C
O
A
FIG. 2.
We shall realise its form more definitely, if we imagine
a semi-circle ACB (Fig. 2) to rotate about its diameter as a
fixed axis. Then, as the semi-circumference revolves, it
generates the surface of a sphere. *
H. S. G. I.-III. P 2

10 - APPENDIX ON SOLID FIGURES.
Now since all points on the semi-circumference are in
all positions at a constant distance from its centre O, we
see that all points on the surface of a sphere are at a constant
distance from a fixed point within it, namely the centre.
This constant distance is the radius of the sphere. Thus
all straight lines through the centre terminated both ways by
the surface are equal: such lines are diameters.
Ex. 11. We have seen that on the curved surfaces of a cylinder
and come it is possible (in certain ways only) to rule straight lines.
Is there any direction in which we can rule a straight line on the
surface of a sphere 7 .
Ex. 12. Again we have cut out a plane figure that could be
wrapped round the curved surface of a cylinder without folding,
creasing, or stretching. The same has been done for the curved
surface of a cone. Can a flat piece of paper be wrapped about a
sphere so as to fit all over the surface without creasing? .
Ex. 13. Suppose you were to cut a sphere straight through the
centre into two parts, in such a way that the new surfaces (made by
cutting) are plane, these parts would be in every way alike. The
parts into which a sphere is divided by a plane central section are
called hemispheres. Of what shape is the line in which the plane
surface meets the curved surface 7 If the section were plane but
not central, can you tell what the meeting line of the two surfaces
would be 7 --
Ex. 14. If a cylinder were cut by a plane parallel to the base,
of what shape would the new rim be? -
Ex. 15. . If a cone were cut by a plane parallel to the base,
what would be the form of the section ?
ANSWERS TO NUMERICAL EXERCISES.
Since the wtmost care cannot ensure absolute accuracy in graphical work, results so
obtained are likely to be only approximate. The answers here given are those found by
calculation, and being true so far as they go, furnish a standard by which the student
may test the correctness of his drawing and measurement. Results within one per cent.
of those given in the Answers may wavally be considered satisfactory.
Exercises. Page 15.
30°; 126°; 261°; 85°. 11 min. ; 37 min.
1123°; 155°; 5 hrs. 45 min. 3. 50°; 8 hrs. 40 min.
4. (i) 145°, 35°, 145°. (ii) 55°, 55°. 86°, 94°.
:
Exercises. Page 27.
1. 68°, 37°, 75° v. nearly. 2, 6-0 em. 4. 22", 50°, 73° nearly.
5, 37 ft. 6. 101 metres. 7. 27 ft. 8. 424 yds., nearly; N.W.
9. 281 yds., 155 yds., 153 yds. 10. 214 yds.
Exercises. Page 41.
1. 125°, 55°, 125°. 12. 15 secs., 30 secs.
Exercises. Page 43. -
3. 21°. 4. 27°. 5. 92°, 46°. 6. 67°, 62°.
Exercises. Page 45.
1. 30°, 60°, 90°. 2. (i) 36°, 72°, 72°; (ii) 20°, 80°, 80°.
3. 40°. 4, 51°, 111°, 18°. 5. (i) 34°; (ii) 107°.
6. 68°. . 7. 120°. 8. 36°, 72°, 108°, 144°.
9. 165°. 11. 5, 15.
- Exercises. Page 47.
2. (i) 45°; (ii) 36°. 3. (i) 12; (ii) 15.
Exercises. Page 54.
4. (i) 81° ; c. (ii) 55°.

Degrees | 15° 30° 45° 60° | 75°
10.
Cm. 4-1 || 4-6 5-7 8-0 |lsº
a Degrees || 0 || 30 60 | 90' | 120 | 150 | 180°
Cm. 1-0 || 2:0 || 3-6 || 5-0 || 6-1 || 6-8 || 7-0
12. 37 ft. 13. 112 ft. 14. 346 yds. 693 yds.
H. S. G.
H
GEOMETRY.
Exercises. Page 61.
14. 54°, 72°, 54°. 15. 36°. 16, 4.
18. (i) 16; (ii) 45°; (iii) 113° per sec.
Exercises. Page 68.
2, 6-80 cm. 3, 2.24”. 4. O'39. 5. 2'54. 8, 10-6 cm,
9. 3-35". 10, 20 miles; 12.6 km.
11. 147 miles; 235 km. 1 cm. represents 22 km.
12. 1" represents 15 mi. ; 1" represents 20 mi.
Exercises. Page 79.
3. 0-53 in. 4. 1 3 cm. 5. 2'4”.
Exercises. Page 84.
1. 4-3 cm., 5-2 cm., 6-1 cm. 2. 1 10. 3, 200 yards,
4, 65°, 77 m., 61 m., 56 m. 5. 6.04 knots. S, 15° E, nearly,
6. Results equal. 9 cm. . 7. 4.3 cm.; 9.8 cm., 60°; 120°.
8. (i) One solution ; (ii) two ; (iii) one, right-angled ; (iv) impossible.
9. 380 yds. 10. 6-5 cm. 11. 6-9 cm.
12. Two solutions; 10.4 cm. or 4-5 cm. 16. 2.8 cm., 4-5 cm., 53 cm.
18, 5.8 cm., 4:2 cm. 19. 7 cm., 8 cm.
Exercises. Page 89.
l, 60°, 120°. 2. 3-54". 3. 2' 12". 4. 4'4 cm.
5. 16:4 cm., 3.4%. 6. 90°. 7. (i) 4'25"; (ii) B = D = 90°.
Exercises. Page 102.
i. 6 sq. in. 2, 6 sq. in. 3. 280 sq. in. 4, 3:50 sq. in.
5. 3-30 sq. in. 6. 336 sq. in. 7. 198 sq. m. 8. 42 sq. ft.
9. 10,000 sq. m. 10. 110 sq. ft. 11. 5 cm. 12, 2-6 in.
14, 900 sq. yds. ; 48 yds. ; 4'8". 15. 11700 sq. m. i
16. 1 cm. = 10 yds. 17, 1 6". 18. 600 sq. ft. 19. 1152 sq. ft.
20. 100 sq. ft. 21. 156 sq. ft. 22. 110 sq. ft.
23. 288 sq. ft. 24. 72 sq. ft. 25. 75 sq. ft.
Exercises. Page 105.
1. (i) 22 cm.; (ii) 36". 2, 3-4 sq. in. 3. 574-5 sq. in.
4, 1°5". 5. 193", 75°.
ANSWERS. º hiſ
:
1.
l
1
3.
10.
13.
16.
17.
18.
19.
Exercises. Page 107.
(i) 180 sq. ft. ; (ii) 8.4 sq. in. ; 1 hectare.
(i) 13:44 sq. om. ; (ii) 15:40 sq. cm.; (iii) 20:50 sq. cm.
15 sq. om. - 4, 6-3 sq. in.
(i) 8”; (ii) 13 cm. 6. 3-36 sq. in.
Exercises. Page 110.
11400 sq. yds. 2. 6312 sq. m.
2-4 cm.; 5:1 cm. 4. 2.04"; 2:20".
Angle | 0 || 30° 60° 90° | 120° 150° | 180°

Area in sq. cm. || 0 || 7-5 13-0 | 15-0 || 13-0 || 7-5 || 0
Exercises. Page 111.
66 sq. ft. 2. 84 sq. yds. 8, 126 sq. m.
132 sq. cm. 5. 180 sq. ft. 6. 306 sq. m.
Exercises. Page 113.
6 sq. in. 2. 170 sq. ft. 8. 615 sq. m. 4. 8-4 sq. in,
31 2 sq. cm. 6. 5:20 sq. in. 7. 24 sq. cm,
Exercises. Page 115.
(i) 25.5 sq. cm.; (ii) 15-6 sq. cm.
(i) 8-95 sq. in. ; (ii) 9-5 sq. in. 3. 12500 sq. m.
Exercises. Page 116.
3.3 sq. in. 5. 7-5 cm. 6. 3-6 sq. in.
Bxercises. Page 121.
(i) 5 cm.; (ii) 6-5 cm.; (iii) 3.7". 2. (i) l'6"; (ii) 2.8 cm.
41 ft. 4. 65 miles. 5. 6-1 km. 6, 16 ft.
48 m. 8, 25 miles. 9. T3 m. 10. 62 ft.
- Exercises. Page 123.
(i) and (iii). 11. 2'83". 12, 4:24 cm.; 18 sq. cm
70-71 sq. m. 14, p=6'93 cm.
(i) 20 cm.; 15 cm. (ii) 40 cm.; 39 cm.
35 cm.; 12 cm.; 306 sq. cm.
(i) 36 sq. in, ; (ii) 90 sq. ft.; (iii) 126 sq. cm.; (iv) 240 sq. yds.
5°l cm. nearly.
GEOMETRY.
Exercises. Page 127.
7-1 cm. 4. 4-0 cm. 5. I-6". 6. 3-1 cm.; 15.6 sq. cm.
Exercises. Page 130.
23.90 sq. cm. 2. 8:40 sq. in.
27.52 sq. cm. 4. 129800 sq. m.
Exercises. Page 134.
(i) (8, 5); (ii) (10, 10).
(i) (4, 5); (ii) (4, 5); (iii) (–4, -5); (iv) (–4, -5).
(6, 5), (12, 10). 6. (5, 8). .
(i) 17; (ii) 17; (iii) 2.5"; 2-5". -
(i) and (ii) 5; (iii) and (iv) 17; (v) and (vi) 37. 9, 10.
(0, 0). (7, 5). 15. 13; (9, 6).
A straight line passing through the points (4,0), (0, - 4).
117 units of area in each case. 18. A square. 2 sq. in. 1 sq. in.
Each =70 units of area. 20. 9 units of area. 31°, 71°, 78°.
(i) 96; (ii) 80; (iii) 120; (iv) 104. º r
(i) 50; (ii) 60; (iii) 120; (iv) 132.
Sides 5, 13; area 63. 24, (i) 27; (ii) 21; (iii) 30; (iv) 27°5.
(i) 50 ; (ii) 65-5; (iii) 21 ; (iv) 83.5.
Each side 13; area 120. 27. 13, 10, 15, 8:24, 42, 30.
AB= 10, BC=9, CD=17, DA= 12-7. Area = 130-5.
10, 13, 5, 5, 3. Area=60. 30. 160,000 sq. yds. 1000 yds. 320 yds.
Side = 15:23; area=232 units of area.
Exercises. Page 145.
5 cm. 2, 24”. 3. 0-6", O'8". 4. N7=2.6 cm.
1 ft. 6. 0-6 sq. in. 7. 0-8".
- Exercises. Page 149. -
1.7" - 2. 3M2=42 cm. 3. 28/3=3-5 cm.
17". 6. 5 cm.
Exercises. Page 151.
4 cm. 7. 1.3%.
Exercises. Page 153.
I '85'. 3. 1-62”. 5. 0.85"; (2.1", 2.1"); 2.97°.
ANSWERS.
w
l-
:
(i) 28-3 cm.; (ii) 628-3 cm.
11:31 cm.; 10:18 sq. cm.
30.5 sq. cm. 8. 8'9".
4. 56 sq. cm.
Exercises. Page 155.
51”, 6. 1-6"; 1'5", 0.6".
Exercises. Page 157.
(8) 11). 5. 17; 10; (0, -8).
Exercises. Page 161.
. 74°, 148°, 16°, 2, 115°, 230°. 3. 55°, 8°, 47°.
Exercises. Page 177.
8-0 cm. 2. 0-6". 3. 8-7 cm. 4. 12", 67°. 5 2.5".
Exercises. Page 179.
3 cm. and 17 cm.
Exercises. Page 181.
72°, 108°, 108°. &
Exercises. Page 187.
1-6". 3. 17”. 4, 198", 1-6".
Exercises. Page 198.
2-3 cm., 4-6 cm., 6.9 cm. 3. 1 39".
6-9 cm.; 20-78 sq. cm. 7. 3-2 cm.
Exercises. Page 199.
2:12", 450 sq. in. * 4, 8.5 cm. 5. 2.0°.
Exercises. Page 200.
128}”; 173".
Exercises. Page 201.
3:46"; 4:00". 2. 259.8 sq. cm.
(i) 41-57 sq. cm.; (ii) 77°25 sq. cm.
Exercises. Page 205.
(i) 16-62 sq. in.; (ii) 352-99 sq. in
5. 43-98 sq. in.
4”; 3". 10. 12:57 sq. in.
Circumferences, 4'4", 6'3". Areas, l'54 sq. in., 3-14 sq. in.
.***
*
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* - sº. 37° Nº ſº
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* * * *
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