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PLANE
GEOMETRY

ABRrDOEDA^in APPLIED
JLLLuii FRi,r Ai\L\TORY

s^ \ VS<i'\^

GIFT or

zed by Google

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PLANE GEOMETRY

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Digitized by Google I

LIPPINCOTT'S SCHOOL TEXT SERIES

EDITED BY WILLIAM F. RUSSELL, Ph.D.

DEAN, COLLSGE OF EDUCATION, STATE UNITEBSITT OF IOWA

PLANE GEOMETRY

I. ABRIDGED AND APPLIED
II. COLLEGE PREPARATORY

BY
MATILDA AUERBACH

SUPERVISOR OP HATHRMATICS IN THE ETHICAL CTTLTURI BIOH SCHOOL, NEW YORK CITY

AND
CHARLES BURTON WALSH

PBINCIPAL or THE FRIENDS* CENTBAL SCHOOL, PHILADKLPHIA

PHILADELPHIA, LONDON, CHICAGO
J. B. LIPPINCOTT COMPANY

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COPTRIOHT, 1920, BT J. B. LIPPINCOTT COMPANY

PBINTED BT J. B. LIPPINCOTT COMPANY

AT THE WASHINGTON SQUARE FBESS

PHILADELPHIA, U. S. A.

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fljO

PREFACE

In separating this text book on Plane Geometry into two parts,
the Authors have followed what appears to them to be a normal
and logical requirement essential to a proper presentation of the
subject, and the most appropriate reference to these divisions
would seem to be a designation of them as a First Study and a
Second Study. In the former, the objective is to afford a general
view of the subject, with emphasis on applications, the study being
intended for the use of all high school pupils, and the material is
so presented as to make it available also for use in junior high
schools. The Second Study is devoted to a more intensive treat-
ment of Plane Geometry, with special emphasis on theoretical
work, and is addressed particularly to Regents' and college entrance
requirements. In the entire text the inductive method is followed
as far as practicable, and simplicity is gained rather by the means
of scientific accuracy than at its expense.

In view of the fact that the purpose and method of the two parts
of the book differ somewhat, a separate consideration of each
of these divisions seems desirable.

Part I

The Authors feel confident that the First Study will serve a
fourfold piupose.

First: That it will contribute to a solution of the question as to
how much mathematics shall be required of high school pupils
who do not intend to enter college, and with this objective, an
effort has been made to plan a course adapted to the needs and
interests of pupils meeting minimum requirements in mathematics.
The Authors believe that it has been customary in many schools
to meet this situation by a study of only a portion — ^three or four
books of the geometry in the required course, thus giving students
merely an intensive knowledge of a part of the subject, instead of
a broadly comprehensive view. The course here outlined covers

459936 °'^--^^^^S^^

yi PREFACE

not only all that is really imports-nt in the five books — although
the syllabus contains fewer propositions, than are now comprised
in the first three books of most available texts — but also work in
the application of three of the trigonometric functions.

Second: That it will suggest a course in elementary geometry
so thoroughly adapted to the mental development of pupils in the
ninth or tenth school year that it may be profitably used to super-
sede the conventional course in formal geometry.

The com^e here outlined will not in any respect restnct the
preparation of the student for college; on the contrary, will find
him much more ready and willing to proceed to the collegiate
preparatory work with his knowledge of the subject enriched by
application and vitalized with interest.

Third: That it will open the eyes of the pupil to the relation of
geometry to the activities and necessities of every-day life, and
emphasize the practical application of the science, both in specific
professions and trades, and in the affairs of daily life.

Fourth: That it will arouse in the pupil a conception of the
dignity and power of the subject. To this end the Authors have
treated it scientifically, endeavoring to develop gradually in the
mind of the pupil a natural assumption of this treatment; and it
•has been their purpose, by departing from formal methods, to lead
the pupil to reason rather than merely to remember.

The following means have served in the attainment of the ends
just stated:

Revision of the Syllabus

In the First Study the number of propositions has been
reduced to approximately half of those given in the standard texts.
Propositions have been retained or selected on three bases:

First: Those that are rich in application.

Second: Those of peculiar interest to the young student.

Third: Those essential to the sequence of the study.

The wording of the propositions retained has departed materially
from the traditional phraseology in an effort to avoid the
formidable and stilted qualities of the latter, while retaining
its scientific correctness. In fact, the language of the students in
the classroom has suggested many of these changes: e.^. "Three

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PREFACE vii

sides determine a triangle " to replace '* Two triangles are congruent
if three sides of one are respectively equal to three sides of the
other." The Authors feel that some of these changes in wording
are desirable scientifically as well as from a practical standpoint:
e.g.f "If the ratio of the sides of one triangle to those of another is
constant, the triangles are similar," to replace "Two triangles are
similar if the sides of one are respectively proportional to the sides
of the other;" and "An angle whose vertex is outside a circle is
measured by half the difference of the intercepted arcs," to replace
three statements.

A transition, more gradual than wsr^aZ, from the geometry of the
grade school to the more scientific work of the secondary school has
been secured. The proofs given in the first section would be wholly
convincing but the forms and detail less conventional than those
demanded in more rigorous demonstrations, some of which are
preceded by explanations. This does not mean that the text is one of
Concrete or Observatumal Geometry, The bare essentials are retained
from the outset, and subsequently there is a gradual introduction
and demand for all the rigor obtainable in secondary school work.
This gradual transition tends to prevent the discouragement so
often manifested during the beginning of the study.

Minor Details.

First: No authorities are required for au^liary construction — con-
sistent reference to them makes a proof unduly formidable — ^no
other reference should be permitted.

Second: Nothing is introdv4ied in the text until it is required — thus
avoiding long lists of definitions, axioms, postulates, and the like.

Third: Throughout the text emphasis is laid upon the idea of
classification; and by means of proper grouping of definitions,
postulates, and developed facts, the student is trained to regard
the subject, not as a miscellany of isolated facts, but as a. frame-
work of interrelated sub-topics. A few reference books are men-
tioned. The use of these has been found so exceedingly helpful
and inspiring in the classes which have been led by the Authors
that they feel that the omission of some such lists would be a great
calamity. The Authors recall many instances in which new life

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viii PREFACE

has been given to the subject through such references; in fact there
are instances where a pupil who might never have discovered him-
self mathematically has developed true mathematical enthusiasm
and ability by browsing through suggested supplementary reading.

In using the text, the Authors earnestly suggest that adtial
writing of proofs be deferred imtil the class is quite ready to fall in
naturally with a more or less set form of proof.

Let the need of a form become apparent to the student who is
trying to write a proof imaided by conventions before insisting
upon the adoption of one in written work. Indeed the Authors
feel that the entire first chapter of this text could well be developed
before the necessity arises for a single written proof from the stu-
dent. Sufficient material for written work will be available from
the exercises during this period.

Part II

In addition to a review of the First Study, the Authors desire
to direct attention to some details of this Second Division of the
work which seem worthy of special mention. .

First: The size of the syllabus. The number of propositions
developed in the First Study has been considerably augmented to
include all the demands of College Entrance and Regents' Examina-
tions. This has been accomplished by inserting the additional
theorems between the theorems of the original syllabus, thus
preserving the sequence of the former Study. A separate syllabus
of construction problems is given in the chapter entitled " Methods
of Attacking Problems " (page 306).

Second : The grouping of the syllabus. To facilitate the reten-
tion of the frame-work of the subject, the propositions are collected
in groups by topics, so far as the sequence permits, and such
grouping has necessitated certain departures from the traditional
arrangement by books.

Third : The type of exercises. In this part emphasis is placed
on theoretical exercises, as contrasted wiih special reference to prac-
tical exercises in the First Study; and a large collection of college
entrance papers, together with a still larger selection of isolated
exercises of this character, is included.

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PREFACE ix

Foubth: a chapter on methods of proof. More scientific habits
of work are fostered through a discussion and a careful classification
of methods of proof with illustrative exercises.

Fifth : A chapter on methods of attacking problems. This chapter
is largely similar in purpose to the chapter just mentioned, except
that it deals with construction work. A special syllabus of what
might be termed fundamental constructions is given, including
those presented in standard texts, and those propositions are of
such character as to form a necessary and sufficient basis for the
work required by colleges. This chapter groups many typical
constructions and methods employed, and definite reference
throughout the book is made to this, as well as to the chapter on
methods of proof.

Sixth: Siiggestions for clvb or other additional work. The chap-
ter entitled "Suggestions," and exercises preceded by the letter
"d" — ^frequent in the book — may be omitted (not constituting a
requirement for college preparation) without impairing the integ-
rity of the course. They are included to give additional interest
and breadth to the subject where time and the ability of the student
permit. This chapter contains in the suggestions for club work a
list of topics suitable for discussion by students or teacher in a
mathematics club of high school grade, and appended to this list
will be found a bibliography of appropriate references.

In summary, then, it may be said that this Second Study is
intended to enlarge upon the course outlined in the First Study,
not only in that it answers the requirements of college entrance
examinations, but in that it also makes possible at the same time
a richer and fuller course for those students whose interest and
ability warrant it.

In closing, the Authors desire to acknowledge three distinct
sources of assistance.

Realizing the present eclectic tendency of teachers in the matter
of exercises, as evidenced by the general use of typewritten lists
of problems, the Authors have availed themselves frequently of
many of the standard texts in the selection of material of this kind.
To Mr. Lewi Tonks, a former pupil of the Authors, who read the
proof and criticised the contents of the book, the Authors fee)

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X PREFACE

especially indebted, and wish at the same time to acknowledge the
assistance of Mr. H. W. Smith, of the Ethical Culture School, and
of Mr. P. S. Clarke, of Pratt Institute, Brooklyn, whose suggestions
for revision of the English of the text were of value.

Finally, the Authors take pleasure in this opportunity to record
their deep appreciation of the endorsement and encouragement
they have received from the authorities of the Ethical Culture
School — Prof. Felix Adler, Rector; Franklin C. Lewis, Superin-
tendent; and Dr. Henry A. Kelly, High School Principal. Their
approval made possible the experimental work in the School which
has developed and justified the course here presented, and the
Authors feel that their kindly S3ntnpathy and intelligent cooperation
in the growth of the experiment have contributed in large measure
to its success.

The Authobs.

OCTOBEB, 1919.

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SYMBOLS AND ABBREVIATIONS

is equal, or equivalent, to

is not equal, or equivalent, to

is similar to

is congruent to

approaches as a limit

varies as

is measured by

is greater than

is not greater than

is less than

is not less than

plus, or increased by

—

minus, or diminished by

• >

divided by

x,(),-

multiplied by

a parallel, or is parallel to

not a parallel, or is not parallel to

ll<

parallels

a perpendicular, or is perpendicu-
lar to

not a perpendicular, or is not per-
. pendicular to

±8

perpendiculars

/-N

arc

ilfi

straight line AB

o®

circle, circles

A, A

triangle, triangles

OylU

parallelogram, parallelograms

<,<'

angle, angles

• •

since

• •

therefore

and so on

(),{Ti]r

the nth root of

3.14159

adj. adjacent

alt. alternate

ap. apothem

approx. approximately

ax. axiom

circf. circumference

oomp. complement, com-
plementary

con. conclusion

oong. congruent

const. construction

cor. corollary

oorres. corresponding

def. definition

diff. difference

ex.

exercise

ext.

exterior

fig.

figure

ht.

height, or altitude

hom.

homologous

hy.

h3rpotenu8e

hyp.

hypothesis

int.

interior

isos.

isosceles

lat.

lateral

peri.

perimeter

pl.

plane

pt., pts.

point, points

n-gon

polygon of n sides

post.

postulate

prob.

problem

proj.

projection

prop.

proposition

rect.

rectangle

reg.

regular

rt.

right

sec.

sector

sq.

St.

subst.
sup.

sym.

th.

vert.

square
straight
substitute
supplement, sup-

plementaiy
synmietrical
theorem
vertical

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CONTENTS

PAGE

PSEFACB V

List of Symbols and Abbreviations xi

PART ONE— FIRST STUDY

CHAPTER I
Introduction

A. A Few Facts Concerning the Early Development of Geometry 3

Summary 10

Bibliography 10

Exercises. Set I. Based Upon Historic Facts 11

B, A Few IixusTRAnoNS of Geobietric Form 12

Exercises. Set II. Illustrations of Geometric Form. . r 13

C Meaning of Geometric Form 14

Exercises. Set IH . Meaning of Geometric Form 15

D, SiroQEsnoNs of a Few Uses of Geobietry 16

Exercises. Set IV. Mensuration 17

Exercises. Set V. Constructions. Designing 17

Exercises. Set VI. Some Other Uses of Geometry 24

E, The Basic Principles of Geometry 24

Axioms 25

Exercises. Set VII. Illustrations of Postulates 26

Postulates of the Straight Line 27

Definitions 27

Exercises. Set VIII. Sums of Angles in Pairs 28

Exercises. Set DC. Measurement of Angles 29

Postulates of the Angle 31

Exercises. Set X. Belative Position of Angles 31

Exercises. Set XI. Instruments for Measuring Angles 36

F, The Discovery of Sobib Facts and Their Informal Proof.. . . 36

I. Classification Based upon Sides 37

II. Classification Based upon Angles 37

Experiments. Theorems 37

Exercises. Set XII. Meaning of Congruence and Classi-
fication of Triangles 38

Exercises. Set XIII^ Application of Congruence of

Triangles 39

III. Some Properties of the Isosceles Triangle 42

Experiment. Theorem 42

a. Some Properties of the Equilateral Triangle 43

Exercises. Set XIV. Equilateral Triangles 43

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xiv CONTENTS

PAGE

IV. Further Discussion of Congruence of Triangles 44

Experiment. Theorem 44

Exercises. Set XV. Further Applications 44

Summary of Chapter 47

CHAPTER II

The Perpendicular, the Right Triangle, and Parallels

A . The Perpendicular 48

Postulate. Theorem 48

Exercises. Set XVI. Distance from a Point to a Line 48

Axioms of Inequality 49

Exercises. Set XVII. Numerical Inequality 49

Exercises. Set XVIII. Inequality of Sects 50

B. The Right Triangle 50

^Ixperiment Theorems 50

C. Parallels 51

Exercises. Set XIX. Parallels 52

Postulate of Parallels 52

Exercises. Set XX. Relative Position of Angles 53

Exercises. Set XXI. Construction of Parallels 55

Exercises. Set XXII. Related Statements 66

Exercises. Set XXIII. Applications of Parallelism 58

Summary of Chapter 59

CHAPTER III
Angles op Polygons and Properties of Parallelogram i

A. Angles op Polygons 60

Exercises. Set XXIV. Sum of Angles of a Triangle 60

Exercises. Set XXV. Sums of Angles oi Polygons 62

Exercises. Set XXVI. Sides and Angles of a Triangle 66

B. Parallelograms 67

Exercises. Set XXVII. Parallelograms 68

Exercises. Set XXVIII. Parallels 73

Smnmary of Chapter 73

CHAPTER IV -

Areas

A, Introduction. Review op Fractions 74

Underiying Principles 74

Exercises. Set XXIX. Fractions 75

B. Areas. Development op Formulas 77

Exercises. Set XXX. Comparison of Sects 78

Exercises. Set XXXI. Areas of Rectangles • • • .. r 80

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CONTENTS XV

PAGE

Exercises. Set XXXII. Areas of Parallelograms 81

Exercises. Set XXXIII. Altitudes of Triangles 81

Exercises. Set XXXIV. Areas of Triangles 82

Exercises. Set XXXV. Areas of Trapezoids 83

Summary of Chapter 86

CHAPTER V
Algebra as an Instrument for Use in Applied Mathematics

A, Logarithms 86

I. Introduction 86

Exercises. Set XXXVI. Meaning of Logarithms 86

Exercises. Set XXXVII. Applications of Laws of Expo-
nents 88

Exercises. Set XXXVIII. Use of Tables of Powers 90

Historical Note 90

II. Principles of Common Logarithms 91

Exercises. Set XXXIX. Common Logarithms 92

HI. Fundamental Theorems 93

IV. Use of the Table of Common Logarithms 94

Exercises. Set XL. Use of Table 95

Exercises. Set XU. Computation by Logarithms 98

B. Ratio, Proportion, Variation 105

I. Ratio and Proportion 105

Exercises. Set XUL Ratio 106

Exercises. Set XLHI. Proportion 106

Exercises. Set XLIV. Applications of Proportion 108

n. Variation 110

Exercises. Set XLV. Applications of Variation Ill

Summary of Chapter 114

CHAPTER VI
Similarity

A. Introductory Experiments and Theorems 1 15

Exercises. Set XLVI. Proportional Sects 116

B. Idea op Simlarity 119

Exercises. Set XLVII. Meaning of Similarity 121

C. Similarity op Triangles 122

Exercises. Set XLVIII. Similarity of Triangles 124

D. PERIBiBTERS AND ArEAS OP SIMILAR TrIANGLES 132

Exercises. Set XLIX. Areas of Similar Triangles 133

E. Applications of Similar Trllnqles 133

Exercises. Set L. Projections. Pythagorean Relation 135

Exercises. Set U. Trigonometric Ratios 140

Summaiy of Chapter 148

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xvi CONTENTS

CHAPTER Vn
The Locus

PAGE

A. Review op Algebraic Idea op Locus 149

Exercises. Set LII. Location of Points 149

Exercises. Set LIU. Applied Problems in Graphic Mathematics 150

Exercises. Set LIV. Graphic Solution of Equations 151

Exercises. Set LV. The Equation aa the Statement of a Locus' 152

B, Peculiarity op the Proop op a Locus Proposition 155

Exercises. Set LVI. Related Statements — Direct, Converse,

Opposite 155

Theorems 156

Exercises. Set LVII. Applications of Locus 157

Summary of Chapter 159

CHAPTER Vm

The Circle

Definitions 160

A, Preliminart Theorems 161

Exercises. Set LVIII. Circle as a Locus 161

B, Straight Line and Circle 162

Exercises. Set LIX. Congruence of Curvilinear Figures 163

Exercises. Set LX. Constructions Based Upon Circles '. . 164

Exercises. Set LXI. Equal Chords 165

Exercises. Set LXII. Tangent and Circle 167

Exercises. Set LXIII. Tangent Circles 168

C. The Angle and Its Measurement 172

Exercises. Set LXIV. Secant and Circle 173

Exercises. Set LXV. Circles 174

Exercises. Set LXVI. Inscribed Angles 175

Exercises. Set LXVII. Measurement of Angles 177

Exercises. Set LXVIII. Tangent and Secant 179

D. Mensuration op the Circle 180

Exercises. Set LXIX. Regular Polygons and Circles 181

Historical Note 183

Postulates of Limits 184

Exercises. Set LXX. Perimeters of Regular Polygons 187

Axioms of Variables 187

Historical Note 188

Exercises. Set LXXI. Value of t 188

Exercises. Set LXXII. Circimiference 189

Exercises. Set LXXIIL Area of Circle 192

Summary of Chapter 196

Miscellaneous Exercises. Set LXXIV 197

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PART I
FIRST STUDY

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PLANE GEOMETRYrr; ::::v

CHAPTER I

INTRODUCTION

Note. — ^For pupils who have had no mtuitional geometry
section A of the Introduction may be postponed until the work in
Areas has been completed — p. 77,

A. A FEW FACTS CONCERNING THE EARLY DEVELOP-
MENT OF GEOMETRY

In a cabinet in the British Museimi there is a piece of clay some-
what over an inch thick and perhaps fifteen inches square which
might be referred to as the first book about geometry. Near it,
on a roll of papyrus, yellowed by age, is a collection of notes con-
taining instructions for finding the contents of areas and solids.
When we reflect that this clay tablet and the manuscript are con-
siderably over thirty-five hundred years old, we can see that the
study of geometry is by no means a modern development.

The tablet and the manuscript represent respectively the earliest
available records of the geometric knowledge of the Babylonians
and the Egyptians. Centuries ago these two races found it neces-
sary to devise some means for accomplishing what today seems a
very simple imdertaking. Perhaps the necessity was forced upon
the Egyptians for a reason that does not seem very apparent at
first. The River Nile, as we know, rises twice a year and inundates
the coimtry bordering on it for many miles. Naturally this flood
produces changes in the line of the river banks, and new turns and
curves give the adjacent land a very different appearance on each
occasion. A farmer whose land bordered the river might therefore
find himself one year in possession of a good deal of property, and
the next year with much less. This condition, we are told by the
historian Herodotus, caused Rameses II, who was king of Egypt
about 1350 b.c, to declare a law: "This king divided the land
among all Egyptians so as to give each a quadrangle of equal size,

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4 PLANE GEOMETRY

and to 3raiv:fi561]i»each his revenues by imposing a tax to be levied
3feafly ; -bjit eyecyone from whose part the river tore away anything
ha*(f tcr* ge Wtuiil aad notify hina of what had happened; he then
sent the overseers who had to measure out by how much the land
had become smaller, in order that the owner might pay on what
was left in proportion to the entire tax imposed." As a result of
this it became necessary for the Egyptians to employ surveyors
who should determine the areas of the land lost or gained, and these
surveyors put into practical use such rudimentary knowledge as
was then available.

Restoration of the Great Hall of Karnak

Seven centuries before this Egyptian kings had imdertaken large
construction operations, the very nature of which showed that
their contractors and builders imderstood the elementary prin-
ciples of the science of mensuration. Menes, the first Egyptian
king, built a large reservoir and two temples at Phthah and Mem-
phis, the ruins of which are still in existence, and under Amenembat
III, a later king, the Egyptians designed and constructed a very
large irrigating system covering considerable territory and requir-
ing a careful calculation of areas, water flow, gradients, etc. Stu-
dents of Egyptian art and religion find frequent evidence that this
race had a crude knowledge of geometrical principles. The pave-
ments of the temples show designs of triangles, squares, five-
pointed stars, and rectangles, and the locations of the buildings
themselves show geometric knowledge, for their temples were
supposed to be constructed with reference to a certain fixed point.
The Egyptians were sun-worshippers, and their temples were

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INTRODUCTION 5

designed to receive sunlight through the doorways at certain times
of the day, as a part of the religious ceremonies. It is interesting
to note that the movement of the North star has been in a measure
demonstrated to our later-day astronomers by the fact that Egyp-
tian temples, built three or four thousand years ago and designed to
face the North star, are no longer in the perpendicular to it. The
Egyptians were astronomers, and in locating their temples used
the sun and the North star to establish base lines. The surveyors,
called the "harpedonaptae" or "rope-stretchers,"
fixed the right angle to the north-south line by
stretching a rope knotted in three places around
pegs. The distances between the knots were
in the ratio of 3-4-5, showing that they knew
this to be the ratio of sides of a right triangle.
Our present day surveyors are still following the same method
and have improved upon the method of the Egyptians only by
substituting a steel tape for the rope.

We mentioned the ancient manuscript of the Egyptians now in
the British Museum. The writer of this manuscript was called
Aah-mesu (The Moon-born), an Egyptian scribe commonly called
Ahmes. The original from which he copied it was probably in
existence about 2300 B.C., but has never been discovered. The
commercial value of the document is shown by the fact that it
contained rules and formulas for finding the capacity of the wheat
warehouses constructed in ancient Egypt, as well as a treatise of
considerable length on a crude algebraic system. A temple built
for the worship of the god Horus on the island of Edf u has on its
walls hieroglyphics describing the land which the priests of the
temple owned, and the formulas for finding the areas of these plots.
Less is known about the Babylonians in these par-
ticulars, but so far as we can learn, their geometrical
knowledge was used more in thearts than for practical
purposes. Their moniunents, found in the ruins of
Babylon, show geometrical designs, such as a regular
hexagon in a circle, and the pictures of their chariots show the
wheels divided into sixths. The Babylonians appear to have fol-
lowed this division into sixths in their arrangement of the calendar,

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PLANE GEOMETRY

for their year consisted of three hundred sixty days, and they
divided the circle into three hundred sixty degrees, on the theory
that each degree represented the supposed revolution of the sun
round the earth.

Although the geometric knowledge of the Egyptians and Baby-
lonians may seem to us somewhat crude and simple, we must
remember that, as compared with the savage races which sur-
rounded them, these people represented the greatest advancement
in civilization and scientific knowledge. We see that much of this
was due to the very necessities of life; that to build public works,
levy taxes, determine boundaries, required a knowledge of the
science of mensuration. In the case of the Egyptians, their require-
ments, so far as we are able to estimate them, were even broader
than those of the Babylonians. The construction of the pyramids

shows clearly a
geometrical design,
executed scientifi-
cally, and this work,
y7N/^£~^y ^~;^^ ^^^ l&^--;^^^^- "^X. as well as the erec-

rfC-_^5:55^ic-:zrI^^^SK^^^^S^?^^ tion of wheat ware-

^:v5^'^:;?%^^^:^ :^^ir: houses and storage
^^^!^B^:^■SS'^s^^■■^#^.-'''^■v^^^:^J■- ::■■ reservoirs, necessi-
tated what was

Egyptian Pyramids , , . , ^ ,

doubtless to them
a very advanced conception of the principles of solid geometry.
Strangely enough, however, we are indebted to neither of these
races for the development of this knowledge into a science, but to
a race whose place in history is much later. The Greeks in the
ancient world occupied a position in some respects similar to that
which America has held in the modern world. They were a people
much given to exploitation and expansion, as well as to scientific
and philosophical pursuits, and in addition to this, prided them-
selves on their high degree of adaptability. Plato said, " Whatever
we Greeks receive we improve and perfect." They did not origin-
ate ideas so much as they adopted those of other races and improved
upon them to a degree which causes history to associate the Greeks
themselves with the original conception. The Greeks were travelers

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t^^mi^'r:

INTRODUCTION

and traders, interested in the arts and
sciences, and a distinguishing character-
istic of the race was their desire to learn
and experiment with new things. Seven
hundred years before Christ, Greek mer-
chants began sending their ships across
the Mediterranean to Egypt. Travelers
began to bring back accounts of this other
great nation, and the Greeks were imme-
diately interested in the reports of what
the Egyptians had done and were doing.

Thales (640-^546 B.C.), a merchant of
Miletus, was among those who became
acquainted with the Egyptians as a result

of commercial intercourse. Thales was at heart a student, and the
geometrical theories and practices of the Egyptians interested
him. In later hf e he terminated his business activities and opened
a school in his native city, Miletus, where he began teaching the
principles of geometrical science as it was then known. The
problems with which his pupils concerned themselves would
seem elementary to us. They dealt merely with finding the
heights of objects or the distances of ships from the shore, but
his school, which has come down to us under the name of the
Ionic school (so called from the Greek province in which Miletus
was situated), was the first intelligent effort to systematize the
study of geometry.

One of the students in the school of Thales was a noteworthy
successor of the first Greek geometrician. Pythagoras (580-501
B.C.) founded a school of mathematics at Crotona in the southern
part of Italy. His plan was much more elaborate than that of
Thales. Pythagoras felt that the study and character of the school
would create a deeper impression if it were organized as a secret
society. The watchword was ** Silence," and its members were
pledged to secrecy as to the nature of the work which was done.
The Greek government felt that the secret methods of the school
might be used to conceal harmful activities, and finally ordered
the institution closed. This circumstance, and the pledge of secrecy

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PLANE GEOMETRY

imposed upon the members of the school, have prevented our
learning much about it. The Pythagorean proposition, which states

that the square on the hypothenuse of
a right triangle is equal to the sum of
the squares on the other two sides, bears
the name of the school, although the
fact was known for a longtime before
Pythagoras proved it to be true. "Py-
thagoras changed the study of geometry
into the form of a liberal education,
for he examined its principles to the
bottom and investigated its pro-
Pythagoras positions in an immaterial and intel-

lectual manner."

Archytas (430-365 B.C.), who followed Pythagoras, was not so
much interested in speculative or geometrical subjects as he was
in the application of the science to practical uses. He invented
several mechanical toys operated on geometrical principles. Very
few sailors reaUze that the ability of one
man to move a tremendous weight of
sail was made possible by the discovery
of this Greek mathematician who lived
over twenty centuries ago, for it was
Archytas whp worked out and applied
the principles of the pulley. He is be-
lieved to have been the first student to
find a solution of the problem called
the "duplication of the cube," that is,
to find the dimensions of a cube the
volume of which shall be twice that of
a given cube.

Plato (429-348 b.c.) was a contem-
porary of Archytas, and his name is
associated even more generally with
geometrical science than that of his
compatriot. Plato called his school the
"Academy," and the underlying prin- piato

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INTRODUCTION 9

ciple of his theory was the abstract and systematic development of
geometric science. Plato insisted that the only instruments needed
for the study of the subject were the straight edge and the com-
passes, and the history of the science has demonstrated the ac-
curacy of his conclusions, as they are the only scientific tools
needed for all elementary work in the science.

Eudoxus (408-355 B.C.) studied imder Plato for a time, and
subsequently did some independent
work in the science. He directed his
attention chiefly to the principles of
proportion and certain methods of
proof, to which we shall make refer-
ence later. He was the first scientist
to begin to put into book form the
mathematical knowledge of his time,
and may properly be considered the
logical forerunner of the mathemati-
cian Euclid. Euclid, who was a teacher (S*.*^.''^*-/
in a school of mathematics in Alex-
andria, Egypt, about 300 B.C., was the author of what is probably
the most famous book on geometry. He collected and arranged all
the knowledge of the science down to his own time, and his book
still stands today in many respects as a final authority and the
background of the entire science. Despite the fact that the Egyp-
tians and Babylonians first developed a crude knowledge of the
subject, it is perhaps appropriate that the Greeks, whose generations
of scientists, culminating in Euclid, gave so much study to it, should
have furnished the name by which we call it. The Greek word
"ge" meaning the earth, and "metron" to measure, are the roots
which formed our name for the science

From its first crude beginning in the necessity for measuring the
destruction wrought by an ancient river, its instruments, crude
pegs and a knotted rope, developed and applied as a science by the
mathematicians of five centuries before the Christian era, supple-
mented and enlarged by the observations and discoveries of nearly
twenty centuries of research, the science by which the surveyors of
Egypt located their boundaries is today the method used for deter-
mining the power of a battleship or the contents of a mountain range.

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10 PLANE GEOMETRY

SUMMARY
I . Geometry Among the Babylonians and Egyptians

A. Derivation of the word '^geobietry."

B. Evidences op knowledge op geometry.

1. Among the Babylonians.

o. Documentary evidence.
(I) Clay tablets.
(II) Talismans.

(III) Monuments.

2. Among the Egyptians.

a. Evidences in practical life.
(I) Surveying.
(II) Reservoirs.
(ni) Irrigation.

(IV) Pavements.

h. Evidences in religious life.

(I) Orientation of temples.
(II) Pyramids.
c. Documentary evidence.
(I) Ahmes Papyrus.
(II) Hieroglyphics
II. Geometry Among the Greeks.

A, Source.

B, Schools that contributed to the development op geometry.

1. Ionic School, a. Thales (640-546 B.C.).

o T> ^u a u 1 ( ^- Pythagoras (580-501 b.c).

2. Pythagorean School j^^ ^^^^ ^^^^^^^ ^^ .

Q T>w • a k 1 !«• Pla*o (429-348 b.c).

3. Platonic School -{it^, />irkooercr \

(6. Eudoxus (408-355 b.c.)

^ ^ ., fa. Hippocrates (C. 440 B.C.).

C, Compilers \ , ^ ^tT: ,^ ^ JL . ^

^ (6. Euclid (C. 300 B.C.).

BIBLIOGRAPHY

Allman, G. J.: "Greek Geometry from Thales to Euclid": pp. 2, 3, 5, 7, 15,

16, 22, 29, 41, 139-140, 143 (footnote) and 154.
Ball, W. W. R.: "A Primer of the History of Mathematics": pp. 3-6, 8-14,

32, 42-48.
Ball, W. W. R.: "A Short History of Mathematics": pp. 5-8, 15-16, 25-27,

40, 54-63, 69-72.
Fink, K.: "Brief History of Mathematics": pp. 190-214 (with omissions).
Gow, J.: "A History of Greek Mathematics": pp. 176-178.
BoYER, J. F.: "Histoiredes Mathematiques": Chapters I-V.
Cantor, Moritz: "Vorlesungen tiber Geschichte der Mathematik": Vol. I,

pp 17-52, 90-114. 134-146, 170-187, 201-277. ,^^^ ,^ GoOgk

INTRODUCTION 11

EXERCISES. SET I. BASED UPON HISTORIC FACTS

1. From the derivation of the word geometry, can you suggest
any studies or professions in which geometry may be applied?

2. What gave rise in the first place to the art and eventually to
the science of geometry?

3. Prom the little told you in the foregoing paragraphs and any
references you may have read, what would you judge to be the
essential difference between the geometry of the Egyptians and
the geometry of the Greeks?

4. Judging from the character of the Roman, would you expect
him to do much to advance the science of geometry?

5. Among the formulas given in Ahmes Papyrus for determining
areas are the following: I. The area of an isosceles triangle equals
half the product of the base and one of the equal sides. II. The
area of an isosceles trapezoid equals half the product of the sum
of the bases, and one of the equal sides.

a. Using 6, 6i, for the bases, and s for each of the equal sides
write an algebraic formula for each of these areas, A.

b. What is the error in each of these formulas?

c. Draw figures to show that at times this error would not
matter much.

d. Draw figures showing cases where the error would make a
considerable difference. ^^^^

e. In the accompanying dia- y
gram find just what error is made «'=3o/ \^-30

(correct to tenths) by using the ^

Egyptian formula. ^=^o

6. Another formula given in Ahmes Papyrus is that for finding
the area of a circle. It tells you to multiply the square of the radius
by 1%. What value must the Egyptian then have assigned to ir?

7. What is meant by saying that a 3-4-5 triangle is a right
triangle?

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12 PLANE GEOMETRY

8. Show by knotting a piece of cord so that the parts have the
ratio 3 to 4 to 5 how the Egyptian "ropenstretchers" obtained their
east-west Une. (Stretch the cord around pins on a board and after it
is in place test the accuracy of the method with your right triangle.)

9. Plato and his school interested themselves in the so-called
Pythagorean numbers. Such numbers are those that would repre-
sent the lengths of the sides of a right triangle.
In this kind of triangle they must be such that
a^+b^=c\ The school of Plato found that
t(^)2+i]2=n^+[(^)2_U2,and hence that(^)^
+ 1, n, and (^)^ — 1 were Pythagorean numbers.

a. Verify the statement.

6. Find ten sets of Pythagorean numbers.

10. Pythagoras himself found that n, J^(n2 -1), and mn^+1)
were numbers such as described in the last exercise. Verify this
statement.

IL Bramagupta, a Hindu writer of the seventh century, gave
Pf }4{^+^f and 3^(^— q) as Pythagorean numbers,
a. Give various values to p and q to test his statement.
6. Verify his statement.

12, In the Culvasutras, a Hindu manuscript, directions for con-
structing a right angle are as follows: Divide a rope by a knot
into parts 15 and 39 units in length respectively, and fasten the
ends to a piece 36 units in length.

a. Draw a diagram to show what is meant by this.
6. Check to see whether these are Pythagorean numbers.
c. Is it true that all numbers having the ratio of these three are
Pythagorean numbers?

13. Archimedes proved that the value of t lay between 3}/j and
31^1. How does this compare with the value we use to-day?

B. A FEW ILLUSTRATIONS OF GEOMETRIC FORM

Before we begin a systematic study of geometry, let us see if we
can find any illustrations of the kind of forms about which we hope
to learn something. Do we not find such forms in nature? We

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INTRODUCTION

recall sjnnmetrical trees and conical mountains; we think of a
circular moon, spherical raindrops and crystals of many forms.

Crystals
EXERCISES. SET II. ILLUSTRATIONS OF GEOMETRIC FORMS
14. Make a list of some geometric forms you have found in
nature, under the following heads: spherical, conical, cylindrical,
prismatic, circular, etc.

Aside from natural objects, geometric forms continually appear
in the works of man. The building and room in which we are, the
furniture, windows, doorways, all are geometric in form. The
familiar objects of our daily life — coins, boxes, cylindrical tubes,
balls — all illustrate the application of geometrical principles.

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14 PLANE GEOMETRY

16. As in the preceding exercise, make a list of some geometric
forms you can find in the works of man.

C. MEANING OF GEOMETRIC FORMS

If we note these forms carefully, we see that they are various
combinations of the simple elements — ^points, lines, surfaces and
solids. At the outset, therefore, we should be sure that our ideas
about these elements are correct.

First let us consider a geometric solid. We have seen cones made
of wood, and others of ice cream; we have looked into a well and
said it was cylindrical; we have watched a soap bubble and called
it spherical. So we see that it is the shape or form, and not the
substance of which an object is made, to which we refer in speaking
of a geometric solid. When we mention a sphere, we mean the
space which it occupies or its shape without reference to its physical
properties or the material of which it is made.

If, as we have just noted, a solid is a limited portion of space, what
limits it? How is a solid separated from the rest of space? The
boundaries of a solid are surfaces, and since a solid is identified,
not by its material, but only by its shape, so must its boundary
be identified by its shape. A chalk box, for instance, is in the
form of a prism, i.e., the space it occupies is a geometric prism.
The boundaries of this prism are called surfaces, and they divide
it from the rest of space.

The surfaces meet and form lines. The edges of the chalk box
are referred to as the union of its sides. Now if we think of the
geometric prism — ^the space occupied by the box — ^the intersection
of the surfaces are lines.

It is evident, then, that lines crossing form points. A limited
portion of space is called a solid, the boundaries of a solid are called
surfaces, the intersections of surfaces a/re called lines, and the places
where lines cross are called points.

We see thus that a point is a place or position, and can, therefore,
have no length, breadth, or thickness. For convenience, we repre-
sent a point by a dot of lead, ink, or chalk. Such a dot is obviously
not a point, because it has some size, however small it may be, but
it marks a location, which is the real point.

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INTRODUCTION 15

If we could imagine a point to move, we would call its path a
line. A line, then, would have no width, but it would have length.
If we could now imagine a line to move, not along itself, we would
say it generated a surface. Again, the surface would have only
length and width, but, of course, no depth. If we consider a surface
to move, not along itself, it would form a solid, which would have
three dimensions.

In our study of geometry we shall have to deal with straight and
curved lines. Let us note, then, that a straight line is one which
is fixed {or determined) by any two of its points, and a curved line
is one no part of which is straight. Throughout the book, it is ito
be understood that when the word line is used without a qualifying
adjective, a straight line is designated. A line is of indefinite
length, so that when we wish to refer to a limited portion of a line
we shall call it a sect. All the facts with which we shall be concerned
for a time will be those relating to a single plane. A plane is a
surface such that if any two points in it be connected by a straight
line, thai line lies wholly within the surface.

EXERCISES. SET lU. MEANING OF GEOMETRIC FORMS

16. If a series of 600 points were put within an inch would they
form a line?

17. A machine has been manufactured which will rule 10,000 dis-
tinct lines within the space of one inch. Are these lines geometric?

18. Fold over a piece of paper. What will the crease represent?

19. If oil is poured on water, of what material is the surface
formed?

20. Put your foot in a heap of snow and quickly withdraw it.
Is the impression that is left a physical or geometric solid?

21. If I place a piece of red paper on a blue one, what is the
color of the surface between them?

22. If 1000 geometric surfaces were placed one on top of the
other would a geometric solid be formed?

23. Is a cake of ice a geometric solid?

24. (a) Make a list of some things in life which are referred to
as points. (6) How many of these are geometric points?

26. (a) Make a list of some things in life which are referred to

as lines. (6) How many of these are geometric lines? ,

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PLANE GEOMETRY

D. SUGGESTIONS OF A FEW USES OF GEOMETRY

We have reviewed briefly the historical development of geometry
and have called to mind illustrations of geometric forms, both in
nature and in manufactured articles and have clarified our ideas
of these forms.

Let us now consider a few of the uses of geometry. The subject
grew out of the need of land-measuring. Hence, historically at
least, surveying is the first known use of geometry. The following

illustrations show some
of the things that we
ought soon to be able to
do. Laying out bounda-
ries of property so that
the owner shall have his
just share, or finding the
areas of pieces of ground,
Diagram I are problems requiring

practical mensuration

Diagram III

Transit

about which we shall soon study. Finding the distances between
inaccessible points, as from Z to F in diagram I, across rivers and

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INTRODUCTION 17

over swamps, as in Exercises 111-113, and heights of objects as m
diagrams II and III, are all geometric problems such as are included
in surveying. We have all seen men in our streets with transits
on tripods. (A transit is an instrument for measuring angles.)
Geometry is necessary to solve the problems for which they are
getting the data.

EXERCISES. SET IV. MENSURATION

26. Make a Ust of mensuration formulas with which you are
already familiar.

27. Find the area of the following piece of property (three lots).
The measurements taken by a surveyor are noted on the diagram.

Other surveying problems

will be found later in the book. ^..-^"^1^?^^^ — ~

We do not yet know enough ^^y^C^^^^^ ^'

geometry to solve many such ^^ Pvt: IwU n/j

problems. ^' ^^'

Another use of geometry that quickly comes to the mind is
designing. We mark the use of geometric design in parquet floors,
linoleums, tilings, wall and ceiling papers, grill-work, stained-glass
windows, arches, and in similar objects. By learning how to
make five fundamental constructions (Exercises 28-32, inc.),
we shall be able to combine them into many geometric designs, and
thus get a clearer idea of one of the uses of geometry. The reasons
why these constructions are correct, and more elaborate work
in design, must be postponed until later in the text.

EXERCISES. SET V. CONSTRUCTIONS— DESIGNING
All the constructions in these exercises are to be made with the
use of compasses and unmarked straight edge only. The pupil is
reminded that unfamiliar technical terms will be found by refer-
ring to the index.

In general, in geometry, auxiliary lines (those needed only as
aids) are indicated by dotted lines, preferably light.

28. From a given point on a given straight line required to draw
a perpendicular to the line.

Let AB be the given line and P be the given point.

It is required to draw from P a line perpendicular to AB,

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PLANE GEOMETRY

/ B

With P as center and any convenient radius strike arcs cutting

AS at X and y.

With X as center and XF as radius
strike an arc, and with Y as center
and the same radius strike another
arc, and call one intersection of the
arcs C.

With a straight edge draw a line

+—5 through P and C, and this will be the

perpendicular required.
29. From a given point outside a given straight line required
to let fall a perpendicular to the line.

Let AB be the given straight line and P be the given point.
It is required to draw from P a line perpendicular to AB.
With P as center and any convenient
radius describe an arc cutting AB at X
andr.

With X as center and any convenient \X I Y /

radius describe an arc, and with Y as
center and the same radius describe an-
other arc, and call one intersection of the
arcs, C

With a straight edge draw a straight wk

line through P and C, and this will be

the perpendicular required.

It is interesting to test the results in Ex-
ercises 28 and 29 by cutting the paper and
fitting the angles together.

30. Required to bisect a given sect.
Let AB be the given sect.
— It is required to bisect AB.

With A as center and AB as radius de-
scribe an arc, and with B as center and the
same radius describe another arc.

Call the two intersections of the arcs X
andr.

Draw the straight line XY.
Then XY bisects the sect AB at the point of intersection M.

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INTRODUCTION

31.

From a given point on a given line required to draw a line
making an angle equal to a given angle.
Let P be the given point on the given
line PQ, and let angle AOS be the given
angle.
What is now required?
With as center and any radius de-
scribe an arc cutting AO at C and BO at D.
With P as center and OC as radius de- ^^B

scribe an arc cutting PQ at M,

With M as center and CD as radius de-
scribe an arc cutting the arc just drawn at
JV, and draw PN.

Then angle MPN is the required angle.

32. Required to bisect a given
angle.
Let AOB be the given angle.
It is required to bisect the angle
AOB.

With as center and any conve-
nient radius strike an arc cutting
OA at X and OB at Y.
With X as center and the sect ZF as radius strike an arc, and
with Y as center and the same radius strike an arc, and call one
point of intersection of the arcs P.
Draw the straight line OP.
Then OP is the required bisector.
33. Make constructions similar to the following:

Suggestion: AB=OB.

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PLANE GEOMETRY

34. Draw the fol-
lowing figure.

36. Make a con-
struction similar to a.

d36.* Copy 6.

a b

37. From a study of Exercise 28, suggest how to erect a per-
pendicular at the end of a sect.

38. On a given sect construct a square.

39. These figures show a parquet floor design, and one of the
imits of the design enlarged. Construct figures similar to these.

ABCD is a square,
and X, Y, Z, and W are
the mid-points of the
semidiameters OEy OF,
OG, OH, respectively.

A E B

40. Make a construction similar to the ad-
joining figure.

The vertices of the square are used as centers
for four of the arcs.

The radius equals one side of the square.

42. Make a
construction
similar to b.

41. Make a
construction
similar to a.

* As here, d will be prefixed to any exercise which the student is likely to
find difficult at this stage.

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INTRODUCTION

d43. Make a construction similar to the following:
The figure is based on an equilateral triangle, the centers of the
interior arcs being the midpoints of radii
drawn to the vertices of the equilateral tri-
angle inscribed in a circle (i.e., having its ver-
tices on the circle).

Note: See Fig. 2, exercise 33.

N. B. — ^The design shown in Mable Sykes, "Source-
Eook of Problems for Geometry," page 160, II, 5 (Fig.
138a), shows a good application of exercise 43.

44. Make a construction
similar to a.

d46. Make an ornamental
design similar to 6. The circle
is divided into how many
equal arcs? How many de-
grees in each central angle?
a "^

What kind of triangle is formed by
two consecutive radii and the sect
joining their ends? What other method
does this suggest of dividing a circle
into six equal arcs?

46. Make an ornamental drawing
similar to the one in the accompanying
figure. Describe the construction.

Suggestion: First draw an equilateral poly-
gon with six sides in a circle.

47. Construct a six-
-f- ^.--Tir--^ —I- pointed star.

48. Bisect each of
the four right angles
formed by two Imes
intersecting each other
at right angles.

49. Make construc-
tions similar to the
following:

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22 PLANE GEOMETRY

60. Make constructions similar to the following:

In such figures artistic patterns may be made by coloring vari-
ous portions of the drawings.

In this way designs are made for stained-glass
windows, pil-cloth, colored tiles, and other dec-
orations.

51. Draw a sect of any convenient length,
and upon it construct a design similar to the
one in the figure.

62. On a line LM take a sect AB, Divide it

into 8 equal parts. With your compasses make

an ornamental scroll as shown

in the diagram.

53. Using the hints given,
make a copy of the accompany-
ing outline drawing of a Gothic
window. The arc BC is drawn

with A as center and AB as radius. The
small arches are described with A, D,
and B as centers and AD as radius.
The center P is found by taking A and
B as centers and AE as radius. How
may the points D, E, and F be found?
64. In many different machines, such

as the sewing machine, printing press,

A F D E B etc., there is a wheel called a cam, which
is used to modify the motion of the machinery. Cams are con-
structed in various shapes and dimensions, depending upon the use
for which they are designed. The figure shows the method of draw-
ing the pattern of a heart-shaped or " uniform-motion " cam. Let

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INTRODUCTION 23

the " throw " be AB and the center 0. Divide AB into eight equal
parts at C, Z>, etc. Through A,C,Dy . . . , fi, draw circles with
centers at 0. Draw sects dividing the
angular magnitude around into six-
teen equal parts. Beginning at A,
mark the points where the consecutive
circles and consecutive sects intersect,
and through these points draw a smooth
curve, as in the figure.

Draw such a cam with AS equal to a
given sect m, and OA equal to a given
sect n.

(Taken with modifications from
Stone-Milhs, Elementary Plane Geometry.)

56. Select and copy som6 geometric design.

66. Make an original design based on the fundamental construc-
tions given in exercises 28-32.

Pupils particularly interested in this part of the work are
referred to: Sykes, Mabel, "Source Book of Problems in Geo-
metry." (Pub. Allyn and Bacon.)

Geometry is used in architecture. Whether the architect is
drawing the plans for an ordinary dwelling-house or a massive
cathedral, he is constantly concerned with geometric forms and
constructions. Consider for a moment what problems of this
character must have confronted the architect of some large building
in our community.

The list of the direct uses of geometry would be very long if
complete. In a few sentences let us,, therefore, simply enumerate
a few more miscellaneous suggestions for its uses. Problems
scattered throughout this book show more concretely how geometry
is used in the cases enumerated. In making all kinds of diagrams,
reducing and enlarging maps, the principles of geometryare apphed.
In engineering, geometry is needed for such matters as laying out
railroads, and planning the constructions of machines, bridges,
and tunnels; and in astronomy, ascertaining the altitude of stars
and similar problems require geometric principles.

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24 PLANE GEOMETRY

EXERCISES. SET VI. SOME OTHER USES OF GEOMETRY
57. State any other uses of geometry which you know.
68. At the entrance to New York Harbor is a gun having a

range of 12 mi. Draw a Une inclosing the range of fire, using any

convenient scale.

59. Two forts are placed on opposite sides of a harbor entrance,
13 mi. apart. Each has a gun having a range of 10 mi. Draw a
plan showing the area exposed to the fire of both guns, using any
convenient scale.

60. Make an accurate diagram of a tennis court or a foot-ball
field noting all lime lines.

d61. Draw to a convenient scale a plan of the ground floor of
your school building.

E. THE BASIC PRINCIPLES OF GEOMETRY

As in all scientific work of an exact nature, the discoveries in
geometry rest upon a few basic principles. These may be classified
under three heads: definitions, axioms, and postulates.

You all can probably recall having heard people argue most
heatedly about some question and reach no conclusion at all. This
is often the case simply because when two people argue, they very
often do so without having clearly in mind the conditions about
which they are arguing. In all debates or discussions it is essential
that we start with the same premises, and our work in geometry
should help us to learn to collect our premises in orderly fashion.

The premises upon which the early parts of geometry rest are
to a great extent definitions, and it is therefore very necessary that
we have a clear image and definition of each new technical term
we meet. The wording of our definitions may differ, but the con-
tent must be the same. Every good definition should include all
that may fall under a particular class, and clearly exclude all that
does not fall under that class. Suppose, for instance, we want to
define the word botany. We might say, to begin with, that it is
a science — but we have not differentiated it from the physical
sciences, so we say it is a natural science. But, then, so is zoology.
Hence it is necessary to differentiate still further, and say it is the
natural science which deals with plant life. Now have we fully

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INTRODUCTION 25

and finally defined it? Can you possibly think of any science —
now that it is thus defined — ^with which it can be confused? Make
all possible tests, and if you find that it cannot be confused with
any other science, well and good, — ^then we have found an accept-
able definition.

Throughout this book no word will be defined until we are ready
to make use of it, but then it will be our duty to see the word in
its full meaning.

From the knowledge we have of algebra, we already know what
some of the axioms are, and some uses to which they may be put,
though we may not have defined the word "axiom." Some of the
axioms for which we shall have immediate use are:

1. The sums of equals added to equals are equal,

Elxample If 5=5

and a=h
then 5+a=5+6.

2. The remainders of equals subtracted from equals are equal.

Example If a^c

and b=d
then a—b=c—d

3. The products of equals multiplied by equals are equal.

Example li a=x

and b=y
then ab=xy.

4. The quotients of equals divided by equals are equal.

Example If x=y

and m=p

then ^^^
m p

Cases in which the divisor is zero will not be considered in this
text.

5. A quantity may be substituted for its equal in a statement of
equality or inequality.

Example If x s 5 If a = 6

and x+y^7 and 2a+5>a+2

then by substitution 5+y=7 then 26 + 5 > a + 2

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26 PLANE GEOMETRY

6. Two quarditiea which are equal to equal quarditieSy are equal to
each other.

Example If a^b

b^c

and csd,

then a=d

7. The whole is equal to the sum of its parts.

-n 1 a , a , a

Example 2"*"3"*"6~^

Note: "part" is here ueed in the sense of a common or vulgar
fraction.

Other axioms will be stated as we need them- Thus we see that
an axiom is the statemerU of a general maihematical truth which is
granted withovl any proof.

A postulate is simply a geometric axiom. That is, it is the state-
ment of a geometric truth which is granted without any proof.

We shall now attempt to formulate a few such truths.

EXERCISES. SET VII. ILLUSTRATIONS OF POSTULATES

62. Why is it shorter to cut across a field than to go around it?

g63.* How many pairs of roots are there when two simultaneous
linear equations are solved? What is the graphic explanation of
this?

g64. Why is it that in making the graph of a linear equation,
such as x+y = 13, we need to plot but two points, and that a third
point may be used to check the correctness of our work?

66. Why is it that the Panama Canal is a great advantage over
the route formerly used to reach a point on the western coast of
South America Irom the West Indies?

66. Why is it that in putting up a croquet set all one needs to
do to get the wickets in line with the stakes is to tie a string tightly
to one stake and stretch it and fasten it to the other stake?

* As here, g will be prefixed to any exercise in this text which presupposes
an acquaintance with the graph. The pupil is here referred to M. Auerbach,
An Elementary Course in Graphic Mathematics (Alljm and Bacon), pp. 29-31,
for review, and previous pages in the same if the subject is new to him.

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INTRODUCTION 27

67. What does the bricklayer do to get a row of bricks in a
straight line? Why? Does the gardener do anything similar to
this?

68. Point out which of the following postulates upholds each
of your answers to exercises 62-67.

POSTULATES OF THE STRAIGHT LINE NEEDED IN PROOFS

1. Two intersecting straight lines determine a point.

2. Two points determine a straight line.

3. A straight line is the shortest distance between two points.

69. How often can two straight lines intersect?

70. Use your answer to the last question to state postulate 1 in
another way.

71. How many straight lines can be drawn between two points?

72. Use your answer to the last question to state postulate 2 in
another way.

73. Give at least one good illustration of how each of the three
postulates mentioned may be used in practical life.

We have used several other postulates in making some of the
constructions on pages 13 to 18. They are: (1) A sect may be
produced indefinitely. (2) A circle may be described with any point
as center and any sect as radius. (3) A point and direction determine
a straight line. But these are so exceedingly obvious that we shall
not feel obliged to quote them.

DEFINITIONS

An angle is the opening between two lines. The lines are called
the sides' of the angle, and the point at which they mset the vertex.

An angle may be named in any one of three ways as ^A in
Fig. 1 where there is no danger of confusion, or as in Fig. 2 ^ABC,
^ABX, ^XBC where there are several angles (the vertex always
being read second), or again as in Fig. 3 ^ca, ^cdy -^ad.

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28 PLANE GEOMETRY

1. Kinds of angles defined according to individual size.

A straight angle is one whose sides run in opposite directions so

./^^l^ , ds to form a straight line.

'^ '^AOB is a straight angle.

A right an^h is one^haJf a straight
tfiffe.
^AOB and ^BOC are right

-2

An acute angle is less than a right
angle.
-^ '^RST is an acute angle.

An obtuse angle is greater than a right angle but less than a straight
angle.

-^XOY is an obtuse angle. x

2. Kinds of angles defined according to sums.
Complementary angles are two whose sum is a right angle.
Supplementary angles are two whose sum is a straight angle.

EXERCISES. SET Vm. SUMS OF ANGLES IN PAIRS
74. Construct: (a) two complementary angles whose ratio is 1
to 3; 3 to 5.

(6) Two supplementary angles whose ratio is 1 to 3; 3 to 5.

MEASUREMENT OF ANGLES

There are three systems of measurement of angles. The one
probably known to most of us is the sexagesimal system, and was
mentioned on page 6. It divides the entire angular magnitude
about a point into 360 parts, each of which is called a degree; these
again are divided into sixtieths, each of which is called a minute,
and each minute is again divided into sixtieths, each of which is
called a second.

The size of an angle then depends upon the amount of opening
between its sides. The amount of opening depends upon the

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INTRODUCTION

amount that one side has to revolve to bring it into the position
of the other, and the greater that amount the greater the angle.
Thus in these com-
passes the first angle
is smaller than the
second, which is also
smaller than the third.
The length of the sides
has nothing to do with
the size of the angle.

A special instrument, called a protractor, is frequently used
for angle measurements. The figure given below represents
one form of the protractor. By joining the notch O of the
protractor to each graduation mark we obtain a set of angles
at 0, each usually representing an angle of one degree.

To measure a given angle with the protractor, place the notch
of the protractor at the vertex of the angle, and the base line along

one side of the angle. The
other side of the angle then
indicates on the protractor
the number of degrees in
the angle. Thus the angle
AOB contains 50°.

To draw an angle of a
given number of degrees,
-^ place the base of the pro-
tractor along a straight line and mark on the line the position of the
notch 0. Then place the pencil at the required graduation mark,
and (after removing the protractor) join the point so marked to 0.

EXERCISES. SET IX. MEASUREMENT OF ANGLES

76. Show how a fan can be used to illustrate the idea of angular
magnitude.

76. What kinds of angles are formed by the hands of a clock at
(1) two o'clock, (2) three o'clock, (3) four o'clock, (4) six o'clock?

77. What kind of angle is equal to (1) its complement, (2) its
supplement?

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30 PLANE GEOMETRY

78. What kind of angle is less than its supplement?

79. What angle is 7° less than one-third its complement?

d80. State as a formula (1) the number of degrees in the com-
plement of the supplement of any angle a, (2) the number of
degrees in the supplement of the complement of angle a.

81. Two supplementary angles are in the ratio of 7 to 2. Find
the number of degrees in each.

82. If four lines a, 6, c, d, are drawn from a point in the order
given, so that a is perpendicular to c, and b is perpendicular to d,
find ^ad if ^bc is 60°. (See definition, page 31.)

83. Three angles together make up the angular magnitude about
a point. The first is 10° greater than the second, and the second
is 17° greater than one-half the third. How many degrees in each?

84. Find that angle whose supplement is eight times its comple-
ment. Is it possible to find one whose supplement is one-eighth
of its complement?

86. How many degrees in the angle which exceeds one-third its
complement by 15°?

86. Find the number of degrees in the angle whose excess over
its complement is one-fourth the difference between its complement
and itself.

87. Construct two straight angles. Cut one out and place it
on the other. What can you say of them? Do you think it is
true of all straight angles?

88. Construct two right angles, (a) How is each related to a
straight angle? (6) How are they related to each other? Why?

89. Using a protractor (a) construct three angles each of which
is the complement of 20°. (b) Construct two angles each of 30°.
Construct their complements, (c) What conclusions can you
draw from (a) and (6)? (d) Why?

90. Do the same as you did in exercise 89 for the supplements
of 50° and 60°.

Corollary. A truth that is directly derived from another is called
a corollary. The conclusions drawn in exercises 88-90 are called
corollaries, since they are all directly derived from the conclusion
drawn in exercise 87.

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INTRODUCTION

POSTULATES OF THE ANGLES

1. All straight angles are equal.

Cor. 1. All right angles are equal. Why?

Cor. 2. Complement^ of the same angle or equal angles are

equal. Why?
Cor. 3. Supplements of the same angle or equal angles are

equal. Why?
Perpendicular. A perpendicular is a line that meets another at
right angles,

3. SlUkIs of angles defined according to relative position.

Adjacent angles are two that have a ^o^

common vertex and a common side lying y^ ^^^B

between them, ^^^^^-^ —

^AOB and ^BOC are adjacent. o"^ A

EXERCISES. SET X. RELATIVE POSITION OF ANGLES

91. Why is it that <A05 and ^AOC are not adjacent angles?

92. Can you name other angles in the diagram which are not
adjacent?

93. Tycho Brahe (1546-1601), a Danish nobleman who built
and operated the first astronomical observatory, in his earliest

observations used a quadrant
for measuring the altitudes
of stars, or their angular dis-
tances above the horizon.
Show that when the instru-
ment was held in a vertical
plane, and the sights A and
B aligned with the star 5, the
altitude of the star was de-
termined by observing the
angle CAT),

(Taken from Stone Millis,
Elementary Plane Geometry.)

94. Make such a quadrant
of cardboard or wood and

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PLANE GEOMETRY

use the method of exercise 93 to find the elevations of objects in
the neighborhood such as trees, hills, steeples, telephone-poles, etc.

Vertical angles are those which
are so jjj^aced that the sides of each
are the prolongations of the sides
of the other, OX is the prolonga-
tion of BOy and F is the prolon-
gation of AO. Therefore < YOX
and ^AOB are vertical angles.

EXERCISES. SET X (continued)

96. Name two other angles in the preceding figure which are
vertical, and tell why they are vertical.

96. Classify angles according to (1) individual size, (2) relative
size, (3) relative position.

97. (a) Draw two complemen-
tary-adjacent angles.

(6) Draw two angles of the same
size as those in (a) but not adjacent.

(d) Draw two non-adjacent sup-
plementary angles.

98. In the accompanying diag-
ram select those angles which are
straight, right, acute, obtuse, com-
plementary, supplementary, adja-
cent, and vertical.

99. Read the angle which is equal
to:

(a) ^AOB+^BOC.
(6) ^AOC+^COD.

(d) ^AOC+^COD-^BOD.

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INTRODUCTION

100. Construct an angle equal to:

(a) The sum of these three angles.

(6) The sum of ^PQR and ^XYZ less ^STV

101. Show that sects bisecting two complementary-adjacent
angles form an angle of 45°.

102. What kind of angle do sects bisecting two supplementary-
adjacent angles form? Prove your answer.

103. The following will illustrate the unreliability of observation
and the need of logical proof.

(Subdivisions (a) through (/) were taken fromWentworth-Smith,
Plane Geometry, and {g) through (i) from Hart and Feldman,
jv ^jj Plane Geometry.)

^ * (a) Estimate which is the longer sect,

Xn ' ^^ ZB orXF , and how much longer. Then

test your estimate by measuring with the ^ q B

compasses or with a piece of paper carefully
marked.

(6) Estimate which is the longer sect, AB
or CD J and how much longer. Then test your
estimate by measuring as in (a).

AB and CD are both straight

ines. If one is not straight,
which one is it? Test your an-
swer by using a ruler or the
folded edge of a piece of paper.

(d) Look at this figure and state Ay/MW/////AW/B
whether TB and CD are the same dis- y7777/7777777777777/
tance apart at A and C as at B and D. C ^^^^^^^^^^P
Then test your answer as in (a).

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PLANE GEOMETRY

{e) Look at this figure and state whether AB will, if prolonged,
lie on CD, Also state whether WX will, if prolonged, lie on YZ.
Then test your answer by laying a ruler along the lines.

A (/) Look at this figure and state which
of the three lower lines is AB prolonged.
Then test your answer by laying a ruler
along AB,

(g) In the figures below, are the lines
everywhere the same distance apart? Test your answer by using
a ruler or a sUp of paper.

/////////////
WW WW

/////////////

(h) In the diagrams given below, tell which* sect of each pair is
the longer, a or 6, and test your answer by careful measurement.

(i) In the figures below, tell which lines are prolongations
of other lines. Test your • ^

:z:

answers.

104. (a) Draw two un-
equal supplementary-adja-
cent angles.

(6) Extend the common side of these angles through the vertex,
and call the angles thus formed a, 0, y, 5,

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INTRODUCTION

(d) What relation exists between /3 and 7?

(/) Do the same, using angles j8, 7, h.

(g) Do the same, using angles 7, 5, a.

(h) What conclusion can you draw?

106. If a plumb-line is fastened to a
horizontal wire nail at the vertex of the
angle of a quadrant, and the quadrant
is turned so that the plumb-line falls
along 90® (here indicated by OA), by
noting where the shadow of the nail
strikes the quadrant the angular alti-
tude of the sun may be obtained. Ex-
plain why OC in the diagram gives the
angular altitude of the sun.

INSTRUMENTS FOR MEASURING ANGLES

Surveyors and engineers em-
ploy for measuring angles costly
instruments called theodolites.*
An inexpensive substitute for a
theodoUte is shown in the accom-
panying figure 1. It consists of

Fia. 1

* The identical theodolite with which the historic Mason-Dixon line, be-
tween Maryland and Pennsylvania was run, settling a controversy of a century
growing out of the overlapping charters of Charles I to Lord Baltimore and
Charles II to William Penn, has lately become the possession of the RoyaJ Geo-
graphical Society, of London, through Edward Dixon, descendant of Jeremiah
Dixon, who used it. Mason and Dixon had observed, for the Royal Society,
the transit of Venus at the Cape of Good Hope in 1761, and did their American
work two years later. When the Une was resurveyed 150 years later, by the
Coast and Geodetic OflBce of Washington, it was proved to be exceptionally
accurate, with no errors of latitude of more than two or three seconds— cer-
tainly a creditable result for the time and the primitive instrument with which
the work was done. The Mason-Dixon theodolite has two sights, a large com-
pass in the cenfer of its horizontal plate, and is adapted for measuring either
horizontal circles or magnetic bearings. The graduated ch-de is twelve inches
in diameter, divided into five minutes, and read by a single vernier.

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PLANE GEOMETRY

Fio. 2

two pieces of wood shaped like rulers mounted on a vertical axis,

by a pin driven through their exact centers. The vertical needles

inserted near the end of the
rulers are used for sighting.
In place of the needle near-
est the eye, it is better to
employ a thin strip of wood.
Ay having a fine vertical slit;
and in place of the other
needle, a vertical wire fixed
in a light frame, JS. By the

help of this instrument, and a protractor, one can measure with

considerable accuracy an angle on the ground; for instance, the

angle MON (figure 2). The following is

a simple substitute for the theodolite.

By means of it angles may be measured

in both horizontal and vertical planes.

The vertical rod MM' is free to revolve

in the socket at M, carrying a horizontal

pointer which indicates readings on the

horizontal circle divided into degrees.

These divisions must be marked. The

pointer at M' is provided with sights

and is free to move in a vertical circle

around M\ By sighting along this pointer, vertical angles may

be measured on the quadrant.
(This instrument was suggested in Betz and Webb, Plan e Geometry.)

EXERCISES. SET XI. INSTRUMENTS FOR MEASURING ANGLES

106. Construct an instrument such as that shown in figure 1 of

the precedin g section or an astrolabe or a good substitute by means of

which angles may be measured in vertical and horizontal planes.

F. THE DISCOVERY OF SOME FACTS AND THEIR
INFORMAL PROOF

A theorem is the statement of a fact which is to be proved.

The fact which you discovered if you worked exercise 104 and
applied in 105 is one which was known to Thales about 600 B.C.

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INTRODUCTION 37

It is very important, so we shall attempt to prove it again. It is
the first theorem in our syllabus.

Theorem' 1. Vertical angles are equal.

In the accompanying .diagram, what angle is the supplement
of ^BOZ?

What other angle is the sup-
plement of ^BOZ^

What postulate can you
quote to prove the equality of
these two angles which are the
supplements of '^BOZt

In similar fashion prove that ^BOZ^ ^AOY.

A plane polygon is a portion of a plane whose boundaries are
straight lines. The lines which bound the polygon are called its sides ;
the intersections of its sides are the vertices of the polygon; and the
angles formed by its sides, its angles.

A triangle is a polygon of three sides. Triangles may be classified
in at least two ways.

1. CLASSIFICATION BASED UPON SIDES
A scalene triangle is one having no two sides equal.
An isosceles iriangle is one having two equal sides.
An equilateral triangle is one having all three sides equal.

2. CLASSIFICATION BASED UPON ANGLES

An acute triangle is one in which all the angles are acute.

A right triangle is one which has one of its angles a right angle.

An obtuse triangle is one which has one of its angles an obtuse angle.

The following experiment will enable us to solve such problems
as the one in Ex. 110, in which we would like to find the distance
between two points separated by marshland, but for which we
lack the necessary information.

EXPERIMENT

(a) Construct a triangle <?

having two of its sides equal 5 ^^
tosectsft and c, andtheangle ^^.^^f^
whose sides they form equal to ^A. ^.^'"""'"K"^ *»
Construct a second such triangle. ^^— 1 1

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38 PLANE GEOMETRY

(6) Cut out and tear oflf part of one triangle as indicated by the
ragged dotted line between Q and P in the diagram, and place it

yX^,^ upon the other so that the equal
;/r angles coincide, and so that the

equal sides 61* and 62 fall along

/ each other.

^--ss^H |_^ (c) Describe what happens.

^' - -^ (d) What parts of these tri-

•vP^^'' angles are you sure will fit on
each other to start with?

^i^ g^ T^o placed them as you were told to?

(f) What is it that finally de-
termines that the triangles may be made to coincide throughout?
{q) Can you quote a postulate to uphold your statement in (/)?
(A) Repeat the process of constructing, cutting, and placing
when the given ^A is a right angle.

(i) Repeat again when the given ^A is an obtuse angle,
(j) What conclusion can you draw?

Such triangles as A\XY and AJPQ in the figure^ which may be
made to coincide throughout are said to be congruent.

Congruent polygons are those which may be made to coincide
throughovt. Hence they have not only the same shape but the
same size. The symbol used to denote congruence is ^ , which is
simply the sign of equality in addition to the initial letter of the
word simUis (Latin for similar) thrown down on its side. •

Superposition Postulate. It may be noted that in the preceding
experiment we have assumed the following fact. Any geometric
figure may be moved about without changing its size or shape.

EXERCISES. SET XII. MEANING OF CONGRUENCE AND
CLASSIFICATION OF TRIANGLES

107. Draw a polygon having ^ IL

the same area as the accom-
panying figure, but not the same
shape.

♦Read "b sub-one," "b sub-two," indicating that these sects equal the
original sect b. ,g.^.^^^ ^^ GoOglc

INTRODUCTION

108. Draw a polygon having the same shape as that in exercise
107, but one-fourth its area.

109. Classify triangles 1 through
18 in the following diagram (a),
according to sides and (6) accord-
ing to angles.

Just as no matter how many
straight lines are drawn between
two points they will all coincide
with one another, so, no matter
how many triangles are constructed with two sides aiid the angle
between them, equal each to each, they may be made to coincide.
We said that two points determine a straight line. Likewise
we can state the second theorem of our syllabus as follows :

\\l4\ 9

\'/

\ ^'^ /^N\ /

IsVifi 13 /\

/\^^j /io\

* Theorem 2.
triangle.

Two sides and the included angle determine a

EXERCISE. SET Xm. APPLICATION OF CONGRUENCE OF

TRIANGLES

110. Show how the distance
from A to 5 can be found when
because of some obstruction it
cannot be measured directly.
Suppose a commander of an
army wished to iBnd the dis-
tance across a stream. He would
have a problem different from
that in Ex. 110, and hence that case in the congruence of triangles
wouldn't help him. The following experiment would enable him
to solve the problem, however, as outlined in Ex. 111.

* It is suggested that theorems marked in this way be developed in the class-
room only. The proofs are introduced rather for the sake of letting the pupils
realize that the sequence is a perfect chain than as a test of the pupil's power.
In order to avoid any possible danger of memory work, the authors believe it
wiser to omit entirely the proof of such propositions as appear to be beyond
the power of the pupil to develop alone.

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PLANE GEOMETRY

EXPERIMENT

(a) Construct a figure having two of its angles equal to ^A and
-^C respectively, and the side common to the two angles equal to the

sect 6, as in the ac-
companying dia-
gram. Construct a
second such figure.
(6) Cut one of
themout and place
it upon the other
so that the equal
parts coincide.

(d) What would happen if
thelinessimilarlyplacedinthe
other figure were produced?

(e) What conclusion can
you draw?

\«

(/) Would all triangles having two angles and the side between
them equal each to each be congruent?

(Test for cases where ^C isa. right angle or an obtuse angle.)

If the test is satisfactory we can state our third theorem as
follows:

'^Theorem 3. Two angles and the included side determine a
triangle.

Note. — The discovery of Theorems 2 and 3 is attributed to Thales.

EXERCISES. SET XIII (continued)
111. If it is necessary for a commander of an army to know the
distance from B to the inaccessible point C across a stream, he may

find it as follows: Run a line BD at
right angles to BC, Prolong CS. From
D locate a point A in the prolongation
of CJSsothat ^BDA = ^CDB. Meas-
ure AB. Show that CB=ABj and
hence that BC may be found.

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INTRODUCTION

112. To measure the distance from B to the inaccessible point
Fy nm BD in any convenient direction. Locate C, the mid-point
ofBD, FromD, with an in-
strument, run DE so that
^CDE^^FBCy and lo-
cate-B in line with Cand F.
Show that BF=ED, and
hence may be determined. ^

Such problems as 111 suggest methods that could be used in
spy work, since only crude instruments need be employed.

113. To find the distance ZC, when C is inaccessible, let £ be a
convenient point from which A and C are
visible. Lay out a triangle ABCi making
<3=^land<4=^2. Show that the dis-
tance AC may be found by measuring'ZCi.

This is a method attributed to Thales for
finding the distance of a ship from shore.
114. Thales of Miletus is said to
have invented another way of finding A

the distance of a ship from shore. This
method may have been as follows:

Two rods, m and n, are hinged to-
gether at A. One arm m is held
vertically while the other n is pointed
at the ship S. Then the instrument is
revolved about m as an axis until n
points at some familiar object Si on
the shore. Explain why BSi = BS^

dll6. Tell
what measure-
ments to make
to obtain the
distance be-
tween two in-
a c cess i ble
points, A and
Fig. 2 B, in Rgure 1.

dll6. Explain the method suggested by the diagram in Figure 2
for finding the distance from S to the inaccessible point R. y^

42 PLANE GEOMETRY

SOME PROPERTIES OF THE ISOSCELES TRIANGLE

Before we can solve many more practical problems it will be
necessary for us to collect a number of geometric facts. This we
will do first in a simple fashion — ^by experimenting — and second by
actual proof, for though many of us may already believe them to
be true we must be ready to convince others.

EXPERIMENT

(a) Construct an isosceles triangle having sect b as its base,
and each of its equal sides equal to sect a. Bisect its vertex
angle (i.e., the angle included by the equal sides).

___________^ (6) Cut the triangle

out, crease along the bi-

. __ sector of the vertex angle

and see what happens.

(d) Make a list of three otherfacts you have thus discovered
concerning the isosceles triangle.

By the bisectors of the angles of a triangle, those sects of the
bisectors are indicated which are terminated by the opposite sides.

The facts discovered in the last experiment may be stated as
corollaries to the fact that the bisector of the vertex angle of an
isosceles triangle divides it into two congruent triangles. We find
that the parts of those triangles similarly placed with respect to
the parts known to be equal to begin with, are equal. Such parts
are called homologous. Homologous parts of congruent polygons
are always equal.

Theorem 4. The bisector of the vertex angle of an isosceles
triangle divides it into congruent triangles.

Cor. 1. The angles opposite the equal sides of a triangle are
equal.

Cor. 2. The bisector of the vertex angle of an isosceles tri-
angle bisects the base, and is perpendicular to it.

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INTRODUCTION

SOME PROPERTIES OF THE EQUILATERAL TRIANGLE

' When we study the equilateral triangle we find more corollaries
to theorem 4.

EXERCISES. SET XIV. EQUILATERAL TRIANGLES

117. What fact concerning the angles of an equilateral triangle
can you base on the fact concerning the angles opposite the equal
sides of an isosceles triangle?

118. How would you state a'coroUary concerning the equilateral
triangle corresponding to corollary 2 under theorem 4?

119. Construct an equilateral triangle and
bisect any two of its angles as in the accom-
panying diagram.

120. Can you prove triangles ABD and
CBX congruent?

121. In these triangles what side of CBX
is homologous to AD in ABD?

122. What conclusion can you draw con-
cerning these sects?

123. Try to prove the same fact, using triangles ACD and
CAX.

124. Could you use two other triangles to prove the same fact?
The results of exercises 117-124 may be stated as follows:

Theorem 4.
Cor. 4.

Cor. 3. An equilateral triangle is equiangular.

The bisectors of the angles of an equilateral triangle
bisect the opposite sides and are perpendicular to
them.

Cor. 6. The bisectors of the angles of an equilateral triangle
are equal.

The following experiment will help
us to understand why it is that a
long span of a bridge in which the
truss is made with queen-posts and
diagonal rods, as shown in the dia-
gram, is sufiiciently supported.

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/^^.

•V^

s.$.

"^^

:/.-—

_-h£.

.-^.^^

^^-^ '

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a.N

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44 PLANE GEOMETRY

EXPERIMENT

(a) Using three sects of diflferent lengths construct two triangles,
XYZ and XPZ and place them as in the accompanying diagram
so that the longest sides c are coincident, and XY^a^XP are

next to each oth er.

(6) Draw YP. What kind of
triangles are AXYP and AZYP?
Why?

(d) In (c) what axiom did you
havetoapply to prove '^XYZ^
^XPZt

(e) Now what parts do you
know to be equal in 6XYZ

>^^ and AXPZ?

(/) What conclusion can you draw concerning these triangles?

{g) Test to see whether you could prove AXFZ and AXPZ
congruent by placing XY=a=XP (the shortest sides) together.

(h) If you placed the triangles as suggested in (g) what axiom
would you have to apply to prove two angles of the triangles equal?

(i) Is the conclusion you drew in (/) true here? If so, we may state

Theorem 6. A triangle is determined by its sides.

EXERCISES. SET XV. FURTHER APPLICATIONS OF
CONGRUENCE OF TRIANGLES

126. Why is it that the last case in the congruence of triangles is se-
parated from the first two cases by a theorem on the isosceles triangle?

126. (a) Three iron rods are hinged at
the extremities, as shown in the diagram.
Is the figure rigid? Why?

(6) Four iron rods are hinged, as shown
in the diagram. Is the figure rigid? If
not, how many rods would you add to
make it rigid, and where would you add them?

Note. This experiment can be tried conveniently with the Mecano toy.

127. How many diagonal braces are needed to support a crane?
Why?

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INTRODUCTION

128. How many tie-beams connecting each pair of rafters are
needed to brace a two-sided roof suf-
ficiently? Why?

129. Note how the girders of the
bridge in the accompanying diag-
ram are fastened. Why cannot the
bridge collapse?

130. Why is the long span of
the bridge represented in the
diagram sufficiently supported?

131. Prove by means of con-
gruent triangles that the directions given on page 19 for the dup-
lication of an angle are sound.

132. Could you have duplicated -^AOB in that exercise if
OD>OC?

133. Prove the directions given for bisecting an angle correct.

134. Prove that the directions for erecting a perpendicular to a
line at a given point in it are correct. Can you suggest any varia-
tion in this construction such as is pointed out in exercise 132?

dl36. Prove that the directions for bisecting a sect perpendicu-
larly are correct. Compare this construction with the bisection of
any angle.
136. Prove that the directions for dropping a perpendicular from

a point to a line are correct.
dl37. In the sixteenth cen-
tury, the distance from A to
the inaccessible point B was
3 found by use of an instrument
consisting of a vertical staff
AC, to which was attached a
horizontal cross bar DE that could be moved up and down on the
staff. Sighting from C to J5, DE was lowered or raised until C, E
and B were in a straight line. Then the whole instrument^ was re-
volved, and the point F at which the line of sight CEi stru ck the
ground again was marked, and FA measured. Show that FA ^AB-
(This exercise is taken with modifications from Stone-Millis,
Plane Geometry.)

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46 PLANE GEOMETRY

dl38. Cor. 1, theorem 4, is known as the "Pons asinorum/'
or "Bridge of Asses." Its discovery is attributed to Thales. The
proof suggested in this exercise, however,
is due to Euclid. He produced BA and
BC, the equal sides of the triangle to D and
jB, so that BD=BE. Then he proved (1)
ABCD^ABAE,(2) AACD^ ACAE, and so,
by subtracting -^ACD from -^BCD, and
<^CAE from ^BAE found ^BAC^ ^BCA .
^^ Give the details of the proof.

LIST OF WORDS DEFINED IN CHAPTER I

Solid, surface, line, point ; straight, curved line, sect ; plane . Angle, vertex,
sides; straight, right, acute, obtuse angles; complementary, supplementary
angles; adjacent, vertical angles; perpendicular. Polygon, vertices, sides,
angles; triangle; scalene, isosceles, equilateral triangles; acute, right, obtuse
triangles. Congruent, homologous. Theorem, corollary, axiom, postulate.

SUMMARY OF AXIOMS IN CHAPTER I

1. The simis of equals added to equals are equal.

2. The remainders of equals subtracted from equals are equal.

3. The products of equals multiplied by equals are equal.

4. The quotients of equals divided by equals are equal.

5. A quantity may be substituted for its equal in a statement of equality
or inequality.

6. Two quantities which are equal to equal quantities are equal to each
other.

7. The whole is equal to the siun of its parts.

SUMMARY OF POSTULATES IN CHAPTER I

Straight Line

1. Two intersecting straight lines determine a point.

2. Two points determine a straight line.

3. A straight line is the shortest distance between two points.

Angle

4. All straight angles are equal.

Cor. 1. All right angles are equal.

Cor. 2. Complements of the same angle or equal angles are equal.
Cor. 3. Supplements of the same angle or equal angles are equal.
SuperpasiHon

5. Any geometric figure may be moved about without changing its size or
shape.

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INTRODUCTION 47

SUMMARY OF THEOREMS PROVED IN CHAPTER I

1. Vertical angles are equal.

2. Triangles are determined by two sides and the included angle.

3. Triangles are determined by two angles and the included side.

4. The bisector of the vertex angle of an isosceles triangle divides the
triangle into two congruent triangles.

Cor. 1. The angles opposite the equal sides of a triangle are equal.
Cor. 2. The bisector of the vertex angle of an isosceles triangle

bisects the base, and is perpendicular to it.
Cor. 3. An equilateral triangle is equiangular.
Cor. 4. The bisectors of the angles of an equilateral triangle bisect

the opposite sides and are perpendicular to them.
Cor. 5. The bisectors of the angles of an equilateral triangle are

equal, i

5. Triangles are determined by their sides.

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CHAPTER II

THE PERPENDICULAR, THE RIGHT TRIANGLE AND

PARALLELS
A. THE PERPENDICULAR

We have beenatu(l3ring the congruence of triangles in general, and
as a necessary and interesting part of that topic we have considered
some properties of the isosceles triangle. Now we are to give our
attention to another special kind, the right triangle, but before
doing so, we need to know more than we do about perpendiculars.

Two facts that we should note at the beginning are sufficiently
obvious to permit our accepting them without proof, i.e., postu-
lating them.

Postulates of Perpendiculars. 1. At a paint in a line only one
perpendiadar can be erected to that line.

2.ExQS%apoint outside a line only one perpendicular can be drawn
toOiatJine*

stuse of this property of perpendiculars, by the distance from
)a point to a line is meant the length of a perpendicular from the
point to the line.

-^TEel^we postulate and the more we prove, the more scientific
is our work. Hence, later in our study of geometry, we shall prove
these postulates of perpendiculars.

SET XVI. DISTANCE FROM A POINT TO A LINE
139. Prove the familiar fact that
the image of an object in a mirror
appears to be as far behind the mirror
as the object is in front of it.

Wnts: (a) It is proved in physics that a
ray of light striking a plane surface is reflected
from it at the same angle as it strikes it.
Assume that fact here, (b) i, Af, 72 lie in a
straight line. See the diagram, (c) Prove
triangles congruent.
Before proving another important property of perpendiculars
we must add to our list of axioms. ^ .

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PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 49

Axioms of Inequality. 1. // unequals are operated on by
positive eqiuils in the same way, the results are unequal in the
same order.

In other words, if we add a positive number to each of two
imequal numbers the sums will be unequal in just the same way
that the original numbers were unequal, i.e., the greater num-
ber increased will still be greater than the smaller number
increased; e.g., 7>5 and 7+2>5+2. Would the same be true
if we started with 7> -5, or -7< -5? Give your reason.

If instead of adding equal numbers to the two unequal num-
bers, we had subtracted equal numbers from each, or divided
or multiplied each by equal numbers, the remainders, quotients,
or products would have been unequal in the same order as the
original numbers.

2. // unequals are subtracted jrom equals^ the remainders are
unequal in the reverse orcfer.

Example. 10s 10

and 6>3
.-. 4<7

EXERCISBS. SETXVn. NUMERICAL INEQUALITY

140. State in algebraic symbols the above axioms of inequality.

14L (a) Why is it that if -2< -1, ( -2)2'»>( -1)«"?

(6) Why is this not an exception to our axiom?

We are now ready to prove another important fact about per-
pendiculars which will give us another reason for measuring dis-
tance from a point to a line by means of the perpendicular.

Theorem 6. The perpendicular is the shortest sect that can
be drawn from a point to a line.

To show that from P a point outside
of ABf the shortest sect that we can
draw to AB is the perpendicular PQ,

let us draw any other se^ say P/S to A

AB, and then show that PQ < PS. So
far in our study of geometry we have
no statement which will lead us to this
oonclusion directly. o^Sized by Google

50 PLANE GEOMETRY

The axiom of inequality which w6 have just discussed might
suggest, however, that we can derive the conclusion we wish by
p showing that 2PQ<2PS.

Then, we also recall that we have

V listed one fact concerning the inequality

\ of sects and that is the postulate that a

— ^H r- straight line is the shortest distance be-

j /' tween two points. Thus we are led to

i / make the construction shown in the

i^ diagram, i.e., produce PQ to /J so that

QR =PQ and connect R and S.
mwPQ+QR_{oT2PQ)j<PS+RS. Why?
PS+RS^2PS_^ PS^RS.
Try to prove PS=RS.
How do we prove sects equal? Finish the proof.

EXERCISES, SET XVIII. INEQUALITY OF SECTS

142. Given the point P outside AB and L in the line. Which
is shorter, the distance from P to L or

the distance from P to i4J5? 'P

143. Which is the longest side of a

right triangle? Give a reason for your '—f -^

answer.

B. THE RIGHT TRIANGLE

We are now ready to go on with the study of the congru-
ence of triangles by noting two cases of the congruence of right
triangles.

EXPERIMENT

Construct a right triangle given the side opposite the right angle,
called the hjrpotenuse, and an angle adjacent to it. Let us see if
we have data sufficient to determine the triangle.

(a) What is the only side of the triangle that is known?

(6) Where then will you have to start the construction?

(d) Is the direction of the third side then fixed? Why?
We are thus led to expect:

♦Theorem 7. The hypotenuse and adjacent angle determine
a right triangle. igtized by Google

PERPENDICULAR, RIGHT TRIANGLE, PARALLELS- 51

Let us convince ourselves of the truth or falsity of this conclusion
by actual proof. a

Given: A ABC and AX FZ
with 45 and 2J.F each a rt.
2^ and ^C^^Z and lU^
XZ.

To prove: aABC^aXYZ

Outline of proof : Plaoe

A ABC on aXYZ so that J? ^__

coincides with its equal ZZ and C5 falls along ZF.
do this?

What right have we to

Then AB will fall along XF. Why? Finally B will coincide with F. Why?
The other case of congruence of right triangles is:
Theorem 8. The hypotenuse and another side determine a
right triangle.

Given: a ABC and A-^FZwith AB^rF, AC s^Z and 4.J5 and 4. F each
art.2^.

Toprove: aABC^ A-X^FZ.

Suppose we place AXYZ
next to AABC (as in the fol-
lowing diagram) with ZF
coinciding with 3B. (Do
we know that we can?)
(1) Why will the figure

S\ ^ formed be a triangle?

^^ ^ (2) What kind of triangle will be formed?

(3) Can you now throw this theorem back to the previous one?

C. PARALLEl^

Parallels, or parallel straight lines, are coplanar lines (lines lying
in the same plane) which never meet. Do you see any reason for
emphasizing the fact that the lines must lie in the same plane?
Take two pencils and hold them so that they would neither meet
if continued nor be parallel.

Draw a line on a piece of paper and erect two perpendiculars
to it. Do these perpendiculars appear to be parallel? Since they
lie in the same plane they must either be parallel or meet. Can
they meet? Give a reason for your answer. This leads us to state
another theorem for our syllabus, one which is frequently used in
mechanical drawing. ^.g,^^, .^ Google

PLANE GEOMETRY

Theorem 9. Lines perpendicular to the same line are parallel.

EXERCISES. SET XIX. PARALLELS
144. What principle is a carpeDter using when he lays oflf parallel
lines on a board by moving one arm of his square along a straight
edge of the board, and marking along the other arm?

146. What is the principle involved in the
use of the T-^uare for drawing parallels?
dl46. The accompanying picture shows
a carpenter's plumb-level, the forerunner
of the spirit-level. AE and EB are strips
of wood of equal length. CE=ED and
B is the midpoint of CD. A and B rest on
the points to be levelled, and they are found to be level when EF
passes through 0. Explain.
Before proceeding with our study of parallels, we need the
Postulate of Parallels. Through

a given point only one line can he ^^

drawn parallel to a given line.

Not both lines a and 6 can be .^

parallel to c. Why?

Cor. 1. Lines parallel to the same line are parallel toone another.
Suggestion for proof: If lines a, b, both parallel to c, should meet how would
the postulate of parallels be violated?
♦Theorem 10. A line perpendicular to one of a series of par-
allels is perpendicular to the others.

Given: AXB \\ CTb\. ^EXYT ± IXB.
To prove: EF±CD,
Suggestions: Draw YK±EF.

What relation will exist between YK and
ABl

Why wiU YK and CD coincide?

What relation exists between CD and
EF? Why?

Why is it necessary to consider only two
parallels?

♦ When naming a Une by more than two of its points it is necessary to use
a bar over the letters. In the case of two points it is immaterial. Why?
t How do the statements given show that EF cuts AB and CD?

.__

r--

^^-K

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PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 53

EXERCISE. SET XIX (continued)

147. Representing a series of lines by p, g, r, construct the

following figures, stating in each case all possible relative positions
of the first and last line of the series:

(a) p\\q, p II r. (6) p\\q, q \\ r. (c) p±q, q±r. (d) p±q,
q\\r. (e) p±q,q\\r,r±s. (/) p |k, g±r, r || s.

ANGLES FURTHER DEFINED ACCORDING TO RELATIVE POSITION

If, as i n the accompan ying dia gram,
ABC and DBF are cut by GBEH, which
is called a transversal (since it cuts
across), certain sets of angles are
formed to which for brevity we give
the following names :

^ABE and ^FEB are called alter-
nate-interior angles.

^GBA and ^HEF are called alternate-exterior angles.

^EBC and ^FEB are called consecutive-interior angles.

^GBA and ^DEH are called consecutive-exterior angles.

^CBG and ^FEB are called corresponding angles.

EXERCISES. SET XX. RELATIVE POSITION OF ANGLES

148. Only one pair of each kind of angles is mentioned in the
last paragraph, though there are two pairs of each except the last
kind. Name the second pair in each case, and three remaining
pairs of corresponding angles.

149. Explain the meaning of alternate as used here.

160. Explain the meaning of consecutive as used here.

161. Explain the meaning of interior as used here.

162. Explain the meaning of exterior as used here.

163. What kind of angles with regard to relative position are
formed in the letter Z? In the letter il? In the letter £7? H? N?

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PLANE GEOMETRY

164. In the accompanying diagram select angles under each
class you know (including adjacent and
vertical) using:
(a) c and d with a as transversal.
(6) c and d with b as transversal.

(d) a and 6 with d as transversal.

PROPERTIES OF PARALLELS
Theorem 11. If, when lines are cut by a transversal, the alter-
nate-interior angles are equal,
the lines thus cut are parallel.

Given: IBC, XYZ cut by BY.

2S^ABY^2^BYZ.
To prove: ABC\\XYZ.

Proof
Take Q in BF so that QB^QY,

Draw QP±AB cutting AB at P.
Extend PQ to i2 in XZ.

Then in APQB and ARQY,
(1) QB^QY

(2)
(3)
(4)

2^.PBQ^2i.RYQ
T^PQB^T^RQY
:. APQB ^ ARQY

(5) :.2^.BPQ^2^.YRQ

(6)
(7)
(8)

(9) .
(10)

But PQR±AB
.'.^BPQisart.^.
;.4Fi2Q is a rt.4

;. P^SXXSZ

:. ASCII xy2

(1) Construction.

(2) Data.

(3) Vertical angles.

(4) Two angles and the included side
equal each to each.

(5) Homologous parts of congruent
triangles.

(6) Construction.

(7) Definition of perpendicular.

(8) Quantities equal to the same quan-
tity are equal to each other.

(9) Definition of perpendicular.

(10) lines perpendicular to the same
line are parallel.

The student's attention is called to the form and arrangement of
this demonstration, as it is the first formal proof given in the text.
Note that after the general statement of the theorem following the
words ** given'' and *Ho prove/'* specific statements are given

* In place of the word "given" either "data" or "hypothesis" is frequently
used, and in place of "prove" the word "conclusion."

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PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 55

referring to the particular diagram drawn. These statements
should be as brief as possible, and such, that were the diagrams
erased, it could be reconstructed. The steps of the proof and the
reasons for them are arranged in parallel' colimms. The con-
venience of such an arrangement is at once apparent if it be com-
pared with a proof written in essay form. Write the proof that
way and draw your own conclusions as to which you would prefer
to use, giving your reasons.

Show that the following corollaries are true by showing that a
pair of alternate-interior angles are equal.
Cor. 1. If the alternate-exterior angles or corresponding angles
are equal when lines are cut by a transversal, the lines
thus cut are parallel.
Cor. 2. If either the consecutive-interior angles or the con-
secutive-exterior angles are supplementary when lines
are cut by a transversal, the lines thus cut are parallel.

EXERCISES. SET XXI. CONSTRUCTION OF PARALLELS

166. Parallels may be constructed by using a T-square and a
triangle. Explain.

166. Draw three pairs of parallel lines using successively each
of the three sides of one of your triangles against a ruler.

167. By using your knowledge of corresponding angles, draw a
line through a given point, and parallel to a given line.

158. (a) Practice drawing parallels with compasses and ruler
until you can draw them accurately. (6) Test your work by draw-
ing any transversal, and measuring a pair of angles that should be
equal, (c) Which is more likely to be in error, your drawing or
your test?

159. The diagram suggests a ^ ^

method of running one line parallel ^\^

to another when you are on a field y^s.

without a transit. Explain and p q

justify the procedure.

Presently we shall prove a theorem which is closely related to
theorem 11. Before doing so, however, we shall want to see why
it need be proved at all.

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56 PLANE GEOMETRY

EXERCISES. SET XXII. RELATED STATEMENTS

160. (a) Is it true that if a triangle is equilateral, it is also
isosceles?

(b) Is it trae that if a triangle is isosceles, it is also equilateral?

161. (a) Is it true that if two angles are right angles, they are
equal?

(b) Is it true that if two angles are equal, they are right angles?

162. (a) Is it true that all men are bipeds?
(&) Is it true that all bipeds are men?

163. (a) Is it true that if a man lives in Cincinnati, he lives in
Ohio?

(6) Is it true that if a man lives in Ohio, he lives in Cincinnati?

164. Explain how each of the statements (a) and (6) in each of
the four preceding exercises is related to the other: that is, how
can (a) in each case be formed from (6), and how can (6) be formed
from (a)?

166. Make a statement related to each of the following as (6)
is related to (a) in each of exercises 160 to 163, and tell whether
or not your statements are true.

(a) If a man lives, he breathes.

(6) If a polygon is a triangle, it has three sides.

166. From exercises 160 to 163 and 165 what can you conclude
as to the truth or falsity of two statements related in this way?
(a) May both of them be true? (6) May one of thfem be true and
the other false?

167. If, then, we have proved that a statement is true, is it neces-
sary to prove a statement related to it as the second is to the first
in each of those exercises, or may we take it for granted that the
second will be true without proof?

Statements related as those in the last set of exercises, are called
converse statements. We saw that each of two converse state-
ments could be formed from the other by interchanging the data or
hypotheses with the conclusion or conclusions; that is, the two state-
ments were so related that what was given in each was what was
supposed to follow in the other. From the fact that it is so much

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PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 67

easier to make a statement whose converse is absurd than one
whose converse is true, it appears that we should never claim that
the converse of a theorem in geometry is true without having
proved it so.

EXERCISES. SET XXII (oontinued)

168. Make a statement of something in life which you know to
be true, but whose converse is false.

189. Make a statement of something in life which you know to
be true, but whose converse is true.

170. Do the same as you were requested to do in the last two
exercises, but take the statements from geometry.

171. Select from the theorems already proved two that are
converse statements.

172. The converse of a definition is always true. Test your
definitions of Chapter I from this point of view. (See lists at end
of Chapter I, page 46.)

173. State what was given us in theorem 11.

174. State what we proved in theorem 11.

176. State what would be given in the converse of theorem 11.

178. State what would have to be proved in the converse of
theorem 11.

Theorem 12. Parallels cut by a transversal form equal alter-
nate-interior angles. \

Pill in all the blank spaces ^
and answer the questions in the
following:

Given: p o^s! R

To prove:

Pboof

(1) If through the mid-point of
YQ we were to draw a line perpendic-
ular to XZ what other fact would we
know about that line?

(2) What parts have we then
equal in the triangles thus formed?

(3) By what method could we
then prove the fact we wish to prove?

(1) Why?

(2) How do you know each of
these pairs are so related?

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zed by Google

PLANE GEOMETRY

Cor. 1. Parallels cut by a transversal form equal corresponding
and equal alternate-exterior angles.

Cor. 2. Parallels cut by a transversal form supplementary con-
secutive-interior, and supplementary consecutive-ex-
terior angles.

EXERCISES. SET XXIII. APPLICATIONS OF PARALLELISM

177. The accompanying diagram
suggests a convenient method of
measuring the distance from B to
an inaccessible point F. Explain.

Note DE II FB.

How can this be done on the ground?

178. The "square network" shown in the figure is used in
designing for drawing a great variety of patterns. The patterns
A and B drawn upon it are examples of Arabian frets. The best
way to rule the square network is to
draw a horizontal line MN and mark
ofif equal divisions on it. At each
point of division, by use of the tri-
angle, draw two lines, such as PQ
and PR, each making an angle of 45®
with MN.

Draw such a network, then upon it construct a pattern, either
an original design or a copy of these Arabian frets.
(Taken from Stone-Millis, Plane Geometry.)

179. In the annexed dia-
gram if Xy||PQand/2>S|| FT
how many angles would you
need to know in order to find
the remaining angles?

180. If the side AC of a
triangle ABC is extended,
as in the annexed diagram,
how could a line be drawn
through C to make an angle

Why?

equal to the <^B?

a D

Would any other angles be equal?

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PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 59

LIST OF WORDS DEFINED IN CHAPTER n

Distance (point to a line). Hypotenuse. Parallels, coplanar, transversal;
alternate-interior, alternate-exterior, consecutive-interior, consecutive-ex-
terior, corresponding angles. Converse.

SUMMARY OF AXIOMS IN CHAPTER H

Inequality

1. If unequals are operated upon by positive equals in the same way, the
results are unequal in the same order.

2. If unequals are subtracted from equals, the remainders are unequal in
the reverse order.

SUMMARY OF POSTULATES IN CHAPTER H

PerpendiciLlars

1. At a point in a line only one perpendicular can be erected to that line.

2. From a point outside a line only one perpendicular can be drawn to
that line.

Parallels

3. Through a given point only one line can be drawn parallel to a given
line.

Cor. 1. Lines parallel to the same line are parallel to one another.

SUMMARY OF THEOREMS IN CHAPTER H

6. A perpendicular is the shortest sect that can be drawn from a point
to aline.

7. The hypotenuse and an adjacent angle determine a right triangle.

8. The hypotenuse and another side determine a right triangle.

9. Lines perpendicular to the same line aje parallel.

10. A line perpendicular to one of a series of parallels is perpendicular to
the others.

11. If when lines are cut by a transversal the alternate-interior angles are
equal the lines thus cut are parallel.

Cor. 1. If the alternate-exterior angles or corresponding angles are
equal when lines are cut by a transversal, the lines thus
cut are parallel.

Cor. 2. If either the consecutive-interior angles or the consecu-
tive-exterior angles are supplementary when lines are cut
by a transversal, the lines thus cut are parallel.

12. Parallels cut by a transversal form equal alternate-interior angles.

Cor. 1. Parallels cut by a transversal form equal corresponding,
and equal alternate-exterior angles.

Cor. 2. Parallels cut by a transversal form supplementary consecu-
tive interior, and supplementary consecutive-exterior
angles.

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CHAPTER III

ANGLES OF POLYGONS AND PROPERTIES OF
PARALLELOGRAMS

A. ANGLES OF POLYGONS

In attempting to develop a formula for the sum of the angles of
a polygon, it is best for us to begin with the simplest polygon,
the triangle.

EXERCISE. SET XXIV. SUM OF ANGLES OF A TRIANGLE.

181. By reproducing the angles of a given triangle, place these
angles adjacent to one another. What does the sum of the angles
appear to be in this case?

Theorem 13. The sum of the angles of a triangle is a straight angle.

Prove this proposition, using the hints given by the following
j5 diagram and notes:

Produce CA and draw
AP II CB.

_v2^x^^ What is the sum of

a' ^' — i2 <M, 2, and3?
What is the relation of ^2 to ^2i? Of ^3 to ^3i?

Cor. 1. A triangle can have but one right or one obtuse angle.

Cor. 2. Triangles having two angles mutually equal are mutu-
ally equiangular.

Cor. 3. A triangle is determined by a side and any two homolo-
gous angles.

An exterior angle of a polyyon is one formed by a side of the
polygon and the prolongation of an adjacent side.
In the preceding diagram ^ RAB is an exterior angle of the AABC.

Cor. 4. An exterior angle of a triangle is equal to the sum of
the non-adjacent interior angles.

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POLYGONS AND PARALLELOGRAMS 61

EXERCISES. SET XXIV (continued)

182. (a) The theorem that the sum of the angles of a triangle
equals a straight angle may be proved by drawing a line through
a vertex parallel to the opposite ^
side. Give proof.

Note. — This proof is attributed to
the Pjrthagoreans.

(6) Prove the same fact by ^

drawing a sect from a vertex parallel to the opposite side.

C (c) Prove the same theorem by

y^ ^^"^^^.^^^^ drawing through any point on one

^ parallel to

the other sides of the triangle.

183. Prove this same fact another ^
way by erecting perpendiculars to one side at its extremities and
dropping a perpendicular to the same side from the opposite vertex.

184. Show how the following procedure may be used to test accu-
racy with which you measure angles with an instrument. Select the
three positions not in a straight line; call them stations -4, B, and
C From station A measure the angle between the directions to
B and C; at B, measure the angle between the directions to C
and A; at C, the angle between the directions to A and B. Would
your measurements be accurate, and if not, what error would
there be, if you found the angles to be respectively : IW 27', 23° 13',
and 56° 23'?

186. If one angle of an isosceles triangle is 60°, find the other
angles.

186. (a) If the vertex angle of an isosceles triangle is t;° write
an expression for each of the other angles.

(6) If a base angle of an isosceles triangle is 6° write an expres-
sion for each of the other angles.

187. What angle do the bisectors of the acute angles of a right
triangle form?

188. Construct the following angles: 60°,30°,120°,75°,150°,105°.
The pupil is again reminded that the use of instruments is

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PLANE GEOMETRY

restricted to the straight edge and the compasses in scientific geo-
metric constructions.

dl89. (a) Two mirrors, mi and m2,
are set so as to form an acute angle with
each other. A ray of light is reflected
by mi so as to strike m^. The ray is
again reflected by m^ and crosses its
• first path. Prove that '^rr2=2'^m\m2.
(6) How should these mirrors be placed in order that the ray
in the second case may be parallel to the original ray?

(c) Part (a) of this exercise is the principle underlying several
important optical instruments such as the "optical square' and
the "sextant." A description of
them may be found in '.\ny good
encyclopaedia or in Gillespie's

Surveying, p. 61, and Stone-Millis' Plane Geometry, p. 40, Ex. 14.
Look them up and make a crude optical square either of paste-
board or of wood.

A diagonal of a polygon is a sect connecting any two non-consecutive
vertices.

EXERCISES. SET XXV. SUMS OF ANGLES OF POLYGONS

190. Find the sum of the angles of a quadrilateral. (Can you so
divide it into triangles that the sum of the angles of the triangles
formed will be the sum of the angles of the quadrilateral?)

191. Draw a five-sided polygon.

(a) How many diagonals can be drawn from one vertex?

(6) How many triangles are formed by drawing these diagonals?

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POLYGONS AND PARALLELOGRAMS 63

(d) What is the sum of the angles of a 6-gon?

Give a reason for each of your answers. Check your conclusion
concerning the sum of the angles by the use of the protractor.

A polygon^ said to be convex if each of its angles is less than a
straight angle. Only convex polygons will be considered in the early
study of geometry. If a polygon has four sides it is called a quad-
rilateral; if ^6 sides, a pentagon; six sides, a hexagon; seven sides
a heptagon; eight sides, an octagon; nine sides, a nonagon; ten
sides, a decagon; etc.

Theorem 14. The sum of the angles of a polygon is equal to a
straight angle taken as many times less two as the polygon has
sides.

(1) Draw all the diagonals possible from one vertex of the
polygon.

(2) If the figure has n sides how many triangles will you form
by drawing these diagonals?

Give the proof following the suggestions given in (1) and (2).
The pupil will find the following (1) and (2) also give hints
leading to an equally desirable proof.

(1) Connect any point inside the polygon with each vertex.

(2) If the figure has n sides how many triangles will you thus form?
A regular polygon is one which is both equilateral and equiangular.
Can you draw a polygon which is equilateral and not equi-
angular?

. Can you draw a polygon which is equiangular and not
equilateral?

Cor. 1. Each angle of a regular polygon of n sides equals the
^th part of a straight angle.

If the sides of a polygon are produced in turn forming one exterior
angle at each vertex^ these angles are called the exterior angles of
the polygon.

Could two exterior angles be formed at each vertex? How would
two such angles be related?

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PLANE GEOMETRY

Cor. 2.* The sum of the exterior angles of a polygon is two
straight angles.

(D What would be the sum of the adjacent interior and exterior
angle at each vertex?

(2) What would be the sum of all the interior and exterior angles
together?

Cor. 3. Each exterior angle of a regular polygon of n sides is
equal to the ^th part of a straight angle.

EXERCISES. SET XXV (continued)
192. Fill in the blank spaces in the accompanying table:

No. of sides of
polygon

No. deg. in sum
of int. angles

No.deg.ineach

angle of regular

n-gon

193. How many sides has a regular polygon if each angle is
(a) 150^ (6) 144^ (c) 170^ (d) 175°?

Hints on solution : ^ (180) = 150; or | ( 180) = 30. Solve for n.

194. In surveying an hexagonal field the angles were found to be
118°, 124°, 116°, 129°, 130°, 112°.

(a) What error was made?

(6) Before making a drawing of the survey, the engineer has to
distribute this error proportionately over all the angles so as to
increase or decrease all the angles according as the sum of the angles
measiu^ was too small or too great. Distribute the error correctly
in this case.

* Theorem 14 and this corollary were proved in their general form by
Regiomontanns (143^1476), although the facts were known to earUer mathe-
maticians, and were proved by them for special cases.

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POLYGONS AND PARALLELOGRAMS 65

196. In surveying a pentagonal field, the angles were found to be
4 = 103° 15',B = 110°3r,C=99°45',Z) = 122° 40',and £=102° 16'.
(a) What error was made?
(6) Distribute the error proportionately.

196. (a) Make a list of regular polygons of the same number of
sides that may be used to cover a plane surface with a geometric
design.

(6) For what purposes are such designs used?

(d) Sketch some of these designs.

(e) Why is it these combinations are possible?

197. Make as many constructions as possible of each of the
polygons mentioned, using only rule and compasses for the purpose.

198. Construct accurately a design based upon each kind of
regular polygon or a combination of regular polygons mentioned
inEx/l96.

dl99. Mabel Sykes, Source Book of Problems for Geometry,
p. 16, par. 23, Ex. 2.

d200. Ibid., p. 16, par. 23, Ex. 4.

d201. Ibid., pp. 16-17, par. 24, Ex. 1.

d202. Ibid., pp. 16-17, par. 24, Ex. 4.

203. Prove the proposition concerning the sum of the angles
of a polygon, according to the following suggestions:

(a) By connecting a point on one of the sides of the polygon
with each vertex.

(6) By connecting a point outside of the polygon with each vertex.

The converse of a proposition concerning isosceles triangles has
many interesting and important applications, and we are now ready
to prove it.

Theorem 16. If two angles of a triangle are equal, the sides
opposite them are equal.

Hint: Can you draw a line to cut A ABC
into two triangles in such a way that the con-
struction itself will make an angle and a side
equal respectively in the two triangles?

Cor. 1. Equiangular triangles are
equilateral.

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PLANE GEOMETRY

EXERCISES. Set XXVI. SIDES AND ANGLES OF A TRIANGLE
201. (a) Show that a person may find the distance (AC) at
which he passes an object A, when going in the direction BC, if
he notes when his course makes an angle of
46** with the direction of the object and
again when it is at right angles to the
object taking account of the distance (BC)
which he traversed between the observa-
tions.

(6) How could this method be applied to
the problem of finding at what distance
from a Ughthouse a ship passes it?
206. To ascertain the height of a tree or of the school building,
fold a piece of paper so as to make an angle of 46°. Then walk
back from the tree until the top is seen at an angle of 45® with the
ground (being, therefore, careful to
have the base of the triangle level).
Then the height AC will equal the
base AB, since ABC is isosceles. A
paper protractor may be used for
the same purpose. Can you suggest
a better method than that of meas-
uring from the ground?

206. The distance of a ship at sea may be measured in the
following manner:
Make a large isosceles triangle out of wood, and standing at T,

sight to the ship and along the shore
on a line TA, using the vertex angle
of the triangle. Then go along TA
until a point P is reached, from which
T and S can be seen along the sides
of a base angle of the triangle. Then
TP = TS. By measuring TB, BS can
be found.
207. Distance can easily be meas-
ured by constructing a large equilateral triangle of heavy paste-
board, and standing pins at the vertices for the purpose of sighting.

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POLYGONS AND PABALLELOGRAMS 67

To measure PiC, stand at some convenient point A, and sight
along APC, and also along AB. Then walk along AB until a point
Bi is reached, from which BiC makes with BiA an angle of the tri-
angle (60°). Then prove that AC=ABi. Also, since APi can
be measured, find PiC
C

\ 2^ -

X A ^ -A B Bi

208. Measure the angle CAX, either in degrees with a pro-
tractor, or by sighting along a piece of paper and marking down
the angle. Then go along XA produced until a point B is
reached from which BC makes with BA an angle equal to half of
angle CAX. Then show that AB=AC. /{L

Note. — ^A navigator uses the prin-

•

ciple involved in the foregoing exercises ^^

when he ''doubles the angle on the bow" /^

to find his distance from a lighthouse or ^^ \\

other object. If he is sailing on the — ^^ ' ' — J ^

course ABC and notes a lighthouse L '^

when he is at A, he takes the angle A; and if he notices when the angle that
the lighthouse makes with his course is just twice the angle noted at A, then
BL^AB. He ha3 AB from his log (an instrument that tells how far a ship
goes in a given time), so he knows BL. He has ''doubled the angle on the
bow" to get this distance.

B. PARALLELOGRAMS

A parallelogram is a qvMrilateral whose opposite sides are parallel.

Theorem 16. Either diagonal of a parallelogram bisects it.

Suggestion: In the proof note that the diagonal is a transversal of the
parallel sides.

Cor. 1. The parallel sides of a parallelogram are equal, and
the opposite angles are equal.

The sects of common perpendiculars included by parallels are
called the distances between the parallels.

Con 2. Parallels are everywhere equidistant.

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PLANE GEOMETRY

EXERCISES. SET XXVII. PARALLELOGRAMS

209. One angle of a parallelogram is 20° more than three times
another. Find all the angles.

210. (a) Establish a relation between consecutive angles of a
parallelogram.

(6) How many angles of a parallelogram must be known in
order to determine the others?

211. A stairway inclined 45° to the horizontal leads to a floor
15' above the first. What is the length of the carpet required to

cover it if each step is 10" high? If each
is 12'' high? If each is 9"?

Can this problem be solved without
knowing the height of the steps? Is it
necessary to know that the steps are of
the same height?

(Taken from Slaught and Lennes,
Plane Geometry.)

212. Is the converse of the fact that a diagonal bisects a paral-
lelogram true or false? Give a reason for your answer.

Theorem 17. A quadrilateral whose opposite sides are equal
is a parallelogram.

(1) What fact would we need to know about these opposite sides?

(2) By what method can we obtain this fact?

EXERCISES. SET XXVII (continued)

213. An adjustable bracket such as dentists often use, is out-
lined in the figure, It is fastened to the wall at Ay and carries a
shelf B. Why is it that as the bracket
is moved so that B is raised and low-
ered, the shelf remains horizontal?

(Taken from Stone-Millis, Plane
Geometry.)

214. The accompanying figure
is a diagram of the "parallel
ruler," which is used by designers
for drawing parallel lines.

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POLYGONS AND PARALLELOGRAMS

(a) Upon what principle of parallelograms must its construction
depend?

(6) Make such an instrument.

A rectangle is a parallelogram one of whose angles is a right angle.

Cor. Each angle of a rectangle is a right angle. Why?

EXERCISE. SET XXVII (continued)
216. Prove that a parallelogram whose diagonals are equal is a

rectangle. (This fact is used as a check by carpenters and builders.

Could it be used in laying out a tennis court?)
Theorem 18. A quadrilateral having a pair of sides both equal

and parallel is a parallelogram.

EXERCISES. SET XXVII (continued)

216. Justify the following method used by surveyors for pro-
longing a line beyond an obstacle; that is, show that in the diagram
EF is AB prolonged beyond

0. BC is run at right angles ^« '^

to AB; CD±BC; DE±CD
and DE=CB; EF±DE.

217. The accompanying
diagrams show another way
of extending a line be-
yond an obstacle, (a)
By reference to dia-
gram, state the proce-
dure in words. (6) Show
that it is correct, (c) Compare this method with that of Ex.
216. Note, for example, why two lines (CB and DE) are used
in Ex. 217, and only one (CB) in Ex. 216. Under what con-

2) ff ditions would you use each method?

~ 218. If the vertices of one paral-

lelogram are in the sides of another,
the diagonals of the two parallelo-
grams pass through the same point
(calledthecenterof the parallelograms).

Suggestions: Call the intersection of the diagonals AC and BD, 0. Draw
OE, OF, OGy OH and prove EOG and FOH straight lines.

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PLANE GEOMETRY

d219. Ad interesting outdoor application of the theory of paral-
lelograms is the following: Suppose that you are on the side of
this stream opposite to XY, and wish to measure the length of
XY. Run a line AB along the bank. Then take a carpenter's

square, or even a large book, and
walk along AB until you reach
P, a point from which you can
just see X and B along two sides
of the square. Do the same for
y, thus fixing P and Q. Using
the tape, bisect PQ at M. Then
walk along YM produced until
you reach a point Yi that is ex-
actly in line with M and F, and also with P and X. Then walk
along XM produced until you reach a point Xi that is exactly in
line with M and X, and also with Q and 7. Then measure
YiXi and you have the length of XY. For since YXiA^PQ,
and XYl^^PQ, YXx || XY^. And since PM = MQ, therefore
XM=MXi and YiM=MY. Therefore Y^XiYX is a parallelo-
gram. Give the reason for each of these steps.

* Theorem 19. A parallelogram is determined by two adjacent
sides and an angle, or two parallelograms are congruent if two
adjacent sides and an angle are equal each to each.

Methods of proof.

(1) Congruence of triangles, or

(2) Superposition, or

(3) Properties of parallelograms.

Note. — K (1) or (3) is used it must be shown that the parts proved equal
are arranged in the same order in the two parallelograms.

EXERCISES. SET XXVTI (continued)

220. Construct a parallelogram, given

(a) Two adjacent sides and an included angle.
(6) Two adjacent sides and a non-included angle,
(c) A side, a diagonal, and the angle between them.

221. In physics it is shown that if two forces (such as a push
and a pull) are exerted in different directions upon the same object,
they have the same effect as a single force called their resultant

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POLYGONS AND PARALLELOGRAMS 71

How is this fact illustrated by the bean-shooter?

Referring to the accompanying diagram, if the directions and
magnitudes of two forces working on object A are represented by
the sects AB and AC, the direction
and magnitude of the resultant will
be represented by the sect AD, which
is the diagonal of the parallelogram,
with AB and AC as adjacent sides.
In physics, such a diagram is, for
obvious reasons, called the Paral-
lelogram of Forces.

A force of 50 pounds is exerted upon a body pulling it in one
direction, and at the same time another force of 100 lbs. pulls it
in a direction at an angle of 45° with the first. Show by the
parallelogram of forces the effect on the body.

Note: (o) Use only compasses and ruler in solving. Represent 50 lbs.
by any given sect as unit; draw the forces to scale, and find the resultant.

(&) Use protractor and marked edge in reading result.

222. Two forces, 250 lbs. and 400 lbs. respectively, are exerted
upon a body at right angles with each other. Find their resultant
as in Ex. 221. Check the result by computation. Why was such
a check not available for you in Ex. 221?

223. Find the resultant of two forces exerted upon a body at
an angle of 150° with each other, one of 50 lbs., the other of 60 lbs.

224. When a train is approaching a station at a velocity of 40 ft.
per second, a mail bag is thrown at right angles from the car with
a speed of 20 ft. per second. Find the actual direction and speed
of the moving bag.

(Taken from Stone-Millis, Plane Geometry.)

226. The resultant of three forces may be found by getting the
resultant of two of the original forces and then finding the resultant
of that and the third original force. This process may be continued
to obtain the resultant of several forces.

Find the resultant of three coplanar forces of 200 lbs., 150 lbs.,
and 175 lbs. respectively, acting on the same body at the same

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72 PLANE GEOMETRY

time. The angle between the first and second force is 45°, and that
between the second and third is 60°. Why note that the forces
are coplanar?

226. A force of 100 lbs. makes an angle of 60° with a second
force of 120 lbs. exerted on the same body, and makes an angle of
90° with a third force of 140 lbs., and an angle of 120° with a fourth
force of 160 lbs. If the forces are coplanar and act simultaneously,
find their resultant.

Theorem 20. The diagonals of a parallelogram bisect each
othef.

EXERCISES. SET XXVII (continued)

227. Draw any line through
the intersection of the diagonals
of a parallelogram.

(a) Give a list of the pairs of
. congruent triangles formed.
Give reasons for your asser-
tions.

(6) Give a list of the pairs of
congruent quadrilaterals. Verify your statements.

228. Cut a parallelogram out of cardboard. Placing a pin at the
intersection of the diagonals, try to balance the parallelogram.
Why would you expect it to balance?

The intersection of the diagonals is called the center of gravity
of a parallelogram. Why?

Theorem 21. A quadrilateral whose diagonals bisect each other
is a parallelogram.

EXERCISE. Set XXVH (continued)

229. The same principle is often
used in the construction of iron
gates that was employed in the
making of a parallel ruler used in
the eighteenth century (see dia-
gram). What is the principle?

The student is encouraged to make such an instrument.

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POLYGONS AND PARALLELOGRAMS 73

EXERCISES. SET XXVm. PARALLELS

230. Classify quadrilaterals.

231. Summarize ways of proving:
(a) Sects equal.

(6) Angles equal,
(c) Lines parallel.

LIST OF WORDS DEFINED IN CHAPTER m

Exterior angle and angles of a polygon. Convex polygon, diagonal; quadri-
lateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon; regular
polygon. Parallelogram, rectangle. Distance between parallels.

SUMMARY OF THEOREMS DEVELOPED IN CHAPTER IH

13. The sum of the angles of a triangle is a straight angle.

Cor. 1. A triangle can have but one right or one obtuse angle.

Cor. 2. Triangles having two angles mutually equal are mutually
equiangular.

Cor. 3. A triangle is determined by a side and any two homolo-
gous angles .

Cor. 4. An exterior angle of a triangle is equal to the siun of the
non-adjacent interior angles.

14. The sum of the angles of a polygon is equal to a straight angle taken
as many times less two as the polygon has sides.

Cor. 1. Each angle of a regular polygon of n sides equals the ^ th
part of a straight angle.

Cor. 2. The sum of the exterior angles of a polygon, made by produc-
ing each of its sides in succession, is two straight angles.

Cor. 3. Each exterior angle of a regular polygon is the -th part
of a straight angle.

15. If two angles of a triangle are equal, the sides opposite them are equal.

Cor 1. Equiangular triangles are equilateral.

16. Either diagonal of a parallelogram bisects it.

Cor. 1. The parallel sides of a parallelogram are equal, and the

opposite angles are equal.
Cor 2. Parallels are everywhere equidistant.

17. A quadrilateral whose opposite sides are equal is a parallelogram.

18. A quadrilateral having a pair of sides both equal and parallel is a
parallelogram.

19. A parallelogram is determined by two adjacent sides and an angle.

20. The diagonals of a parallelogram bisect each other.

21. A quadrilateral whose diagonals bisect each other is a parallelogram.

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CHAPTER IV

AREAS

A. INTRODUCTION. REVIEW OF FRACTIONS ♦

In dealing with areas, we are largely concerned with ratios. A
ratio is a fraction, and therefore our work in this chapter presup-
poses a familiarity with fractions. For those of us who need a
review of this topic the following will be helpful.

A fraction is an indicated guotienty the dividend ot which is the
numerator and the divisor the denominator.

Since this is review, let us summarize without discussion the
fundamental facts which we need to recall about fractions.

PRINCIPLES

I. The valve of a fraction is not changed if both numerator and
denominator (i.e., both terms of the fraction) are multiplied or
divided by the same quantity. Why?

This statement includes cancellation, for that is the process of
dividing both numerator and denominator by a common factor.

Illustrations: !• fL — §E — 5? ^' 64q»6 ^ o^

6 "" 66 "^ n6 128a6« "^ 26*

3. a«-4 o-f2

2a2-8aH-8'°2(o-2)

II. Considering as the signs of a fraction that of the numerator,
that of the fraction, and that of the denominalor, any two may be
changed withovi changing the value of the fraction. Why?

Recall that the numerator and denominator of a fraction are
treated as if each were enclosed in a parenthesis.

Illustrations: !• ct „ _ ^;^ ^ £_ ^ _21? ^* Q— 2 ^ 2— o ^

6~ 6 "" — 6 ~* — 6 o— 5""5— o~

3. a6c ^ o(— 6)c _ —abc __

deS^de{-f)^ -def^

* Those who have not studied algebraic fractions are advised to take up this
topic at this time. Any standard algebra text will furnish sufficient material.
74

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AREAS 75

III. Fractions may be added or subtracted by changing them to
equivalent fractions having the same denominators and adding or
subtracting their numerators, jmttiTig the sum or differerux over the
common denominator. Why?

Changing to a same denominator depends for its validity upon
preceding principle I. The same or common denominator may
be found by getting the Lowest Common Multiple of the
denominators.
Illustration: m— 2n m»— 3n* . ^m—n

The L. C. M. of m*+mn+n^, rn^—n*, and m—n is m^—n*.

^. (m—2n)(m—n) (in^—2n^) , (3m— n)(m*+mn4

.'. the sums H

m*— n* wi* — n* m'—n*

m*-3mn+2n*-m«H-3n*+3m»+2m*n+2mn*-n«

in*—n*
2m» +m» +2m«n - 3mn +2mn« +5n« - n»

or ; ;

m*—n*

IV. Fractions may be multiplied by multiplying their numerators
and denominators separately y obtaining thus the numerator and de-
nominator respectively of the product. Why?

Illustration: 5c»-20d» c»-2cd+4d»
c»+8d» ' 25cd*

5(c-f2rf)(c-2d) c«-2c(f-f4d»
~ (c +2d) (c« -2cd-i-4d«) ' 25cd*
c-2d 5

V. Fractions may be divided by inverting the divisor and proceeding
as in multiplication. Why?

Illustration: a»-'7o+12 . o«-16
a-1 • 1-a*
-1

(a-4)(o-3) (l-o)(l-fa) _ (l-fa)(3-a)
^ a-1 •(a-4)(a+4) 4+a

EXERCISES. SET XXIX. FRACTIONS

232. Would the value of a fraction be changed if both numerator
and denominator were squared? Illustrate and give a reason for
your answer.

233. Why in dividing fractions do we invert the divisor and
multiply?

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76 PLANE GEOMETRY

234. Would the result of cancellation ever be zero? Illustrate
and give a reason for your answer.
236. Write in four ways the fraction -^g-

236. Find the difference between

n m n+m n m n-m 3 3

237. Are any or all of the expressions in the following groups
identical?

2'^' 2 ' 2 T~' 2 "^ 2' 2 ^ "^^

Justify your answers.

238. State, with reasons, whether or not the following are
identities:

(a) 1 _ 1 (6) 1 _ 1

(6-a)2 " (a-by (b -a)« "" (a-6)»

239. Add: 3a— b 2 Why is it correct to refer to such a

a^-b^" b—a combination as addition?
Simplify (which means do whatever is indicated by the symbols).

hxy \ 2-x)'x^+^+2x
-a^\ /-aV /aV/-l\*

244.

-2x 2a;+l ' 4x«-l

^■y-l^y-2 y-3-^ (y2-y)(y-2)

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AREAS 77

248. (a) x-\ — T-T • (6) [x-\ — —j-rr

a a+b \ a /a+b

219. a-b-'-^
a—b

250 ^^— 5m— 84

251 (a;^-49)(x^-16a;+63)

' (a:2-14x+49)(x2-2x-63)

252. -^ - (^2+-^') r ^-^'"i ( ^+^ ")
m+2 V ^m -3/ \4 -m^/ Vm^+m -6/

B. AREAS. DEVELOPMENT OF FORMULAS

Just as a sect is measured by finding the number of linear units
it contains, so a surface is measured by finding the number of
square units it contains, or better, its roiio to the unit square y which
is known as the area o/ the surface. A square unit is a square each
side of which is aUnearunit. For example, if we were to measure the
length and the width of this page (taking the inch as a linear unit)
a square inch would be the corresponding square unit, and the area
would be found in square inches. The selection of the unit in
practical measurements is just a matter of convenience. Why, for
instance, select an inch instead of a mile in measuring the length
of this page?

In comparing areas or sects, we compare the abstract numbers
which express the ratio of their measures in terms of a common
unit. When we say that two sects compare as 5 to 4 we refer to the
fact that when measured in terms of the same unit their measures
would compare as 5 to 4 — say one was 5" and the other 4". Sim-
ilarly, when we say two rectangles compare as 6 to 5 we mean that if
the area of one were 6 sq. ft. the area of the other would be 5 sq. ft.

Such quantities as we have just referred to are said to be com-
meiisurable because they can be measured in terms of the same unit.
In eomparing two sects it is sometimes impossible to get a common
unit <rf measure, that is, to select a unit so that it will be contained
an exact number of times in both. Such sects are said to be incom-

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78 , PLANE GEOMETRY

mensurable. It is known from experimental work in mensuration
that a circumference and its diameter are incommensurable, for if
the diameter were 1 inch, the circumference would be 3.14159+
inches (r inches).

EXERCISES. SET XXX. COMPARISON OF SECTS

263. Assuming the fact that the square of the hypotenuse of a
right triangle is equal to the sum of the squares of the other two
sides, show that the diagonal of a square and a side of the square
are incommensurable.

264. When two sects are commensurable a common measure
can be found as follows:

^ X Zi Zj B If the sects are AS

'. ' • ' aiidPQ. Lay off PQ

p Y Z Q ^^ '^^' I* ^^^ ^

'^ ^ ' ^ contained once with

a remainder XB. Lay off XBopi PQ. It will be contained twice
with a remainder ZQ. LayoffZQonZ5. It will be contained just
3 times with no remainder. Then Z2By is a common measure (in-
deed the greatest common measure) of AP and PQ . Any part of
Z2B would also be a conmion measure of AB and PQ. Call ZJB, u.

How many u^s does XB contain?

How many li's does YZ contain?

How many S's does AX contain?

Show that li is a common measure of AB and PQ.

We shall refer to the consecutive sides of a rectangle as its dimen-
sionSy calling one the altitude, and the other the base.

For the sakft of brevity the expressions "the ratio of two sects"
or "the products of two sects" will be used to indicate the ratio
or the product of the abstract numbers expressing their lengths
in terms of the sam^ unit of measure. For the same reason "rec-
tangle," "parallelogram," etc., will be used for the "area of rec-
tangle," "area of parallelogram," etc. Therefore, the expression
"two rectangles compare as the products of their dimensions" is a
conventionally abbreviated form of "the areas of two rectangles
compare as the products of the lengths of their dimensions ex-
pressed in terms of the same linear unit."

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AREAS

Theorem 22."' Rectangles having a dimension of one equal to
that of another compare as their remaining dimensions

PQ,

Q M N 8

Given: Rectangles ABCD (call it R) and PQ&T (call it Ri) with AB*

and EC and Q8 commensimible.
To prove: R^ BC

Suggestions for proof:

Let u be a common measure of BC and QS.

Lay uoff on BC and QS and erect ±8 at points of division F, O, . . ^

M,N, _ _ SC

If ii is contained m times in BC and n times in QS what is the ratio === 7
What kind of figures are ABF^, . . . , PQAf L, . . .? ^^

Are they congruent? Why?
If R and Ri are respectively composed of tn and n congruent parallelograms

R
what is the ratio of -^7
Ri

Theorem 23. Any two rectangles compare as the products of

their dimensions.

Given: Rectangles R and Ri with dimensions h, h, and hi, bi respectively.

R bh
Topn>ve:^^»gy^

Suggestions for proof:

Construct a rectangle (Rt) having as dimensions h and &i.

R ,

Rt^

"Ri '

^Although the proof here suggested applies only to those cases where the
sects are commensurable, the theorem is aJway? valid.

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80 PLANE GEOMETRY

Theorem 24. The area of a rectangle is equal to the product

of its base and altitude.

H Given : Rectangle R with
base b and altitude h.
1 To prove: Area of R=bh.

Suggestions for proof:

b What does area mean?

Call the unit of surface u. u will be a rectangle with base and altitude 1.

^y^ S^n' ^y^ For what does ^ stand?

EXERCISES. SET XXXI. AREAS OF RECTANGLES
266. What per cent of surface is allowed for joints and waste,
if 120 rectangular sheets of tin, 14" by 18", are just sufficient to
cover a roof 165 sq. ft.?

266. A map is drawn to a scale of 1" to 1000 miles. What actual
area would be represented on the map by a foot square?

267. From a rectangular sheet of paper cut a strip H of the sheet
in width. What part of the sheet is left? Then from the same sheet
cut off M of its length. What part of the original sheet is now left?

268. Prove, by letting a and h represent the lengths of two
sects, that {a^hy^w'+h^^2ab.

269. How long would a rectangular strip of paper 1 sq. ft. in
area be if it were .01" wide?

260. Both a square and a rectangle three times as long as it is
wide have a perimeter of 64 ft. Compare their areas.

261. Thucydides (430 b.c), a Greek historian, estimated the
size of the Island of Sicily by the time it took him to sail around it,
knowing how long it took him to sail around a known area. Was
his method correct? Give a reason for your answer.

262. 144 sq. ft. is the area of a square and also of a rectangle
four times as long as wide. How do their perimeters compare?

// one side of a parallelogram is selected as its base, the distance
between U and the opposite side is called its altitude.

Would it make any difference at what point its altitude was
measured? Why?

How many altitudes has a parallelogram?

Is this definition consistent with what we have referred to as
an altitude in the case of the rectangle?

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AREAS

Theorem 26. The area of a parallelogram is equal to the
product of its base and altitude. A h d

Given: OABCD with base b and altitude k.
To prove: Area of OABCDsbh.
Suggestions for proof:

DTs,wAHBXidDHi±BC. What kind
of figure is AHHiD?

What is the area of AHHiD.

Compare A ABH and DCHi,

How do the areas of AHHiD and ABCD compare?

What is the base of each, and what the altitude of each?

Suppose AH cut BC produced, will the theorem still hold true?

EXERCISE. SET XXXII. AREAS OF PARALLELOGRAMS

263. If one side of your parallel ruler is held fixed while the
opposite side is raised and lowered to various positions, will the
areas of the various parallelograms be changing? If so, what will
be the greatest area obtainable, and what the least?

Any side of a triangle may be taken as its base, and the perpeiv-
dicvlar from the opposite vertex to thxit side will be its altitude.
EXERCISE. SET XXXIII. ALTITUDES OF TRIANGLES

264. (a) How many altitudes has a triangle? Illustrate.

(6) Does your answer to (a) hold for a right triangle? Illustrate.

(d) What fact have we already proved about the altitudes of an
equilateral triangle?

Theorem 26. The area of a triangle is equal to half the product

of its base and altitude.
Given: AABC with & as base and

h as altitude.
To prove: Area oiAABC
Suggestions for proof:

Draw BXWCAsjid AY
meeting at P.

^bh,
\CB,

What kind of figure is APBC7

What is the area of APBC?

How is the area of AABC related to that of APBC7

Cor. 1. Any two triangles compare as the products of their bases

and altitudes.
Cor. 2. Triangles having one dimension equal compare as their

remaining dimensions.
6

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PLANE GEOMETRY

EXERCISES. SET XXXIV. AREAS OF TRIANGLES
266. Prove theorem 26, using the suggestion given by the accom-
pan3ring diagram.

266. Calculate the area
of the letter Z shown in

'6 the figure, the dimensions ^ \\ ^' — V

being indicated in centimeters,
^t 267. With a marked edge draw a triangle, and tak-

ing necessary measurements find its area, using in turn its three
bases and altitudes.

268. Where do the vertices of all triangles having the same base
and the same area lie? Give reasotis for your answer.

A trapezoid is a qyMrUateral with one, and only one, pair of par^
aUel sides. The parallel sides are called its basesi and the distance
between them is called its altitude.

Why do the words "and only one" need to be included in the
definition?

Theorem 27. The area of a trapezoid is equal to half the product
of its altitude and the sum of its bases.

Given: Trapezoid ABCD with bases b and 6i

and altitude h.
To prove: Area of trapezoid equals -^h (6+61).
Notes on proof:
This fact may be proved by making con-
structions that will divide the figure into
rectangles, parallelograms, or triangles, or combinations of them. Why?
The following diagrams show some such constructions. The student may

"51

use one of them or suggest another. It is
interesting as an exercise to show that the
same formula for the area of a trapezoid
may be derived from each of the following
diagrams. Which gives the simplest deri-
vation?

2. DrawAB

B D T

3. Draw J5±5t and CDXJP.

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AREAS

P _ M^

4. Draw PQ and MN±AB
produced.

A. B-

--~—f

-D s^a

6. Throuj^ ff, the midpoint of £C,dTaw
MN II £>A cutting AB produced in P.
u4^ B

6. Extend ABhy_DC to ^ and DC 7. Draw CZ
by AB to F.
8. Extend YX to
i2 so 1^ XR=AB.
Draw /gZ || Fil. Bi-
sect AF in Af and
draw_MP II rX cut-
ting RZmQ.

I DA cutting AB produced
inF.

:z^

EXERCISES. SET XXXV. AREAS OF TRAPEZOIDS

269. The diagram shows how the area of an uregular polygon
may be found, if the distance of each vertex
from a given base line, as XYy is known.
These distances AAiy BBiy etc., are called
offsets, and are the bases of trapezoids
whose altitudes are AiBi, Bid, CiDi, etc.
The area ABCDEF may now be found
by the proper additions and subtractions.
On cross-section paper plot the points
whose coordinates are given below, join
them in order, draw F-A, and find the inclosed area in each case;

Case

(a)

5,7

4,5

4,1

7,0

5,3

6,3

(b)

2,7

3,3

6,0

5,2

6,6

8,7

(c)

2,5

3,6

2,4

4,2

6,6

4,7

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PLANE GEOMETRY

270. In order to determine the flow of water in a certain stream,
soundings are taken every 6 ft. on a line AB at right angles to the
current. A diagram may then be made to represent a vertical
cross-section of the stream. If the area of this cross-section and

the speed of the current are known,
it is possible to determine the
amount of water flowing through
the cross-section in a given time.
The required area is often found
approximately by joining the ex-
tremities of the offsets i/o, i/i, 2/2, etc., by straight lines, and finding
the sum of the trapezoids thus formed. That is, the strips between
successive oflFsets are replaced by trapezoids. This gives the Trape-
zoidal Rvle for finding an area. It may be stated as follows: To
half the sum of the first and last offsets add the sum of all inter-
mediate offsets, and multiply this result by the common distance
between the offsets.

(a) State this as an algebraic formula.
(6) Verify the formula.

(c) Find the area of the cross-section of a stream if the soundings
taken at intervals of 6 ft. are respectively 5 ft., 6.5 ft., 11 ft., 14.5
ft., 16 ft., 9 ft., and 6.5 ft.

(d) In the midship section of a vessel the widths taken at
internals of 1 ft. are successively 16, 16.2, 16.3, 16.4, 16.5, 16.7,
16.8, 15, 10, 4, 0, measurements being in feet. Find the area
of the section. (Use the line drawn from the keel ± to the deck
as base line.)

6271. A third rule for finding plane areas, known as Simpson's
Rvlej usually gives a closer result than the Trapezoidal Rule. In
proving Simpson's rule two consecutive
strips are replaced by a rectangle and
two trapezoids as follows: Divide 2d
into three equal parts, erect Jls at the
points of division, and complete the rec-
tangle whose altitude is the middle
offset yij as in the figure. Join the
extremities of y and j/2 to the nearer upper vertex of this rectangle.

T 1

y 1

1
1

%d|%

di%d

*— d— ^

«— d— *

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AREAS 85

Then if the areas of the trapezoids are T and Ti, and if the area
of the rectangle is Rj

.-. r+i2+ri^| d (2/+4i/i+t/2).

If, now, the number of strips is eoeUy and if the offsets are lettered
consecutively t/o, yi, 2/2, , 2/n, the addition of the areas of suc-
cessive double strips, found by the above formula, gives the result :

Area^gd (2/o+42/i+2y2+42/8+ . . .+2yn-2+42/„-i+yn).

In words: To the sum of the first and last offsets add twice the
sum of all the other even offsets and four times the sum of all the
odd offsets, and multiply by one-third the common distance
between the offsets.

(a) Verify this rule.

(6) By Simpson's Rule find the area of the stream in Ex. 270 (c) .

LIST OF WORDS DEFINED IN CHAPTER IV

Ratio, fraction, numerator, denominator, terms of fraction, cancellation,
simplify. Area, commensurable, incommensurable; dimensions, base and
altitude of rectangle, parallelogram, triangle, trapezoid.

SUMMARY OF THEOREMS PROVED IN CHAPTER IV

22. Rectangles having a dimension of one equal to that of another compare
as their remaining dimensions.

23. Any two rectangles compare as the products of their dimensions.

24. The area of a rectangle is equal to the product of its base and altitude.

25. The area of a parallelogram is equal to the product of its base and
altitude.

26. The area of a triangle is equal to half the product of its base and altitude.

Cor. 1. Any two triangles compare as the products of their bases
and altitudes.

Cor. 2. Triangles haying one dimension equal compare as the re-
maining dimensions.

27. The area of a trapezoid is equal to half the product of its altitude and
the siun of its bases.

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CHAPTER V

ALGEBRA AS AN INSTRUMENT FOR USE IN APPLIED
MATHEMATICS

A. LOGARITHMS

Although logarithms are introduced at this point, as a part of
a chapter on Algebra, it is not essential that they be studied at
this time. They might well be omitted until there is a feeling of
necessity on the part of the pupils in the solution of the more diffi-
cult problems based upon similarity and the trigonometric func-
tiDns. In fact, those schools wishing to oinit such problems imder
Ratio, Proportion, Variation and Similarity need not include the
topic at all. For this reason, when logarithms are desirable* in
the solution of problems, this fact has been indicated. The integ-
rity of the course will not be injured in the least by the omission
of the topic of logarithms or any of these exercises.

I. INTRODUCTION

If 8^=64, then 2 is called the logarithm of 64 to the base 8.
This is written log864 = 2. Again 3^=81 or logsSl = 4. Hence we see
that the logarithm of a number is simply the exponent of the power
to which another number (called the base) miLst be raised to obtain
that number. Thus, in the last example 4 is the exponent of the
power to which the base 3 must be raised to obtain the number 81.

EXERCISES. SET XXXVI. MEANING OF LOGARITHMS

272. Write in logarithmic form:

(a) 5»=125 (c) 72=49 (e) 10i-3oio3=20 (g) 103*7712=3000
(6) 2^=128 (d) 10^ = 10 (/) 103 = 1000 (h) 105*771 = 300,000
(i) 10^ = 1,000,000 0*) 100^-^=1000

273. Write in exponential form:

(a) loge216 = 3 (d) logio2 = .3010 (g) logio.Ol = -2

(6) log981 = 2 (e) logiol = (h) logio.001= -3

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ALGEBRA IN APPLIED MATHEMATICS 87

Before studying logarithms, their principles and applications, it
wiU be well for us to recall the laws concerning exponents in multi-
plication, division, involution and evolution. We have learned to
multiply, divide, raise to powers and extract roots of simple
expressions when the exponents have been positive integers, so
that we shall here only summarize what we already know, adding
the statement that the laws governing positive integral exponents
govern all exponents, both fractional and negative

a. PRINCIPLES OF EXPONENTS

1. The exponent of the -product of any number of factors of like
base is equal to the sum of the exponents of the factors.

lUustrations: (1) a^'a^a^^a''-^^-^

(2) J'a^=a^

(3) 6-n~^s6-«-P

(4) 6+*-6"»'^s6*~*^=6*s6*

2. The exponent of the quotient of two quantities of like base is
equal to the exponent of the dividend diminished by that of the divisor.

Illustrations: (1) x«-x-^sxo-fr

(2) x^'^x-h^x''-^^

(3) w ''^m'^m «• -"

3. The exponent of any power of a quantity is equal to the product
of the exponent of the base and the index of the power.

lUustrations: (1) (b'^y^h^

(p\r pr

4. The exponent of any root of a quantity is equal to the quotient
of the exponent of the base and the index of the root when that index
is not zero.

Illustrations : (1) _„j _ _ £

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88 PLANE GEOMETRY

EXERCISES. SET XXXVU. DRILL IN APPLICATION OF LAWS

OF EXPONENTS

Perfonn the operations indicated in the following problems:
274. x*a;-^a;-8 282. 103«82.io78m

276. -a*-^* 283. 10"*10^-5-10^.ViO»

8 6

276. fc"+2-5-3A;»

m n 1

277. p^jp^p ""

284. 10~il0t-f(l0^v^)
286. (10««723)»

-••• y y y f i \(«'-»')«

/ _7,_4\ 286. \^d*+»)

278. (r«<^»)V-g«« V , ■

16^125~»

279. 4* 8* 287. -^^

(Hin t : Express 4 and 8 as

powers of 2 so as to have On/'Qn — l\n

a common base). 288 ^ '

280. 25^.125*

3»+i(3»-i)(9-«)

289.* -^i^lIMlCio!)!!

281. 25 - *-5 -* ^ (10- «)«(10*)>»

* Note : No doubt some of us are by this time curious to know the mean-
ing of the zero, negative and fractional exponent.

(I) Meaning qf the zero exponent. (II) Meaning of the negatwe exponent.

Let ao« Leta-Psa;

Then x-a^^a^-a^

Thena"^-a^sxaP

Why?

Buta-Pa^sl

Why?

.•.»<iPal

Why?

.•..-,.j

Why?

Quote the axiom apphed.
(Ill) Meaning oj the fractional exponent.

The fourth law given in the text expresses the meaning of the fractional

exponent since -C^=a^, but we may make the meaning still clearer by the
following:

Since x*'a:*sT,x* must by definition of square root be the square root (Vx)-

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ALGEBRA IN APPLIED MATHEMATICS 89

Let us see how the annexed table may be used to sunplify cer-
tain calculations.

Table of Powers of 2

21=2 2« =64 211=2048 2" =65536

22=4 2^=128 2" =4096 2" = 131072

2»=8 28 =256 2" =8192 2^=262144

2* = 16 2» =512 2" = 16384 2i» =524288

2* =32 2«> = 1024 2^=32768 2» = 1048576

1. Suppose we wished to multiply 1024 by 512
•/ 1024 = 210 and 512 = 2« and 2io.2» = 2^

.-. 1024. 512 = 21^ = 524288.

2. Suppose we wished to divide 1,048,576 by 32768.
•.• 1048576 = 220 and 32768 = 21^ and 22o^2i^ = 2S
;•. 1048576 -^ 32768 = 2^=32.

3. Suppose we wished to raise 64 to the third power.
•.• 64 = 2« and (2f^y = 2^^ and 218=262144,

.•.64» = (2«)3 = 262144.'

4. Suppose we wished to find the 5th root of 1,048,576.
•.• 10 48576 = 2 ^ and^V^22o=2^'' =2*

.-. \/l048576=\/22o=2^=16

In similar manner a table of the powers of any number may be
computed, and these four operations (multiplication, division,
involution, evolution), reduced to the operations of addition, sub-
traction, multiplication, and division of exponents.

Ill
Similarly x«a;»-x»- to 8 factors

1 +1 +i to « terms. „„ „

sx* • • Why?

sxfax Why?

/.x't^y/z Why?

p 1 i. i
Still more generally, x^^^x^ofi to p factors.

But Vx^ also means ^^ ^^^ixxx- to p factors)^'.

Ill

sa;«-xix^- to p factors.

1 + 1+ Ia to p terms.

ax'

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PLANE GEOMETRY

EXERCISES. SET XXXVUI. USE OF TABLES OF POWERS
Using the given table of 2's, find the values of the following:

290. 512X2048 292. (32)» 296. \/262144

266X16384 293. (512^1 296. V^S6

29L

262144
297,

294. V131072

, 4/ (256)K524288)»
• V (32)«

298. [(2«)*]'

299. 2«**

Table of Powebs of 16

No.

Power

No.

Power

0.00

1.50

0.25

256

2.00

0.60

1024

2.50

0.75

4096

3.00

1.00

65536, etc.

4.00

1.25

EXERCISES. SET XXXVIII (concluded)
From the foregoing table compute the following:

300. 16X4096

301. 8*

303.

/ 65536(256)^
\ 64(1024)

302. 65536-5-4096

304. 16 X

b. HISTORICAL NOTE

3/(65536X1024)^

2562

There is a large amount of computation necessary in the solution of some
of the practical applications of mathematics. The labor of making extensive
and complicated calculations can be greatly lessened by employing a table of
logarithms. About the year 1614 a Scotchman, John Napier (1550-1617),
Baron of Merchiston, invented a system by which multiplication can be per-
formed by addition, division by subtraction, involution by a single multiplica-
tion, and evolution by a single division. From Henry Briggs (1556-1631),
who was a professor at Gresham College, London, and later at Oxford, this
invention received modifications which made it more convenient for ordinary
practical purposes.

Laplace, the great French astronomer, said: "The employment of logar-
ithms by reducing to a few day^ the labors of months, doubles, as it were, the
life of an astronomer, besides freeing him from the errors and disgust insepar-
able from long calculations.''

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ALGEBRA IN APPLIED MATHEMATICS 91

The logarithms now in general u^ are known as common logarithms, or as
Briggs' logarithms, in order to distinguish them from another system^ also a
modified form of Napier's system. The logarithms of this other modified
system are frequently employed in higher mathematios, and are known as
natural or hyperbolic logarithms.

n. PRINCIPLES OF COMMON LOGARITHMS

For practical purposes, the exponents of the powers to which 10,
the base of our decimal system, must be raised to produce various
nimibers are put in table form. That is, the logarithms of numbers
to the base 10 are tabulated. For the positive integral powers of
10 we would need no tables, for those we can find by inspection.
But exponents may be negative and they may be fractional. For
the negative integral powers of 10 as we shall see presently, we
would need no tables either. But fractional exponents or the frac-
tional parts of exponents we cannot readily find, and hence for
them we need tables.

We all know that Similarly it can be shown that

VlO' = 1000, .-.♦log 1000=3. ''10-^ = ^> .-. log.l = -1

V102 = 100, .-. log 100 =2. -.10-2= jL, ... log .01 = -2

•.•101 = 10, .-. log 10 =1. •;10-» = -^,, .-. log .001= -3

•.• 12! =101-1= 100, andl5J = i; ... 100=1, ...log 1 = 0.

4 7 7 12

10-47712 or lOioooooj that is, the one-hundred-thousandth root
of 10-^7712 is nearly 3. .-. log 3 = .47712 nearly.

Although log 3 can never be expressed exactly as a decimal
fraction, it can be found to any required degree of accuracy. In
this book logarithms are given to four decimal places. These are
sufficient for ordinary computations.

* When the base 10 is used the base is not indicated in writing the loga-
rithms of numbers. Thus we write log 3 =.47712, not log io3 = .47712.

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92 PLANE GEOMETRY

EXERCISES. SET XXXIX. CX)MMON LOGARITHMS

306. What are the logarithms of the following to the base 10:

(a) 10000? (c) .0001? (/) \/lO?

^ ' 10000 (e) 10-»? (A) 10^?

306. Between what two consecutive integers does the logarithm
of each of the following numbers lie? Why?

(a) 600 (c) 13 (e) 46923

(6) 5728 (d) 496,287 (/) 9

307. Between what two consecutive negative integers does the
logarithm of each of the following numbers lie? Why?

(a) .06 (c) .0008 (e) .00729 (g) 0.5

(6) .007 (d) .0625 (/) .00084

308. What is meant by saying that:

(a) log 880=2.94448? (6) log 92.12 is 1.96435?

Since 3585 lies between 1000 and 10,000, its logarithm lies
between 3 and 4. It has been calculated as 3.55449. The integral
part 3 is called the characteristic, and the decimal part .55449, the
mantissa of the logarithm.

•••358.5=3585 MO, .*. log 358.5= log (3585 5- 10) = log 3585-
log 10=3.55449 -1 = 2.55449.

That is, since log 3585=3.55449

log 358.5 = 2.55449 and similarly it can be shown
that log 35.85 = 1.55449
log 3.585=0.55449
log .3585 = .55449-1*
log .03585 = .55449 -2.
Thus we see that

(a) The characteristic can he found by inspection in all cases.
'.' the number 589 lies between 100 and 1000, log 589 lies between
2 and 3. .'. log 589=2+some mantissa.

* Log 0.3585 = .55449— 1 may be written in two other ways as follows:
1.55449 or 9.55449—10. The last method is the most practical, aa we shall
flee as we proceed.

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ALGEBRA IN APPLIED MATHEMATICS 93

(6) The mantissa is the same for any given sitccession of digits,
wherever the decimal point may be.

(See last table of numbers with their logarithms.)

EXERCISES. SET XXXIX (continued)

309. What is the characteristic of the logarithm of:

(a) 384? (c) .297? (e) A number of n integral places?
(6) 5286? (d) Any number of millions?

(f) Any decimal fraction whose first significant digit is in the
first decimal place?

(g) In the second decimal place? (h) In the third decimal place?
(i) In the seventh decimal place? (j) In the n'* decimal place?

310. From the last exercise formulate a principle by means of
which the characteristic of the logarithm of any positive number
may be found.

m. THE FUNDAMENTAL THEOREMS OF LOGARITHMS

(a) The logarithm of the product of two numbers equals the sum of
iheir logarithms to the same base.

1. Let as 10**, then log a =Z]

2. Let 6 = 10^ then log 6=fe

3. .-. o6=10''+Sandloga6=Zi+fe=loga+log6.

(6) The logarithm of the quotient of two numbers equals the logarithm
of the dividend minus the logarithm of the divisor ^ all to the same base.

1. Leta=10^ then log a=Zi

2. Let 6=10'*, then log b=h

a 10'* , _ , a

3. .'. 5 = 10^=10' ^ aiid log -=li-k^log a -log b.

1. Leta=10^ then log a=l

2. .'. a'*=10''*, and log a''=nl=n log a.

(d) The logarithm of the n^ root of a number equals ^<a of the
logarithm of the number.

1. Leta=10', then log a=Z

i ' * / 1

2, .*, a'^^lO*, and log a»» = ;is- log a.

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PLANE GEOMETRY

Th. (c) might have been stated mofe generally, so as to include

The proof would be substantially

Th. (d) thus: Loga''=- log a.
the same as in Ths. (c) 'and (d).

EXERCISES. SET XXXIX (concluded)

Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990, log 7 = 0.8451,
and log 514 = 2.7110, find the following:

811- Log 6. 312. Log 14. 313, Log 7^\

316. Log 42. 316. Log 5*. 317. Log 105.

319. Log Vsii. 320. Log 514^. 321. Log 1542.

323. Logl799[=log(i-5147)]. 324. Log\/3*.

326. Show how to find log 5, given log 2.
IV. USE OF THE TABLE

(a) Given a number, to find its logarithm.

In the table on p. 103 only the mantissas are given. For in-
stance, in the row beginning 43, and in columns headed 0, 1, 2,
3, ,9 will be found:

314. Log\/2.
318. Log 1.05.
;. Log 257^
326. Log\/2L

6335

6345

6355

6365

6375

6385

6395

6405

6415

6425

This means that the mantissa of log 430 is .6335, of log 431 is
.6345, and so forth, to log 439.

Therefore log 431 = 2.6345, log 434 = 2.6375, log 43.7 = 1.6405,
log 4.39 = 0.6425, log .438 = .6415 - 1, log .0433 = .6365 -2. Now,
since 437.8 is .8 of the way from 437 to 438, .'. log 437.8 is about
.8 of the way from log 437 to log 438. .*. log 437.8= log 437+. 8
of the diflFerence between log 438 and log 437. .*. log 437.8 =
2.6405+.8 of .0010=2.6405+.0008 = 2.6413.

This process of finding the logarithm of a number lying between
two tabulated numbers is called interpolation. This is not wholly
accurate, since the numbers do not vary as their logarithms, but
it is suflSciently accurate for most practical purposes. If greater
accuracy is desired, tables of five or six or even more places are
used. The mantissas here given are correct to .0001. This will
give a result which is correct to three figures in general, and an
approximation to four figures, which will be sufficiently accurate
for the computations in this book.

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ALGEBRA IN APPLIED MATHEMATICS

EXERCISES. SET XL. USE OF TABLE
Find by using the table:

327. Log 49. 332. Log 14.7.

328. Log 723. 333. Log 14.73.

329. Log 1580. 334. Log 5.93.

330. Log 4285. 335. Log .00432.

331. Log 14.5. 336. Log 1.672.
(6) Given a logarithm to find the corresponding number.

The number corresponding to a given logarithm is called its anti-
Iofi;aiithm. Example: ".• 0.4771 =log 3, .'. antilog 0.4771=3.

337. Log .00002376.

338. Log V29.

339. Log 5. 692».

340. Log \/36.54.
34L Log .0057*.

6335

6345

6355

6365

6375

6385

6395

6405

6415

6426

Here we see that antilog 4.6345=43100, antilog 0.6395=4.36,
antilog 3.6405 = .00437.

Now suppose we wished to find antilog 2.6417.

•/ 2.6417 is .2 the way from 2.6415 to 2.6425.

.*. antilog 2.6417 is about .2 the way from antilog 2.6415 to anti-
log 2.6425.

.-. antilog 2.6417 is about .2 the way from .0438 to .0439.

/. antilog 2.6417 = .04382.

The following form is a good one for use.

Required: Antilog 2.7361. Antilog 2.7364 = 545 2.7361

Antilog 2.7356 = 544 2.7356

Tabular difr.=

8
;. Antilog 2.7361 = 544f = 544.6.

EXERCISES. SET XL. (concluded)

Diflf.=

Find from the table:

342. Antilog 1.9321.

343. Antilog 2.9049.
344: Antilog 2.7813.
346. Antilog 3.0354.

346. Antilog 1 .0354.

347. Antilog 3.1628.

348. Antilog 4.8393.

349. Antilog 10.5843.

360. Antilog 0.6923 -2.

361. Antilog 8.6923-2.

362. Antilog 7.5194-10.

363. Antilog 9.2490-10.

364. Antilog 10.4687-10.
366. Antilog 3.5357.
366. Antilog 0.3471.

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96 PLANE GEOMETRY

Since many errors occur because of failure to arrange work care-
fully, the pupil is advised to arrange all work in as compact and
neat a form as possible. A few examples worked out in full may
be suggestive, therefore we append the following:

1. In how many years will $600 double itself at 3% interest
compounded annually?

Solution: Let the number of years be n.

At the end of one year the amdimt will be 1.03 of $600, at the
end of the second year it will be 1.03 of 1.03 of $600, or l.OS^
of $600, and so forth.

.-. at the end of n years it will be 1.03« of $600.
.-. 1.03«X600=1200 or 1.03«=2

/. n log 1.03 = log 2 and .'. n=, ^^ ^^

log l.Uo

log 2 = .3010] ^.3010

log 1.03 = .0128j"^ .0128

.-. log n = log .3010 -log .0128

log .3010=9.4786 -10

log .0128= 8.1072 -10

.*. log n = 1.3714 antilog 1.3714 = 23.5+

•. n=23.5+, or the sum will double itself in 24 years.

2. Required the value of

Solution:
log 529 = 2.7235
log 528 = 2.7226

\/l754X3292

Tab. diff. =9

5.4.-. log 528^ = 2.7231

log 427=2.6304 .-. log ^^427 = 1.3152

.-. log numerator =4.0383 14.0383 -10

log 1760=3.2455

log 1750 = 3.2430

Tab. diff. = 25
.4

10.0

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ALGEBRA IN APPLIED MATHEMATICS 97

/.log 1754 = 3.2440 .'. log \/i754 = 1.0813;

log numerator = 14.0383-10
log 3300 = 3.5185
log 3290 = 3.5172
Tab. difif.= 13

2.6 or 3. Mog 3292 = 3.5175

.". log denominator = 4.5988 4.5988

antilog 9.4409 - 10 = .276 . ". log fraction (log

numerator— log denominator) =9.4395 - 10
antilog 9.4393-10 = .275 9.4393 - 10

Diflf. for 1 = 16 2

.-. antilog 9.4395 -10 = .275A=.275i.
.'. Fraction = .2751

As soon as the pupil is able to interpolate mentally the left
column may be omitted.

Often in problems involving the process of evolution difficulties
may arise owing to the fact that the characteristic may be negative
and not a multiple of the divisor, while the man tissa is always
positive. For instance, it may be desired to find \/0.03.
log \/a03 = Hog 0.03

= i (2.477 1) . Why is this a decidedly impracticable

form?
= i (8.4771 -10). Why is this an inconvenient

form?
' =i (28.4771 -30). Why is this the form which is

adopted?
= 9.4924-10
/. \/a03= 0.3107
Again, suppose \/0.3* is called for.
log 0.3=1.4771

.-. log VoJ' =1(9.4771 -10) == 37.90^-40 ^^^^ ^^^^ ^^^^
multiple of 10 and 7, hence we change to the equivalent
log ^0J« =67:9^4^=9.7012 -10.
.-. antilog 9.7012 -10= -^/oT*.

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98 PLANE GEOMETRY

In place of a table of logarithms engineers often use an instru-
ment called a "slide rule." This is really a mechanical table of
logarithms arranged ingeniously for rapid practical use. Results
can be obtained with such an instrument far more quickly than
with an ordinary table of logarithms, and that without recording
or even thinking of a single logarithm. A ''slide rule'' ten inches
long gives results correct to three figures. In work requiring greater
accuracy a larger and more elaborate instrument which gives a
five-figure accuracy is used.*

EXERCISES. SET XLI. COMPUTATION BY LOGARITHMS

367. If the hypotenuse of a right triangle and one leg are known,
the other leg may be found by means o f log arithms, for i f h = hypo-
tenuse, ii = one leg, then k=y/h^-'li^=\^(h+li)(h''li).

.-. log k^i [log (A+W+log (h -W].

If the hypotenuse of a right triangle is 587, and one leg is 324,
what is the other leg? «

368. The area of an equilateral triangle whose side is s, is T\/3-

(a) Find the area of an equilateral triangle whose side is 15.38 units.

(b) Find the side of an equilateral triangle whose area is 89.5
square inches.

369. The formula for the area of a triangle in terms of its sides
iBA = Vs(s— o)(s— 6)(s— c) where a, b, c are the sides of the triangle
and s the semi-perimeter. Find the area of a triangle whose sides
are 436, 725.4, 951.8 units respectively.

This is often referred to as Heron of Alexandria's formula, and
will be proved later.

a360.t Show that the amount of P dollars at interest com-
pounded annually for n years is P( 1 + 77:7: ) ; compounded semi-

/ ^ \2n V 100/

annually is P

i'+mj

* It is suggested that pupils will find an inexpensive slide rule of great use
ifi rapid calculation. Such an instrument, called the "Favorite," can be pur-
chased of KeuflFel and Esser, New York, N. Y.

t As here, the letter a preceding the number of an exercise indicates that
algebra beyond the solution of simple linear equations is required to solve the
problem.

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ALGEBRA IN APPLIED MATHEMATICS 99

361. In how many years will $1.00 double itself at
(a) 3% interest compounded annually?

(6) 4% interest compoimded annually?

(d) 6% interest compoimded annually?

362. In how many years will $1.00 double itself at
(a) 3% interest compounded semi-annually?

(6) 4% interest compounded semi-annually?

(d) 6% interest compounded semi-annually?

363. Find the amount at compound interest, compoimded
annually of:

(a) $150 for 7 years at 5|%.: (6) $1850 for 5 years at4%.
(c) $10 for 50 years at 5%.

364. To find the present value Ao of an annuity (a fixed sum of
money, payable at equal intervals of time) of s dollars to continue
for n years at R% compound interest, the formula

Ao^:

'^Wol^'a^^'"'^'^'

Find the present value of an annuity (i.e., the amount which, if
put at compound interest for the given time and rate, will amount
to the given sum).

(a) Of $1000 for 10 years, at 4% compound interest.

(6) Of $1200 for 10 years, at 4% compound interest.

(d) Of $500 for 5 years, at 5% compound interest.

365. What annuity can be purchased for $3000, if it is to run
for 15 years, at 5% compound interest compounded annually?

366. The diameter in inches of a connecting rod depends upon
the diameter D of the engine cylinder, I the length of the connecting
rod, and P the maximum steam pressure i n poun ds per sq. inch,
according to Mark's formula d^0.02758V Ddy/P.

What is d when D = 30, Z=75, and P= 150?

367. If fluid friction be used to retard the motion of a flywheel
making Vo revolutions per minute, the formula F=Fo«~*' gives
the number of revolutions per minute, after the friction has been

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100 PLANE GEOMETRY

applied t seconds. If the constant fc = 0.35, the value of e being
2.718, how long must the friction be applied to reduce the number
of revolutions from 200 to 50 per minute?

a368. The pressure, P, of the atmosphere in pounds per sq. inch,
at a height of z feet, is given approximately by the relation Po=
Pffi "* **, where Po is the pressure at sea level and k is a constant,
the value of e being 2.718. Observations at sea level give Po = 14.72,
and at a height of 1 122 feet, P = 14. 1 1 . What is the value of k1

369. If a body of temperature Ti be surrounded by cooler air
of temperature To , the body will gradually become cooler; and its
temperature, T, after a certain time, say t minutes, is given by
Newton's law of cooling, that is T^To+(Ti -T^e~^y where A; is
a constant and e — 2.718. In an experiment a body of temperature
55° C. was left to itself in air whose temperature was 15° C. After
11 minutes the temperature was found to be 25°. What is the
value of A;?

370. How many ciphers are there between ^e decimal point and
the first significant figure in (0.0504)^°?

371. The loss of energy E through friction of every pound of
water flowing with velocity v through a straight circular pipe of
length I ft. and diameter d ft. is given by 0.0007y'^Z-^<^.

Given t; = 8.5 ft. per sec, Z=3000 ft., d=6 inches, find E.

372. A man bequeaths $500, which is to accumulate at compound
interest until the interest for one year at 5% will amount to at
least $300, after which the yearly interest is to be awarded as a
scholarship. How many years must elapse before the scholarship
becomes available, assuming that the original bequest is made to
earn 5% compoimd interest?

373. In 1624 the Dutch bought Manhattan Island from the
Indians for about $24. Suppose that the Indians had put their
money out at compound interest at 7%, and had added the interest
to the principal each year, how large would be the accumulated
amount in 1910?

Ans, In round niunbers $6,000,000,000. The actual valuation
of Manhattan and Bronx real and personal property in 1908 was
$5,235,399,980.

374 The population of the State of Washington in 1890 was

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ALGEBRA IN APPLIED MATHEAUTI£S IQl

349,400, and in 1900 it was 518,100. What w&s-tfce 3<xcTa'g6 ^e&rfy
rate of increase? Assuming the rate of increase to remain the
same, what should have been the population in 1910?

375. The founder of a new faith makes one new convert each
year, and each new convert makes another convert each year, and
so on. How long would it require to convert the whole earth to the
new faith, assuming that the population of the world is
1,500,000,000?

Ana. Between 30 and 31 years.

376. The combined wealth of the United States and Europe was
estimated (1908) to amount to about $450,000,000,000. Let us
assmne that the entire wealth of the world amounts to $10^^. How
long would it take $1.00 put out at compound interest at 3% to
equal or exceed this amoimt?

Ans, 935 years.

The following problems should be solved by means of five-place
tables:

377. The circumference of a circle is 27rr (r being radius). (Use
ir=3.i416.)

(a) Find the circumference of a circle whose radius is 143.7.
(6) Find the radius of a circle whose circumference is 528.45 units.

378. The area of a circle is irr'^.

(a) Find the area of a circle whose radius is 12.34 '^•
(6) Find the radius of a circle whose area is 243.5 sq. ft.

379. The area of the surface of a sphere is ^itr\

(a) The radius of the earth is 3959 miles. What is its surface?
(6) What is the length of the equator?

380. The volume of a sphere is ^irrK What is the weight in
tons of a solid cast-iron sphere whose radius is 6.343 feet, if the
weight of a cubic foot of water is 62.355 pounds, and the specific
gravity of cast-iron is 7.154?

381. The stretch of a brass wire when a weight is hung at its
free end is given by the relation :

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:ifl2 - ••• • :•■:•- PLANE GEOMETRY

» •»• ••••• • . **■,

'\Aier€fh is 'the height applied, gf = 980, I is the length of the wire,
r is its radius, and fc is a constant. Find k for the following values :
m= 944.2 grams, Z= 219.2 centimeters, r=0.32 centimeters, and
S== 0.060 centimeters.

382. The weight P in pounds which will crush a solid cylindrical
cast-iron column is given by the formula:

P=98,9205L_^,

where d is the diameter in inches, and / the length in feet. What
weight will crush a cast-iron column 6 feet long and 4.3 inches in
diameter?

383. The weight W of one cubic foot of saturated steam depends
upon the pressure in the boiler according to the formula:

PO.941

w=-

330.36'
where P is the pressure in pounds per sq. inch. What is W if the
pressure is 280 pounds per sq. inch?

384. The number, n, of vibrations per second made by a
stretched string is given by the relation:

1 /i^

^. '21 \ m

where I is the length of the string, M the weight used to stretch the
string, m the weight of one centimeter of the string, and gf = 980.
Find n, when M= 6213.6 grams, Z=84.9 centimeters, and m=
0.00670 gram.

385. If p is the pressure and u the volume in cubic feet of 1 lb.
of steam, then from pu^'^^^=479 find u when p is 150.

The practical problems 366-369,[380-384, were taken from Rietz
and Crathorne's College Algebra (Henry Holt and Co.).

The interesting problems 372-376 were taken from White's
Scrapbook of Mathematics (Open Court Pub. Co.).

The student is referred to such texts if his interests or needs
require further work in logarithms.

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ALGEBRA IN APPLIED MATHEMATICS

103

0000

0043

0086

0128

0170

0212

0253

0294

0334

0374

0414

0453.

0492

0531

0569

0607

0645

0682

0719

0755

0792

0828

0864

0899

0934

0969

1004

1038

1072

1106

1139

1173

1206

1239

1271

1303

1335

1367

1399

1430

1461

1492

1523

1553

1584

1614

1644

1673

1703

1732

1761

1790

1818

1847

1875

1903

1931

1959

1987

2014

2041

2068

2095

2122

2148

2175

2201

2227

2253

2279

2304

2330

2355

2380

2405

2430

2455

2480

2504

2529

2553

2577

2601

2625

2648

2672

2695

2718

2742

2765

2788

2810

2833

2856

2878

2900

2923

2945

2967

2989

3010

3032

3054

3075

3096

3118

3139

3160

3181

3201

3222

3243

3263

3284

3304

3324

3345

3365

3385

3404

3424

OAAA

3464

3483

3502

3522

3541

3560

3579

3598

3617

3636

3655

3674

3692

3711

3729

3747

3766

3784

3802

3820

3838

3856

3874

3892

3909

3927

3945

3962

3979

3997

4014

4031

4048

4065

4082

4099

4116

4133

4150

4166

4183

4200

4216

4232

4249

4265

4281

4298

4314

4330

4346

4362

4378

4393

4409

4425

4440

4456

4472

4487

4502

4518

4533

4548

4564

4579

4594

4609

4624

4639

4654

4669

4683

4698

4713

4728

4742

4757

4771

4786

4800

4814

4829

4843

4857

4871

4886

4900

4914

4928

4942

4955

4969

4983

4997

5011

5024

5038

5051

5065

5079

5092

5105

5119

5132

5145

5159

5172

5185

5198

5211

5224

5237

5250

5263

5276

5289

5302

5315

5328

5340

5353

5366

5378

5391

5403

5416

5428

5441

5453

5465

5478

5490

5502

5514

5527

5539

5551

5563

5575

5587

5599

5611

5623

5635

5647

5658

5670

5682

5694

5705

5717

5729

5740

5752

5763

5775

5786

5798

5809

5821

5832

5843

5855

6866

5877

5888

5899

5911

5922

5933

5944

5955

5966

5977

5988

5999

6010

6021

6031

6042

6053

6064

6075

6085

6096

6107

6117

6128

6138

6149

6160

6170

6180

6191

6201

6212

6222

6232

6243

6253

6263

6274

6284

6294

6304

6314

6325

6335

6345

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104

PLANE GEOMETRY

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ALGEBRA IN APPLIED MATHEMATICS 105

B. RATIO, PROPORTION AND VARIATION
I. RATIO AND PROPORTION

On page 74 a ratio was defined as a fraction, and a fraction as an
indicated quotient. Hence the ratio of two numbers is their indi-
ceded quotient. We recognize in this definition that a ratio and a
fraction are identical, i.e., the ratio of 3 to 4 and the fraction | are
the same. Thus we see at the very beginning that we are not deal-
ing with a new subject, but that, having learned how to work with
fractions, we already know a good deal about ratios. The symbols
for expressing ratio we are familiar with as the symbols for expres-
sing division, viz., often used in Europe to express division, — ,
the fractional form which we shall find most convenient, and shall
use exclusively, and -^, really a combination of the two symbols
previously given.

The ratio of concrete numbers can be found only when the quan-
tities are of the same denomination. When a common unit of
measure can be foimd for two such quantities their ratio is the
quotient of the numbers expressing their measures in terms of
that unit. To find the ratio of 10 hours to 5 days, one must express
both in terms of a common unit. Here, calling 5 days 120
hours, the ratio is yW> ^^ ^^. It is not always possible to get
the ratio of two quantities, e.g., 10 minutes and 5 bushels can
have no common measure. Again, it is not always possible to
get an exact ratio, as in the case of the circumference of a circle
and its diameter. As we know, this ratio is called t, and its
value we may express more or less accurately, but can never
calculate exactly.

In a ratio the first number is called the antecedent, and the second
the consequent. It is evident, then, that the antecedent corresponds
to the numerator, and the consequent to the denominator, when
we think of a ratio as a fraction.

Since a ratio is a fraction, one fundamental property of ratios is
apparent, that is, both antecedent and consequent may be multi-
plied or divided by the same number without changing the value
of the ratio. How could you state this fact algebraically? What
principle of fractions verifies it?

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106 PLANE GEOMETRY

EXERCISES. SET XLII. RATIO

386. What is the ratio of 3 ft. 8 in. to 4 in.?

387. What is the ratio of 37i% to 87i%?

388. What is the ratio of 5x^ to (5x)2?

x^ -1

389. Simplify the ratio 7 — r-rrz

390. Simplify the ratio ^^^

391. Simplify the ratio of 7 — t-tto to , , /»

392. Find the ratio of a to 6 if 6a -76 = 3a+46.

393. Does a ratio always remain the same if a constant is added
to both antecedent and consequent? Discuss in detail.

394. What ratio is implied in the statement that the death-rate
of a certain town for the month is 4 out of each 1000?

The statement of the equality of two ratios is called a proportion.
In other words, a proportion is a fractional equation each member

0/ c

of which is a single fraction (or ratio); e.g., 7 = -j is a proportion.

The first antecedent and the second consequent of a proportion are
called the extremes, and the first consequent and the second antecedent

a c
are called the means of a proportion. In r= j, a and d are the

extremes, and b and c the means.

The antecedents and consequents in a proportion are called the

terms of the proportion. // the first consequent equals the second

antecedent, each of those terms is called a mean proportional between

the first and the last terms of the proportion; the whole expression is

referred to as a mean proportion. In the mean proportion r^",

b is said to be a mean proportional between a and c. If we solve
this equation for b two values would be obtained, =t \/ac,

EXERCISES. SET XLIII. PROPORTION

395. Find the value of y in the proportions

^""^ 4 5 • ^^^ 2/ -3 22/+6-

396. Find the mean proportionals between

(a) 16 and 4. (6) ^ and ^ (c) a+b and a —6.

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ALGEBRA IN APPLIED MATHEMATICS 107

397. What number added to each of the numbers 3, 7, 15, and
25 will give results which are in proportion?

398. In sterling silver, the amount of silver is .925 of the entire
weight of the metal, (a) How many ounces of pure silver are
needed to make 500 oz. sterling silver? (5) 500 oz. of pure silver
will make how many ounces of sterling silver? (Form a proportion
and solve this problem by means of it.)

399. The volumes of two similar solids have the same ratio as
the cubes of any two homologous dimensions. The diameter of
the first of two bottles which have the same shape is three times
the diameter of the second. If the first holds 5 ounces, how much
does the second hold?

Many properties of a proportion may be derived from its defini-
tion and the fundamental laws of algebra. We shall now suggest
proofs for three theorems of proportion which are especially im-
portant because of their application to geometry.

Theorem 28. Any proportion may be transformed by alternation^
I.e., the first term is to the third as the second is to the fourth.

Suggestion for proof: By what must we multiply r to get - ?

Theorem 29. In any proportion, the terms may be combined
by addition (usually called composition) ; i.e., the ratio of the sum
of the first and second terms to the second term {or first term)
equals the ratio of the sum of the third and fourth terms to the
fourth term (or third term).

N.B. — ^Addition and sum are used in the algebraic sense.

Given: f^^.

To prove: (1) ^ «

or ^ =

d^c
d

or (2) « «

h^a

d^c

c '

Suggestions for proofs

• (1)

a ^ c

±1.

Authority?

(2)

. b

^? Why?

c-d

' d

why does ^

-^-?'

Complete the proofs.

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108 PLANE GEOMETRY

Theorem 30. In a series of equal ratios, the ratio of the sum of
any number of antecedents to the sum of their consequents
equals the ratio of any antecedent to its consequent.

Proof: Let|sr

/. a^br, csidr, etc.

Finish the proof.

EXERCISES. SET XLIV. APPLICATIONS OF PROPORTION

400. In any proportion the product of the means is equal to the
product of the extremes.

Hint What axiom do we use in clearing an equation of fractions?
State a corollary of this theorem which will express the mean
proportional as a function of the other terms of the proportion.

401. If the product of two numbers is equal to the product of
two other nmnbers, the factors of either product may be made the
means of a proportion of which the factors of the other are the
extremes.

How are Exs. 400 and 401 related?

402. (a) If 7 = J prove that — ^ = — 3.

(6) State this fact as a theorem in proportion (sometimes re-
ferred to as addition and subtraction or composition and division).

403. (a) Prove that itl = % then - = -.

^ b d a c

(6) State this fact as a theorem in proportion. This transforma-
tion is usually referred to as inversion.

404. (a) State a brief way of testing the correctness of a
proportion. g.^^ ^^^

(6) Is the following a true proportion ^-^ = f^?

406. A and B are in business^ and their respective shares of the
business are in the ratio of f . If the profits of a certain year are
$16000, and during the year A draws $1200 and B $1000, at the
end of the year how much of the profits does each receive?

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ALGEBRA IN APPLIED MATHEMATICS 109

406. Is the validity of a proportion impaired by adding the same
number to all the terms? Prove yom* answer to be correct.

407. By the term specific gravity of a substance is meant the ratio
of the weight of a volume of that substance to the weight of an equal
volume of some other substance taken as a standard. In practice
that standard is water. A cubic foot of water weighs 62.4 lbs.

(a) What is the specific gravity of steel if a cubic foot of it
weighs 490 lbs.?

(b) What is the specific gravity of ice if a cubic foot of it weighs
57.5 lbs.?

408. What is the weight in tons of a solid cast-iron sphere whose
radius is 5.433 ft. if the weight of a cubic foot of water is 62.355 lbs.,
and the specific gravity of cast-iron is 7.154? (See Ex. 380.)

409. What number added to a ratio whose antecedent is 6 and
consequent is 8 will give the ratio |?

410. What number added to the terms of the ratio -^j will give
the ratio f ?

Many of the important applications of proportion are made in
physics. The exercises in proportion which follow depend upon
three well-known facts in physics. The facts are:

(1) Law of the Inclined Plane. If F represents the force applied
along an inclined plane, W the weight of the body, h the height of
the plane and I the length of the plane, then i^,

W^V ->^
Illustrative Problem. How heavy a weight could a force of
300 lbs. pull up an incline 75 ft. high and 400 ft. long?
Substituting: iP=300, A =75, ^=400; calculate W.

(2) Boyle's Law. If Pi be the pressure of a gas of volume Vi, and
P2 the pressure of the same gas of volume F2, the temperature re-
maining constant, we have 5^ ==, ^2

¥2" Pi

Illustrative Problem. Keeping the temperature the same, if
200 cu. cm. of gas exerting a pressure of 2 lbs. per sq. cm. be allowed
to expand to a volume of 300 cu. cm. what pressure will it exert?

Substituting: 7i=200, Pi=2, 72=300; calculate K. GooqIc

igi ize y ^ g

no PLANE GEOMETRY

(3) Charleses Law, When the pressure is constant, if Vi is the
volume of a gas of temperature <i+273 centigrade (called absolute
temperature) and V2 the volume of the same gas of temperature
6j+273, then Vi _ <i+273

■F8""fe+273'

Illustrative Problem. If 200 cu. cm. of gas is 300° C. (absolute
temperature) keeping the pressure the same, what will be the
temperature if the gas is expanded to 250 cu. cm.?

Substituting: 7i = 200, 72 = 250, <i=300 -273; calculate fe.

EXERCISES. SET XLIV (concluded)

411. Find the force which must be exerted to draw a sled weigh-
ing 240 lbs. up a hill which is 300 ft. long and 50 ft. high.

412. A bladder holds 40 cu. in. of air under a pressure of 15 lbs.
per sq. in. What is the size of the bladder when the pressure is
reduced to 12 lbs. per sq. in.?

413. A certain mass of gas occupying a volume of 160 cu. cm.
at a temperature of 47° C. is cooled to 17° C. Find the volume at
the lower temperature.

414. A boy is able to exert a maximum force of 80 lbs. How
long an inclined plane must he use to push a truck weighing 320
lbs. up to a doorway which is 3| ft. above the level of the ground?

n. VARIATION

If one variable is a constant number of times a second variable,
the first quantity is said to vary as the second. If x^ky, then x
is said to vary as t/, and is written x ccy. The .circumference of a
circle varies as its diameter because it is equal to a constant (tt)
times its diameter, i.e, C^tD or CccD.

The illustration cited is called direct variation, but of ten fti?o
quantities are said to vary inversely when an increase in the one
causes a proportional decrease in the other , e.g., the time it takes to
go from America to Europe varies inversely as the speed of the

vessel. If we call the time t and the speed 5 we might write ^ oc -^^
Again, we might illustrate this inverse variation by the apparent
height of objects and our distance from them. If a building
appears 6" high when we are 200 feet from it, how high will it
appear when we are only 100 feet from it?

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ALGEBRA IN APPLIED MATHEMATICS 111

A qumdity is said to vary jointly as several others if it is equxil to
a constant number of times the proditct of the others; i.e., a varies
jointly as c, d, and e (accede), if a^kcde. This relation may be
illustrated by the area of a triangle which varies jointly as the base
and altitude. In this case k=^. Again, we might illustrate this
relation by a man's wages which vary jointly as the number of
days he works and the pay he receives for one day's work. The
constant in this as in many other cases is unity.

EXERCISES. SET XLV. APPLICATION OF VARIATION

415. The velocity of a falling body varies directly as the time
during which it falls, (a) State this fact as a formula. (6) If
the velocity of a body is 160 feet per second after falling 5 seconds,
what will the velocity be after 12 seconds?

Notes on Solution: Such problems may be solved by first solving for the
constant or by throwing them into the form of a proportion.

Here: By the first method 160=5fc .*. A = 32 .'. t; = 3212=384,

or by the second method z-rjz = — .'. v= — '-z — =384.
•^ 160 V 5

416. The distance through which a body falls from rest varies
as the square of the time during which it falls.

(a) State this fact as a formula.

(6) If a body falls 576 ft. in 6 sees., how far does it fall in 10 sees.?

(d) How far will a body fall during the twelfth second?

(e) How long will it take a body to fall a mile?

417. The pressure of wind on a flat smf ace varies jointly as the
area of the surface and the square of the wind's velocity.

(a) State this fact as a formula.

(6) The pressure of the wind on 1 sq. ft. is 0.9 lb. when the
velocity of the wind is 15 miles per hour. What is the pressure of
the wind against the side of a house 120 feet deep and 70 feet high
when the wind is blowing 40 miles an hour?

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112 PLANE GEOMETRY

418. The heat one derives from a stove varies approximately
inversely as the square of one's distance from the stove. If I move
my position from 10 feet away from a stove to 35 feet away, what
part of the original heat will I then receive?

419. The law of gravitation states that the weight of a body varies
inversely as the square of its distance from the center of the earth.

(a) If a body weighs 10 lbs. on the surface of the earth what
will it weigh 5 miles above the siuf ace? (Consider the radius of
the earth to be 4000 miles.)

(6) How high would a body have to be raised above the siuf ace
of the earth to lose half its weight?

420. The intensity of light varies inversely as the square of the
distance from its source.

(a) How much farther from a lamp 20 feet away must a piece
of paper be moved to receive half as much light?

(b) What is the relation of the intensity of light 15 feet from an
electric light and 37 feet from the same light.

1421. Kepler proved that the squares of the times of revolution
of the planets about the sun vary as the cubes of their distances
from the sim. The earth is 93,000,000 miles from the sun, and
makes a revolution in approximately 365 days. How far is Venus
from the sim if it makes one revolution in 226 days? (Use logs.)

422. The strings of a musical instrument produce soimds by
vibrating. The number of vibrations in any fixed interval of time
varies directly as the length of the string, if the strings are alike
in other particulars.

A C string 42" long vibrates 256 times per second. A G string,
like the C string except for length, vibrates 384 times per second.
How long must it be?

423. The relation between the time of oscillation of a pendulum
and its length is given by the following formula:

If two pendulums are of lengths L and I respectively and the
number of oscillations per second are T and t respectively, then:

(a) A pendulum which makes 1 oscillation per second is 39. I*'
long. How often will a pendulum 156.4" long vibrate per second?

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ALGEBRA IN APPLIED MATHEMATICS

113

(6) How long would a pendulum have to be to oscillate once a
minute?

1424. The relation between Q, the quantity of water in cubic
feet per second passing over a triangular gauge notch, and H, the
height in feet of the siuf ace of the water above the bottom of the
notch, is given by Q oc H^

When H isl, Q is found to be 2.634. What is the value of Q
when H is 4?

If the area of the reservoir supplying the notch is 80000 sq. ft.,
find the time in which a volume of water 80000 sq. ft. in area and
3 inches in depth will be drawn off when H remains constant and
equal to 4 ft.

(The relation between Q and H may be written Q=kHi where
fc is a constant.)

425. In steam vessels of the same kind it is found that the
relation between H, the horse-power; 7, the speed in knots; and
D, the displacement in tons, is given by

HozVDl

Given H= 35640, 7=23, and D = 23000, find the probable
numerical value of H when V is 24.

426. Some particulars of steam vessels are given. Assuming in
each case the relation H.P. ex V^D^to hold, where H.P. denoted the
horse-power at a speed of V knots and displacement D in tons,
find in each case the probable H. P. necessary to give the indicated
speed.

Name

H.P.

(i) Paris

20000

20.25.

15000

(ii) Teutonic

19.50

13800

(iii) Campania

22.10

19000

(iv) K&iser

22.62

20000

(v) Oceanic

20.50

28500

(vi) Conmiunipaw

23.00

23000

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114 PLANE GEOMETRY

g427.* Assuming that the circumference of a circle is 3| times
its diameter, make a graph showing that the circumference varies
as the diameter.

Some of the numerous applications of Ratio, Proportion, and
Variation to geometry will be given or suggested in the pages of
this book.

LIST OF WORDS DEFINED IN CHAPTER V

Logarithm, base, characteristic, mantissa, interpolation, antilogarithm.
Ratio, proportion, antecedent, consequent, terms, extremes, means, mean
proportional, mean proportion, addition or composition, alternation, subtrac-
tion or division, inversion. Variation, direct, inverse, joint.

SUMMARY OF THEOREMS PROVED IN CHAPTER V

28. Any proportion may be transformed by alternation.

29. In any proportion the terms may be combined by addition.

30. In a series of equal ratios, the ratio of the sum of any number of ante-
cedents to the sum of their consequents equals the ratio of any antecedent
to its consequent.

* As here, "g^' preceding the number of an exercise indicates that its solu-
tion involves a knowledge of graphs.

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CHAPTER VI

SIMILARITY

A. INTRODUCTORY THEOREMS
EXPERIMENT I

1. Construct a scalene triangle.

(a) Divide one side into 4 equal sects, and through the first
point of division construct a line parallel to a second side of the
triangle. Compare the lengths of the sects thus cut on the third
side of the triangle.

(6) Repeat the work of (a), dividing the first side of the triangle
into 5 equal sects.

(c) Repeat the work of (a) , dividing the first side of the triangle
into 9 equal parts and drawing the line through the fourth point
of division.

2. Repeat the work of 1, using ian equilateral triangle.
Theorem 31. A line parallel to one side of a triangle, and

cutting the other sides, divides

them proportionally.

(External division of the

sides will be considered later.)

^en: A ABC, D in BC and ^ in

Jb, so that DE \\CA.
AE CD

(1)

ADEA

ADBE
(2) Similarly it can be shown that
AEDC CV

h'UE

Proof

(1) Triangles having one dimen-
sion equal compare as their remain-
ing dimensions.

ADBE

(3) But^if ADCJS:='-

-, h being

the distance between Z5B and CA,
IDE,,
2

ADEA^'

'DEWTJA.

(3) Data and ||« are everywhere
equidistant.

115

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116

PLANE GEOMETRY

(4) aadAEDC^ADEA.
ADEA AEDC

(5)

(6)

' ADBE ADBE'
EX CD

(4) Quantities equal to the same
quantity equal each other.

(5) Quotients of equals divided by
equals are equal.

(6) See (4).

Eb" Ub

Cor. 1. One side of a triangle is to either of the sects cut off
by a line parallel to a second side, as the third side is
to its homologous sect.

Sugge8tion:K = -=- why would -^---=-

Cor. 2. A series of parallels cuts off
proportional sects on all trans-
versals.

„. XB BC ,W CD —. ..^— -^

^XY^BiCu ZW^CiDu if AXY and BZW are
how drawn?

Cor. 3. Parallels which intercept equal sects on one trans-
versal, do so on all transversals. ^

Cor. 4. A line which bisects one side of
triangle, and is parallel to the
second, bisects the third. B* ^^

EXERCISES. SET XLVI. PROPORTIONAL SECTS

428. A sheet of ruled paper is useful in dividing a given sect into
equal parts, (a) Explain. (6) Make such an instrument by using
a sheet of tracing paper and drawing at least twenty parallels.
(What must you be sure that these parallels do?)

429. Draughtsmen

d^^^-^r'^^'^^^ a^d designers sometimes

b^^^ — \ \ \ divide a given sect into

j ^ X X \ X ^ Tj ^^y required number of

equal parts by the fol-
lowing method: To di-
vide AB into 5 equal

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SIMILARITY

117

parts, draw AC at any convenient angle with AB. DrawBZ)
parallel to AC. Beginning at Aj mark oS on AC five equal sects
a, 6, c, etc., of any convenient length. Beginning at B, mark oflF
on BD five sects equal in length to those on ACy ai, 61, ci, etc.
Join their extremities as in diagram. These lines divide AB
into 5 equal sects. Prove that this is a correct method.

430. Among the applications of the propositions on parallel
lines is an interesting one due to Arab Al-Nairizi (ca. 900 a. d.).
The problem is to divide a sect into any number of equal parts.
He begins with the case of trisecting a sect AB.

Make BQ and AQi perpendicular to AB, and make BP=PQ =

APi = PiQi. Then A XYZ is con-
gruent to A YBP, and also to
AXAPi. Therefore AX=Xy=
YB. In the same way we might
continue to produce BP, imtil it
is made up of n lengths BP, and
so for APiy and by properly join-
ing points we could divide AB into n+l equal parts. In par-
ticular, if we join P and Pi, we bisect; the sect AB. Prove the
truth of these statements.

Divide a sect into seven equal
parts by this method.

431. Find the cost of fencing the
field represented in the diagram.
Field is drawn to scale indicated, ^.
and the fence costs $2.75 per rod. ^ 55 100 200 rods

432. If DE is parallel to BC in triangle ABC, compute the sects
left blank from those given in the following table:

112

418

342

433. Divide a sect 11 units long into parts proportional to 3,
5, 7, and 9.

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118

PLANE GEOMETRY

First compute the lengths of the required sects, then construct
them, and measure the sects so obtained. Compare the results.
Which method is more convenient? Which is more accurate?

C 434. The accompanjdng

diagram suggests a method
for finding the distance from
a point -4 to a second point
By visible from but inacces-
sible to the first point A.
Hints: C is selected, from which both A and B are visible. CB \\ ED.

435. IfDBisparalleltoBCintriangleilBC, prove that ^=1 = ^iz=r.

AE EC

436. Under the same conditions show that

437. State Exs. 435-436 in words.

AD+AE AD

BD+EC BD

EXPERIMENT U

1. Construct a scalene triangle.

(a) Divide two sides of the triangle into 8 equal sects. Join
the corresponding points of division. By comparing certain angles,
establish a relation between the lines just drawn and the third
side of the triangle.

(6) Repeat (a), dividing the sides into 7 equal sects.

2. Repeat 1, using an equilateral triangle.

Theorem 32. A line dividing two sides of a triangle proportion-

ally is parallel to the third side. ^

BD
Given: A ABC and :=^

EE ^

^—^(DmWyEmAB),

To prove: ]5B || CJ.
Proof: Details to be sup- ^.
plied by the student.

Draw CY WW: cutting BXmX, Then := s

Why?

W^EX'
Show why X must coincide with A.

Cor. 1. A line dividing two sides of a triangle so that those
sides bear the same ratio to a pair of homologous sects
is parallel to the third side.

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119

B. IDEA OF SIMILARITY

Earlier in the book it has been noted that two figures are called
similar if they have the same shape. The symbol (<^), due to
Leibnitz, for ''is (or are) similar to," it has been pointed out, is
an S thrown on its side. The S was doubtless used because it is
the initial letter of the word ^^simUis^^ (Latin for ''like"). Before
developing the subject we need, however, a more careful definition
of similarity, for shape is only a vague notion and not a scien-
tifically defined term.

Before defining similar figures we must note the meaning of

similar sets or systems of points. The points Ai, Bi, Ci, and

A^y B2, C2, .... are said to be similar systems if they can be so
placed that all the sects joining corresponding points, AiA2y B1B2,
C1C2, . . . . , pass through the same point, and are divided by that
point into sects having the same ratio.

or D,

Flo. 1.

FiQ. 2.

In Figs. 1 and 2, points Ai, Bi, Ci, Di and A2, B2, C2, D2 are

similar systems. A1A2, B1B2, C1C2, and D1D2 pass through P and

AiP BiP C^P DiP T rv: o u or .u

dT~ — dB" — "BrT — UrT — ^- ^^ -^S- 2 where P hes on the pro-

X ii2 X jD2 ■iiy2 -t i-'2

longation of sects A1A2, B1-B2, etc., it is said to divide these sects
externally. The topic " External Division of a SecV^ will be further
developed in the "Second Study."

The point P is called the center of similitude, and the ratio r is
called the ratio of similitude.

Similar figures are those which can he placed so as to have a center
of similititde.

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120 PLANE GEOMETRY

The following are illustrations of similar figures:

v.^.

7-->0

(a) Triangles.

-~-~^^

We are now in a position to prove our right to the use of the
double symbol ( ^ ) for congruence.

Cor. Congruent figures are similar.

If n-gon ABC .... n^AiBiCi . . . . ni, they may be made
to coincide. Then any point may be taken as the center of

. OA^OB

smulitude, and == = == =

OAi OBi

Oni

= 1.

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SIMILARITY

121

EXBRCISES. SET XLVII. MEANING OF SIMILARITY

438. In (a) wha t is t he ratio of similitude? What is the center
of similitude? If OAi = A^ and OB2 = 0.5", what does BiB
equal?

439. In (a) draw a sect through which will be divided by sides
of triangles in the ratio of OAi to OA2. How many such lines can
be drawn?

44 0. In (c) if 0Bi=5 cm. and 0-62 = 1 dm. what is the ratio of
0-4 1 to 0-4.2? Can you mention any other sects in this figure
having the same ratio?

441. In (d) if OCi is ^ of C1C2 what is the ratio of similitude?
Under the same conditions what is OZ2 if OAi is J"?

Shadows fmnish familiar illustrations of similar figures,
such cases the source of light is the center of similitude.

BAn6 Descartes

The lens of the camera gives a figure similar to the object in
front of it with the image inverted.

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PLANE GEOMETRY

In reducing or enlarging maps we have another fam^Jiar illus-
tration of the application of the principle of similarity.

EXERCISES. SET XLVII (concluded)

442. State any illustrations or applications of similarity with
which you are familiar.

443. If the ratio of similitude is 1, what relation between the
figures exists besides similarity.

C. SIMILARITY OF TRIANGLES

^ * Theorem 33. The homologous angles of similar

triangles are equal, and the homologous sides

Y have a constant

z/ X ratio.

Z Given : A XYZ ^ A
ABC,
Prove: (I) 4.X = 4.ii,

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SIMILARITY

123

Proof

(1) /.AXrZc^A ABC, it maybe
placed in the position of AXiYiZi
with a center of similitude of it
and A ABC.

(1) Data and def. of sim. figs.

(2) Def. of center of similitude.

(3) Why?

(2) :.UXi_OYiOZi

'UA~'1)S"'0C

(3) .-. XTYi II IS

Y^ZiWBC
Z^iWCA

(4) .-. 2S^XiYiO = 2S^ABO (4) Why?

4.071^1 = 4- 05C and

2S^OXiYi^2S^OAB

2S^OXiZi^2S^OAC

(5) /. What two 4? of A XiYiZi (5) Why?
or A XYZ are equal to what two 4?
of A ABC?

(6) Are the third 4! equal? (6) Why?

(7) •.• ^.Z s 4. A how can A XYZ (7) Why?
be placed with respect to A ABC?

(8) What proportion will this (8) Why?
give us?

(9) In how many ways will you
have to superpose A XYZ in order
to prove (II)?

Complete the proof.

♦Theorem 34. Triangles are similar when two angles of one are
equal each to each to two angles of another.

Given: 4.P = 4.A, 4.O = ^B in

APQS and ABC.
Prove: A PQS <^ A ABC.
Suggestions for proof:

Draw PiOi \\AB and equal A^
to PQ.

Draw APiX a nd S^ IZ.

(1) lfj[PjC\\BQ[Z,thenABQiPi
is a O.

(2) a.nd AB^PQ
(3j .'. AP QS ^ AAB C.
If APiX intersects BQ^ at O,

draw PiR \\ AC and intersecting CO
&tR. Bt&wQiR.

(1) Why?

(2) Why?

(3) Why?

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PLANE GEOMETRY

/A^ rru ^0 AO .AO CO
(4) Then^=p^andp^ = ^

^^^ •• QiO RO

(6) Any sect through O cut- ^
ting PiQi in y and AB in r is

TO AO
VO^PiO

(7) .-. APiQiR ^ lABC,

(8) .-. 4.Qis4.Band4Pi3 4-A.

(9) .-. APiQiR^APQS

(10) /. APO>S may be made to
coincide with APiQiR.

divided so that 777\^Tr?\

(4) Why? ^^^^^:^C\\

(5) Why? -^^

(6) Why? X\^

(7) Why? Xf-

(8) Why? 2r

(9) Why?
(10) Def . of congruent figures.

(11) /. A PQS ^ AABC, (11) Def. of similar figures.

Discussion : Consider the instance in which PQ ^AB. Note that the figures
may be so placed that the center of similitude Ues between them. Is a proof
necessary for this case?

EXERCISES. SET XLVIII. SIMILARITY OF TRIANGLES

444. State a necessary and sufficient condition for the similarity
of (a) right triangles, (6) isosceles triangles, and (c) equilateral

triangles.

445. If rays of light from
a tree (TTi) pass through a
hole (H) in a fence (F) and
strike a wall, an inverted
outline (OiO) of the tree will
be seen on the wall.

(a) Explain why this
should be.

(6) If the distance HD is
35 ft. and ^P is 9 ft. and the height of the image (O^) is 8' 8",
find the height of the tree.

d446. The location of the image ili of a point A, formed in a
photographer's camera, is approximately found by drawing a

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SIMILARITY

125

^--1 ^

J LB

JBx

straight line AA\ through the
center of the lens L. If CE is
the position of the photographic
plate, then A\Bx is the image of
AB. How large is A\B\ if AB = 6
ft., LZ) = 12 ft., and LF = 6 in.?

447. Show that an object which
appears of a certain height, will,
when moved twice as far away,
appear to be comparatively of
only one-half the height.

Hint: Show that XS = JZB.

448. To measm-e indirectly from an accessible point A to an
inaccessible point B, construct AD perpendicular to the line of

^^ sight from -4 to B, and ED perpendi-
cular to AD. Let C be the point on
AD which lies in line with E and B.
By measurement, ED is 100 ft., CD
90 ft., and CA 210 ft. What is the
distance across the river?

449. A man is riding in an automobile at the uniform rate of
30 miles an hour on one side of a road, while on a footpath on the
other side a man is walking in the opposite direction. If the dis-
tance between the footpath and the auto track is 44 ft., and a tree
4 ft. from the footpath continually hides the chauffeur from the
pedestrian, does the pedestrian walk at a uniform rate? If so,
at what rate does he walk?

460. A mirror (referred to as a ** speculum") has been used
for crudely measuring the height of objects, such as trees.

In the diagram a mirror is placed hori-
zontally on the ground at M, The obser-
ver takes such a position that the top of
the tree (C) is visible in the mirror. What
distances must he measure to be able to
compute the height of the tree (BC)?

Note. — Light is reflected from the surface of a
mirror at an angle equal to the angle at which it strikes it.

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PLANE GEOMETRY

461. The accompanjring diagrams show a simple device for
measuring heights, using a square (such asABCD), with a plumb-
line (AE) suspended from one corner. BC and DC are divided
into equal parts (say 10, 100, or 1000).

Study the diagrams, and show how to use the square in each case.

452. Look up a description of the hypsometer, and construct one
of wood, stiff pasteboard, or any material that can be used prac-
tically.

(See D. E. Smith, " Teaching of Geometry." Ginn & Company.)

453. The distance from an

accessible point B to an in- , ^,...r^d<SI

accessible one, A, was meas-
ured in the sixteenth century
by the use of drmnheads. On
a drumhead placed at 5 a
sect I was drawn toward -4,
and another m toward an accessible point C, BC wa^ measured.
The drumhead was then placed at C with m in the direction CB.
m

The ratio of ^ was noted.

A sect p was drawn in the direction

CA» Would any further measurements be necessary to make it

possible to compute ABI Explain.
454. To a convenient scale draw
N' a symmetrical roof, pitch 7 inches
to the foot, on MN, which is to repre-
sent 30' 6". The figure suggests the
construction.

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SIMILARITY

127

455. Another way of determining the distance from A to an

inaccessible point B is to align Ay D, and B. Run DE at random.

Rim^lC parallel to DE. Align C,

Ey and B, Measure ofT FA equal

to ED. Measure CF, EF, CA.

Show how to compute AB and

justify the method.

466. The accompanying diagram

shows how coordinate (squared)

paper may be used to divide a given sect into a number of equal

parts (here 9). If the sect AB is laid
off on one of the horizontal lines, the
vertical lines will be perpendicular to it
(CB±AB). Draw AC. At the first
division on CB from C draw FE\\BA.
^CFE^^CBA. (a) Why, then, is

EF = ^? (6) What part olABi^NMt

(AB) in the diagram?
457. The principle of the diagonal scale is the same as that
imderlying the division of a sect by the method of Ex. 456. In the
accompanying diagram of a diagonal scale the unit is marked u.
What is the length of ZB, CD, EF, and MFt

■~~

—

■^

TT*

•h

-J

1L^9 Jl 7 » I ^J

468. Show how by the use of the diagonal scale to measure 0.3w,
0.56ii, 0.75ii, l.Sti, S.lii, 0.35w, 0.82u, 2.67^.

469. Draw a triangle and measure the lengths of the sides to
hundredths of an inch by the use of a diagonal scale in which ti= 1
inch. In measuring adjust the dividers to the side of the triangle,
then apply them to the diagonal scale.

460. Make a diagonal scale on tracing paper, or of stiff card-
board or of wood, and with its aid measure correct to .01 in. (a)

128 PLANE GEOMETRY

the hypotenuse of a right triangle whose legs are 1 in. and 2 in.;
(6) the diagonal of a square of side 1 in.; (c) the altitude of an equi-
lateral triangle of side 2 in.; (d) Verify each measurement by com-
putation, assupiing the Pjrthagorean Theorem.

d461. To find the height of an object AB: Place the rod CD in

^A an upright position. Stand

at X and sight over DtoA.

Move the rod any conve-

nient distance, so that it

It- rf

XG X^C,

takes the position DiCi, and
sight over Di U> A. What

measm^ments are necessary, and how may the height of the

object be determined from them?

Theorem 36. Triangles which have two sides of one propor-
tional to two sides of another and their included angles equal
are similar. yO

Given: A ABC and A AiBid with 4.C3 y^\

^^^^^"z^^W /\ aY V

To prove: A ABC eo y/^ \ y/^ \.

^A,B,Cu / \ ^X As

Suggestionsforproof: '^ ^

PlaceAAiBiCi in position of A2B2C. How do you know this is possible?

Why will A^2 be parallel to ZB?

What foUows about T^B^A^C and 4.BAC?

Can you throw this theorem back to the one immediately preceding?

EXERCISES. Set-XLVIII (continued)
462. Extend yom* arm and point to a distant object, closing
your left eye and sighting across your finger tip with your right
eye. Now keep your finger in the same position and sight with
your left eye. The finger will then seem to be pointing to an object
some distance to the right of the one at which you were pointing.
If you can estimate the distance between these two objects, which
can often be done with a fair degree of accuracy, especially when
there are buildings of which we can judge the width intervening,
then you will be able to tell approximately the distance of your
finger from the objects by the distance between the objects, for
it will be ten times the latter. Find the reason for this.

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SIMILARITY

129

463. Explain how the accom-
panying figure can be used to
find the distance from A to B

on opposite sides of a hill.

CE=\BC, CD=\AC. m is found by measurement to be 125 ft.
What is the distance ABl

464. The accompanying picture shows a pair of proportional
compasses. Note that rods AB and CD are of equal length ^ t>
and pivoted together at 0.
(a) Prove AilOCA^ BOD.

(6) Prove 5=^.
y h

(d) How may such an instrument be used to con-
struct a triangle similar to a given
triangle?

0*466. This picture shows a pair c\-^-^
of sector compasses. It can be used in
much the same way as the proportional
compasses. Show how by means of it to
get any part of a given sect.

Hint: To bisect a sect, open the compasses
so that the distance from 10 to 10 is equal to the
given sect. Then the distance from 5 to 5 equals
one-half the given sect.

What part of the given sect would the distance from 6 to 6 be?

c466. Show how by using the sector compasses to divide a given
sect into 10 equal parts.

c467. The sector compasses may be used to find the fourth
proportional to three given sects as follows: From center on OL

mark off OA equal to a. Open the a

sector imtil the transverse distance ^

at A equals h. Then if OB be marked '

off on OL equal to c, the transverse ^

distance at B is the required fourth proportional. Why?

* c is prefixed to the niunbers of exercises better suited to class dis-
cussions than to written or home work.

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PLANE GEOMETRY

Theorem 36. // the ratio of the sides of one triangle to those
of another is constant, the triangles are similar.

^Av aivt^n? A ARr^rxA X >l,R,/7. with ?5? ~ ^ «^^

a b c
To prove: A ABC ^

Given: A ABC and A AiBiCi with

Ay,

o:b:

AY^AiCi. DrawXF.

ProveAilBC'^ AAXY.
Prove AAXY ^ AAiBiCi.-

Note: — s— = - (given) and - s
di oi ci z

A AiBiCi.
Suggestions for proof:

On AB lay off ZX =

AiBi, and on -AC lay off

h c

i-3- (Why?) butysci.
z y

EXERCISES. SET XLVIII (concluded)

d468. In the figure, £, F, G, and H are the mid-points of the

sides of the square A BCD, and the points d G o

are joined as shown.

Show that the following triangles are
similar:

(a) EBC, ELB, ELY, and EUN; (b)
BLZ,BST,SindBXO; (c) EYZ and EUC;
(d) YLZ, YMB, and BHP; {e) BOY and
BHD; (/) BEZ and ATB; (g) BYZ and
BHT; (h) BYF and BHC.

d469. The following gives a procedure used in surve3dng for
running a line through a given point parallel to a wholly inacces-
sible line. Study the diagram and
notes and then justify the method.
Notes : Take C in sect MB, Select
D at any convenient place. Run MF \ \ DB.
Find E in MF in line with D and C Run
EN II AD, meeting AC at iV. Then AfiV
lUB.

Why is it that we can run paral-
lels through M and D, whereas we cannot run the one through M
directly?

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131

d470. The pantograph, invented in 1603 by Christopher
Scheiner, is an instrument for drawing a plane figure similar to a
given plane figure, and is, hence, useful for enlarging and re-
ducing maps and diagrams.

The pantograph, shown in two positions, consists of four bars so
pivoted at B and E that the opposite bars are parallel. Pencils are
carried at D and F, and A tm-ns upon a fixed pivot. BD and DE may

be so adjusted as to make the ratio of j^^ (and hence ^^and-j-^)

whatever is desired. So if F traces a given figure, D will trace a

AD
similar one, the ratio of similitude being the fixed ratio -t-«.

Fia. 1

Fza. 2

(a) Prove that Aj D, and F are always in the same straight line.
(6) Prove that -j-^ is constant and equal to -j^.

Fia. 3

Fia. 4

Note : Figs. 3 and 4 show interesting elaborations of the pantograph.

471. Summarize the conditions necessary and sufficient to make
triangles similar.

472. How may four sects be proved proportional?

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PLANE GEOMETRY

D. PERIMETERS AND AREAS OF SIMILAR TRIANGLES

Theorem 37. The perimeters of similar triangles are propor-
tional to any two homologous sides, or any
two homologous altitudes.

Given: Aabc c/y AaibiCi with h±b and hxbi.

^ a-\-b-\-c a , b c h

To prove: — , . , = — (or — or — ) s -.

Suggestions for proof:

— s — s -. Why?

Oi Oi Ci

2+h+c
' ai + bi-\-Ci

a
ai

Why?

Prove--- = — by showing A CBX w ACiBiXi,
hi ai

Cor. 1. Any two homologous altitudes of similar triangles have
the same ratio as any two homologous sides.

Theorem 38. The areas of similar triangles compare as the
squares of any two homologous sides.

Given: Aabc </» AaibiCi.
Aabc

a2 / 6« c*\

— I or rr «r -- I

To prove:

AaibiCi

Suggestions for proof:

Aabc hb h b

-rr-r-r- why?

AaibiCi
a_ ^b_

CLl bi
Aabc
AaibiCi

hibi

i = i
Ci hi

hi bi
Why?

— lor — or— J. Why?

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133

EXERCISES. SET XLIX. AREAS OF SIMILAR TRIANGLES

473. Draw a triangle. Construct a second one similar to it,
having an area nine times as great.

474. Connect the mid-points of two adjacent sides of a parallelo-
gram. What part of the area of the whole figure is the triangle
thus formed?

476. Fold a rectangular sheet of paper from one corner as shown.
The successive creases are to be equally distant from each other
and parallel. Prove that the ratio of
the successive areas between creases is
1 to 3 to 5 to 7, etc.

476. When, in forestry, shadows
cannot be used, justify the following
method of getting the height of a tree.
A staff is planted upright in the
ground. A man sights from S to the
top and foot of the tree. His assistant
notes where his Une of sight crosses
the staff.

(a) What measurements does he need to take?

(b) Assume a reasonable set of data and calculate AB.

E. APPLICATIONS OF SIMILAR TRIANGLES

Theorem 39. The altitude upon the hypotenuse of a right
triangle divides the triangle into triangles similar each to each
and to the original.

Given: Aa6cwithaJ.6andA J.c-
To prove: ABCH ^ ACHA o>
h "<^ ABC.

Suggestions for proof:

TrNC- !^^^^ ^B is common to the two

right triangles BCH and ABC,
^A is common to the two right triangles CHA and ABC.

I. PROJECTIONS
The word "projection" has a variety of meanings in general use.
We refer to projecting ourselves into a situation; or to projecting
a picture on a screen by means of a lantern. In geometry the word

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134

PLANE GEOMETRY

has a technical significance which we exemplify in the following
little experiment. Place in the sunlight a table covered with a
white cloth. Hold over the table (parallel to it) a plate of thin
glass on which small figures of dark paper have been pasted.
The sun will project these diagrams on the cloth. Upon
reflection you will note a certain relationship between the point
or sect represented on the glass and its manifestation on the cloth.

To use scientifically the idea of projection we need exact defini-
tions rather than these vague assimiptions. Let us express these
ideas in geometrical terms.

The intersection of a perpendicular to a line with that line is called
the foot of the perpendicular. The projection of a point on a line
is the foot of the perpendicular from the point to the line. The pro-
jection of a sect on a line is the sect cut off by the projections of its
extremities on thai line. For example:

-fe

F is projection of P on AB.

F S

CP is projection of CD on AB. EP is the projection of CD on AB.

Would the projection ever be as long as the original sect? Ever
longer?

Note: These projections are sometimes referred to as orthogonal to dis-
tinguish them from other types met with in higher mathematics.

The notion of projections was originally obtained from that of
shadows. The projection of a circle in one plane on another plane
was its shadow. It is evident that a scientific study of shadows
becomes very complicated. Consider, for instance, the effect on
the shadow caused by the various relative positions of the planes
and the positions of the light. Projective geometry, an advanced

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SIMILARITY

135

study, concerns itself with the more complex phases of the subject.

In elementary geometry, we refer only to one very small instance

of shadow geometry, reduced, as you note, to geometric definitions.

These projections we refer to as right or orthogonal.

Cor. 1. Each side of a right
triangle is a mean propor-
tional between the hypote^
nuse and its projection
upon the hypotenuse.

Note: aBCH eo aCBA. Form a B"
proportion involving BG, BZ, BE.

Cor. 2. The square of the hypotenuse of a right triangle is
equal to the sum of the squares of the other two sides.

I^ote: a*=c.BHy&iidb*=C'HA. Why?

This fact is very important in geometry, and has an interesting
history. The first proof of the theorem is attributed to Pythagoras
about 500 B. c, although the fact was known much earlier.
Reference has already been made to the history of the theorem in
Chapter I. Its later development consists of numerous proofs
worked out by later mathematicians. In the following exercises
specimens of such will be found, and further interesting proofs are
contained in Heath's Monograph, "The Pythagorean Theorem."

EXERCISES. SETL. PROJECTIONS. PYTHAGOREAN RELATION

477. What would be the projection of a sect 10" long on a line
with which it makes an angle of (a) 30°? (6) 45°? (c) 60°?

478. In the sixteenth century the distance from A to the inac-
cessible point B was determined by means of an instrument called

the "squadra." The squadra, hke
a modern carpenter's square, con-
sisted of two metallic arms at right
angles to each other. To measure
AB the squadra was supported, as
in the figure, on a vertical stafiF
AC. One arm was pointed toward
B, and the point D on the ground, at which the other arm

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136

PLANE GEOMETRY

pointed, was noted. By measuring AD and AC, show how AB
may be computed.

d479. Two stakes are set on a hillside whose slope is 20% (i.e.,
20 ft. rise in 100 ft. measured along the slope). The distance
between the stakes, measured along the slope, is 458 ft. What is
the horizontal distance between them?

480. The accompan3dng drawing
represents a plot of land divided as
indicated. Z)£=22'8", £B=100',
BF = 50', FG (alley) = 16', GH = 150',
HI = 66', and IK = 232'6". Find the
length of AB in feet, and the area of
triangle ACD in square rods.
d481. The figure shows a ground plan of a zigzag or "worm"
fence. The rails are 11 feet long, and a lap of 1 foot is allowed
at each corner. Stakes, supporting
the rider rails, are set along the
boundary Une. Find the amoimt
of ground wasted by the construc-
tion of such a fence 100 rods long.

How much more fence is needed in this zigzag fence than in a
straight one?

d482. Let ABC be any right-angled triangle, right-angled at C,
and let the square ABDE be described on the hypotenuse AB,
overlapping the triangle. Prove that the perpendicular from E
upon AC is oi length b, and hence that the area of the triangle
ACE is 16^. Similarly, prove that the area of the triangle BCD
is ^a^. Notice that these two triangles have equal bases c and
total height c. Hence prove that a^+b^=c^.

483. The great Hindu mathematician,
Bhaskara (born 1114 a. d.), proceeds in a
somewhat similar manner. He draws this
figure, but gives no proof. It is evident that
he had in mind this relation:

/i2^4~-|-(6-a)2=a2+62.

Boundary

120°

Lihe

Give a proof.

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SIMILARITY

137

d484. A somewhat similar proof can be based upon the following
figure:

If the four triangles, 1+2+3+4, are taken
away, there remains the square on the hypo-
tenuse. But if we take away the two shaded
rectangles, which equal the four triangles, there
remain the squares on the two sides. Therefore
the square on the hypotenuse equals the sum of these two squares.
Give details of the proof.

d486. This exercise makes the Pythagorean Theorem a special
case of a proposition due to Pappus (fourth century a. d.), relating
to any kind of triangle.

Somewhat simplified, this proposition asserts that if ABC is
any kind of triangle, and MC, NC are parallelograms on AC, BC,

the opposite sides being produced
to meet at P; and if PC is produced
making QR=PC; and if the paral-
lelogram AT 18 constructed, then
AT^MC+NC.

For MC^AP=AR, having equal
altitudes and bases.
Similarly, NC=QT. Adding, MC+NC=AT.
If, now, ABC is a right triangle, and if MC and NC are squares,
it is easy to show that ^T is a square, and the proposition reduces
to the Phythagorean Theorem. Show this.

d486. The Arab writer, Al-Nairizi (died about 922 a.d.), attri-
buted to Thabit ben Korra (826-901 a.d.) a proof substantially
as follows:

The four triangles T can be proved congruent.
Then if we take from the whole figure T and
Ti, we have left the squares on the two sides
of the right angle. If we take away the other
two triangles instead, we have left the square
on the hypotenuse. Therefore the former is
equivalient to the latter. Give details of proof.

d487. A proof attributed to the great artist, Leonardo da Vinci
(1452-1519), is as follows:

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PLANE GEOMETRY

The construction of the following figure is evident. It is easily-
shown that the four quadrilaterals
ABMX, XNCA, SBCP, smdSRQP
are congruent.

.-. ABMXNCA equBlsSBCPQRS
but is not congruent to it, the
^if congruent quadrilaterals being dif-
ferently arranged.

Subtract the congruent triangles
MXN, ABC, RAQ, and the pro-
position is proved.
Give details of proof.

d488. A proof attributed to President Garfield is
suggested by the accompanying diagram. Work it out .

Note: asai, h^hi, csci. il BDC is a trapezoid. What is
the altitude of the trapezoid? Its bases? Its area? How
else may the area of the trapezoid be found?

d489. Show that if AB=a (in Fig. 3),

(6) SL^^x/B. {d)LY^^y/b. {j)ZBJ^V2.{h)ZY^-^Vb.

490. In the middle of a pond 10 ft.
square grew a reed. The reed projected
1 ft. above the surface of the water.
When blown aside by the wind, its top
part reached to the mid-point of a side
of the pond. How deep was the pond?
(Old Chinese problem.)

491. Show that the following dia-
grams illustrate methods of representing

Moorish Design, from Mabel the squarc roots 01 integers.

Sykes* Source Book of Problems

for Geometry. B 1 C

AB=BC=1. _ '

AC=BD = y/2.
AD = BE =V3, etc.

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SIMILARITY

139

AC is a square.
BD=BG=\/2.
AG=AE^VZ, etc.

Note: Such methods would not be
practicial for large numbers Why? JB

The first of these methods is used on the "line of squares" on
the sector compasses.

492. The Hindus said that triangles having the following sides
are right triangles. How is the assumption they apparently made
related to Theorem 40, Cor. 2? (a) 5, 12, 13. (6) 15, 36, 39.
(c) 8, 15, 17. (d) 12, 35, 37.

n. TRIGONOMETRIC RATIOS

Let ABC be an acute angle. Drop a
series of perpendiculars to BA from
any points on BC-

It will readily be noted that the right
triangles formed, BPiRi, BP2R2, etc.,
E^ E^ R^ R^ Ri, ^are similar, having the angle B in com-
mon. The equality of the following ratios will result:

,^x ^./j-. i , _ RJP2 _RzPz __ ^^t- * __

BPx BP2 BPi

R4PA

BP,

^^=l^and(3) 1^=

RiPi _
BRi'

JLtsx 5 ,fy. BR\ BR2 BRz

R2P2 RzPz R^Pa RbPb

Why?

BPs BRi BR2 BRz BRi BRs

If we think of ^ABC as being generated by sect BC revolving
counterclockwise from the position BA to BC we may call BA the
initial side and BC the terminal side. These perpendiculars from
points on the terminal side may be thought of as projectors form-
ing on the initial side the projections of sects of the terminal side.
We may then summarize the facts given as ratios in what preceded
as follows: For any acute angle if perpendiculars be dropped to the
initial side from any points on the terminal side the ratios (1) of
the projector to the sect of the terminal side, (2) of the projection to the
sect of the terminal aide, and (3) of the projector to the projection of a
sect of the terminal upon the initial side are constants. These ratios

are given the names sine, cosine, and tangent, respectively. ,

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140

PLANE GEOMETRY

IniJtial Side

C Thus (1) the sine of angle ABC is=^-

■DJ

(2) The cosine of angle ABC is ==•

-^ (3) The tangent of angle ABC is =•

These are referred to as trigonometric ratios of an angle.

If in the fixed A ABC, right angled
at C, the side opposite A is called o,
the side adjacent, a, and the hypote-
nuse, h, fill in the following:
"C sin -4 = ? cos -4 = ? tan -4 = ?

EXERCISES. SET LI. TRIGONOMETRIC RATIOS

493. Make a table of the values of the
sines, cosines, and tangents of angles of 30°,
46^ and 60^

494. Showbymeans of similiar trianglesthat
the sine of 60° is the same as the cosine of 30°.

d496. When a wagon stands upon an incline, its weight is
resolved into two forces, one the pressure against the incline, the

other tending to make it run down the
incline. Show that the force along the
incline is to the weight of the wagon as
the height of the incline is to its length.
If the incline makes a 30° angle with the
horizontal, with what force does a loaded
wagon, weighing three tons, tend to run down the incline, i.e., disre-
garding friction, what force must a team exert to pull it up the slope?
496. Fill out the following table:

Polygon

Dimensions

Perimeter

Area

Parallelogram

Base 18
Angle 60°

300

Rectangle

Base 18

300

Rhombus

Angle 60°

300

Square

300

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SIMILARITY

141

497. To measure the height of an object AB by drawing to scale:
Measure a distance CD towards A. Measure the angle ACB

and the angle J.Z)B. Then draw a plan thus: Representing, say,

100 ft. by a sect an inch long, draw

EF to represent CD in the plan,

and draw the angle HFG equal to

the angle ADB, and the angle

FEG equal to the angle ACB, and

draw GH at right angles to EF

prolonged. Measure GH.

Show how to compute -4B.

(a) If CD = 175 ft., the
angle ADB = 45'', and
angle DCB =30'', compute ^ ^
AB.

(6) Draw a diagram to scale and compare the result with that
obtained from yom* calculations.

498. If a survey is made, using a 100 ft. tape, and on a hill,
the lower chainman holds his end of the tape 2 ft. too low:

(a) What error will be caused in one tape length?
(6) If the distance between two stations on the hillside is
recorded as 862 ft., what is the actual distance?

(c) If the problem is to layoff adistanceof
900 ft., what is the actual distance laid off?
499. A ladder 30 ft. long leans against
the side of a building, its foot being 15 ft.
from the building. What angle does the
ladder make with the ground?

600. In order to
find the width of a
river, a distance AB
was measured along
the bank, the point A being directly opposite
a tree C on the other side. If the angle ABC
was observed to be GO'', and AB 100 ft., find
the width of the river.

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PLANE GEOMETRY

601. A house 30 ft. wide has a gable roof whose rafters are 20
ft. long. What is the pitch of the roof? (The pitch is the angle
between a rafter and the horizontal.)

602. A barn 60 ft. wide has a gable roof whose rafters are 30\/2
ft. long. What is the pitch of the roof, and how far above the

eaves is the ridgepole?

The angle of elevation is the angle between
the ray of light from the object to the eye and
the horizontal line in the same planej when the
object is above the horizontal line. When the
object observed is below the horizontal line^ the
angle is called the angle of depressioxi*

For example:

Eye

Horizontal

Bye

Object

Horizontal

EXERCISES. SET LI (continued)

603. At a point 200 ft. in a horizontal line from the foot of a
tower the angle of elevation of the top of the tower is observed to
be 60**. Find the height of the tower.

604. The vertical central pole of a circular tent is 20 ft. high,
and its top is fastened by ropes 40 ft. long to stakes set in the
ground. How far are the stakes from the foot of the pole, and
what is the inclination of the ropes to the ground?

606. At a point midway between two towers on a horizontal
plane the angles of elevation of their tops are 30° and 60° respec-
tively. Show that one tower is three times as high as the other.

606. A flagstaff 25 ft. high stands on the top of a house. From
a point on the plane on which the house stands, the angles of
elevation of the top and the bottom of the flagstaff are observed
to be 60° and 45° respectively. Find the height of the house.

607. A man walking on a straight road observes at one mile-
stone a house in a direction making an angle of 30° with the road,
and at the next milestone the angle is 60°. How far is the house
from the road?

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SIMILARITY

143

608. Find the number of square feet of pavement required for
the shaded portion of the streets
shown in the figure, all the streets
being 60 ft. wide.

It is not possible to determine the
trigonometric ratios of angles other
than 30°, 45°, and 60° by elementary
plane geometry. By the use of the protractor any acute angle can
be drawn, and with a ruled edge the sects needed may be meas-
lu^ and approximations may be made for the ratios. The values
correct to many decimal places have been scientifically worked
out and tabulated. A table correct to four places follows for
use in subsequent problems. Corrections for fractions of min-

utes may

be made as in the case of logarithmic

tables

Deg.

Sine

Cosine

Tangent

Deg.

Sine

Cosine

Tangent

Deg.

Sine

Cosine

Tangent

.0175

.9998

.0176

.5150

.8572

.6009

.8746

.4848

1.8040

.0349

.9994

.0349

.5299

.8480

.6249

.8829

.4695

1.8807

.0523

.9986

.0524

.5446

.8387

.6494

.8910

.4540

1.9626

.0698

.9976

.0699

.5592

.8290

.6745

.8988

.4384

2.0503

.0872

.9962

.0875

.5736

.8192

.7002

.9063

.4226

2.1445

.1045

.9945

.1051

.6878

.8090

.7265

.9135

.4067

2-2460

.1219

.9925

.1228

.6018

.7986

.7536

.9205

.3907

2.3559

.1392

.9903

.1405

.6157

.7880

.7813

.9272

.3746

2.4751

.1564

.9877

.1584

.6293

.7771

.8098

.9336

.3584

2.6051

.1736

.9848

.1763

.6428

.7660

.8391

.9397

.3420

2.7475

.1908

.9816

.1944

.6561

.7547

.8693

.9455

.3256

2.9042

.2079

.9781

.2126

.6691

.7431

.9004

.9511

.3090

3.0777

.2250

.9744

.2309

.6820

.7314

.9325

.9563

.2924

3.2709

.2419

.9703

.2493

.6947

.7193

.9657

.9613

.2756

3.4874

.2588

.9659

.2679

.7071

1.0000

.9659

.2588

3.7321

.2756

.9613

.2867

.7193

.6947

1.0355

.9703

.2419

4.0108

.2924

.9563

.3057

.7314

.6820

1.0724

.9744

.2250

4.3315

.3090

.9511

.3249

.7431

.6691

1.1106

.9781

.2079

4.7046

.3256

.9455

.3443

.7547

.6561

1.1504

.9816

.1908

5.1446

.3420

.9397

.3640

.7660

.6428

1.1918

.9848

.1736

5.6713

.3584

.9336

.3839

.7771

.6293

1.2349

.9877

.1564

6.3138

.3746

.9272

.4040

.7880

.6157

1.2799

.9903

.1392

7.1154

.3907

.9205

.4245

.7986

.6018

1.3270

.9925

.1219

8.1443

.4067

.9135

.4452

.8090

.5878

1.3764

.9945

.1045

9.5144

.4226

.9063

.4663

.8192

.5736

1.4281

.9962

.0872

11.4301

.4384

.8988

.4877

.8290

.5592

1.4826

.9976

.0698

14.3006

.4540

.8910

.5095

.8387

.5446

1.5399

.9986

.0523

19.0811

.4695

.8829

.5317

.8480

.5299

1.6003

.9994

.0349

28.6363

.4848

.8746

.5543

.8572

.5150

1.6643

.9998

.0175

57.2900

.6000

.8660

.5774

.8660

.5000

1.7321

1.0000

.0000

144 PLANE GEOMETRY

In trigonometry a more extended study of these ratios will be
given. The ratios of angles of more than 90° will be considered,
and other ratios which are constant will be developed.

Work involving calculations with the trigonometric ratios is
often simplified by the use of tables of logarithmic functions. For
this purpose, and for greater facility in the use of logarithms in
general, and in the use of the natural functions, it would be to the
pupil's advantage to procure a compact volume of tables. An
excellent book for this purpose, costing only twenty cents, is
Prof. A. Adler, Fiinfstellige Logarithmen (Sammlung Goschen).

EXERCISES. SET LI (concluded)

609. The sect AB 15 inches long makes an angle of 35° with the
line OX. Find its projection on OX. Find its projection on the
line UY perpendicular to OX and in the same plane as OX and AB»

510. What is the angle of the sun's altitude if the shadow of a
telegraph pole 30 ft. high is 40 ft. long?

611. A tower is 6 15 ft. high. How large an angle does it subtend
at a point which is 1^ mi. away and on the same horizontal plane
as its base?

612. A mariner finds that the angle of elevation of the top of a
cliff is 1&*, He knows from the location of a buoy that his distance
from the foot of the cliff is half a mile. How high is the cliff?

613. At 40 ft. from the base of a fir tree the angle of elevation
of the top is 75°. Find the height of the tree.

614. A flagstaff 75 ft. high casts a shadow 40 ft. long. Find the
angle of elevation of the sun above the horizon.

616. To find the distance across a lake from a point A to a
point B, a man measured 100 rods to a point
C on a line perpendicular to the line AB,
and foimd that the angle ABC was 50°.
How could he find the distance across the
lake? What is the distance?

616. What is the angle of slope of a road
bed that has a grade of 5 per cent? One with
a grade of 25 hundredths per cent? (By "a grade of 5 per cent"
is meant a rise of five feet in a horizontal distance of one hun-

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SIMILARITY

145

dred feet. By "the angle of the slope'' of such a grade is meant
the angle whose tangent is 0.05.)

617. A steamer is moving in a southeasterly direction at the
rate of 25 miles an hour. How fast is it moving in an easterly
direction? In a southerly direction?

518. A balloon of diameter 50 ft. is directly above an observer
and subtends a visual angle of 4*^. What is the height of the
balloon?

d619. The angle of elevation of a balloon from a point due south
of it is 60°, and from another point 1 mile due west of the former,
the angle of elevation is 45°. Find the height of the balloon.

620. Wishing to determine the width of a river, I observed a
tree standing directly across the bank. The angle of elevation
of the top of the tree was 32°; at 150 ft. back from this point, and
in the same direction from the tree, the angle of elevation of the
top of the tree was 21°. Find the width of the river.

621. A tree is standing on a bluff on the opposite side of the
river from the observer. Its foot is at an elevation of 45°, and its
top at 60°. (a) Compare the height of the bluff with that of the
tree (i.e., find the ratio), (b) What measurement would you use
to find the height of the tree? (c) The height of the bluff? (d)
The width of the river?

d622; Two men are lifting a stone by
means of ropes. As the stone leaves the
ground one man is pulling with a force of
85 lbs. in a direction 25° from the vertical,
while the other man is pulling at an angle
of 40° from the vertical. Determine the
weight of the stone.

623. A 60 ft. pole stands on the top of
a mound. The angles of elevation of the
top and the bottom of the pole are
respectively 35° and 62°. Find the height of the mound.

624. From the top of a mountain 1050 ft. high two buildings
are seen on a level plane, and in a direct fine from the foot of the
mountain. The angle of depression of the first is 35°, and of the
second is 21°. Find the distance between the two buildings.

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PLANE GEOMETRY

626. Certain lots in a city are laid out by lines perpendicular

to B Street and running
through to A Street, as
shown in the figure. Find
the widths of the lots on A
Street if the angle between
the streets is 28*" 40'.

626. In surveying
around an obstacle meas-
urements were taken as
shown in the figure.
Find the distance on
a straight line from
A to E. (Use log- ^.
arithmic tables.)

627. With data,
find the length and bearing of DF, a proposed street. (Use
logarithmic tables.) jy

li4.62

1^ 28-80-66.2 E

d628. Look up and explain the principle of the Vernier. (Lock
and Child, Trigonometry for Beginners (Macmillan), is a good
reference for this point — pp. 120-126).

629. The shadow of a vertical 10 ft. pole is 14 ft. long. What
is the angle of elevation of the sun?

630. The tread of a step on a certain stairway is 10" wide; the
step rises 7" above the next lower step. Find the angle at which
the stairway rises.

631. The width of the gable of a house is 34 ft. The height of
the house above the eaves is 15 ft. Find the length of the rafters
and the angle of inclination of the roof.

632. Find the angle between the rafter and horizontal in the
following pitch of roof: two-thirds, one-half, one-third, one-fourth.

533. Two trees M and N are on opposite sides of a river. A line
NP at right angles to MN is 432.7 ft. long, and the angle NPM is
52° 20'. What is the distance from ikf to iV? (Use logarithmic tables.)

SIMILARITY 147

684. In an isosceles triangle one of the base angles is 48® 20', and
the base is 18". Find the legs, the vertical angle, and the altitude
drawn to the base.

636. To find the height of a tower, a distance of 311.2 ft. was
measiu'ed from the foot of the tower, and the angle of elevation of
the tower was found to be 40*^ 67'. Find the height of the tower.
(Use logarithmic tables.)

636. Find the shorter altitude and the area of a parallelogram
whose sides are 10' and 25' when the angle between the sides is 74*^ 33'.

d637. The angle of elevation of the top of a spire from the third
floor of a building was 35° 12'. The angle of elevation from a
point directly above, on the fifth floor of the same building, was
25*^ 33'. What is the height of the tower and its horizontal distance
from the place of observation, if the distance between consecutive
floors is 12 ft., and the first floor rests on a basement 6 ft. above
the level of the street?

638. (a) What size target at 33' from the eye subtends the same
angle as a target 3' in diameter at 987 yds.?

(6) Find the angle it subtends.

639. The summit of a mountain, known to be 14,450 feet high,
is seen at an angle of elevation of 29° 15' from a camp located at
an altitude of 6935 feet. Compute the air-line distance from the
camp to the summit of the mountain. (Use logarithmic tables.)

d640. Two towns, A and 5, of which B is 25 miles northeast of
A, are to be connected by a new road. Ten miles of the road is
constructed from A in the direction N. 23° E. What must be the
length and direction of the remainder of the road, assiuning that it
follows a straight line?

641. A car track rims from -4 to -B, a horizontal distance of 1275'
at an incUne of 7° 45', and then from -B to C, a distance of 1586'.
C is known to be 509' above 4. What is the average inclination of
the track from B to C? (Use logarithmic tables.)

4642. On a map on which 1" represents 1000', contour lines
are drawn for differences of 100' in altitude. What is the actual
inclination of the surface represented by that portion of the map
at which the contour lines are J" apart?

d643. The description in a deed runs as follows: Beginning at a
stone (A), at the N. W. corner of lot 401; thence east 112' to a

148 PLANE GEOMETRY

stone (J5); thence S. Se.S*" W. 100'; thence west parallel with AS to
the west line of said lot 401; thence north on the west line of said
lot to the place of beginning. Find the area of the land described.

LIST OF WORDS DEFINED IN CHAPTER VI

Similar sjrstems (or sets) of points, center of similitude, ratio of similitude,
similar figures. Projection of a point, and of a sect, on a line; projector. Initial
and terminal side of an angle. Trigonometric ratios; sine, cosine, tangent of
an angle. Angle of elevation, angle of depression.

SUMMARY OF THEOREMS PROVED IN CHAPTER VI

31. A line parallel to one side of a triangle, and cutting the other sides,
divides them proportionally.

Cor. 1. One side of a triangle is to either of the sects cut off by a line
parallel to a second side, as the third side is to its homo-
logous sect.
Cor. 2. A series of parallels cuts off proportional sects on all trans*

Cor. 3. Parallels which intercept equal sects on one transversal,

do so on all transversals.
Cor. 4. A line which bisects one. side of a triangle, and is parallel to

the second, bisects the third.

32. A line dividing two sides of a triangle proportionally is parallel to the
third side.

Cor. 1. A line dividing two sides of a triangle so that those sides
bear the same ratio to a pair of homologous sects is paral-
lel to the third side.

33. The homologous sides of similar triangles have a constant ratio, and
their homologous angles are equal.

34. Triangles are similar when two angles of one are equal each to each
to two angles of another.

35. Triangles which have two sides of one proportional to two sides of
another and the included angles equal are similar.

36. If the ratio of the sides of one triangle to those of another is constant,
the triangles are similar.

37. The perimeters of similar triangles are proportional to any two homolo-
gous sides, or any two homologous altitudes.

Cor. 1. Homologous altitudes of similar triangles have the same ratio
as homologous sides.

38. The areas of similar triangles compare as the squares of any two homol-
ogous sides.

39. The altitude on the hypotenuse of a right triangle divides the triangle
into triangles similar to each other and to the original.

Cor. 1. Each side of a right triangle is a mean proportional between
the hypotenuse and its projection upon the hypotenuse.

Cor, 2, The square of the hyx)otenuse of a right triangle is equal to
the sum of the squares of the other two sides. ^

CHAPTER VII

THE LOCUS '

A. REVIEW OF THE IDEA OF LOCUS AS MET WITH IN

ALGEBRA

L REVIEW AND SUMMARY OF ESSENTIAL POINTS IN THE
INTRODUCTION TO GRAPHIC MATHEMATICS

a. Location of Points.

In locating places on a map we are accustomed to noting their
longitude and latitude, which means that we refer to their distances
north or south of the equator, and east Or west of some meridian.
So we may locate points on a piece of paper by stating their dis-
tance up or down from some fixed line of reference, and to the right
or left of some other line of reference at right angles to the first.

These lines of reference are called the axes, the distances up or
down are called the ordinates, and those to the right and left the
abscissas of the points. The ordinaie and the abscissa of a point
are together called its coordinates. Paper ruled off in squares is
used for convenience in counting, and in locating points. Such
paper is called coordinate paper.

The abscissa of a point is given first, followed by the ordinate.
Plus or minus are used in the case of the abscissa to denote distance
to the right or left of the so-called y-axis; plus or minus, in the
case of the ordinate, denote distance above or below the so-called
X-axis. The intersection of the axes is called their origin.

The coordinates of a point are written in a parenthesis with a
comma between them; e.g. (5,-2) refers to a point 5 imits to the
right of the y-axis and 2 imits below the x-axis.

EXERCISES. SET LII. LOCATION OF POINTS
644. With reference to a single pair of axes, plot the following
points on a sheet of coordinate paper:
(4,5),(-2,6), (-2,-6), (5,-2).

646. On the same sheet plot also the points: (3, i), (—3, — «),
(6, -I), (-3, -i).
646. Locate the points: (2, 0), (-5, 0), (0, 5), (0, -|), (0, 0).

140

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ISO PLANE GEOMETRY

647. (a) All the points on the x-axis have what ordinate?
(6) All the points on the y-axis have what abscissa?

648. Construct the triangle whose vertices are (1, 1), (2, -2), (3,2).

649. Construct the quadrilateral whose vertices are (2, -1),
--4,-3), (-3,5), (3,4).

660. Construct the rectangle whose vertices are (—3, 4), (4, 4),
-3, -2), (4, -2), and find its area.

661. Construct the triangle whose vertices are (—3, -4),
(—1, 3), (2, -4), and find its area.

b. The Graph and its Applications.

The line connecting a series of points plotted as explained is called a
graph. Graphs are useful for giving information quickly, in making
estimates, and in the solution of many problems such as those involv-
ing time and distance. We have all seen the charts of trained
nurses, and newspaper and magazine reports given in graphic form.

Among the numerous applications of the graph, then, we may
list (1) records of statistics, (2) ready reckoners which furnish
bases of interpolation or give convenient diagrams, (3) repre-
sentations of formulas which make quick approximations possible.

EXERCISES. SET LIII. APPLIED PROBLEMS IN GRAPHIC
MATHEMATICS

662. Observe the readings of the same thermometer at the same
hours daily for a week, and record the results of your observations
graphically.

663. A boy who can throw a stone from a sling shot with a

velocity of 80 ft. per second is experimenting. He finds that

when he throws it in a direction making an angle of 16*^ with the

ground it pitches 35 yds. away. This and other results are given

in the table below:

Angle (in degrees) 16 24 32 40 48 56 64 72 80

Distance (in yards) 35 .50 60 65 66 61 52 39 22

(a) Draw a graph to represent these facts.

(jb) Find (1) how far he can throw when the angle is 60®.

(2) what angle will produce a throw of 57 yds.

(3) what is the greatest distance the boy can throw,

and what angle will produce this.

664. The vertices of a pentagonal field are located by the fol-
lowing points, A = (-20, 15), 5= (10, 20), C = (23, -20), D =

(-10, -30), J?=(-30, -10). Digitized by Google

THE LOCUS 151

(a) Draw the outline of the field.

(b) Give new values io A, B, C, D, E, so that the area shall
remain the same, but the diagram lie wholly in the first quadrant,
with E on the north-south axis, and D on the east-west axis.

666. The boiling-point of water on a Centigrade thermometer
is marked 100°, and on a Fahrenheit 212°. The freezing-point on
the Centigrade is zero, and on the Fahrenheit is 32°. Conse-
quently a degree on one is not equal to a degree on the other.

(a) Show that the correct relation is expressed by the equation
C=l (F-32), where C represents degrees Centigrade, and F
degrees Fahrenheit.

(b) Construct a graph of this equation. Can you, by means
of this graph, express a Centigrade reading in degrees Fahrenheit,
and vice versa?

(c) By means of the graph express the following Centigrade
readings in Fahrenheit readings, and vice versa: (1) 60° C;
(2) 150° F.; (3) -20° C; (4) -30° F.

(d) What reading means the same temperatiu'e on both scales?

c. The Graph of Equations.

If in such an equation as x+y = 10 various values of z are taken
as abscissas of points whose ordinates are the corresponding values
of y, and the points are joined, we have what is known as the graph
of the equation. It is a fact which is proved in more advanced
mathematics that the graph of an equation of the first degree is
always a straight line.

Thus, if we represent graphically such a system of equations as
x+y = 10, and x-y=4:, we have two straight lines. The coordinates
of their intersection will give the solution of the equation. Why?

EXERCISES. SETLIV. GRAPHIC SOLUTION OF EQUATIONS

Solve the following systems of equations graphically.

666. x+4y = 1 1 669. 2x-9y = 23 662. 2x -3t/ = 7

2x-T/ = 4 5x+y= -IZ 5x-7t/ = 14

657. 2x+3y = 19 660. x+5y = 663. 6x-3t/ = 16

7x-2y=4: 3x+9y= -6 2x+7y = ^5

668. x+5y= -3 661. 7x+2y==U

2x^3y = 20 5a: -32/= -21 „^,,,,,^GoOgle

152 PLANE GEOMETRY

11. APPLICATION OP ELEMENTARY GRAPHIC MATHEMATICS TO

GEOMETRY

Since we have studied the graphic solution of simultaneous
equations, the idea of locus (plural, loci) is not an entirely new one.
In our graphic work we found that the locus, obeying the law
expressed by a linear equation, was a straight line.*

In our graphic work we find that all points +2 units from the
a;-axis are to be found on a line parallel to the x-axis and 2 units

above it; likewise we found that
all points in this line, no matter
how far it may be extended, will
be +2 units from the a:-axis.
Another way of expressing these

Xi \. a* -axis « facts is to say that the path of

all points the ^/-value of which is
2, is the line parallel to the x-axis
y^ and 2 units above it, and next,

that the y-value of every point in
the line parallel to the a:-axis and 2 units above it is +2. This is
stated algebraically by means of the equation 2/ = 2.

EXERCISES. SET LV. THE EQUATION AS THE STATEMENT OF

A LOCUS

564. Where are all points -2 units from the x-axis to be found?
(Answer in a complete sentence.)

666. What can you say of all points in the line described in
your answer to the last question? (Answer in a complete sentence.)

666. State the law which is obeyed by the line described in
exercise 664 by means of an algebraic equation.

667. Where are all points + 10 imits from the 2/-axis to be found?

668. What can you say of all points in the line described in
yoiu" answer to the last question?

669. State this law by means of an algebraic equation.

670. Answer the last three questions, inserting the following
words in place of "+10 units from the T/-axis":

* In this work, those for whom it is not review will find Auerbach, An Ele-
mentary Course in Graphic Mathematics, Chapter I and Chapter III, pp.
22, 23, and 28-31, helpful.

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THE LOCUS 163

(a) -15 units from the 2/-axis. (6) +12 units from the x-axis.
(c) -18 units from the a:-axis. (d) -7 units from the 2/-axis.
(e) +13xuiitsfrom they-axis.

671. What is the y-vahie of every point in the line parallel to
the a;-axis and +6 units from it? -6 units from it?

672. What is the x-value of every point in the line parallel to
the 2/-axis and +17 units from it? -17 units from it?

673. What is the path of every point whose 2/-value is -f-18?
-18? +20? -3?

674. What is the path of every point whose x-value is 30? -27?
+16? -8?

676. Make a list of the equations expressing the facts stated,
in order, in the last four questions.

676. (a) What is the 2/-value of every point in the x-axis?
(6) What is the path of every point whose 2/-value is this?
(c) What, then, is the equation of the x-axis?

677. a: = 19 expresses algebraically what two facts?

678. What is the equation of the 2/-axis? Why?

679. What is the equation of the parallel to the 2/-axis through
the point ( -5, 7)?

The arrangement of points that completely fulfills a given geometric
condition is called the locus of that condition. This arrangement
usually gives rise to a line or group of lines either straight or curved.
For instance, the locus of the condition expressed by the equation
a; = 7 is the line drawn parallel to the T/-axis at a distance 7 units
to the right of it. This is a brief way of saying that (1) all points
in this line are 7 xmits to the right of the y-axts and (2) all points
7 units to the right of the y-axis he in this line.

Because of the idea of motion involved, another acceptable
definition of the word locus would be : The complete path of a
point that moves in accordance with some specified geometric condition.
For instance, the complete path of a point that moves so that its
distance from the y-axis is -7 is the line 7 units to the left of the
2/-axis and parallel to it. Hence this line is called the locus of the
point which moves so as to remain constantly 7 imits to the left
of the y-axis.

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154 PLANE GEOMETRY

EXERCISES. SET LV (concluded)

680. What is the locus of points:

(a) 3 units from the x-axis? (6) -5 units from the y-suds?

581. What two facts do you imply in the answer to each of the
parts in the last question?

582. Give the equation expressing the condition which deter-
mined each of the loci in exercise 580.

683. What is the locus of the condition expressed in each of the
following equations?

(a) x=15 (b) y= -9 (c) x= -12 (d) y = 20

684. What locus is represented by the equation x^ = 25?
686. What locus is represented by the equation

(a) a:=y? (d) x-t/=10? (g) 2x+3;y = 20?

(6) x= -y? (e) x= -3t/? (h) 3x--5y = 12?

686. Give the equation of the locus of a point:
(a) Just as far from the x-axis as from the ^-axis.

(6) Three times as far from the x-axis as from the T/-axis.

(d) Minus five times as far from the T/-axis as from the x-axis.

(e) Minus seven times as far from the x-axis as from the t/-axis.
(J) Such that the sum of its distances from the axes is -11.

(g) Such that three times its distance from the x-axis increased
by 9 times its distance from the y-axis is 26.

(A) Such that five times its distance from the x-axis diminished
by twice its distance from the 2/-axis is 7.

(i) Such that the simi of the squares of its distances from the
axes is 49.

(j) Such that four times the square of its distance from the
X-axis increased by the square of its distance from the T/-axis is 144.

Check the answer to each part of the last two questions by
plotting the equation.

687. The theorem of Pythagoras is employed to find the ''equa-
tion of a circle'' about the origin as a center.

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THE LOCUS 155

Take any point P in a circle about the origin 0. D r aw th e ordi-
nate MP. Let OM=x, and MP^y. Then OAP+MP^^OP^.
K the radius OPsr, this becomes x^+y^
^r\ This equation holds for the co-
ordinates of any point on the circle, and
is called the equation of the circley r being
any known number.

Form the equation of the circle with
the origin as center and (a) 7 as radius,
(b) 7\/2 as radius.

B. THE PECULIARITY OF THE PROOF OF A LOCUS
PROPOSITION

When we say that the locus of points on this page just one inch
from its right edge is a line parallel to that edge, and one inch in
from it, we really imply three facts. First, that any point on that
line is one inch from that edge; second, that any point which is
one inch from that edge and on the page is on that line; and third,
that any point not on that line and on the page is not one inch
from that edge. If the first of these three facts be called the
direct statement, we already know that the second is its converse.
The third is known as its opposite.

In symbols, three theorems so related might be stated as follows:
Direct theorem. If a = 6, then c = d.

Converse theorem If c=d, then a =6.

Opposite theorem. H aj^h, then c^d.

Hence we see that while the converse of a fact simply inter-
changes its data or what is given with its conclusion, the opposite
of a fact negates both the daia and conclusion.

Now let us discover what we can as to the* truth or f aMty of
converse and opposite theorems when the direct theorem is true.
EXERCISES. SET LVI. DIRECT, CONVERSE, OPPOSITE

The following exercises will help us in this task:

688. Form (1) the converse, and (2) the opposite of each of the
following facts:

(a) If a man lives in Boston, he lives in Massachusetts.

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156 PLANE GEOMETRY

(6) If it rains, the ground is wet.

(d) All vertical angles are equal.

(e) The supplements of equal angles are equal, or.

If two angles are equal, their supplements are equal.
(/) All men are bipeds.

689. (a) Of which of the six facts mentioned in the last exercise
are the converse facts true?

(6) Of which of them are the opposite facts true?

(d) Of which of them are the opposite facts either false or not
necessarily true?

690. (a) Can you draw any conclusion as to the truth or falsity
of converse and opposite theorems?

(6) Test this conclusion with several more instances.
Though a statement is true:

(1) Its converse may or may not be true.

(2) If its converse is true, its opposite is also true.

(3) If its converse is false, its opposite is also false.
The proof of this fact follows:

Given: That when a = 6, c = rf, and when c = rf, o = 6.
Prove: That when a?^6, c^d.
Proof : Suppose c = d.

Then what follows? Why?

What conclusion can you draw?

Let us see how these conclusions help us to decide just how much must be
proved in order to establish the truth of a locus proposition.

Since the converse of a fact is not necessarily true, in order to prove a
line a required locus, not only must we prove (1) that any point in it fulfills
the required conditions, but also (2) that any point that fulfills the required
conditions is in the line. It is, however, unnecessary to prove the opposite,
since if the converse has been proved true, we know the opposite is also true.
In short, if we know (1) and (2) are true, we know without further proof that
any point not in the line does not fulfill the required conditions.
Suppose we wished to prove

Theorem 40. The locus of points equidistant from the ends
of a sect isjhe perpendicular bisector of the sect

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THE LOCUS 157

Given: XO?±AOE so that AO^UE.

Prove: XF is the locus of points equidistant from A and B,

If we proved fimst that any point P in -^

XY is equidistant from A and B, and
second that point Pi which is equidistant
from A and B lies in line XF, what would
we know about any point not in 3T? y"^

.'. (1) Prove: PA sP5, given: Z^sOB y^

and P in XY±AB at 0; and (2) prove: ^^ J ^^

OPi±AB, given: P^^T^S&ndlO^OB.
To prove (1):

What parts of A AOP and A BOP do
you know are equal? ^

Are the A congruent?
To prove (2): ^

What parts of A AOPi and A BOPi do you know are equal?

Are the A congruent?

What fact must you prove to show that OPi±AB?

Is there any other converse which we might have proved in place of the
one here proved? Why are there two converses in this case?

Cor. 1. Tiuo points each equidistant from the ends of a sect
fix its perpendicular bisector.

How many points determine a straight line?

EXERCISES. SET LVII. APPLICATIONS OF LOCUS

691. Show why a circle may be defined as the locus of points
at a fixed distance from a given point.

Describe without proof:

692. The locus of the tip of the hand of a watch.

693. The locus of a point on this page and just 3" from the upper
right comer.

694. The locus of the center of a hoop as it rolls along the floor
in a straight line.

696. The locus of the edges of the pages of a book as it is opened.

696. The locus of the handle of a door as it is opened.

697. The locus of the end of a swinging pendulum.

698. The locus of places described as 1 mile from where you are
standing.

699. The locus of points 1' above a given shelf.
dCOOj The locus of points 1' from a given sbelf .

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158 PLANE GEOMETRY

601. The locus of the center of a circle as it rolls around another
circle, its circumference just touching that of the other circle.

602. The locus of the center of a ball as it rolls around another
ball, its surface just touching that of the other ball.

603. The locus of one side of a rectangle as it revolves about
the opposite side as axis.

604. The locus of the entire rectangle in the last exercise.
606. The locus of a point at 3" from a fixed point P.

606. The locus of a point 3" from a given Une.

607. The locus of a point equidistant from two parallel lines.

608. The locus of a point equidistant from two given points.

609. The locus of a point equidistant from two intersecting lines.

610. The locus of a point the distance <-d— >• from a given line l.

611. The locus of a point the distance <—d -> from a given point P.

612. The locus of a point the same distance from the center and
the circumference of the circle c.

We shall now prove one more veryimportant fact concerning loci.

Theorem 41. The locus of points equi-
distant from the sides of an angle is the
bisector of the angle.

I. Given: 4CBA, EXso that Z^CBX s 4XBA.
P anypoint in BX, PR±lB cutting AB at R.
PS±BC cutting^C at S,
To prove: Pi2sP5.

(Proof left to the student).

II. Given; Pi a point within the ^CBA, so
that PiR (the ± t o AB) =P^ (the ± to Bc).

To prove: PiB bisects 4CBA.
(Proof left to the student.)

Cor. 1. The locus of a point equidistant from two intersecting

lines is a pair of lines bisecting the angles.

Hint: At what angle do the bisectors of any two adjacent angles formed
by the pair of intersecting lines meet each other? At what angle, Sien, do the
bisectors of the vertical angles meet each other?

EXERCISES. SET LVII (concluded)

613. If a gardener is told to plant a bush 10' from the north fence
doesheknowexactly where to plant it? If not, state another direction
which might be given him that he may know just where to plant it.

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THE LOCUS 169

614. Is the second direction given the gardener the only one
you could give him to have the bush definitely located? If not,
state other directions which might have answered.

616. How many loci are needed to locate a point on the floor of
the room? On any one of its walls? On the ceiling?

616. (a) Where would all points that are two feet from the
floor of a room lie?

(6) Where would all points 3' from the front wall lie?

(d) Suppose a point was described as being 2' from the floor,
3' from the front wall, and 4' from a side wall. On how many loci
is it? Exactly where is it?

(e) How many conditions are needed to fix a point in a room?

617. A man wants to build his home at the same distance from
two railroad stations, (a) Is the location of his home fixed?
(6) If at the same time he wishes to build a half mile from the
bank of a river which runs parallel to and four miles from the
road connecting the stations, is the location of his home fixed?
Make an accurate construction showing how many locations
answer the description.

618. Prove theorems 40 and 41 by means of direct and opposite.

619. What is the locus of the vertices of triangles which have a
common base and equal areas?

620. What is the locus of points dividing sects which connect
a given point and a given line in the ratio of 5 to 7?

621. What is the locus of the vertices of triangles resting on a
common base and having fixed areas in the ratio of 5 to 7?

LIST OF WORDS DEFINED IN CHAPTER VH
Locus, opposite.

SUMMARY OF THEOREMS PROVED IN CHAPTER VH

40. The locus of points equidistant from the ends of a sect is the perpen-
dicular bisector of the sect.

Cor. 1. Two points each equidistant from the ends of a sect fix its
perpendicular bisector.
4L The locus of points equidistant from the sides of an angle is the bisector
of the angle.

Cor. 1. The locus of points equidistant from two intersecting lines
is a pair of lines bisecting the angles.

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CHAPTER VIII

THE CIRCLE

We have not only used the word "circle" very freely throughout
the text, but have also used our compasses for the construction of
the circle, or any part of it, whenever necessity arose. This has
been due to the fact that the idea of a circle seems to be one with
which all of us have grown up. But we have now reached the
time to consider the idea scientifically, and add to our stock
of facts concerning it.

Up to the present we have thought of the circle as a portion of
a plane, and the curve bounding it as its circumference. This is
not the sense in which the word circle is used as we advance in
mathematics, so we shall have to revise our notion of it. The word
circle is used to refer to both the portion of a plane and the curve
which bounds it — the one to which it refers being determined by
the context, but the definition covers only the boundary. When
there is any danger of ambiguity the word circumference will be
used in this text.

A circle is a plane curve which contains all points at a given
distance from a fixed point in the planej and no other points* Thus
we see that a locus definition may be given as a corollary to this
one; namely, a circle is the locus of points at a given distance from a
fixed point* The fixed point is called the center, and the given dis-
tance (the distance from the center to any point on the circle)
is called its radius (plural, radii). The sect through the center and
terminated by the circle is called its diameter.

At this point we may state several corollaries to these definitions.
It will be left to the student to verify them.

'^ Cor. 1. All radii of equal circles are equal.
V Cor. 2. Circles of equal radii are equal.
v;^Cor. 3. All diameters of equal circles are equal.

* Refer to Exs. 585 (i), 586 (i), 588, 591, and 611 for previous illustrations
of this definition.

160

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THE CIRCLE 161

Cor. 4. A point is inside, on, or outside a circle, according as
its distance from the center is less than, equal to, or
\ greater than the radius.

Cor. 6. A point is at a distance less than, equal to, or greater
than the radius from the center according as it is inside,
on, or outside the circle.

A. PRELIMINARY THEOREMS
Theorem 42. Three points not in a
straight line fix a circle.
Given: Points A, B, D, not in a straight line.
To prove: (1) A circle can be passed through A,
B,D.
(2) Only one circle can be passed through A,
B,D,

Suggestions for proof: What is the locus of the

centers of all circles passing through A and D? Why? Through B and D?

By means of the transversal XF in the diagram, show that PY and QX

are not parallel, and that hence point C exists
Why can no second point such as Ci exist?

EXERCISE. SET LVllI. THE CIRCLE AS A LOCUS
622. An amusement park is to be located at the same distance
from each of three villages 7i, 72, and Vz. V2 is 5 miles from 7i,
Vz is 6 miles from 7i, and Vz is 8 miles from V2. Show by an accur-
ate construction the actual location of the park. Can you think
of any locations of 7i, 72, and at the same time Vz which would
make it impossible to have a park so located? (Use sect V2 V3 to

represent 8 mi.) 72 — ■ Vz

In the accompanying diagram, ^BOA is known as a central
angle. Define such an angle.

Any portion of a circle^ such as BA is known
as an arc. When referring to any definite arc,
[^ such as BA or CD, we write it thus: BA, Cb.
1| Any two points on a circle (unless they are
(^ the ends of a diameter) are the ends of two
arcs known as minor and major arcs.^^^or
instance, (^XD is the minor CD, and DYC is
the major DC. The shorter of two arcs cut off by any two points on

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162 PLANE GEOMETRY

a circle is known as the minor arc, and the longer is known as the
major. When not otherwise stated, the minor arc is referred to.

The sect CD is known as a chord . Define the word chord. The
diameters of a circle are simply chords. Why?

Angles in a circle are said to intercept arcs, and chords are said
to subtend arcs. Intercept comes from the Latin "inter," meaning
"between," and "capio," meaning "to take," hence the angle
intercepts or "takes the arc between its sides." Subtend comes
from the Latin "sub," meaning "imder," and "tendere," meaning
"to stretch." Thus, as we see, the chord is the straight line which
"stretches imder the arc.'* Arcs are considered passive, and are
referred to as being "intercepted by" angles, and "subtended by"
chords, although in engineering the expression "the central angle
subtended by an arc of w® " is very common. If the expression \S
used in this text it will only be when referring to engineering
problems.

^ Theorem 43. In equal circles equal central angles intercept equal

arcs, and conversely.

I. Direct.
Given: OC s OCi, ^ACB^

Prove: AB^fY.

II. Converse.
Given: OCs oCi, ABsjfy.
Prove: ^ACB^iS^XCiY.

Method of proof, superposition. In superposing, which parts will you
make coincide in proving the direct? Which in proving the converse?

B. THE STRAIGHT LINE AND THE CIRCLE

- Theorem 44. In equal circles, equal arcs are subtended by equal
chords, and conversely.

Suggestion: Prove A ACB^
APCiQ.
What means have you of
proving the triangles con-
gruent in the direct?
What in the converse?

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EXERCISES. SET LIX. CONGRUENCE OF CURVILINEAR
FIGURES

623. Prove that the curved figures
BDEFsmdCFGH Biecongrnent. The
figure is based on a network of equi-
lateral triangles. The vertices are the
centers, and the sides the radii for the
arcs.

624. In the accompanying figure ^ ^

prove that the curved triangles ABC,

CHD, etc., are congruent. Also BCDEFG and DHML.

V- — ^^'

\l \t

/ ^

-A A

/ ^ ' \ \

L. ^^ ^

/\7\

''K — 7\

y Theorem 45. A diameter perpendicular to a chord bisects
it and its subtended arcs.

Given: 00, diameter EF ± chord AB at D.
Prove: ZDsDB. I^s^. 5^^/^.
Proof: ^^Sh if what angles are equal?

These angles are equal if what triangles are con-
gruent? Write a complete proof.
7^ By means of what angles can you prove Bp a fXf

Cor. 1. A radius which bisects a chord is
perpendicular to it.

Which of the methods of proving lines perpendicular can be applied here?

Cor. 2. The perpendicular bisector of a chord passes through
the center of the circle.

Hints: Draw the radius which bisects the chord and prove the given
bisector coincident with it, or treat as a locus.

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PLANE GEOMETRY

EXERCISES. SET LX. CONSTRUCTIONS BASED UPON CIRCLES

626. Give the con-
struction for Fig. 1.

From Carlisle Cathedral.
(Figs. 1 and 2 applied.)

626. Inscribe Fig. 1 in a given circle. (See Fig. 2.)

627. Give the construction for the design shown in Fig. 3.

Suggestion : Construct the equilateral AABC. Divide each side into three
equal parts and join the points as indicated. The intersections are the centers,
and AM is the radius for the arcs drawn as indicated.

Fig. 4.

From Exeter Cathedral.
(Fig. 3 applied.)

628. Construct Fig. IV.

Suggestion: Construct the
equilateral AABC, A, B,
and C are the centers for CjB,
CAf and AB. The semi-
circles are constructed on the
sides of the linear triangle
as diameters.

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165

629. A civil engineer wishes to continue the circular track AB
for some distance. Suggest how he
can do it.

630. From the measurements of a A^ ^b
piece of broken wheel a new wheel is to
be cast of the same size. Show how
to find the radius of the new wheel.

y Theorem 46. In equal circles, equal chords are equidistant

from the center, and
conversely.

I. In the direct what
parts are known to be
equal in AOBX and
AOiDY='!

II. State and prove
the converse.

EXERCISE. SET LXI. EQUAL CHORDS

631. The following method of locating points on an arc of a
circle that is too large to be described by a tape is used by engineers.
If part of the curve APE is known, ^
take P as the mid-point. Then
stretch the tape from A to B and
draw Pilf perpendicular to it. Then
swing the length AM about P, and
PM about Bj until they meet at L,
and stretch the length AB along PL to Q. This fixes the point Q.

In the same way fix the point C. Points

on a curve can thus be fixed as near to-

^ /^ \ gether as we wish. Why is this method

. correct?
.0

A straight line is said to be tangent to a
circle whe n it touches it once, and only once.
Thus XY is tangent to ©C if it touches QC
at pt. P, and nowjiere else.

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PLANE GEOMETRY

'^Theorem 47. A line perpendiadar
to a radius at its outer extremity is
tangent to the circle.

Given: OO, AB ± radius OC at C.
Prove: AB tangent to ©0.

Pboop

(1) Given.

(2) Authorities left to the student to
insert.

(1) AB^ radius at C.

(2) .-. AB touches OO at C.

(3) It remains to prove that no other

point such as P in AB touches
O0._/. Draw sect OP.

(4) OC ± AB^_ (4)
(6)/.0P/-AB. (5)

(6) .-. OP > OC. (6)

(7) .-. P lies outside of OO. (7)

(8) .-. Cand only C in' ABib in QO. (8)

(9) r.ABia tangent to QO, (9)

The first three corollaries following are partial converses of this
theorem.

We are not yet ready to prove the converse of Theorem 6.
It will be proved later under the topic of inequalities. For the
present, then, we shall add it as a postulate.
Postulate, the third postulate of perpendiculars:
The sfiartest distance from a point to a line is the perpendicular
to that line.

Cor. 1. A tangent to a circle is perpendicular to^ the radius
^ drawn to the point of contact.

Given: Xf tangent to OO at P, radius OP.

Prove: Xy±OP.

Suggestions forproof: Draw OQ to any point QinXY.

Where does Q lie?

How does OQ compare with OP^

What conclusion can be _= p — ^

drawn with respect to OP^ ^

Cor. 2. The perpendicular to a tangent at
the point of contact passes through
the center of the circle.
Hint: Show that KP coincides with the
radius CP^ and .'. C lies on KP. ,Qoglc

THE CIRCLE 167

^ Cor. 3. A radius perpendicular to a tangent passes through the

point of contact.

Hint: Draw the radius to the point of con-
tact P, and show that CR coincides with CP.

Cor. 4. Only one tangent can be
drawn to a circle at a given
. point on it.
Hint: How many ±8 can be erected to CR ^
&tR?

EXERCISES. SET LXII. THE TANGENT AND THE CIRCLE
632. When a ray of light strikes a spherical mirror (represented
in cross section by the arc of a circle), the angle of incidence is
found by drawing a tangent to the circle at the
point of incidence, and erecting a perpendicular to
the tangent at that point. In this case the perpen-
dicular (called the normal) is a diameter. Why?
d633. The line of propagation of a sound wave
also follows the law of reflection of a ray of light,
namely, that the angle of incidence is equal to the angle of reflection.
The circular gallery in the dome of St. Paul's in London is known
as a whispering gallery, for the reason that a faint sound produced
at a point near the wall can be heard aroimd the gallery near the
wall, but not elsewhere. The sound is reflected along the circular
wall in a series of equal chords. Explain why these chords are equal.
634. What is the locus of the centers of a nimiber of hoops of
different sizes (one inside the other) tied together at one point?

636. What is the locus of the centers of all circles tangent to a
gjven line at a given point?

1B86. What is the locus of the center of a wheel as it rolls straight
ahead along level ground? Prove this fact.

637. What is the locus of the centers of all circles tangent to
both sides of an angle? •

638. Two straight roads of different
width meet at right angles. A is the nar-
rower, B the wider. It is desired to join
them by a road the sides of which are
arcs of circles tangent to the sides of the
straight roads. What construction lines
are necessary? Draw a figure.

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PLANE GEOMETRY

How would you construct a tangent to a given circle at a
X given point P on the circle?

'7aheorem48. Sects of tan-
gents from the same point
to a circle are equal.

Why are A OPX "and OPY
congruent?

Circles are said to he tangent to each other when they are tangent
to the same line at the same point. Why is OCi tangent to 0C2?
Why to 0C3? Why is 0C2 tangent to 0C3?

Circles are externally
tangent when their centers
lie on opposite sides of the
common tangent. Name
two pairs of circles that
are externally tangent
in the diagram.

Circles are internally
tangent when their centers lie on the same side of the common tangent.
Which circles in the diagram are internally tangent?

EXERCISES. SET LXIII. TANGENT CIRCLES
640. (a) How are the hoops mentioned in Ex. 634 related to one
another?

(fe) How is the line in which their centers lie related to their
common tangent? Why?

mesh? Why?

The line of centers of
two circles is the sect con-
necting their centers.

Theorem 49. The line
of centers of two tangent
circles passes through
their point of contact.

(For suggestions see Ex. 634, p. 167, and Ex. 640, p^68.) ,

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THE CIRCLE 169

The common chord of two intersecting circles is the se^ connecting
their points of intersection.

EXERCISES. SET LXIII (concluded)

641. How is the line of centers of two intersecting circles related
to their common chord ? Prove your answer.

642. If two cog-wheels mesh, show that the point where they
mesh is in a straight line with the centers of the wheels.

643. (a) Show how to construct an
equilateral Gothic arch. (See the accom-
panying diagram.)

Suggestion : Construct the equilateral tri-
angle ABC. With A and B as centers and AB
as radius, construct the arcs BC and AC,

(6) If E, F, and D are the mid-points
of the Unes AC, CB, and AB, respec-
tively, prove that equal equilateral tri-
angles are formed.

Suggestion: Points D, E, C, F, B, and A are the centers and AD is the
radius for the arcs drawn as indicated.

d644. An application of geometry to engineering is seen in cases
where two parallel streets or lines of track are to be connected by a

"reversed curve." If the lines
are AB and CD, and the con-
nection is to be madf from
B to C, as shown, we may pro-
ceed as follows: Draw BC
and bisect it at M. Erect
PO the perpendicular bisector of BM, and BO perpendicular to
AB. Then is one center of curvature. In the same way fix Oi.
The curves may now be drawn, and they will be tangent to AB, to
CD, and to each other. Prove that the curve BMC is a reversed
curve tangent to AB and CD] i.e., prove (a) BM tangent to AB
at B,^M tangent to CD at C; (6) ^ tangent to 6m at M]
(c) BM = CM.

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PLANE GEOMETRY

646. The figure represents a Persian arch. Triangles ABC and
DEF are congruent and equilateral. The centers of the upper
arcs MC and NC are respectively D and F;
while the lower arcs are drawn with the cen-
ter E, Prove that the area of the arch equals
the area of triangle ABC.

646. The trefoil ADBF, etc., is constructed
from circles described on the semisides of
A ABC. The points D, E, and F are the
centers for the arcs which are tangent to the sides of AABC, and
which form the trefoil HYKZGX. If PD and RF in Fig. 1 are
radii for the arcs ZK and YK, prove that PD = FR.

Fig. 1.

FiQ. 1. applied.

d647. The semicircle AGHB in Fig. 2 is constructed on AB as
diameter, and CD is perpendicular to AB at its mid-point.

(a) Construct arcs CH and
CG tangent to the line CD at
point C, and to the semicircle.

Suggestion : Make KC ^ AD.
Draw NE, the perpendicular bisector
of KD, meeting CK extended at E.
E is the center for the arc CH. What
general problem in construction of
circles is involved here?

D B

Three-centered ogee arches.

(6) If the arcs CH, drawn fw. 2.

with ^ as a center, and HB drawn with D as a center, are tangent
at H, prove that the points D, -ff , and E are coUinear.

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171

Suggestion: Join C and H, and B and C, and prove that each is parallel
to DK,

(d) UAB^Sy and CZ)=8, find the length of CE.

Suggestion: i?D=4\/5. Compare the sides of the similar triangles
KCDmdKNE. Find iC^ and hence C^.

(e) UAB=s and CD=A, find the length of CK.

648. What must be the relation between ry^
the length and width of the rectangle ^
ABCD in order that the tangent circles ^
may be inscribed as shown. ^-^

649. ABCD is a parallelogram with four tangent circles inscribed.

a (a) If the lines AC and BD

are supposed to be indefinite in
extent, show how to construct
circle tangent to the lines -4 C,
AB, and BD, and circle X tan-
gent to lines AC and BD and to
circle 0.

(6) If Sis the mid-point of AB,
and and X are the centers of
the circles, prove that the points
Ej 0, and X are coUinear.

660. Fig. 1 shows a trefoil

formed of the three circles X, F, and Z tangent to each other at

the points T, S, and R. It is ^

inscribed in the circle as shown,
(a) Show how to construct the

figure.

Solution : Circumscribe an equilateral
triangle about the circle. Connect each
vertex with the center. Inscribe a circle
in each of the triangles FOG, GOE, and
EFO.

(6) Prove that the small cu-cles
are tangent to the large circle and

to each other. Fiq. l. Trefoil formed of tangent circles-

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PLANE GEOMETRY

651. (a) Construct the quadrifoil of tangent circles inscribed
in a square so that each circle is tangent to two sides of the square
and to two other circles.

Inlaid tile design.
(Fia. 2 applied.)

(6) Prove that the lines joining the centers of the circles form
a square.

C. THE ANGLE AND ITS MEASUREMENT
"^ Theorem 50. In equal circles central angles have the same
ratio as their intercepted arcs.

Given: OCsOCi, ^ACBand
4 XCiY, AB and XY com-
mensurable.

Proxe; ^^CB ^Xb

Suggestions for proof: Select a
unit of measure for AB and

Divide these arcs into such units and connect the points of division with
the centers C and Ci.

What can you say of all the central angles thus formed?

How do l^sBCA and YCiX compare?

How does BA compare with YX''
Cor. 1. A central angle is measured by its intercepted arc.
Suppose 2$. a were the angular unit of measure, and a the circular unit of
measure in a circle of radius r.

What would be the numerical measure of ^JCCiYt ^^ — \a

What would be the numerical measure of XF?
How do these numerical measures compare?

©'

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173

Up to the present time we have emphasized the fact that a
magnitude can be measured by a unit of the same kind only. We
must then justify a statement such as that given in Cor. 1, Theorem
50. This corollary should be stated as follows: The ratio of a
central angle of a circle to the angular imit is equal to the ratio of
its intercepted arc to the circular imit; or: The numerical measure
of a central angle of a circle is equal to the numerical measure of
its intercepted arc. But since this fact is one frequently referred
to, and the correct statement of it is so lengthy, mathematicians
have agreed to the abbreviated statement given in Cor. 1. The
s3anbol " 29 " is used for "is measured by." It suggests the ideas
of both equality and variation.

A secant is a straight line which intersects the circle.

EXERCISES. SET LXIV. SECANT AND CIRCLE

g662. (a) Show by a graphic solution of the equationso:'^ 4-2/2 =49,
and x=3, that a secant cuts a circle in two points.

(6) What numbers would have to replace 3 in the second equa-
tion to change the equation to one of a tangent?

Y Theorem 61. Parallels intercept equal arcs on a circle.

Case I. When the parallels are a secant and a tangent. (Fig. 1.)
Given: PQ tangent to circle C at T, secant RS \\ PQ, cutting C at jB and S.
Prove: ST^TR.
Suggestions for proof : Draw diameter through T, cutting 0C in M.

What relation exists between TCM and PQ?

Then what relation exists between TCM and RS*(

What follows as to iS? and 5^?

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174 PLANE GEOMETRY

Case II. When both parallels are secants. (Fig. 2.)
(Proof left to the student.) Suggestion: Draw the diameter perpendicular
to one secant.

Case III. When both parallels are tangents. (Fig. 3.)
Draw the diameter through T, and give the proof in full.
EXERCISES. SET LXV. CIRCLES

663. Does it make any difference in what order the cases under
Theorem 51 are proved, if the proofs are given as suggested?

664. Can you suggest methods of proof for Cases II and III
which depend upon Case I? Which methods do you prefer? Why?

666. Construct a diagram consisting of:

(a) Two concentric circles whose radii are in the ratio of 1 to 3.

(6) Six circles lying between them, tangent to them, and each
tangent to two others.

666. Make a diagram oi the mariner s compass, putting in six-
teen points of the compass.

Angles such as ABC in Figs. 1, 2, 3, Theorem 62, are called
inscribed angles. Their vertices are not only in the cirdej hut their
sides are chords. An angle is said to be inscribed in a semicircle
when its sides intercept a semicircle, in less than a semicircle when
its sides intercept a major arc, and in more than a semicircle when
its sides intercept a minor arc.

Since measurement is but a numeric relation, axioms may be
converted into authorities for statements concerning measure-
ment by simple changes such as those illustrated by the following:

The measure of the sum of two magnitudes of the same kind is

If X 99 d )
equal to the siun of their measures. E.g. : , ^j^^x+y££a+b.

The measure of any multiple of a magnitude is equal to that
multiple of its measure. E.g.: If a 29 a, then

/ ^y'yr \ v^eorem 62. An inscribed angle or one
^Lr^-y^ ] formed by a tangent and a chord is measured

\ / / ^ one-half its intercepted arc.

\ / / Given: OO, 4ABC so that B is on the circle.

^V.^ / To prove: ^ABC^h its intercepted arc.

Fig 1 Suggestions: Fig. 1. Compare 4^5Cwith 2^.^00.

What is the measure of ^ABCi

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175

Figs. 2 and 3. By drawing the diameter through B, reduce these cases

to that of Fi g. I.
Fig. 4. DT2iwZAY\\DC.

What is the measure of 2^ YAB?
What arc may be substituted for YB*l Why?
What, then, is the measure of 2^DBA? Of its supplement ^ABC?

Fia. 2.

Fig. 4.

Show that ABC is

EXERCISES. SET LXVI. INSCRIBED ANGLES

■•

667. What kind of angle is inscribed in a semicircle? In less
than a semicircle? In more than a semicircle?

668. In carpentry, circular pieces of molding for door panels,
etc., are sometimes turned out in the form of rings on a lathe;
then these are cut into pieces according
to the places in which they are to be used.
To cut such a ring into two equal parts,
place a carpenter's square upon it with
the heel at the edge of the ring, and
mark the points A and C where the
arms of the square cross the edge of the ring.

half of the ring.

669. Pattern-makers and others use
the carpenter's square as follows to
determine if the "half-round hole" is
a true semicircle: The square is placed
as in the figure. If the heel of the square
touches the bottom of the hole in all
positions of the square, while the sides rest against the edges of
the hole, the hole is a true semicircle. Justify this test.

660. Prove that the semicircles intersect in pairs on the sides
of the triangle in Ex. 628.

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PLANE GEOMirrRY

'11

661. In practical work a line EF is sometimes drawn parallel

j ^-t— <r to a given line-AB, as shown in

~ " the figure. Explain the con-

struction, and prove that
EF\\AB.
Y^ / P 662. A ship is steered past

^^n,^^^" a known region of danger as

follows : A chart is made in which a circle is drawn through two
points A and B, which can be seen from the
ship, and with sufiicient radius to inclose the
danger region. The inscribed angle ACB is
measured. Observations of A and Bare made
from the ship from time to time, and the course
of the ship directed so that the angle between
the directions to A and B never becon;es
greater than -^ACB. Justify this method.

663. Show by a diagram how you can use a protractor with a
plumb-line attached to determine the horizontal line AC while
sighting the top B of a building.

664. The diagram shows how the latitude
of a place may be determined by observa-
tion of the pole star. Let EEi represent
the equator, ilili the axis of the earth, P the
place whose latitude is to be found, PF a
plane (the horizon) tangent to the earth at
Pj and PS the line of observation of the
pole star. Then ^ a represents the latitude,
and -^ot is called the altitude of the pole star.
For practical purposes we may assume that
AAi II PS. Prove that <^ai= ^a.

666. Suppose XY to be the edge of a
sidewalk, and P a point in the street from
which we wish to lay a gas pipe

perpendicular to the walk. From -2r :

P swing a cord or tape, say 50 feet long, until it meets XY at A.
Then take Af, the mid-point ©f PA, and swing MP about M, to
meetXFat-B. Then J5 is the foot of the perpendicular. Verify this.

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177

666. By driving two nails into a board at A and B, and taking
an angle P made of rigid material, a pencil
placed at P will generate an arc of a circle i
if the arms slide along A and B, Why?
Try the experiment.

d667. Circle O is tangent to the sides of
square ABCD at points E, F, G, and H,
and cuts the diagonals at points iC, L, Af , and N, Point X on OE

2) O Q is so chosen that iS^XsXS. OX

=OY=OZ=OW, and the points
are joined as indicated.

B r— -->— >*0--P 1 F

//
1/

Y^-J5^

\ \ \ /

\ N ' / /

/ / 1 \ \

A5iJ^A\

1 1

SC 1 ^

^^st>>^ ^^^

6 r"-^^^I^--<-^

®®f^

i^;^

v$7^

'TTT^

(M)

ffv^^v ^

: '■ ;

tWM

•^ ^ -^ Parquet flooring. Arabic and Roman.

(a) Prove that KELFM, etc., is a regular octagon, and that
KELX, LYME, and MGNZ are congruent rhombuses.
(6) Prove that XO^AK.
Suggestion: Compare Ai^^O and Ai^i^. 2(.45 2(.55i rt.2(..

'^Theorem 63. An angle whose vertex is

Q inside the circle is measured by half the sum

of the arcs intercepted by it arid by its vertical.

Given: OC, chords AD and BE intersecting at O.

Prove: ^AOBgc

Ib+Ge

Hint: How is ^AOB related to 4? A and E?
EXERCISES. SET LXVII. MEASUREMENT OF ANGLES
668. Fill out the blank spaces in the table:

AND

35^

40°

80°

48°

50°

216°

40^*

60°

54°

190°

45°

90°
108°

180°

34°

164°

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PLANE GEOMETRY

d669, (g) Prove that (1)AB=
AK^KC^CM, etc, (2)^KCM
^^MEP, etc. (3) ^AKC=
^CME, etc.

(6) Find the number of de-
grees in each angle mentioned
in (a).

X Theorem 54. An angle whose
vertex is outside the circle is
measured by half the difference
of its intercepted arcs.

Given: ^ABC with vertex outside OO, inter-
cepting^ in Fig. 1 ^and ED; in Fig. 2 Ad
and EA ; in Fig. SAC and CA . ^_^

Prove: ^.ABCgc AC-Sb • p. j AC-SI
2 ^* ' 2

in Fig. 2, and ^^^"^ in Fig. 3.

Hint: In Figs.
>B 1 and 2 draw
CX||3B,and
reduce case
3 to case 2
by drawing
any secant
BSR,

Fig. 3.

^ Theorem 66. A tangent"^ is the mean
proportional between any secant and
its external sect, when drawn from
the same point.

Given: OC, AB tangent at A and secant BP

cutting OC at Q and P.

-. 5P J5

Prove: z=:r= ■=

AB BQ

Suggestion: Prove AABQ^APBA.

Fio. 2.

* By "tangent" in such cases is meant the sect from the point to the point
of tangency.

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EXERCISES. SET LXVIII. TANGENT AND SECANT

670. (a) Assuming that the diameter of the earth is 8000 mi.,
how far can a man see from the top of a building 200 ft. high?
(The height of the building is measured along the prolongation of
the diameter.)

(6) How far can one see from the top of a moimtain 1000 ft.
high? 2000 ft. high? 3000 ft. high? 4000 ft. high?

(c) How far can one see from a balloon 1500 ft. above the sea?
d671. Galileo (1564-1642) measured the heights of the moim-

tains on the moon, some of which are as much ^ jj

as 6 mi. high, as follows: AC B was the illu- ^
minated half of the moon just as the peak of
the moimtain M caught the beam SM of the
rising or setting sun. He measured the dis-
tance AM from the half-moon's straight edge
AB to the mountain peak M. Then by using
the known diameter of the moon, show how he was able to com-
pute the height of the mountain.

d672. Since the earth is smaller than the sun, it casts a conical
shadow in space (umbra), from within which one can see no por-

p tion of the sun's disk.

If S is the center of the
sun, Ethe center of the
earth, and V the end or
vertex of the shadow,
prove that the length of
ES X EB
the shadow, VE=-r— — -. Approximately, JFS = 92,900,000

miles, SD= 433,000 miles, and EB =4,000 miles. Compute VE,
673. Look up the terms ''umbra'' and '* penumbra," and answer
the following questions:

(a) How are the umbra and penumbra afiFected by a change in
the distance apart of the lumi-
nous and opaque bodies?

(b) If a golf ball is held be-
tween the eye and the sun, is
there any penumbra? Explain.

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180

PLANE GEOMETRY

the opaque body are of the same size?
'^i:— - — ^^ s. (d) What is the shape of the umbra
K^^><C^( ) if the luminous body is larger than
the opaque body, as in the case of the
Sim and the earth?
(Exs. 672-673 are taken from Betz and Webb, Plane Geometry.)

D. MENSURATION OF THE CIRCLE

We now come to a very important section of our plane geometry
— ^the section which develops the formula for ih^e length of the
circle and for the area enclosed by the circle. The latter is known
as the area of the circle.

A polygon is said to be cir-
cumscribed abovi a circle when
each of its sides is tangent to the
circle, and to be inscribed in a
circle when each of its sides is
a chord of the circle. In the
first case, the circle is said to be inscribed in the polygon, and in the
second case, the circle is said to be circumscribed aboiU the polygon.

Theorem 66. A circle may he circumscribed about, and a circle
may he inscribed in, any regular polygon.

Given: The regular n-gon PQRST ...

To prove: 1. A circle may be circumscribed about PQRST , . .

2. A circle may be inscribed in PQRST . . .
Hints for proof 1:

Let be the center of the circle | Three non-collinear points determine

a circle.

determined by P, Q, and R.

Fio. 2.

Draw 5P, 05, OS, OS.

By means of ^ OPQ and
0725 prove 0*S=OP.

To what triangle can /^OST
be proved congruent?

Would it make any differ-
ence how many sides the
original polygon had?

Give the details of the
proof.

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Hint for proof 2: If is the center of the circumscribed circle, what are PQ,
QR, RS, etc.?

Cor. 1. An equilateral polygon inscribed in a circle is regular.
Hint: In Fig. 1 what is the measure of each of the angles P, Q, R, etc.?
Cor. 2. An equiangular polygon circumscribed about a circle is
regular.

Suggestion: In Fig. 2 connect consecutive points of contact B, C, D, etc.
Prove ABRC, ACSD, etc., congruent isosceles triangles.

What can then be said of the sums of any two of the equal sides of the
triangles? (Such as RC+CS, SD+DT, etc.)

An angle such as -^POQ is called a central angle of the regular
polygon.

EXERCISES. SET LXIX. REGULAR POLYGONS AND CIRCLES

674. Make a design for tiling or linoleum pat-
terns baseduponinscribedequilateralhexag(^

676. Make a copy of the accompanying

design of arosewindowof six lobes. (Fig. 1.)

676. Make an accurate construction

of the accompanying design representing

a Gothic win-
dow. (Fig. 2.)
d677. Fig. 3
shows a star in- rig. i.

scribed in a given square with all of its
vertices on the sides of the square.

(a) Construct a figure in which points
KyL,MjNy etc., shall be the mid-points
of AE, EB, BF, FC, etc.
(6) Show that a circle can be circum-
scribed about the star constructed as in
(a) and find its radius, if AB = a.

678. If a series of equal chords are laid
off in succession on a circle, what relation
exists between:

(o) the arcs subtended by the chords?
(6) the central angles intercepting the arcs?
(c) the inscribed angles formed by any two successive chords?

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Fig. 3.

182

PLANE GEOMETRY

679. If the series of equal chords mentioned in the last exercise
were such as finally to form an inscribed polygon, what kind of
polygon would it be? Why?

680. How are the central angles of a regular polygon related?

681. Make a table showing the number of degrees in the central
angle of a regular inscribed polygon of 3, 4, 20 sides.

682. Write a formula by means of which we can obtain the
number of degrees in the central angle of any regular polygon.

683. Which of the angles foimd for the table in Ex. 681 can
we construct by means of compasses and straight edge?

684. Inscribe in a given circle each of the regular polygons you
can by the means mentioned in the last exercise. (Do not go
beyond the sixteen-sided polygon.)

686. Express as a formula the number of sides of the regular
polygons you can inscribe in a circle up to this point.

Theorem 67. If a circle is divided into any number of equal

arcs, the chords joining the successive points of division fortn a

' regular inscribed polygon; and the tangents drawn at the points of

division form a regular circumscribed polygon.

Given: QO with A^^Bt;^ ^fl,

(1) Chords i4J5, BC FA,

(2) GH, HKy etc., tangent to OO at il, B, ....
F respectively.

To prove: (1) ABC F a regular inscribed polygon.

(2) GHK i\r a regular circumscribed polygon.

Hints: To prove (1) use Cor. 1, Theorem 56.

To prove (2) use Cor. 2, Theorem 56.

Cor. 1. Tangents to a circle at the vertices of a regular inscribed
polygon form a regular circumscribed polygon of the
same number of sides.

How do the vertices of the regular inscribed polygon divide the circle?

Cor. 2. tines drawn from each vertex of a regular polygon to
the mid-points of the adjacent arcs subtended by the
sides of the polygon form a regular inscribed polygon
of double the number of sides.

Hint : Show that the polygon is equilateral. Why ? Or throw the corol-
lary back to the proposition.

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THE CIRCLE 183

Cor. 3. Tangents at the mid-points of the arcs between con-
secutive points of contact of the sides of a regular
circumscribed polygon form a regular circumscribed
polygon of double the number of sides.

How does the corollary rest on the theorem?

Cor. 4. The perimeter of a regular inscribed polygon is less
than that of a regulat inscribed polygon of double the
number of sides; and the perimeter of a regular cir-
cumscribed polygon is greater than that of a regular
circumscribed polygon of double the number of sides.

(Proof left to the student.)

Historical Note. — ^This theorem presupposes the possibility of dividing
the circle into any number of equal arcs. In the table found in answer to
Exercise^681 it was probably seen that the number of equal arcs into which
we are at this time able to divide tke circle is very Umited, but we have not
yet learned to divide the circle exactly in as many ways as possible by means
of elementary geometry. Some other methods will be discussed later.

As early as Euclid's time it was known that the angular magnitude about
a point (and hence a circle) could be divided into 2», 2»-3, 2»-5, and 2»-15 •
equal parts. In 1796 it was discovered by Karl Friedrich Gauss, then only
sixteen years of age, that 2»*'17 equal parts of the circle could be found by the
use of only the straight edge and compasses. Gauss also showed that in
general it is possible to construct all regular polygons having (2»+l) sides,
when n is an integer and (2^+1) is a prime number. He went still further and
proved that regular polygons, having a number of sides equal to the product
of two or more different numbers of this series, can be constructed.

EXERCISES. SET LXIX (contmued)

686. Show that according to Gauss's formula, regular polygons
of 3, 5, 17, and 257 sides can be constructed.

687. Inscribe a square in a circle, and by means of it a regular
octagon, a regular 16-sided, and a regular 32-sided polygon.

688. What is the perimeter of the square in terms of the diameter
of the circle in the last exercise? How do the perimeters of the
octagon, hexadecagon, and 32-sided polygons compare with it?
With the circle?

689. Repeat exercises 687 and 688 with respect to the inscribed
equilateral triangle.

690. Repeat exercises 687 and 680 with respect to the regular
circumscribed square and triangle. C^ooale

184 PLANE GEOMETRY

691. Between what two values will the length of the circle
always lie?

692, Between what two values will the area of the circle
always lie?

From Cor. 4, Theorem 57, and Exercises 688-692, it is seen that
though the perimeter of the inscribed polygons increases as the
nimiber of its sides increases, it i^ always less than the length of
the circle; and that while the perimeter of the circumscribed
polygon decreases imder the same conditions, it is alwajrs greater
than the length of the circle.

In the first case the perimeter and likewise the area of the
regular inscribed polygon are increasing variables. They are
always less than the perimeter and the area of the circle, which
are fixed or constant

In the second case, the perimeter and the area of the regular
circumscribed polygon are decreasing variables which are always
greater than the perimeter and the area of the circle.

In the first case we say that the perimeter and the area approach
as superior limits the circle and its area, while in the second we
say that the perimeter and the area approach as inferior limits
the circle and its area.

Thus, if Pn represents the perimeter of a regular inscribed
polygon of n sides, a„, its area, jP„ the perimeter of a regular cir-
cumscribed polygon of the same number of sides. An, its area,
and c the circle in and about which they are inscribed and cir-
cumscribed, and C its area, we say that as the number n U in-
creased, Pn approaches c as its limit, Pn approaches c as its limit,
ttn approaches C as its limit, and An approaches C as its limit.
These statements are briefly written as follows:

Pn=C, Pn=C, ^ an = C, A^^C.

POSTULATES OF LIMITS

1. The circle is the limit which the perimeters of regular inscribed
and drcmnscribed polygons approach if the number of sides of the
polygon is indefinitely increased,

2. The area of the circle is the limit which the areas of regular
inscribed and circumscribed polygons approach as the number of
sides of the polygon is increased, ^ ,

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Theorem 68. A regular polygon the number of whose sides
is S-2" mag be inscribed in a circle.

(Proof left to the student.)

EXERCISE. SET LXIX (continued)
693. Inscribe a regular polygon of 3-2^ sides.

Theorem 69. If i^ represent the side of a regular inscribed

polygon of n sides, iin ih e side of one of 2n sides, and r the

radius of the circle, i^n—^^^ - r^4r^ - g.

Given: OO of radius r, AB {ji^ the side of a regular in- ^

scribed i»-gon, AC (iin) the side of the regular inscribed
2ii-gon. j^

Prove: imM \/2r«-r\/4r2-i„«.

Proof

Draw radius OC cutting AB at Z>.
Draw^.

(1) Then OCXAB and AZ>«^

(2) .-. 3C«-(i,n)*^(^)VcD'

(3) But CD ^r-O D.

(« .: (i.)-y+(r-^H-(|)')'
sr(2r-\/4r2~i„0

(l)-Why?

(2) Why?

(3) Why? (Note AOAD.)

(5) /. tt» = \/2r-rV4r2-i^«

EXERCISES. SET LXIX (continued)
694. Given a circle of radius 1 unit, compute:
(o) The length of a side of the inscribed square.
(6) The length of a side of the inscribed regular octagon.

(d) The perimeters of each of these polygons.

696. Do the same as you did in the last exercise for the regular
hexagon, dodecagon, and 24-sided polygon.

696. Do the same as you did in Ex. 694, using a diameter of
1 unit.

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PLANE GEOMETRY

697. Do the same as you did in Ex. 695, using a diameter of
1 unit.

698. (a) Which computation is simpler — that using a radius
of 1 unit or a diameter of 1 unit?

(6) If the radius is one unit what must be done to the perimeter
found for each of the polygons in order that it be expressed in
terms of the diameter?

Theorem 60. If i„ represent the side of a regular inscribed
polygon of n sides, c^ that of a regular circumscribed polygon of

2rin
n sides, and r the radius of the circle, c„ s ^Aft^i i *

Given: AB, a side Cn of a regular circum-
scribed polygon of n sides, tangent to 00
of radius r at point C; in the side of a
regular inscribed polygon of n sides.

Prove: c»= / . , . , .

V4r*-i,i*

Suggestions for proof: Draw AO, SO, CO.

LetilO cut OO at P,andBO cut it at

12, and CO cut 7S at Q.

Show that PJS^in.

Note that OC is an altitude of AAOB and OQ of APOiJ.

Why is AAO-B «« APOR'f

(1) Then ^ •

"A"

(2) In AOP0,OQs^r'-(|y
Substitute (2) in (1) and complete the proof.

(1) WhyT

(2) WhyT

EXERCISE. SET LXIX (concluded)

699. Repeat the computations made in Exs. 694 and 695, or 696
and 697 for the sides and perimeters of regular circumscribed
polygons.

The radiua of its circumscribed circle is called the radius of a
regular polygon, and the radius of its inscribed circle is called the
apothem of a regular polygon.

Cor. The apothem of a regular polygon is perpendicular to
its side.

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Theorem 61. The perimeters of regular polygons of the same

rtumber of sides compare as their radii, and also as their apothems.

Given: Polygons ABCD

•'N&adAiBiCiDi''

Ni, regular, and each

of n sides, and with

perimeters p and pi,

oenters O and Oi, radii

r and n, and apothems

a and ai.

To prove: (1) -^ s 1 and (2)

Pi Ti pi

Suggestions: Proof of (1)

Show that AAOB ^ AAiOiBi.

IfS^JlwhyisP -^?
AiBi n pi fi

Proof of (2)

What are a and ai in AAOB and AAiOiBif

EXERCISE. SET LXX. PERIMETERS OF REGULAR POLYGONS

700, Construct a regular hexagon whose perimeter is f of the
perimeter of a given regular hexagon.

In the proofs that follow, two more assumptions are made,
namely:

AXIOMS OF VARIABLES

1. If any variable approaches a limit, any part of that variable

avproaches the same part of its limit. That is, if Vi = ?i, then *=—

"' n.

2. If two variables are always equal, the limits which they approach
are equal. That is, if v\= h, and vt^Uy and t;i=t;2 then li=l%*

Theorem 62. Cir-
cumferences have the
same ratio as their
radii.

Given: Circles c and ci of
radii r and ri.

Prove: — aa-.

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PLANE GEOMETRY

Proof
Inscribe in each circle a reg-
ular polygon of n sides, and let
p and pi be their perimeters.

(1) Then !^ .
Pi

S — .

(2)or?-H^.

Let

n increase indefinitely.

(3) Thenp=<

;andpi =

= ci and

(4)

r r n

~ri'

(6)

" r n*

(6)

Why?

2-r-2f; ^^^

(1) The perimeters of regular polygons of the
same number of sides compare as their radii.

(2) By alternation.

(3) Postulate of limits for inscribed poly-
gons.

(4) If any variable approaches a limit, any
part of that variable approaches the same
part of its limit as limit.

(5) If two variables are always equal, the
limits which they approach are equal.

(6) By alternation.

Cor. 1. The ratio of any circle to its diameter is constant

ci~2ri'
Cor. 2. Since the constant ratio ^ is denoted by the Greek

tetter w* (which is the initial letter of the Greek word
"periphery") in any circle, c=27rr.

EXERCISE. SET LXXI. VALUE OF 7t
701. Show that the second value given to tt by Brahmagupta
is another form of that given by Ptolemy. (See historical note.)

Theorem 63. The value of t is approximately 3.14169.

If Pn stands for the perimeter of the regular inscribed polygon,
and Pn for the perimeter of the regular circumscribed polygon in

* Historical Note. — "Although this is a Greek letter, it was not used
by the Greeks to represent this ratio. Indeed, it was not until 1706 that an
English writer, William Jones, used it in this way."

Professor D. E. Smith further tells us that "probably the earliest approxi-
mation of the value of x was 3." In I Kings, vii, 23, we read : "And he made a
molten sea, ten cubits from one brim to the other; it was round all about —

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THE CIRCLE 189

a circle of diameter 1, the following table can be derived by using
the formulas of the last two theorems:

(1) iin=^2r* -rV-4r*-i»» and (2) c„s /^=— in which r, ijn,

in, and c„ retain their original meanings.

(The student might verify a few of the values in the table by
using logarithms wherever possible, and for convenience basing
the calculations on a circle of diameter 2.)

No. of sides. pn < c < Pn

6 3.0000000 3.4641016

12 3.1058285 3.2153903

24 3.1326286 3.1596599

48 3.1393502 3.1460862

96 3.1410319 3.1427146

192 3.1414524 3.1418730

384 3.1415576 3.1416627

768 3.1415838 3.1416101

1536 3.1415904 3.1415970

EXERCISES. SET LXXII. CIRCUMFERENCE
g702. Construct a graph by means of which the circumference

of a circle of any given diameter may be obtained.

703. Find the size of the largest square timber which can be

cut from a log 24 in. in diameter.

and a line of thirty cubits did compass it round about." Again in ii Chronicles
iv, 2, we read a similar sentence. And again in the Talmud is found the sen-
tence: "What is three hand-breadth around is one hand-breadth through."
The following list of various other values given to x may be of interest to
some of us.

Value Attributed to

3.106 Ahmes (c. 1700 B.C.)

3W<»-<3J Archimedes (287-212 B.C.)

3^ Ptolemy of Alexandria (87-165 aj).)

UU^ or 3.1416 Aryabhatta (c. 500 a.d.)

Jl§3 and jm or VlO Brahmagupta (c. 600 a.d.)

Hi Metius of Holland (1571-1635)

Computed to the equivalent of over

30 decimal places Ludolph von Ceulen (1540-1610)

To 140 decimal places Vega (1756-1802)

To 200 decimal places Dase (1824-1861)

To 500 decimal places Richter (1854)

To 707 decimal places Shanks (1854)

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PLANE GEOMETRY

704. The following problems illustrate the t3rpe of problem that
is suggested by books on carpentry.

To lay off an octagon on the end of a square

piece of timber -4J5C2>, draw diagonals AC and

BD. With radius EF (the apothem of the

square) draw arc cutting BD at G. Square out

i y^ \ A fromG. Make a similar construction at each of

X^ ^^X the other corners of the square. Justify the rule.

.r^ \^l.f-r N,> 706. How much belting does it require to

^TTXT^TTT

make a belt to run over two pulleys, each 30
in. in diameter, the distance between their centers being 18 ft.?

706. The annexed j&gure repre-
sents a small wire fence used to
protect flower beds. How many
feet of wire are needed per run-
ning foot of fence if AB = 1 ft.,
B2) = CD = 9in., and DE=S in.?
d707. Using 4000 miles as the
radius of the earth, find the num-
ber of feet in the length of one minute at the equator. Use loga-
rithms.

(This distance is conmionly called a "knot.")
708. (a) If a cable were laid around the earth at the equator,
how many feet would have to be added if the cable were raised
10 ft. above the surface of the earth?

(6) If the same were done around a gas-tank whose diameter
is 100 ft.?

d709. Carpenters and other tradesmen
frequently wish to know the circiunference
of a circle of given radius. The accompany-
ing graphic method is given in some of the
self-education books as a substitute for com-
putation:

Draw radii AO and BO at right angles.
Draw chord AB and line OE perpendicular
to AB, meeting circle at E and chord at Z).

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To 6 times the radius add the sect DE. The resulting sect is
approximately the length of the circumference.

Compute the approximate per cent of error, using 7r = 3.14159.

710. The central angle whose arc is equal to the radius is often
used as the unit of measure of angles. It is called a radian. Find
the niunber of degrees in a radian.

711. Many attempts have been made
to construct a sect equal in length to a
circle. The following approximate con-
struction is one of the simplest. It is
due to Kochansky (1685). At the ex-
tremity A of the diameter AB of a given
circle of radius r draw a tangent CD, making <^ COA =30° and
CD = 3r. Taking BD as semi-circumference is equivalent to taking
what value for tt? Carry work to four decimal places.

712. A running track consists of two parallel straight stretches,
each a quarter of a mile long, and two semi-circular ends, each a
quarter of a mile long at the inner curb. If two athletes run,
one 5 ft. from the inner curb and the other 10 ft. from it, by how
much is the second man handicapped?

713. Show how to go into the field and lay out a running track
of the dimensions given in the last exercise.

d714. A conduit for carrying water is
circular in form and is 10' in diameter.

(a) Fmd the length ABCD of the por-
tion of the circular outline which is wet
when the water reaches AD and ^AED
is 120°.

(6) Whatisthe length of AfiCD if <^AEZ>
is 60°?

This so-called "wetted perimeter" is of
the greatest importance in determining
friction, and therefore the resistance of the pipe to the water
flow.

d716. Find the length of the forty-second parallel of latitude,
assuming the radius of the earth to be 4000 miles.

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PLANE GEOMETRY

Theorem 64. The area of a circle is equal
to one-half the product of its radius and its
circumference.

Given: G of radius r, area C, and perimeter c.
Prove: C^icr,

Proof

Circumscribe about QO a regular Write out the proof, giving all
polygon of n sides and let P be its authorities,
perimeter and A its area.

(1) Show that A sirP.
Let n increase indefinitely.

(2) Then A =C and P=c.

(3) .-. irP = irc.

(4) .'. Csirc.

Cor. 1. The area of a circle is equal to n times the square of

its radius.
(Substitution left to the student.)
Cor. 2. The areas of circles compare as the squares of their

rudii.
(Proof left to the student.)

A portion of the area of a circle enclosed by two radii and their
intercepted arc is a sector.

EXERCISES. SET LXXIII. AREA OF CIRCLE

g716. Construct a graph by means of which the area of a circle
of any given diameter may be obtained.

717. Show that the area of a circle is equal to that of a triangle,
whose base equals the length of the circumference of the circle
and altitude equals the radius. This was proved by Archimedes.

718. Fill in the blanks in the following table:

Perimeter Pi

Area Ai

Perimeter Pj

Area At

Square

300

Circle

300

719. Show by means of a carpenter's square how to find the
diameter of a circle having the same area as the sum of the areas
of two given circles.

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193

d720. If a one-inch pipe will empty 2 barrels in 15 min., how
many barrels will an 8-in. pipe empty in 24 hrs.? (Make no allow-
ance for friction.)

d721. In putting up blower pipes, two cir-
cular pipes 11 in. and 14 in. in diameter re-
spectively join and continue as a rectangular
pipe 14 in. in width. Find the length of the
cross-section of the rectangular pipe.

722. A convenient formula used in practical
work for finding the area of a '^hollow circle"

or nng is:

^_irt(D+d)

Establish this formula.

723. A horse tied by a rope 25 ft. long at the corner of a lot
50 ft. square, grazes over as much of the lot as possible. The next
day he is tied at the next corner, the third day at the third comer,
and the fourth day at the fourth corner. Draw a plan showing
the arcs over which he has grazed during the four days, using a
scale of \ inch to 5 feet. Calculate the area.

724. Justify the following rule used by
sheet-metal workers or show the per cent of
error if it is incorrect:

Draw radius AO±OB. Extend each one-
fourth its own length to C and D. Then the
sect CD is the side of the
square of the same area
as that of the circle.
d726. Construct a square with a sides. With
the vertices as centers, and s as radius, construct
arcs as in the figure. Find
the perimeter and the area
of the shaded portion boimded by the arcs.

d726. In the design shown in this figure,
the side of the square is s. The inscribed
semicircles are tangent to the diagonals. Find
the perimeter and the area of the shaded por-
tion of the figure.

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PLANE GEOMETRY

727. Construct an equilateral triangle. With each vertex as
center, and with one-half a side as radius, describe arcs as indicated
in the diagrams. Let 2r represent the length
0^ of a side of the equi-

lateral triangle. Find
the perimeter and the
area of the figure
bounded by the arcs.

728. Modify the pre-
ceding exercise by using
the mid-point of the sides as centers, as indi-
cated in the diagrams.

729. Inscribe an equi-
lateral triangle in a circle
of radius 2r. Using the
mid-point of each radius
of the triangle as center
and r as radius, describe circles. Find the perimeter and the area of
the trefoil and of the shaded part of the resulting
sjonmetric pattern. (Using S for the area of
the shaded portion, C for that of the large circle,
A for that of the small circle, T for that of the
triangle, give the formula for S in terms of r,
r, C, and A.)
730. In the papyrus of Ahmes, an Egjrptian,
the area of a circle was found by subtracting from the diameter one-
ninth of its length and squaring the remainder This was equi-
valent to using what value of x?

731. In the Sulvasutras, early semi-theological writings of the
Hindus, it is said: "Divide the diameter into 15 parts and ta'ke
away 2; the remainder is approximately the side of the square,
equal to the circle." From this compute their value of tt.

d732. The proposition of the so-called
lunesof Hippocrates (ca. 470 b.c.) proved
a theorem that asserts in somewhat more
general form, that if three semicircles be
described on the sides of a right triangle as

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THE CIRCLE 195

diameter, the liines L and Li as shown in the diagram are together
equivalent to the triangle T, Prove it.

d733. A problem of interest is one that Napoleon is said to have
suggested to his stafif on his voyage to Egypt: To divide a circle
into four equal parts by the use of circles aJone.

LIST OF WORDS DEFINED IN CHAPTER Vm

Circle, center, radius, diameter; central angle, arc (minor, major); chord,
intercept, subtend; tangent to a circle, secant, tangent circles (internally,
externally); line of centers,- common chord, inscribed angle; inscribed, circimi-
scribed regular polygons; center, central angles, radius, apothem of regular
polygon; sector. Constant, variables (increasing, decreasing); limits (inferior,
superior).

AXIOMS OF VARIABLES IN CHAPTER Vm

1. If any variable approaches a limit, any part of that variable approaches
the same part of its limit, as limit.

2. If two variables are always equal, the limits which they approach are equal.

POSTULATE OF PERPENDICULARS (third) IN CHAPTER Vm

1. The shortest distance from a point to a line is the perpendicular to that
line.

SUMMARY OF THEOREMS PROVED IN CHAPTER Vm

42. Three points not in a straight line fix a circle.

43. In equal circles equal central angles intercept equal arcs, and
conversely.

44. In equal circles, equal arcs are subtended by equal chords, and
conversely.

45. A diameter perpendicular to a chord bisects it and its subtended arcs.

Cor. 1. A radius which bisects a chord is perpendicular to it
Cor. 2. The perpendicular bisector of a chord passes through the
center.

46. In equal circles, equal chords are equidistant from the center, and
conversely.

47. A line perpendicular to a radius at its outer extremity is tangent to
the circle.

Cor. 1. A tangent to a circle is perpendicular to a radius drawn

to the point of contact.
C<m:. 2. The perpendicular to a tangent at the point of contact

passes through the center of the circle.
Cor. 3. A radius perpendicular to a tangent passes through the point

of contact.
Cor. 4. Only one tangent can be drawn to a circle at a given point

on it.

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196 PLANE GEOMETRY

48. Sects of tangents from the same point to a circle are equal.

49. The line of centers of two tangent circles passes through their point
of contact.

50. In equal circles central angles have the same ratio as their intercepted
arcs.

Cor. 1. A central angle is measured by its intercepted arc.

51. Parallels intercept equal arcs on a circle.

52. An inscribed angle or one formed by a tangent and a chord is measured
by one-half its intercepted arc.

Cor. 1. An angle inscribed in a semicircle is a right angle.

53. An angle whose vertex is inside the circle is measured by half the sum
of the arc intercepted by it and by its vertical angle.

54. An angle whose vertex is outside the circle is measured by half the
difference of the intercepted arcs.

55. A tangent is the mean proportional between any secant and its external
sect, when drawn from the same point.

56. A circle may be circumscribed about, and a circle may be inscribed in,
any regular polygon.

Cor. 1. An equilateral polygon inscribed in a circle is regular.
Cor. 2. An equiangular polygon circumscribed about a circle is
reg^ular.

57. If a circle is divided into any nimiber of equal arcs, the chords joining
the successive points of division form a regular inscribed polygon, and the tan-
gents drawn at the points of division form a regular circumscribed polygon.

Cor. 1. Tangents to a circle at the vertices of a regular inscribed
polygon form a regular circumscribed polygon of the
same number of sides.

Cor. 2. Lines drawn from each vertex of a regular polygon to the
mid-points of the adjacent arcs subtended by the sides of
the polygon form a regular inscribed polygon of double
the number of sides.

Cor. 3. Tangents at the mid-points of the arcs between consecutive
points of contact of the sides of a regular circumscribed
polygon form a regular circimiscribed polygon of double
the number of sides.

Cor. 4. The perimeter of a regular inscribed polygon is less than
that of a regular inscribed polygon of double the num-
ber of sides, and the perimeter of a regular circumscribed
polygon is greater than that of a regular circumscribed
polygon of double the number of sides.

58. A regular polygon the number of whose sides is 3*2^ may be inscribed
in a circle.

59. If in represent the side of a regular inscribed polygon of n sides, and i^n
thesideof oneof 2n sides, and r the radiusof the circle, ijn = V 2r* — r\/4r*— in**

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60. If in represent the side of a regular inscribed polygon of n sides, Cn
that of a regular circumscribed polygon of n sides, and r the radius of the circle,

2rin

61. The perimeters of regular polygons of the same number of sides com-
pare as their radii, and also as their apothems.

62. Circumferences have the same ratio as their radii.

Cor. 1. The ratio of any circle to its diameter is constant.
Cor. 2. In any circle c=2irr.

63. The value of tt is approximately 3.14159.

64. The area of a circle is equal to half the product of its radius and its
circumference.

Cor. 1. The area of a circle is equal to ir times the square of its

radius.
Cor. 2. The areas of circles compare as the squares of their radii.

EXERCISES. SET LXXIV. MISCELLANEOUS
The problems in this set are miscellaneous in that their solution
may depend upon any part of the text; but they are arranged, in
general, in the order of difficulty. Those problems requiring the
application of trigonometric ratios are preceded by "t."

734. Find the num- 78 ft.

ber of feet of lime-line I I U-^ ^

of a tennis-court, as
represented. ^

736, Knd the num- s;
ber of yards of lime-
line for a football field,
which is 300 ft. by 160
ft., including all the ten-yard lines. How long would it take a run-
ner to cover the total distance if he can make 40 feet in 12 seconds?
736. Construct an accurate diagram of a rectangular garden
with a border inside it one-fourth the width of the gi^den.

1^ 737. A designer, in making

a pennant, must make one in

B the same proportion as a given

one, but larger. Find AB, if

D CD = 36", AiBi = 43", and

Ci2)i = 20".

21ft.

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738. By use of the steel square lay out an angle of 45**.

739. By use of steel squares lay out an angle of 60°.

740. A straight railroad AB strikes a mountain at C, and a

tunnel is to be dri ven at C in
the direction ABC. It is de-
sired to commence the work at
the other side of the mountain
at the same time as the work
is progressing from C. -^ABD
is made 150°, BD = 3 miles, and

^ ^Z) = 60°. How far from Din
DE must the tunnel be driven, and in what direction?

741. An instrument for leveling A ,^ '^

consists of a rectangular frame ABCD.
E and F are the mid-points of -4 Sand
DC, respectively. A plumb-line is
suspended from E. Show that when
the plumb-line coincides with the
mark F, DC is horizontal.

This instrument is shown in French books.

jgt 742. The accompanying

diagrams represent an in-
strument for locating the
center of circular discs.
Three pieces of metal are so
joined that AB bisects the
angle formed by BD and BE,
which two sects are equal.
Prove that AB passes through the center of the circle.

743. A carpenter bisects an angle by the
following rule: Lay off AB=AC. Place
a steel square so that BD=CD as shown
in the diagram. Draw the line AD. Is
this method correct? Give proof. Would
this method be correct if the steel instru-
ment did not have a right angle at D?

744. Answer the following questions without proof:

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(o) Are all equilateral polygons equiangular?
(6) Are the diagonals of a parallelograni equal?

(d) Do the diagonals of a parallelogram bisect its angles?

(e) Under what circumstances do two chords bisect each other?

(f) An arc of 30° is subtended by a chord of 6". In the same
circle will an arc of 60° be subtended by a chord equal to, ^eater
than, or less than 5"?

(g) Is it possible to inscribe a parallelogram in a circle?

746. In an ideal honeycomb the cells are 6 sided. Why is this?
In what other regular form might they be built and yet fit snugly?

746. The wheel of an automobile makes 110 revolutions per
minute. If it measures 2 ft. 6 in. in diameter, find the speed of the
machine.

747. Walking along a straight road a traveler noticed at one
milestone that a house was 30° off to the right. At the next mile-
stone the house was 46° off to the right. How far was the house
from the road? Is there more than one solution?

748. Calculate the diameter of the circle of water visible to an
observer at sea, (a) when seated in a small boat, his eyes being 4 ft.
above the surface of the sea, (6) when on the bridge of a steamer
25 ft. above the surface, (c) when at the masthead 60 ft. above the
surface, (d) when on the top of a mountain 3000 ft. above sea level.
(e) How far above sea level does the elevation of the observer
begin to make a perceptible difference?

a749. In finding the diameter of a wrought-iron shaft that will

transmit 90 horse-power when the number of revolutions is 100

per minute, using a factor of safety of 8, we have to find the

3/___90_

diameter d from the formula d= 68.5 \ / 50000 . Rnd d.

' Y lOOX^'o

750. Draw any quadrilateral ABCD. Take such measurements
of your figure as you consider necessary and sufficient, and from
your measurements construct the quadrilateral a second time.
State what measurements you make, and how you draw the second
figure. Cut out the two figures and fit one upon the other.

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PLANE GEOMETRY

(6) Discuss a number of other sets of measurements that you
could use to reproduce the quadrilateral ABCD. How many meas-
urements must there be in every set?

761. Find the nimiber of square
feet in the floor of the room shown
in the accompanying plan.

762. Find the area of the ac-
companying polygon by filling out

the following table, assuming reasonable values for necessary
dimensions. (Fig. 1).

Parts of
Polygon

Factors

Products

Bases, or sums of
parallel sides

Altitudes

AABBi

BBi=6.8

ABi=5.Q

38.08

2) '

Folygon ABCDEFGH =

753. Show how to find the area of
polygon ABCDEFGHK, assuming the
shaded portion to be inaccessible.

Fig. 2.

754. Using goods 20 in. wide, how many strips will it take,
cut on true bias, to put a band 12 in. wide around a skirt 3 yds.
wide?

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766. In the accompanying diagram how do (a) the perimeters
and (6) the areas of the circle and the curvi-
linear figures ADCFBX compare? (c) Use
this figure as a suggestion and show how,
by means of arcs, to divide a circle into any
nimiber of equal areas.
766. This is the cross-
section of a foot-stool,
in which the width of
the top is to be 12 in., d 8 in., e 12 in., and
the lengths of the legs 8 in. In making the
stool, angle a and angle P are first laid out
on paper. Show how, from the required di-
mensions, to lay out these angles on paper.
757. To find the diagonal of a square, multiply the side by 10,
take away 1% of this product, and divide the remainder by 7.
Test the accuracy of this rule of thmnb used by some carpenters.

768. Construct a perpendicular at the
end of a sect without producing the sect.

Hints: Let AB be the given sect. With any
point C, between A and B, but outside the sect, as
center, and radius CB, describe a major arc inter-
secting AB at E. Draw the diameter EK, KB is
the required perpendicular.

Prove that this construction is correct.

769. The resultant of two forces acting upon a body is 400 lbs.
One of the forces is 250 lbs. What are the limiting values for the
other force?

760. A man decided to buy some nmnerals and make the face
of a grandfather's clock, but when he came to divide the face into
minutes he found that he was not able to do it without gues-
sing. How could it be done accurately?

t761. A kite string is 250 ft. long, and makes an angle of 40*^
with the level ground. Find (approximately) the height of the
kite above the ground, disregarding the sag in the string.

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762. Fill out the blank spaces in the table by referring to the
diagram. (Fig. 1.)

i^4

Xnd

35°

40°

80**

30^

30°

50°

190°

105**

763. The accompanying drawing is one

of the earliest of Gothic O

tracery windows. The

archil CB is based on

an equilateral trian- ^

gle. A is the center ^

and AB theradius

for the arc CB,

andB thecen- a

ter and AB

the radius for ' ^^'^

arc AC. D is the midrpoint of span AB. The arches AED
and DFB are drawn on the half span by similar construction.
Find the center and the radius of the circle with center that
shall be tangent to the four arcs, DE, DF, AC, CB.

Suggestions: With A as center and AH as radius cut the altitude CD at 0.
H is the mid-point of DB,

764. How many degrees are there in each of the angles of the
Pythagorean badge?

766. From a strip of metal 2^" wide
it is desired to cut off a rectangle from
which two circular disks 2" in diameter
can be cut. What length AB must be

cut off? A B

766. In taking soundings to make a chart of a harbor it is
necessary at each sounding to determine the position of the boat.
This is sometimes done by measming with a sextant the angles
between lines from the observer to three range poles on the shore.
When a chart is made, the position on the chart of each sounding
is sometimes found as follows: The points A, B, and C represent
the positions of the three range poles. Suppose the angles read

:V:

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THE CIRCLE 203

by the sextant were 50°, and 35° (50° between lines to A and B,
and 35° between lines to B and C). On the chart lay oflF at A and
B, angles BAM and ABM, each equal to 40°, and find the point
M. At B and C, lay off angles CBN and BCN each equal to 55°
and find N. With M and N as centers, draw circles passing through
B. The other common point of these circles is the position of the
sounding. Prove the correctness of this construction. Make the
construction to scale.

767. The resistance offered by the air to the passage of a bullet
through it varies jointly as the square of its diameter and the
square of its velocity. If the resistance to a bullet whose diameter
is .32 in., and whose velocity is 1562.5 ft. per second, is 67.5 oz.,
what will be the resistance to a bullet whose diameteris .5 in., and
whose velocity is 1300 ft. per second?

768. Roman surveyors, called (igrimensores, are said to have
used the following method of measuring the width of a stream:
A and B were points on opposite sides of the stream, in plain
view from each other. The distance AD was then taken at right
angles to AB, and bisected at E, Then the distance DF was taken
at right angles to AD, such that the points B, E, and F were in a
straight line. Make a drawing illustrating the above method, show
what measurement affords a solution, and prove that this is so.

769. A student lamp and a gas jet illuminate a screen equally
when it is placed 12 ft. from the former and 20 ft. from the latter.
Compare the relative intensities of the two lights.

770. An endless knife runs on pulleys 48" in diameter at the
rate of 180 revolutions per minute. If the pulleys are decreased
18" in diameter, how many revolutions per minute will they have
to make to keep the knife traveling at the same speed?

771. In surveying, to determine
a line from the inaccessible point P
perpendicular to AB, lay off FE
perpendicular to AB at an arbit-
rary point and of any length. ^
Make EH=EF. Obtain point D ^^
in lines PF and AB; next, point N in HP and AB; next, K in DH
and FN. PK is perpendicular to AB. Why?

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PLANE GEOMETRY

772. Make the construction shown in the diagram, and -4fi will
be approximately the quadrant of the
circle. Find the per cent by which it
i differs from the correct value.

773. The steel square may be tested by

'measuring across from the 9-inch point

of the tongue to the 12-inch point of the

blade. If this distance is exactly 15

inches, the square is true. Why?

774. Four of the largest possible equal sized pipes are enclosed

in a box of square cross-section 18 in. on an edge. What part of

the space do the pipes occupy?

dt776. A boy pulls a sled with a force of twenty-five pounds by
means of a rope ten feet long, and with his hands three feet from the
ground. Find the component of his force effective in pulling the
sled forward.

776. A camp kettle weighing 20 lbs. is suspended on a wire
from two trees 10 feet apart. The wire is 20 feet long, and the
kettle is suspended on it at a point midway between the trees.
Find the tension on each strand of the wire.

777. A running track having two parallel sides and two semi-
circular ends, each equal to one of the parallel sides, measures
exactly a mile at the inner curb. Two athletes run, one at the
inner curb, and the other 10 ft. from this curb. By how much is
the second man handicapped?

778. The last row of seats in a circular tent is 30 ft. away from
the central pole, which is 20 feet high, and which is to be fastened
by ropes from its top to stakes driven in the ground. How long
must these ropes be in order that they may be 6 feet above the
ground over the last row of seats, and at what distance from the
center must the stakes be driven?

779. A wheelwright is given a part of a
broken wheel to make a duplicate. To do
this he needs the diameter. He measures
the chord of the arc given him, 24"; the
height of the segment is 4". How large is
the wheel?

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THE CIRCLE 205

dt780. The width of the gable of a house v/^'^N.

is 35 ft. The height of the house above the , >^ fe, x. -
eaves is 15 ft. Find the length of the rafters ^^ t^"^^"" nJ
and the angle of inclination of the roof. 1 j

781. An aeroplane travels 1000 ft. upwards, Sj miles due west,
and 2\ miles due north. Find its distance from the starting-point.

782. The sides of a floor are 10 ft. and 6 ft. A man wishes to
tile it with tiles in the shape of a hexagon whose sides are 6 in.
How many tiles will it take?

783. A flat circular sheet of metal is to be stamped into the
form of a spherical segment with a flange. The figure shows a
cross-section of the resulting piece of metal, a being the width of the

^^-«;;j^^ spherical segment, 6 the depth or altitude

//^U^^^^^^ of ^he segment, and c the outer diameter

r — 'k —a—^ y^^ of the flange. The problem is to determine

^ ^ the size to cut the sheet metal in order

that when stamped the piece may have these dimensions. Show
that the required radius of the circular sheet equals e in the figure.

784. To lay off the length of a brace with the steel square:
Suppose that the post is 4 ft. and the beam 3 ft. Apply 3 times
to the timber from which the brace is to be cut, the distance across
the square from the 12 in. point of the tongue to the 16 in. point
of the blade. Why will this give the required length of the brace?

t786. In railroad construction and mining the material is
sometimes hauled in a tram pulled by a horse. If the pull of the
tram in the direction of the track is, say, 200 lbs., and if the horse
walks at the side of track so that its pull is exerted at an angle of
25° with the track, what pull must the horse exert?

786. The force that the wind exerts normally (per-
pendicularly) on the sail of a boat is resolved into
two components: one useless, in pushing

the boat sidewise in the water in '^ — ^Z^ ^ ^ — ^% — j^**^
spite of the keels; and one useful com-
ponent driving the boat directly forward. If, as in the following
diagram, the sail is at an angle of 30° to the keel, and a force of
wind of 100 lbs. acts on the sail (considered as applied at one point),
find the effective value of the two components mentioned above.

Note. — ^A similar problem can be applied to the aeroplane. ^ j

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787. An angle inscribed in a circle varies directly as the inter-
cepted arc. Show that in this case k=^.

788. A street-car track is 12' from the curb (GF^BC =12").

In passing the comer of two streets
which deflect through an angle of
60° the rail must be 5' (DS=50
from the comer, (a) Find the
radius of the curve. (6) Find the
length of the tangents from G and
C to their point of intersection,
(c) Find the length of arc GC; also
the length of the outer arc. The

width of the track is 4'&^". (All these curves are arcs of circles.)

789. Where two straight streets intersect, each comer is usually
"rounded oflf." Show that the problem of laying out the comer
arc is a simple one where the streets made an angle of 90°. Show
by a plan how to lay out a comer where the angle is 45°, and the
radius for the curbing is to be 20 ft. Find the total length of
curved curbing needed.

dt790. A given regular polygon has n sides. How many meas-
urements of the figure are both necessary and suflScient to deter-
mine it in size and shape?

dt791. A flagstaff is seen in a direction due north of a station
il at an elevation of 17°, and from a station B 120 ft.- due east of
A the flagstaff bears 23° west of north. The two stations and the
foot of the flagstaff being at the same level, determine the height
of the flagstaff.

792. A straight street intersects
one which is curved (with a large
radius, such as 200 ft.), and the
comer is to have a small radius,
say 15 ft. Show in a plan how to
find the center for the corner arc
by means of the intersection of two loci. (Fig. 1.)

How would you get, in actual field work, the curved locus?
(Note its large radius.) (Fig. 2.)

793. Solve in another plan the corresponding problem where
both streets are curved.

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207

•^ ^J3<'1-

794. The "gear" of a bicycle is the diameter (in inches) of a
wheel whose circumference would equal the distance gone forward
with one revolution of the pedals. Find the gear if the diameter
of the rear wheel is 28'^ and the front and rear sprockets have 22
and 8 teeth, respectively.

796. A belt runs over two pulleys, one of which is 4 ft. in dia-
meter, and driven by an engine at the rate of 100 revolutions a
minute. What must be the diameter of the other pulley if it is
to turn a fan at the rate of 400 revolutions a minute?

796. The figure is the diagram of a part

of the side of a bridge. The point C must

be located on AB and on DJS?, where the

holes must be bored to fasten the brace

AB to the upright DE, Required to find

the lengths of AB^ x, y, and 2, in order to

locate C.

Suggestion: After finding AB, compare
triangles ACF and ABO.

797. It is desired to construct a subway

under a river. The bank on one side has

a 30% slope from the edge of the river to

the river bed. The maximimi effective

grade of a subway is 5%. On the surface

it is five hundred feet from the bank of

the river to the bed of the river. Deter-
mine the necessary length of the subway
from the bank to the river bed.

798. WW is a wall with a round cor-
ner of dimensions as given in the figure
from A to fi, on which a molding, gutter,
or cornice isto be placed. Find theradius
of thecircleof which thearciliVBisapart.

799. il, B are two beacons on a coast-
line; jS is a shoal off the shore, and the
angle ASB is known to be 120®. Show

that a vessel V sailing along the coast will keep outside the shoal
if the angle AVB is always less than 110°.

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208 PLANE GEOMETRY

Prove a property of the circle that you used in proving that the

ship will clear the shoal.

. 800. The method given by Galileo for finding the strongest

rectangular beam that can be cut from a round log is as follows:
Let the circle ABCD represent the end of the
log, and let AC be a diameter. Divide the
diameter into three equal parts at the points
M and N, and from these points erect perpen-
diculars intersecting the circumference at
points D and B. Draw AD, DC, CB, and BA .
The rectangle thus formed is the cross-section
of the strongest rectangular beam. Show

that the dimensions of the rectangle are in the ratio of 1 to \/2.
dtSOl. The resultant of two forces is 300 lbs. One of the

forces acting at an angle of 37"* with the resultant is 100 lbs.

Find the other force, if the forces act at an angle of 65° with

each other.

dt802. The radius of a circle is 7 ft. What angle will a chord of

the circle 10 ft. long subtend at the center?

803. To prolong a Hne through an obstacle and to measiu-e the
distance along the line through the obstacle.

A and B are two points on

the given line. Take X any — ^

point. Measure AX. Take

XC—AX and in line with AX.

Mark Y on the mid-point of

XC. Make DX=BX and in

UnewithBX MakeZF=F2). ^

Make ZE^ZD and in line with ZD. Make ZF=ZC and in line

with ZC. Prove FE in line with AB. What line of the figure

equals BF?

804. The cross-section of the train-shed of a railroad station is
to have the form of a pointed arch, made of two circular arcs, the
centers of which are on the ground. The radius of each arc equals
the width of the shed, or 210 ft. How long must the supporting
posts be made which are to reach from the ground to the dome of
the roof?

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g806. Show by a graph that the area of a triangle having a
fixed altitude varies as the base, and that one having a fixed
base varies as the altitudp.

g806. Construct a graph showing the relation between the areas
and the sides of equilateral triangles.

g807. Construct a graph showing the relation between a side
and the area of a regular hexagon. By means of it find the area of
a regular hexagon whose sides are 6. Compare with the computed
area.

g808. If a side of a given regular polygon is a and its perimeter
p, graph the relation of the perimeters and the corresponding sides
of polygons similar to the given polygon.

g809. Given a polygon with area A and a side a. Construct a
graph showing the relation between the areas and the sides corre-
sponding to a in polygons similar to the one given.

dtSlO, The figure shows a
method for determining the
horizontal distance PR, and
the difference of level QR be-
tween two points P and Q. A
rod with fixed marks A and
B upon it is held vertical at
Q, and the elevation of these
points ACD ( = a) and BCD
( = /5) are read by a telescope and divided circle at C, the axis
of the telescope being at a distance CP ( = a) above the ground
at P. If .QA = 6 and AB^s, write down expressions for PR
{=z)mdQR{=y).
Find X and y when a =6^ 10' and /3=7° 36', the values
of a, 6, and « being 5 ft., 2^ ft., and 5 ft.
respectively.,

dtSll. Find the radius of a parallel of lati-
tude passing through Portland, Me. (43** 40^
N. lat.), if the radius of the earth is taken as
4000 mi.

(Note that in the figure <x equals <y.
Why?)

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812. The oval in the figure is a design used in the construction
of sewers. It is constructed as follows: In the circle let CD,

the perpendicular bisector of AB, meet
AB at Oi. Arcs AM and BN are drawn
with AB as radius and with centers
B and A respectively.

The chords BOi and AOi meet these
arcs in M and N respectively.

The arc MDN has the center Oi and
radius OiM.

(a) Make an accurate construction
of the design.

(6) Is arc ACB tangent to arc AM
and arc BN at A and B respectively?
Why?

(d) If AB equals 8 ft., find BOi, and hence OiM, and finally CD.
That is, if the sewer is 8 ft. wide, what is its depth?

d (a) If the width of the sewer is a ft., show that its depth is

|(4-V2).
d (/) If the depth of the sewer is d ft., show that its width is

d (g) Compute to two places of decimals the width of a sewer
whose depth is 12 ft.

813. The circumference of the earth is approximately 25000
miles. Suppose an iron band 25000 miles long fits tightly around
it. If you cut the band and put in three feet, will the band then be
raised any appreciable distance from the earth?
As much as ^ inch, for example? yj^ inch?
Consider the band, when enlarged, to be raised
an equal distance from the earth at all points.

814. In this design, the side of the square
upon which it is constructed is 4a. Find the
area of the shaded portion.

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211

dt816. In constructing a sail, the amount of surface of canvas

ABCD is known, and the lengths of AB, AD, and DC are given.

The angle A is a right angle. Show how to

construct the angle between DC and DA.
dt816. Find the number of square yards

of cloth in a conical tent with a circular ^

base and the vertex angle 72®, the center

pole being 12 ft. high.
817. A girder to carry a bridge is in the

form of a circular arc. The length of the ^

span is 120 ft., and the height of the arc is 25

ft. Find the radius of the circle.

dtSlS. In the side of a hill which slopes
upward at an angle of 32®, a tunnel is bored
sloping downwards at an angle of 12® 15' with
the horizontal. How far below the surface of
the hill is a point 115 ft. down the tunnel?

819. In constructing a gas engine the piston
D, which is in the form of an inverted cup, is
5 in. in inside diameter; the crank ABia&in.
between the centers of the pivots, and the con-
necting rod AC is 17 in. between the centers
of the pivots. How far from the mouth of
the cup must the pin C be adjusted in order
that the connecting rod may just clear the

edge of the cup at E and F, the diameter of AC being 1 in.?

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PART II

SECOND STUDY

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CONTENTS

PART TWO— SECO>fD STUDY p^^.

SUOQESTIONS POR A REVIEW OP THE FiRST SXUDY 219

Plates: Family Trees 220, 221, 222, 223, 224

CHAPTER I

Fundamentals. Rectilinear Figures

Exercises. Set LXXV. Triangles 229

Exercises. Set LXXVI. Perpendiculars, Parallels, Sums of Angles

of Polygons 233

Exercises. Set LXXVII. Parallelograms ...235

Exercises. Set LXXVIIl. Inequalities 237

Sttmmary op Terms Introduced 238

CHAPTER II

Areas op Rectilinear Figures

Exercises. Set LXXIX. Areas 241

Summary op Terms Introduced 242

CHAPTER III

Similarity

Division op a Sect 244

Exercises. Set LXXX. Ratio, Proportion, Parallels 245

Exercises. Set LXXXI. Similarity of Triangles 250

Exercises. Set LXXXII. Similarity of Polygons 256

Exercises. Set LXXXIII. Metric Relations 262

CHAPTER IV
Locus

Concurrence 267

Exercises. Set LXXXIV. Locus ^ 269

CHAPTER V

The Circle

Exercises. Set LXXXV. The Straight Line and the Circle 274

Exercises. Set LXXXVI. Measurement of Angles 279

Exercises. Set LXXXVIL Metric Relations 286

Exercises. Set LXXXVIII. Mensuration of the Circle 291

216

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216 CONTENTS

CHAPTER VI
Methods of Proof page

A. DiBECT 297

Exercises. Set LXXXIX. Synthetic Methods of Proof :

I. Geometric 297

II. Algebraic 298

B. Indirbct 299

Exercises. Set XC. Proof by the Method of Exclusion 300

Exercises. Set XCI. Proof by Reduction to an Absurdity 301

Exercises. Set XCII. Analytic Method of Proof 304

Suggestions as to Method of Procedure 305

CHAPTER VII
Constructions. Methods of Attacking Problems

Exercises. Set XCIII. Synthetic Solutions 309

Discussion of a Problem ^ 311

Exercises. Set XCIV. Intersection of Loci 312

Formal Analysis of a Problem 318

Exercises. Set XCV. Problems CaUing for Analysis r . . 321

CHAPTER VIII
Summaries and Applications

A, Stllabus of Theorems 324

B, Syllabus of Constructions 334

C, Summary of Formulas 336

D, Summary of Methods of Proof 338

Exercises. Set XCVI. Congruence of Triangles 339

Exercises. Set XCVII. Equality of Sects 340

Exercises. Set XCVIII. Equality of Angles 340

Exercises. Set XCIX. Parallelism of Lines 341

Exercises. Set C. Perpendicularity of Lines 342

Exercises. Set CI. InequaUty of Sects 343

Exercises. Set CXI. Inequality of Angles '. 344

Exercises. Set CHI. Similarity of Triangles 345

Exercises. Set CIV. Proportionality of Sects 346

Exercises. Set CV. Equality of Products of Sects 346

Exercises. Set CVl. Miscellaneous Exercises 347

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CONTENTS 217

CHAPTER IX

College Entrance Examinations page

Chicago 362

Harvard 363

CHAPTER X

Suggestions

A. List of Topics Suitable for Students' Discussion:

General 374

Arithmetic 375

Algebraic 375

Geometric 376

B. Topics with Definite References:

Geometric Fallacies 376

Number Curiosities 376

Pythagorean Proposition 376

C. List of Books Suttablb for Students' Reading:

History 876

Recreations 377

Practical 377

General 378

Index of Definitions 379

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SUGGESTIONS FOR A REVIEW OF THE FIRST
STUDY OF PLANE GEOMETRY

Before beginning our Second Study of Plane Geometry, it might be well
for us to review the First Study. The following material furnishes a brief,
suggestive outline for such a review.

A. Congruence — ^Theorems 1-8
I. Define: *

Sect, polygon, axiom, postulate', corollary, adjacent angles, congruent,
homologous, perpendicular.
II. Summarize:

a. Conditions under which triangles are congruent in general;

b. Special conditions under which right triangles are congruent;

c. Facts about perpendiculars.
III. Family Trees of Propoaitiona:

To trace a proposition back to its sources, that is, back to the
definitions, postulates, and axioms upon which it rests, will be found
an interesting and profitable form of review. A convenient arrange-
ment is to make a "family tree" of a theorem, the branches of which
are the authorities quoted in proving the proposition. Each branch
should be followed down as in the main proposition until it ends in
a postulate, an axiom, or a definition.

Such a tree of Theorem 6 is given as an iUustration (Plate 1).
The student is advised to make a tree of Theorem 5.

B. Parallels — ^Theorems 9-12
I. Define:

Parallels, transversal, alternate-interior angles,
n. Classify angles according to :
a. Individual size;
&. Belative size;
c. Relative position.

III. Summarize:

a. Conditions under which lines are parallel;

b. Methods of proving sects equal;

c. Methods of proving angles equal.

IV. FamHy Tree:

A family tree of Theorem 11, Cor. 2, is appended (Plate 2).
The student is advised to study it and make one of Theorem 12.

♦ Consult the First Study only when necessary.

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220

PLANE GEOMETRY

111!

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REVIEW OF FIRST STUDY

221

a 1 2 1^-

a a 5 '"•

S I is
^ ? ?

¥:

a ^

i 5.

•^ Hi

rlfl

-I-

Sll — I
a

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J
d

L^D

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222 PLANE GEOMETRY

C. Sums of Anqlbs — ^Thbobbms 13-15
I. Define:

Exterior angle.
n. Summarue:

The facts about the sums of the angles, interior and exterior, of
an n-gon.
III. Make & family tree of Proposition 14.

D. PARALLELOORAM&— ThEOBKMS 16^21

I. Define:

Parallelogram, diagonal
U. Classify qttadrilaterala.

Is it possible, to classify them m more than one way? If so, is there
any preference?

III. Summarize:

a. Conditions under which a quadrilateral is a parallelogram;
h. Properties of a parallelogram.

IV. Make a family tree of Proposition 17.

E. Areas — Theorbius 22-27
I. Define:

Commensurable, measurement, area.
II. Summarize the facts about areas by giving formulas for the areas of

figures studied,
in. Complete the family tree of Theorem 27 as appended (Plate 3).

F. SiMiLARrrT— Theorems 28-39
I. Define.

Ratio, proportion, center of similitude, similar figures.
II. Summarize:

a. Properties arising from similarity of triangles;
h. Conditions under which triangles are similar.
ni. A family tree of Theorem 39, Cor. 2, is begun (Plate 4). The student is
advised to make a tree of Proposition 36.

G. Locusr— Theorems 40-41
I. Define:

Locus of a point.
II. State the facts of which proofs are necessary and sufficient to establish

a locus theorem.
m. State the two locus theorems proved in the syllabus.

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REVIEW OF FIRST STUDY

223

PLATE 4

P
ft?

J-S

The altitude upon the hypotenuse
of a right triangle divides it into
triangles similar to each other
and to the original.

Compe. of equal 2^ are equal.

{ Triangles are similar if two angles
of one are equal to those of
another.

The homologous sides of similar
triangles have a constant ratio.

The products of equals multiplied
by equals are equal.

The sums of equals added to

equals are equal.
The whole is equal to the simi of

its parts.

The homologous angles of similar!
triangles are equal. <

Post, of superposition.

If corresponding angles are equal J
lines cut by a -transversal are]
parallel. [

A line parallel to the base of a
triangle cuts the remaining sides
so that they are proportional to
either pair of homologous sects/

Quantities equal to the same
quantity are equal to each other.

(To be completed by the student.)

H. The Cibcle and Straight Line — ^Theobems 42-49

I. Define:

Circle, chord, secant, tangent.

II. Summarize the facts proved in the syllabus about:

a. Chords;

b. Tangents;

c. Arcs.

III. Make a family tree of Proposition 48.

I. The Circle and Angle Measurement — ^Theorems 50-55

I. Define:

Central angle, inscribed angle.

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224

PLANE GEOMETRY

II. State the method of measuring:
a. Central angles;
6. Inscribed angles;

c. Angles formed by a tangent and a chord;

d. Angles with vertices outside the circle;

e. Angles with vertices inside the circle.

111. A family tree of Proposition 55 is begun (Plate 5).
make one of Theorem 53.

The student should

J. Mensuration of the Circle — ^THEOREiis 56-64

.1. Define:

Sector, apotbem.
II. State formulas for:

a. Area of a regular polygon;

6. Circumference of a circle;

c. Area of a circle.
III. Summarize the methods now known to you of proving:

a. Sects equal;

&. Angles equal;

c. lines perpendicular;

d. lines parallel;

e. Triangles congruent;
/. Triangles similar;

g. Arcs equal;

i. Chords equal. •

PLATE 5

Th. 55. A tangent from a point to a circle is the mean
proportional between any secant and its external sect, from
the same point to the circle.

Any two quanti-
ties of the same
kind compare as
their numeric

C. Homologous
sides of similar
triangles have a
constant ratio.

A, An inscribed
angle, or one
formed by a tan-
gent and a chord is
measured by one-
half its intercepted
arc.

Note. — If the student is in need of further review, he is advised to trace
branches A, B, and C of this family tree back to their sources.

B. Triangles are
similar when two
angles of one are
equal each to each
to two angles of
another.

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CHAPTER I

FUNDAMENTALS. RECTILINEAR FIGURES

Theorem 1. Vertical angles are equal

The pupil is earnestly advised not to refer to the First Study*
for suggestions as to proofs of the theorems there taken up if it
is possible for him to work them out without any help in the review.
For this reason those theorems are stated in this part.

Theorem 2. Two sides and the included angle determine a
triangle.

Given: AABC and ADEF; ABs
Prove: AABC ^ ADEF.

■DE; ^Bs^^; and BC^EF.

(1) Place ADEF on AABC so that
^E coincides with :^B and ED fails
along BA.

(2) Then EF falls along BC.

V ED=^BA, Z> will faU on A.
VEF = BC,/?'willfaUonC.

(3) /. FD coincides with CA.

(4) /. ADEF ^ AABC.

Proof

(1) Superposition post.

(2) Data.

(3) Two points fix a straight line.

(4) Def . of congruence.

•Throughout "A Second Study of Plane Geometry" reference will be
made to "A First Study of Plane Geometry" as "First Study," and no proofs
will be given in this part of propositions contained in the former except in the
case of the congruence of triangles. In this instance, one proof by the method
of superposition will be given in full, since cutting and pasting were used
in the First Study owing to the difficulty of the usual proof for the beginner.
16 .226 g\e

226

PLANE GEOMETRY

Theorem 3. Two angles and the included side determine a
triangle.

Apply the method of proof used in Theorem 2.

What parts of the triangles will you first make coincident in superposing
in this case?
What postulate is needed in order to clinch this proof?

Theorem 4. The bisector of the vertex angle of an isosceles
triangle divides it into two congruent triangles.

Cor. 1. The angles opposite the equal sides of an isosceles

triangle are equal
Cor. 2. The bisector of the vertex angle of an isosceles tri-

angle bisects the base and is perpendicular to it.
Cor. 3. An equilateral triangle is equiangular.
Cor. 4. The bisectors of the angles of an equilateral triangle

bisect the opposite sides and are perpendicular to them.
Cor. 6. The bisectors of the angles of an equilateral triangle

are equal.

Theorem 6. A triangle is determined by its sides.

Theorem 6a. Only one perpendicular can be
drawn through a given point to a given line.

Givto: (I) Point P in AB; (II) Pomt P outside AB,

Prove: Only one line through P is perpendicular n

to AB.
Suggestion: (I) When P is in ZB, in how many
positions of PD will the st. 2^.APB be di-
vided into two right 2^.8?

Pboop (II)

(1) Suppose P25 meets ZB at rt. 2^«
at P, and PC is a line drawn to any
other point C in AE,

(2) Extend PZ> to Pi so that PiD
^PD.

(3) PCPi is not a straight line. (3) Why?

(4) /. 4.PCP1 is not a st.4.. (4) Why?

(5) APCD^APiCD. (5) Why?

(6) .-. 4.PCD = 2^PiCD. (6) Why?

(7) .'. 2S^PCD is not a rt. ^. (7) Why?

(8) .". PC±AB. I (8) Why?

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FUNDAMENTALS. RECTILINEAR FIGURES 227

Theorem 6b. Two sects drawn from a point in a perpendicular
to a given line, cutting off on the given line equal sects from
the foot of the perpendicular, are ^

equal and make equal angles with
the perperulicular.

Given: PQ±AB at Q; P any point in

PQ; QC^QD; C and D in AB.
Prove: PC^PD; ^CPQ^^DPQ, / X?

(Proof left to the student.) A C

In our first study, propositions dealing with inequalities were
with one exception omitted. The following group of propositions
supplies this omission, and before proving them it will be necessary
to study a set of axioms dealing with inequalities.

1. If unequah are operated on in the same way by positive equals,
the results are unequal in the same order.

e.g., If a>b and c^dy where c and d are positive quantities, a+c>

6+d, a -c>b -d, ac>bdy -> 3; while a^>b^ and Va>\/6 under

c a

certain conditions which do not afifect work in elementary geometry.

Test these statements, substituting for the sjmibols the following
values:

(1) a = 5, 6=3, c=d=4. (2) a=^, 6=^, c=d=4. -

(3) a=Z, 6= -Z, c=d=4. (4) a=i, 6= -^, c=d=3,
(5) a=-32V, 6=-i,c=d=3.

What conclusions can you draw?

What meaning is attached to the phrase "in the same order"?

2. The sums of unequals added to unequals in the same order (or
the same sense) are unequal in the same order.

e.g., If a>b, and c>d, then a+c>b+d.
Show why this statement could not be made for subtraction of
unequals.

3. The differences of unequals subtracted from equals are unequal
in the reverse order.

Illustrate this fact.

4. // the first of three quantities is greater than the second, which
in turn is greater than the third, all the more then is the first greater
than the third.

Illustrate this fact. r^ ^ ^ ^i ^

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228

PLANE GEOMETRY

Theorem 6c.

The sum of two sects drawn from any point
inside a triangle to the ends of one of
its sides is less than the sum of its
remaining sides.

Given: AABC; D any point inside.
Prove: DA+DCKBA+BC.

Proop
Extend AD to ^ in BC.

(1) DA-\-DE<whsit part of AB+ (1) Why?
BC? De<Z)^+whatpartofA5+BC?

(2) .-. DA+DE+DCKwhkt? (2) Why?

(3) .'.DA+DCKBA+BC. (3) Why?

Theorem 5d. Two sects drawn from a point in a perpendicular
to a given line and cutting off
unequal distances from the foot of
the perpendicular are unequal in
the same order as those distances,
and conversely.

I. Direct,
Given: PC ±XQUEB; CR>CQ.
Prove: PR>PQ.

Suggestions: Where will d Ke with res-
pect to C and fi if CQi=CQ? Why?

Why is it that whatever you prove true of PQi is true of PQ?

Why is it that whatever is true of PQi+PiQi and PR+PiR is true of

PQi and PR?
Write a complete proof of this theorem.

II. Converse.

Given: PC±AQCRB; PR>PQ.

Prove: CR>CQ.

Suggestions: Why can CR not be less than CQ?

Why can CR not equal CQ?

What remains for the relation of CR to CQ?

For conveniencje PQi+QiPi is at times referred to as PQiPi and
is called a broken line. Siich a line is always composed of two or
more sects.

The method of proof here outlined is known as the method of
exclusion or elimination. Any two quantities of the same kind
(a and b) must bear one of the following relations to each other:

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FUNDAMENTALS. RECTILINEAR FIGURES 229

a>6, a=6, or a<6. If any two of these relations can be shown to
be false, the remaining relation must therefore be true. For
further discussion and illustration see p. 299.

Cor. 1. AU possible obliques from a point to a line are equal
in pairs, and each pair cuts off equal sects from the foot
of the perpendicular from that point to the line.

(I) Under what condition will obliques from a point to a line be equal 7

Why, then will all possible obliques from a point to a line be equal in
pairs?

(II) Use the method of exclusion to prove the second part of the
corollary.

Theorem 6. The perpendicular is the shortest sect from a
point to a line.

Suggestion: Use Theorem 5d to give a much simpler proof than was
possible in the First Study.

Theorem 6a. The shortest sect from a point to a line is perpen-
dicular to it.

Hint: Show that this is a special case of Theorem 5d (converse)

Theorem 7. The hypotenuse and adjacent angle determine a
right triangle.

Theorems. The hypotenuse and another side determine a
right triangle.

EXERCISES. SET LXXV. TRIANGLES
Numeric

820. The perimeter of an isosceles triangle is 13, and the ratio
of one of the equal sides to the base is If. Find the three sides.

Theoretic

821. In proving triangles congruent, two methods have been
used.

(a) When what elements are given equal can superposition be
used?

(6) When must juxtaposition be used? Answer in a single brief
sentence.

822. The sects of any bisector of a given sect cut off by per-
pendiculars erected at the ends of the given sect are equal.

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230 PLANE GEOMETRY

823. The sects of a perpendicular to the bisector of an angle at
any point in it and limited by the sides of the angle are equal.

824. Sects joining any point in the bisector of an angle to points
on the sides equidistant from the vertex are equal.

826. If in the accompanying diagram,
^BAC = ^BCA, smd^BAY^^BCX, sect
XC = sect YA.

826. If the sides of an equilateral triangle
are prolonged in turn by equal lengths, and
the extremities of these sects are joined,
another equilateral triangle is formed.

827. Two isosceles triangles are congru-
ent if one of the equal sides and the
altitude upon that side are equal each to each.

828. Triangles are congruent if two sides and the altitudes upon
one of them are equal each to each.

829. A triangle is determined by a side and the median* and
the altitude to that side.

d830. If in the accompanying diagram
^A is a right angle, YX==AC, CP=PYy
show that PB+PX>CB+CA.

831. The sum of the distances from

any point inside a triangle to the vertices -^ X

is greater than its semiperimeter, but less than its perimeter.

832. The simi of the diagonals of a quadrilateral is greater than
the sum of either pair of opposite sides.

833. The sum of the diagonals of a quadrilateral is less than its
perimeter, but greater than its semiperimeter.

834. The sum of the medians of a triangle is less than one and a
half times its perimeter. (Prove this and the following exercise
without assuming the concurrence of the medians).

836. The simi of the medians of a triangle is greater than its
semiperimeter.

836. Each altitude of a triangle is less than half the sum of the
adjacent sides.

*A median is the sect between any vertex of a A and the midpoint of
the side opposite that vertex.

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FUNDAMENTALS. RECTILINEAR FIGURES 231

837. The sum of the altitudes of a triangle is less than the
perimeter.

838. Show that the bisector of the vertex angle of an isosceles
triangle is coincident with the altitude to its base.

Constrvction *

839. Bisect a reflex angle.

d840. To construct a triangle, having given a side, the median
and the altitude to a second side.

d841. To construct a triangle having given one side, the corre-
sponding median, and the altitude to another side.

Theorem 9. Lines perpendicular to the same line areparallel.

Theorem 10. A line perpendicular to one of a series of parallels
is perpendicular to the others.

Theorem 11. If when lines are cut by a transversal the alter-
nate-interior angles are equal, the lines thus cut are parallel

Cor. 1. // the alternate-exterior angles or corresponding angles
are equal when lines are cut by a transversal, the lines
thus cut are parallel.

Cor. 2. // either the consecutive-interior angles or the consecu-
tive-exterior angles are supplementary when lines are
cut by a transversal, the lines thus cut are parallel.

Theorem 12. Parallels cut by a transversal form equal alter-
nate-interior angles.

Cor. 1. Parallels cut by a transversal form equal corresponding
angles and equal alternate-exterior angles.

Cor. 2. Parallels cut by a transversal form supplementary
consecutive-interior angles and supplementary con-
secutive-exterior angles.

♦ For a discussion of methods of attacking problems in construction see
Chapter VII, p. 306, where additional exercises will also be found. At this
point, for instance, the topics, *'The Synthetic Method of Attacking a Problem"
(p. 309) and '*The Formal Analysis of a Problem" (p. 318) should be studied.

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PLANE GEOMETRY

Theorem 12a. Tuh> angles whose sides are parallel each to each
or perpendicular each to each are either equal or supplementary.

Given: Q) X^YX \\ BA] 2^72

\\BC.

(n) :fiI^X5A;ZiI\Zi
LBC.
Prove: (I) ^XYZ^^ABC
s^XtYZt; 4.ZrX,si80*'-
4ABCs4.Z2rX.
(II) ^XiYiZi^^ABC^
^Xt YiZi ; ^Zi YiXt =
180^-4.^BCs4Z,yxXi.
Suggestions: Using the con-
struction lines prove (I).
(II) If YiCi \\ BC andYiAiW BAt what relation exists between ^ABC

and ^AiYiCi?
How are YiCi and YiZi related? How YiAi and YiXil
Then how are 4' AiYiCi and ZiYiAi related?
Then how are ^ XiYtZi and ZiYiAi related?
Draw conclusions.

Theorem 13. The sum of the angles of a triangle is a straight
angle.

Cor. 1. A triangle can have but one right or one obtuse angle.

Cor. 2. Triangles having two angles mutually equal are mutu-
ally equiangular.

Cor. 3. A triangle is determined by a side and any two homolo-
gous angles.

Cor. 4. An exterior angle of a triangle is equal to the sum of
the non-adjacent interior angles.

Theorem 14. The sum of the angles of a polygon is equal to a
straight angle taken as many times less two as the polygon has
sides.

Cor. 1. Each angle of an equiangular polygon of n sides equals

the th part of a straight angle.

Cor. 2. The sum of the exterior angles of a polygon is two

straight angles.
Cor. 3. Each exterior angle of an equiangular polygon of n

sides is equal to the -th part of a straight angle.

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FUNDAMENTALS. RECTILINEAR FIGURES 233

Theorem 16. If two angles of a triangle are equal, the sides
opposite them are equal.
Cor. 1. Equiangular triangles are equilateral

EXERCISES. SET LXXVI. PERPENDICULARS. PARALLELS.
SUMS OF ANGLES OF POLYGONS

Numeric

842. Find the angles of an isosceles triangle if a base angle is
double the vertex angle.

843. If the vertex angle of the isosceles triangle is 30^; find the
angle formed by the bisectors of the base angles. If the vertex
angle is fi?

844. The bisector of the base angle of an isosceles triangle makes
with the opposite leg an angle of 53® 17'. Find the angles of the
triangle.

846. Find the angles of an isosceles triangle if the altitude is
one-half the base.

846. If the angle at the vertex of an isosceles triangle is 36®,
the bisector of a base angle divides the triangle into two isosceles
triangles. Find the lengths of all the sects in the diagram if a leg
of the given triangle is a and the base is 6.

847. What is the sum of the angles of (a) a hexagon, (6) a hepta-
gon, (c) an octagon, (d) a nonagon, (e) a decagon, (/) a polygon of
18 sides, (g) a polygon of 24 sides, (h) of 30 sides?

848. Find each angle of an equiangular (a) hexagon, (b) hepta-
gon, (c) octagon, (d) nonagon, (e) decagon.

849. In what polygon is the sum of the angles three times as
great as in a pentagon.

860. How many sides has a polygon if

(a) the sum of the interior angles equals 4 rt.-^*? 3 st. '^»?
6 rt. ^«? 8 St. ^.? 20 rt. ^«?

(6) the sum of the interior angles is 2 (or 3, or 4, or 5, or 6)
times as large as the simi of the exterior angles?

(d) the ratio of each interior angle to its adjacent exterior
angle is 2 to 1? 3 to 2? 5 to 1? a to 6? May a and 6 have any
values whatever?

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234 PLANE GEOMETRY

(e) each exterior angle contains 40°? 30°? 20°? 120°?
(/) each interior angle is^st. ^? fst. ^? a rt. ^? | rt. ^?
frt.<? fst.^?

Only what kind of polygon is considered in (d), (e),
and (/)?

d861. If the sides of a polygon are extended until they intersect,
a star polygon results. A five-pointed star is called a pentagram^
and is of historic interest as the badge chosen
by the followers of Pythagoras. Star polygons
may also be formed by chords of circles or
by certain combinations of polygons. What
is the sum of the vertex angles of a five-
pointed star? A six-pointed star? An n-
pointed star? (Make your solution general.)
d862. It A, B, Cy denote the angles of a triangle, ha, hb, the
altitudes upon BC, and AC, ta, h, the bisectors of A and B, find:

(a) ^tah. (h) ^hahb. (C) <tjla.

853. The number of all diagonals of a polygon of n sides is
^ Test this statement in several instances.

, Theoretic

864. The bisectors of a pair of consecutive-interior angles of
parallels cut by a transversal are perpendicular to each other.

866. The bisectors of a pair of alternate-interior angles of paral-
lels cut by a transversal are parallel to each other.

866. The bisector of an exterior angle at the vertex of an isosceles
triangle is parallel to the base, and conversely.

867. An exterior angle at the vertex of an isosceles triangle is
double a base angle.

868. State and prove the converse of Ex. 857

d869. If a leg of an isosceles triangle is produced through the
vertex by its own length, and its extremity joined to the extremity
of the base, the joining line is perpendicular to the base.

d860. The bisectors of the angles of a quadrilateral form a
quadrilateral the sum of whose opposite angles is equal to two
right angles.

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FUNDAMENTALS. RECTILINEAR FIGURES 235

Theorem 16. Either diagonal of a parallelogram bisects it.

Cor. 1. The parallel sides of a parallelogram are equal, and
the opposite angles are equal.

Cor. 2. Parallels are everywhere equidistant.

Theorem 17. A quadrilateral whose opposite sides are equal
is a parallelogram.

Theorem 18. A quadrilateral having a pair of sides both equal
and parallel is a parallelogram.

Theorem 19. A parallelogram is determined by two adjacent
sides and an angle; or parallelograms are congruent if two adjacent
sides and an angle are equal each to each.

Theorem 20. The diagonals of a parallelogram bisect each
other.

Theorem 21. A quadrilateral whose diagonals bisect each other
is a parallelogram.

EXERCISES. SET LXXVII. PARALLELOGRAMS
Theoretic

861. If in the accompanying diagram, DE \\FG\\ HK \\ BC and
KL\\OH\\EF\\AB.

(a) What is the sum of AD, DE, EF, FG, GH, HK, KL,
andLC?

(6) How, if at all, does
the length of AC afifect the
solution?

number of parallels affect the

solution? ^ ^ a K C

d862. The sum of the perpendiculars from any point in the base
of an isosceles triangle to the legs is constant, and equal to the
altitude upon a leg.

863. The sum of the perpendiculars from any point inside an
equilateral triangle to the three sides is equal to the altitude.

The following group of six propositions further supplies facts
relating to inequaUties of sects and angles omitted in the First
Study.

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236

PLANE GEOMETRY

Theorem 21a. The difference between any two sides of a
triangle is less than the third side.

Wnt: If c Ka-^-b, how can we express a in terms of h and c?

Theorem 21b. If two sides of a triangle are unequal, the angles
opposite them are unequal in the same order.

Suggestions: Make ACi-AC. Justify this con-
struction.
What relation exists between 2^1 and 2^2?
What relation exists between 2^2 and 43?
B^^ * ^ ^ O What relation exists between 43 and 4C?

Theorem 21c. // two angles of a triangle are unequal, the sides
opposite them are unequal in the
same order.

Suggestions: Make 41s 4B. Why?

What relation exists between AB,

AAi-^AiC, and AC?

Note. — Since Prop. 21c is the converse

of Prop. 21& an indirect proof would have

been i)06sible. Why? Can you see any

. advantage in giving a direct proof where

convenient?

Theorem 21d. If two triangles have two sides equal each to each

but the included angles unequal,
their third sides are unequal in the
same order as those angles.

Suggestions: Why is it desirable and why
possible that the triangles should be
placed together as suggested by the
accompanying diagram with AB (<AC)
in coincidence in the two triangles?
If we could break the sect BC into two
sects such as BP-\-PCi it would be
evident that BP'\-PCi>BCi.
Then our problem is to so locate P that PCi -PC.
What kind of triangle, therefore, is CPCi?
How, then, shall we draw CiP?

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FUNDAMENTALS. RECTILINEAR FIGURES 237

Theorem 21e. If two triangles have two sides equal each to
each, but the third sides unequal, the
angles opposite those sides are unequal
in the same order.
Suggestions: Use method of exclusion.

Theorem 21/. If one acute angle of a
right triangle is double the other, the
hypotenuse is double the shorter leg, and
conversely.

Suggestions: (a) Direct.

What part of a straight angle is 2a?
What kind of triangle is ACCi?
How are the sects CBy CCi, and AC related?
(b) Converse.
How is ACi related to AC?
What kind of triangle is ACCi?
How are the angles CiAC, BAC, and ACCi
related?

/
/

' Vo

/
/
/

EXERCISES. SET LXXVIII. INEQUALITIES
Numeric
864. Two sides of a triangle are 10" and 13". Between what
limits must the third side lie?

866. If two angles of a triangle are respectively 65** and
65®, which is the longest, and which the shortest side of the
triangle?

866. If one angle of a triangle is one-third of a straight angle,
and a non-adjacent exterior angle of the triangle is five-eighths of a
straight angle, which side of the triangle is the longest and which
is the shortest?

Theoretic

867. Either leg of an isosceles triangle is greater than a sect
connecting the vertex with any point in the base.

868. The sect joining the vertex of an isosceles triangle to any
point in the prolongation of its base is greater than either leg.

869. If one leg AB of an isosceles triangle ABC is pro-
duced beyond the base BC to a point 2), then ^ACD is greater
than ^D.

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238 PLANE GEOMETRY

870. If PQ^QRin the accompanying diagram, prove EP>ER.

871. If no median of a triangle is
perpendicular to the side to which it is
drawn, the triangle is not isosceles.

872. If any point in the prolongation
of a leg produced through the vertex of
an isosceles triangle whose base is shorter
than a leg is joined to the extremity of
the base, a scalene triangle will be formed.

873. If a vertex of an equilateral triangle is joined to any point
in the prolongation of the opposite side, a scalene triangle is formed.
Which is the longest and which the shortest side of e^h of the
triangles thus formed?

d874. Prove that the sect joining an extremity of the base of an
isosceles triangle to any point in the opposite leg is greater than
one sect cut off on that leg. Is it ever greater than either sect?
Under what condition is it less than one of the sects?

876. The diagonals of a rhomboid intersect obliquely, and the
greater angle formed by them lies opposite the greater side of the
parallelogram. (A rhomboid is an oblique O in which the ad-
jacent sides are imequal.)

876. If a triangle is not isosceles the median to any side is not
perpendicular to it, and the larger angle which it forms with that
side lies opposite the greater of the remaining two sides.

d877. Any point not on the perpendicular bisector of a sect is
unequally distant from the ends of the sect.

LIST OF WORDS DEFINED IN CHAPTER I

Exclusion, elimination, broken line.

SUMMARY OF AXIOMS IN CHAPTER I

1. If unequals are operated on in the same way by positive equals, the
results are unequal in the same order.

2. The sums of unequals added to unequals in the same order (or the
same sense) are unequal in the same order.

3. The differences of unequals subtracted from equals are unequal in the
reverse order.

4. If the first of three quantities is greater than the second, which in turn
is greater than the third, all the more then is the first greater than the third.

(For a simimary of theorems see Chapter VIII, p. 324.) C^r\r\o]f>

CHAPTER II

AREAS OF RECTILINEAR FIGURES

Theorem 22. Rectangles having a dimension of one equal to
that of another compare as their remaining dimensions.

Theorem 23. Any two rectangles compare as the products of
their dimensions.

Theorem 24. The area of a rectangle is equal to the product
of its base and altitude.

Theorem 26. The area of a parallelogram is equal to the
product of its base and altitude.
Cor. 1. Any two parallelograms compare as the products of

their bases and altitudes.
In proving Core. 1, 2, 3 of this theorem and the next, express
the areas as algebraic formulas and apply axioms. Why is such a
procedure both convenient and natural?
Cor. 2. Parallelograms having one dimension equal compare

as the rerrudning dimensions.
Cor. 3. Parallelograms having equal bases and equal altitudes

are equal.
Theorem 26. The area of a triangle is equal to half the product
of its base and its altitude.
Cor. 1. Any two triangles compare as the products of their

bases and altitudes.
Cor. 2. Triangles having one dimension equal compare as

their remairiing dimensions.
Cor. 3. Triangles having equal bases and equal altitudes are

equal.
Theorem 26a. The square on the hypotenuse of a right tri-
angle equals the sum of the squares on the two legs.

Note. — In this text the conventional phraseology will be followed, and the
usual distinction between such expressions as "square on" and "square of"
will be observed. Square on wUl mean the area of the square constructed on the
given sect, and square of, the square of the numeric measure of the given sect.

239

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240

PLANE GEOMETRY

Given: AABC; ^^.BCA art. 4.; squares
CBFG, ABDE, and CAKH.

Prove: Sq. ABDE ^&q, C5TO+Sq.
CAKH.

Proof: (Some steps which the student
can readily supply have been pur-
posely omitted.)

Draw CM±EMB.
Draw CD and AF.

(1) :.CM\\BDajidAE.

(2) ACG is a st. line.

(3) AABFs}4 (BF.CB).

(4) AABF^yi (Sq. CBFG),
(6) LMDB is a rect.

(6) ACDB^^ (BDLB).

(7) ACDBai^ (Rect. LMDB)

(8) BD^AB,CBsBF.

(9) 4.DBCS4FBA.

(10) /. AABF^ACBD.

(11) /. Sq. CF=Rect. LD.

(1) Why?

(2) Why?

(3) Why?

(4) Why?

(5) Why?

(6) Why?

(7) Why?

(8) Why?

(9) Why?
(10) Why?

(11) Why?

Draw the necessary construction lines and show that sq. AH equals rect. AM,
Finish the proof.

Theorem 26b. The areas of two triangles having an angle of one
equal to an angle of the other are to each other as the products of the
sides including those angles. Xk

Given: AABC and AAiXY with 4A a 4^1 1.
AABC IB'AU

Suggestions: Proof being left to the student.

Place AABC in the position of AAiQR, Draw QY.
AAiRQ _AiR , AAiQY AjQ
AAiYQ^AiY' ^^ AAiXY^AiX'

Could the case arise in which QR intersects XY between X and Y7
If 80, show that this proof still holds or give one that does hold.

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AREAS OF RECTILINEAR FIGURES 241

Theorem 27. The area of a trapezoid is equal to half the product
of its altitude and the sum of its bases.

EXERCISES. SET LXXIX. AREAS
Numeric

878. Find the expense of paving a path 4' wide inside a square
piece of ground the side of which is 50' if the price is 18 cents per
square yard. What would be the cost if the path were outside
the piece of ground?

a879.* Find the dimensions of a rectangle given:

(a) Area 216 sq. ft., perimeter 60 ft.

(6) Area 600 sq. ft., difference of the sides 10 ft.

(d) Area 340 sq. ft., sum of squares of two consecutive sides
689 sq.ft.

880. Find the change in the area of a triangle of base a and
altitude h in the following cases :

(a) If a and h are increased by m and n, respectively.

(b) If a and h are diminished by m and n, respectively.

(d) If a is diminished by m, and h increased by n.

881. A line of division is drawn between two sides of a triangle,
(lividing it into a triangle and a quadrilateral. What parts are
these two figures, respectively, of the entire triangle if the line of
division cuts off the following parts of the two sides, reckoned
from the intersection of the sides?

(a) i and i. (b) | and |. (c) ^ and i.

(d) i and i. (6) i and ^ (/) \ and I

882. Find the area of a rhombus, given:
(a) The diagonals 18 and 12 units.

(6) The smn of the diagonals 12, and their ratio f .

* As here, *'a" precedes a problem which calls for the solution of an affected
quadratic equation.

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242 PLANE GEOMETRY

Theoretic

883. (a) Prove, geometrically, the algebraic formula,

a(6+c)=a6+ac.
(fr) Prove, geometrically, the algebraic formula,

a{b-'c)^ab -ac.
(c) Prove, geometrically, the algebraic formula,

a2-62=(a+6)(a-6).

884. The area of a rhombus is equal to half the product of its
diagonals.

886. If lines are drawn from any point inside a parallelogram to
the four vertices, the sum of either pair of triangles with parallel
bases is equal to the sum of the other pair.

886. The accompanjdng figures show easy methods of trans-
forming (a) a triangle into a parallelogram, (6) a parallelogram into
a triangle, (c) a trapezoid into a parallelogram. Explain. Can
you give more than one explanation? If so, upon what does your
explanation depend?

A F D

For problems in construction based upon this chapter see
Chapters VII and VIII.

TERMS DEFINED IN CHAPTER H

Square on (a sect)t square of (a quantity), medians of a triangle, rhomboid.

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CHAPTER III

SIMILARITY

The following are terms with which the student should now be
familiar:

• ArUecederUf eonseqtient, extremes, means, m^ean proportion, mean
proportional, continued proportion, inversion, composition, division,
composition and division.*

Theorem 28. Any proportion may be transformed by alternation.
Theorem 29. In any proportion the terms may be combined by
composition or division.

Theorem 30. In a series of equal ratios, the ratio of the sum of
any number of antecedents to the sum of their consequents equals
the ratio of any antecedent to its consequent.

Theorem 31. A line parallel to one side of a triangle divides
the other sides proportionally.
Cor. 1. One side of a triangle is to either of the sects cut off

by a line parallel to a second side as the third side is to

its homologous sect.
Cor. 2* Parallels cut off proportional sects on all transversals.
Cot. 3. Parallels which intercept equal sects on one transversal

do so on all transversals.
Cor. 4. A line which bisects one side of a triangle, and is

parallel to the second, bisects the third.
Cor. 6. A sect which bisects two

sides of a triangle is par-
allel to the third Iside and

equal to half of it.

Suggestions: What means have you for
proving lines parallel?
How may one sect be proved half of ^

another?
What kind of figure would you like to have?

* The student will find an alphabetical index of definitions on p. 379.

248)Ogl^

244 PLANE GEOMETRY

The sect joining the mid'j>oint8 of the non'j>araUel sides of a trape-
toid is called its median.

Cor. 6. The median of a trape-
zoid is parallel to the
W bases and equal to one-

half their sum.

P">

Suggestions: Why draw a diagonal?

** ^ Show that a line through M \\W

will bisect AC and therefore DCf and henoe coincide with MMu

Cor. 7. The area of a trapezoid equals the product of its median
and altitude.

What numerical relation exists between the median and the bases of a
trapezoid?

Theorem 32. A line dividing two sides of a triangle proportion-
ally is parallel to the third side.
Cor. 1. A line dividing two sides of a triangle so that these

sides bear the same ratio to a pair of homologous sects

is parallel to the third side.

DIVISION OF A SECT

A point in a sect is said to divide it internally, and a pointin the
prolongation of a sect is said to divide it externally, and in bo^ cases
the divisions of the sect are reckoned from one extremity to the
point of division and from that point to the other extremity of the
original sect,

A _J B E

Al
I divides sect AB internally in the ratio ==.

AE
E divides sect AB externally in the ratio — .

Note. — ^When neither ''internal'' nor "extemal'' is used to qualify the
division, internal is understood.

Is there any ratio into which a sect cannot be divided externally?
A sect is said to he divided harmonically when it is divided intervr
ally and externally in the same ratio.

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SIMILARITY

245

In the foregoing illustration AB is divided harmonically if

HJae
IB m

Is there any ratio into which a sect cannot be divided
harmonically?

Discuss Theorems 31 and 32 from the point of view of external
division of a sect.

Theorem 32a. The bisector of an angle of a triangle divides ,
the opposite side into sects which are proportional to the adjacent
sides.

Given: AABC; D on AB and Z^ACD s 4DCB.
AD AC

To prove: 55=^.

Suggestions : Compare AADC and
ABDC in two ways. A

Theorem 32b. The bisector of an exterior angle of a triangle
divides the opposite side externally into sects which are propor-
tional to the adjacent sides.

^jy Given: AABC with exterior 4
BCD; EmAB produced; ^BCE'-
^ECD.

AS AC
To prove:—--.

' ; Suggestions: Com-
pare AAEC first with
A5^C, and second with ABtEC

How should Bi be taken so that ABiEC may be substituted for ABEC^
Consider special cases where asb,a>h.

Cor. 1. The bisectors of an adjacent interior and exterior
angle of a triangle divide the opposite side harmonically.
(Proof left to the student. Discuss special cases.)

EXERCISES. SET LXXX. RATIO. PROPORTION. PARALLELS

Numeric

887. Find the value of n if (a) -=1, (6) - = -.

^ ' n 8' ^ n c

T- a« 216 ^ , a

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246 PLANE GEOMETRY

889. If ^=1-, find;-.

890. k:|?=:^, find |.

91. Form all possible proportions involving a, b, p, and q if
ab^pq.

892. Find the mean proportionals between:

(a) 2 and 50. (c) 3 and 21. (e) a and b.

lb) 2 and 75. (d) 3 and 19. (/) a+b and a -6.

893. Find the third proportional to (a) 5 and 6, (6) a and 6.

894. Transform the proportion t=-j so that b becomes the third
term.

896. If -7— = c> what is the ratio of a to 6?

®^- ^* 6~d"/""n' ^"^ HPd+/

897. Find a fourth proportional to:

(a) a, 06, c. (6) a2, 2a6, db\ (c) a:«, rcy, 5x22/.

898. Find a third proportional to:

(a) a%, 06. (6) x\ 2x^. (c) 3a:, 6x2^. (d) 1, x.

899. Find mean proportionals between:

(a) a2, b2. (6) 2x3, &J.. (c) i2ax2^ 3^3. (rf) 27a^b^, 36.

900. If a, 6, c, be in mean proportion, show that:

(a) 4-=?Lz.^. (6) (62+6c+c2) (ac-6c+c2)s64+ac8+c*.

d901.* If ^-^, prove that: (a) ^^—^^^5—,.

a ,6 c ,d -+T -+3

(d) £-^^P_«. (^) "^ = -:^'

^ ^ a2+b2 c^ + ep

* For guidance in proofs such as this exercise requires see p. 304.

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SIMILARITY 247

a902. Solve the equations:

^^^6x^7 Qx+lO* ^^2/+3 52/ -13 14'

,.x^-2x+3 x^-Sx+5 ,j. 2x-l x+4

2x-3 3a: -5 ' '^ a:2+2x-l x2+x+4*

Hint: Transform. In doing so do you lose any roots?

903. Prove that a, 6, c, d are in proportion if

(a+6 -3c -3d) (2a+26 -c+d) = (2a+26 -c -d) (a -6 -3c+3d).

904. If 6 is a mean proportional between a and c, show that
4a2 -962 is to 462 ^9^2 i^ ^^^^ r^tio of a^ to 6«.

906. If a, 6, c, d are in continued proportion, i.e., if vs s -*

prove that 6+c is a mean proportional between a+h and c+d.

d906. Krr^^lr^, prove that asc, or a+6+c+dsO.
o+c a+a

d907. If r s -=- prove that each of these ratios is equal to
s/ 2a2c+3c»c+462c
\ 2fe2d+3d8c+4/2d

908. Two numbers are in the ratio of f , and if 7 be subtracted
from each, the remainders are in the ratio of |; find them.

909. What number must be taken from each term of the ratio
a that it may become f ?

910. What number must be added to each term of the ratio
fj that it may become ^?

911. Ur^ s-i-s-I-, show that p+q+r^O.

b-c c-a a-o'

912. If x^ =-T- ^ — V» show that x -y+z=0.

b+c c+a a-b'

6913. If^=^=|,showthaty^

-2c^e+3aVe2 ace

'2d'f+3¥cd^e'^bdf
914. In the accompanying diagram,
(a) If a=3, 6=4, c=7, findd.
(6) If a=5, c=9, d=10, find 6.
(c) Find each sect in terms of the ^'
other three.

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248

PLANE GEOMETRY

916. To measure indirectly the distance from an accessible
point A to an inaccessible point B, run CK through C, a point

from which A and B are
both visible, making ^KCB
^^BCA. Sight D in line
with K and C and also in line
with A and B. What sects
must now be measured in
order to compute AB?
916. Find the area of a
trapezoid, given the median m, and the altitude h.

Theoretic

917. The perpendiculars dropped from the mid-points of two
sides of a triangle to the third side are equal.

918. Lines joining the mid-points
of two opposite sides of a paral-
lelogram to the ends of a diagonal
trisect the other diagonal.

919. A line bisecting one of the
non-parallel sides of a trapezoid and parallel to the base bisects
the other non-parallel side.

920. The sects joining the mid-points of the consecutive sides
of any quadrilateral form a parallelogram.

d Consider whether there are any modifications of this fact in
the case of (a) the parallelogram, (6) the rectangle, (c) the rhombus,
(d) the square.

d921. The mid-points of two opposite sides of a quadrilateral
and the mid-points of the diagonals determine the vertices of a
parallelogram.

d922. The sects joining the mid-points of the opposite sides of a
quadrilateral and the sects joining the mid-points of the diagonals
are concurrent.

d923. If perpendiculars are drawn from the four vertices of a
parallelogram to any line outside the parallelogram, the simi of the
perpendiculars from one pair of opposite vertices equals the sum
of those from the other pair.

921. The triangle formed by two lines drawii from the mid-point
of either of the non-parallel sides of a trapezoid to the opposite
vertices is equivalent to half the trapezoid. ^

SIMILARITY

249

926. State and prove the converse of proposition 32a.

926. State and prove the converse of proposition 326.

927. If a sect PQ is divided harmonically at R and S, then sect
RS is divided harmonically at P and Q.

Construction
92& Construct a third proportional to two given sects.

929. Construct a fourth proportional to three given sects.

930. If a, 6, c are given sects, construct (a) sect d, so that r= ,
ab c

(6) ds?2.

931. Divide a sect (a) internally into sects proportional to two
given sects, (6) externally into sects proportional to two given sects,
(c) harmonically in the ratio of two given sects.

932. Divide a given sect (a) internally in a given ratio without
the use of parallels, (6) externally, (c) harmonically.

d933. Construct two sects given (a) their simi and their ratio,
(6) their difference and their ratio.

d934. Through a given point P draw a line meeting tiie sides of
an angle A in the points B and C so that (a)AB= AC, (6) BC = 2AC.

Theorem 33. The homologous angles of similar triangles are
equal, and their homologous sides have a constant ratio.

Theorem 34. Triangles are similar when two angles of one are
equal, each to each, to two angles of another.

Cor. 1. Triangles which have their sides parallel or perpen-
dicular each to each are similar.

Given: AiBilAB: AiBt || AB; BiCi±BC; B^Ct \\ BC; CiAi±CA; CiAm || CA,
Prove: AAiBiCi </> AABC; AAiB^t co AABC.
Suggestions : Show that :

(1) Three angles of A^
one triangle cannot be
supplementary to three
angles of the other.

(2) Two angles of
one triangle cannot be
supplementary to two
of the other.

(3) What, then, is the
fact?

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250 PLANE GEOMETRY

Theorem 36. Triangles which have two sides of one propor-
tional to ttoo sides of another and the included angles equal are
similar.

Theorem 36. If the ratio of the sides of one triangle to those of
another is constant, the triangles are similar.

EXERCISES. SET LXXXI. SIMILARITY OF TRIANGLES
Numeric
936. If the sides of a triangle are 3, 7, and 8, find the sides of a
similar triangle in which the side homologous to 7 is 9.

936. If the sides of a triangle are a, b^ and c, find the sides of a
rimilar triangle in which the side homologous to a is p.

Construction

937. The sides of a triangle are 5, 6, and 7. Construct a triangle
similar to the original, having the ratio of similitude 3 to 2.

938. Construct a triangle
similar to the accompanying
triangle with the ratio of
similitude equal to that of
the two given sects a and b.

Theoretic

939. Two isosceles triangles are similar if an angle of one is
equal to the homologous angle of the other.

940. Prove that the altitudes of a triangle are inversely propor-
tional to the sides to which they are drawn.

941. If the altitudes AD and BE in AABC are drawn, prove that

-5^s=p^. Are the altitudes directly or

inversely proportional to DC and EC1

942. If a spider, in making its web,
makes AA \\AB, BiCi \\ BC, CiDi \\ CD,
DiEi II DE, and EiFi \\ EF, and then
runs a Une from Fi\\ FA, will it strike
the point Ai? Prove your answer.

943. If D is taken in the leg AB of an isosceles triangle ABC,

so that CDsAC (the base), then AC^^AD-AB.

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SIMILARITY

251

944. Isosceles or right triangles ABC and PQR are i^milar if

ha^AB

K PQ

946. The diagonals of a trapezoid divide each other pro-
portionally.

946. If in triangle ABC altitudes AD and BE meet at 0, then:
(a) 55 • DUsDO • ifD; (6) BD • ACsBO^'AD.

947. If in a parallelogram PQRSyB, sect QT is drawn cutting the
diagonal PR in V, the side i2/S in L and the prolongation of PS
inr,then7Q'=7L- Ff.

948. In similar triangles homologous angle bisectors are directly
proportional to the sides of the triangle.

949. In a quadrilateral ABCD, right-angled at B and Z), per-
pendicidars PE and PF from any point P in AC to the sides BC

and AD are such that =ir-+==-s 1.
AB CD

d960. Every straight line cutting the sides of a triangle (pro-
duced when necessary) determines upon the sides six sects, such
that the product of three non-consecutive sects is equal to the
product of the other three.

The line XYZ must cut either (a) two sides of the triangle and
the third side
produced (Fig.
1), -or (6) all
three sides pro-
duced (Fig. 2).
The proof in
both instances
is the same.

DrawCDIIAB.

Fig.1

From the similar triangles

i?=i? and — sS
CD CZ^ CY CD'

AJ^^BY AZ'BX

Fis.2

therefore _, ^-
whence AX-E7-

UZ-W
CZ^IZ-EI'CT.

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Google

252

PLANE GEOMETRY

This theorem was discovered by Menelaus of Alexandria about
80B.C.

d951. Prove the converse of the last theorem.
Let XY produced cut AC produced in a point P.
Then IX • BF.- CP =AP -^ • CY. _
But, by hypothesis, AX • B? • CZ=AZ • BX • CY\

whence

AP-

-CP

-Vz-

= —

AC AZ
/.AP^AZ;
*** that is, P coincides with Z.

d952. Lines drawn through the ver-
tices of a triangle, and passing through
a common point, determine upon the
sides six sects, the product of three
non-consecutive sects being equal to
the product of the other three.

The common point may lie either inside or outside the triangle
(Figs. 1 and 2). In both cases apply Ex. 950 to the AACD and
sect BOF and to the ABCD and sect AOE, then multiply the
results.

Fig. 2

Fig. 1

Fig. 2

This theorem was first discovered by Ceva of Milan, in 1678.

d963. Conversely, if three lines drawn through the vertices of a
triangle determine upon the sides six sects, such that the produjct

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SIMILARITY

253

^:>0

of three non-consecutive sects is equal to the product of the other
three, the lines pass through the same point.

The proof is similar to that of Ex. 951.

Theorem 36a. The homologous angles of similar polygons are
equal,and their homologous
sides have a constant ratio.
Suggestions: Why is it that
the n-gons may be placed as in
the accompanying diagram? To
prove the second part use the
similar triangles thus formed and
show that the ratio of the homo-
logous sides is equal to the ratio
of simihtude of the polygons.
Give the complete proof.

Cor. 1. If the ratio of similitude of polygons is unity, they are

congruent.
Cor. 2. The homologous diagonals drawn from a single vertex

of similar polygons divide the polygons into triangles

similar each to each.

Why is A AiBiCi a> A
ABC7

Having proved this can
you prove A ^i^iDi eo
A ACD, and so forth?

Why is it that only these
two sets of A need be proved
similar?

Write the proof in full.

Theorem 36b. Polygons whose homologous angles are equal
and whose homologous

— ^<r^

sides have a constant

ratio are similar.

Given: Polygons -ABC. . .
N and AiBiCi, . .Ni in
which ^^ilis^A, :^Bi
= 4B, . . 2^Niss^N,Sind
AiBi Bj^ NiAi

, AB^EU^ ^ NA.
Prove: AiBiCi , . . Ni </>
ABC. . .N,

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254

PLANE GEOMETRY

(1)

Place A iBiCt ...Nieain the
diagram s o that AfBj\\AB,
and draw Kli^ and ESiZ.

(2) K ilB-iGFi then ABC.
N^AJBiCi,..Nr.

(3) If lB>A^i, then EC>E^i
etc. and ABBtAt is not a O.

(4) and AAtX and ^BsZ intersect
at some point O.

(6) AAOB^AA/)Bi.

(7) •.•2j.ilJ?/)=4.ABO.

(8) and ^AsBsCts^-ABC

(9) /. 2S^OB^iS2^0BC

(10) andBjC.IIBC.

(11) Draw CiKS" cutting BSiZ in Oi

(12) Then ABOiC <^ AB/)iC%.

'5c " So,

4.B,CiOi = 4.BCOi

, ^ 'B^
• So"SOi

,\SiO^B7fi, and Oi coin-
cides with 0.

(18) .-. BOi
through 0.

(19) Any ORL cutting sides of the
polygons in R and L respectively is

divided so that — = -j^

(13)
(14)
(16)

(16)
(17)

iBO and CCt passes

(1) Data.

(2) CJongruent polygons are similar.

(3) The opposite sides of a paral-
lelogram are equal, and conversely.

(4) Non-parallel coplanar lines in-
tersect.

(5) Two angles equal each to each,
or sides respectively parallel.

(6) Homologous sides of similar
triangles have a constant ratio.

(7) Corresponding angles of par-
allels.

(8) Data.

(9) The differences of equals less
equals are equal.

(10) Corresponding 4.' equal.

(11) See (1), (2), (3), and (4).

(12) See (5).

(13) See (6).

(14) See (7).

(15) Data.

(16) Quantities equal to equal quan-
tities are equal to each other.

(17) By division and the products of
equals multiplied by equals are equal

(18) Two points determine a
straight line.

(19) See (6).

In the same way it may be proved that DDif . . . NNt, pass through and
therefore AiBiCi, . .Ni«^ABC, . JV by the definition of similar figures.

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SIMILARITY 256

Cor. 1. If homologous diagonals drawn from a single t;er-
tex of two polygons divide them into triangles similar
each to each and similarly arranged, the polygons
are similar.

(Proof is left to the student.)

Theorem 37. The perimeters of similar triangles are propor-
tional to any two homologous sides, or any two homologous
altitudes.

Cor. 1. Homologous altitudes of similar triangles have the

same ratio as homologous sides.
Cor. 2. The perimeters of similar polygons have the same ratio

as any pair of homologous sides or diagonals.

Apply propositions 360, 37, and the appropriate law concemmg equal
ratios.

Theorem 38. The areas of similar triangles compare as the
squares of any two homologous sides.

Cor. 1. The areas of similar polygons compare as the squares
of any two homologous sides or diagonals.

Proof similar to that of Theorem 37, Cor. 2. Give it in detail.

Cor. 2. Homologous sides or diagonals of similar polygons
have the same ratio as the square roots of their
areas.

Apply suitable axioms to the results obtained in Theorem 38, Cor. 1.

Concurrent lines are those which pass through a common point.

Theorem 38a. If two parallels are
cut by concurrent transversals, the
ratio of homologous sects of the-
parallels is constant. ^^^ ^ P/d^l fW

(Proof left to the student.)

Discuss the case where lies between the parallels.

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256

PLANE GEOMETRY

Theorem 38&. If the ratio of honuy-
logons sects of two parallels cut by
three or more transversals is constant,
the transversals are either parattd or
concurrent.

Given: P n Pi cut by traiuveTBals (, h, h, in
X, Y, Z and Q, R, V lespectively, so that

^ WW

^ Prove: t, ti, it concurrent.

Proop

(1) If XYsQR, t\\ti and similarly

XY YE ..

since g= B=, ti II ti, etc.

(2) KXy<Qfi, fJffiandwiUinter-
aect it at some point 0.

.ov ,™. ^y OY

i (3) Why?

(4) Suppose OZ cuts Pi in Vu

(4) Why?

YI OY
EFi OR

'"■^'S-S-

\ (6) Why?

'•>-i-^

(6) Why?

Complete the proof.

(1) Why?

(2) Why?

EXERCISES. SET ]LXXXII. SIMILARITY OF POLYGONS
Numeric

964. The corresponding bases of two similar triangles are 11 in.
and 13 in. The altitude of the first is 6 in. Find the corresponding
altitude of the second.

965. The perimeter of an equilateral triangle is 51 in. Find the
side of an equilateral triangle of half the altitude.

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SIMILARITY 257

956. The bases of a trapezoid are 20 in. and 12 in., and the alti-
tude is 4 in. Find the altitudes of the triangles formed by pro-
ducing the sides until they meet.

957. The perimeters of two similar polygons are 76 and 69.
If a side of the first polygon is 4, find the homologous side of the
other.

958. The sides of a polygon are 4, 5, 6, 7, and 8, respectively.
Find the perimeter of a similar polygon if the side corresponding
to 5 is 7.

959. The sides of a polygon are 2 in., 2^ in., 3^ in., 3 in., and
5 in. Find the perimeter of a similar polygon whose longest side
is 7 in.

960. The diameter of the moon is approximately 3500 miles,
and it is approximately 250,000 miles from the earth. At what
distance from the eye will a two-inch disk exactly obscure the moon?

Conatmction

d961. Through a given point Pi draw a line such that its dis-
tances from two given points P2 and Pt shall be in the ratio of
(o) 3 to 5, (6) sect 6 to sect c.

d962. In the prolongation of the side AB of triangle ABC, find
point P so that AP'BP^CP\

963. Construct a triangle similar to a given triangle and having
a given altitude.

964. From a given rectangle cut off a similar rectangle by a line
drawn parallel to one of its sides.

965. Construct a triangle having given:
(a) a, b, and the ratio of & to c.

(6) a, and the ratios of a to 6, and a to c.

(d) a, 6+c, and the ratio of & to c.
d{e) B, the ratio of a to c, and he.

966. Given two sects AB and CD, and a point P. Draw a line
XY through P — ^without producing AB and CD to meet — such
that ABy XY, and CD would be concurrent if produced.

967. To draw a parallel to one side of a triangle, cutting off
another triangle of given perimeter.

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258 PLANE GEOMETRY

Theoretic

968. The line joining the mid-points of the bases of a trapezoid
is concurrent with the legs of the trapezoid.

969. Two triangles are similar if an angle of one is equal to an
angle of the other and the altitudes upon the including sides are
proportional.

970. The line bisecting the bases of a trapezoid passes through
the intersection of its diagonals.

Suggestion: Prove that line coincident with the line joining the mid-point
of one base and the intersection of the diagonals.

d971. Any sect drawn through the mid-point of one side of a
triangle and limited by the parallel to that side through the
opposite vertex, is divided harmonically by the second side and the
prolongation of the third side.

972. If two triangles have equal bases on one of two parallel
lines, and their vertices on the other, the sides of the triangles
intercept equal sects on any line parallel to these lines and Ijring
between them.

Reword the theorem (a) when one base is half the other. (6)
When the bases have any given ratio.

Theorem 39. The altitude upon the hypotenuse of a right tri-
angle divides it into triangles simitar to each other and to the
original.

Cor. 1. Each leg of a right triangle is a mean proportional
between the hypotenuse and its projection upon the
hypotenuse.

Cor. 2. The square of the hypotenuse of a right triangle is
equal to the sum of the squares of the other two
sides.

Cor. 3. The diagonal and the side of a square are incommen-
surable.

Cor. 4. The altitude upon the hypotenuse of a right triangle is
a mean proportional between the sects it cuts off on the
hypotenuse.

(The proof is left to the student.)

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SIMILARITY

259

Theorem 39a. In any triangle, the square of the side opposite
an acute angle is equal to the sum of the squares of the other two
sides diminished by twice the product of one of those sides by the
projection of the other upon it.

Fia. 2.

Given: In AABC, :^A <90'*, AD spa,, the projection of c upon b.
Prove: o»»6*+c*— 26p«6.

Pboop

(1) Data and def . of projection.

(2) Why?

(3) Why?

(1) ABCDandABDaiert^A.

(2) /. in ABCD, a^mhb^+DC*.

(3) Butfe»3c»-p(**in AABD,

(4) and in Fig. 1, Sc*s(6-pc6)« (4) Why?

while in Fig. 2, DC =(p(*-6)« which
are identical.

(5) .•.o«=c«-pc6«+(6-pc6)» (6) Why?

=c*— p(**+&*— 26-p(*+p<*'
«c«+6«-26-p(*.

Theorem 39b. In an obtuse triangle, the square of the side
opposite the obtuse angle is equal to the sum of the squares of the
other two sides increased by twice the product of one of those
sides by the projection of the
other upon it.

Given: In AABC, 2^A>90^ 5Zs

Pcbf the projection of c upon b.
Prove: a*s6*+c*+26-p(*.
(Proof left to the student.)

Theorem 39c. L The sum of the squares of two sides of a
triangle is equal to twice the square of half the third side, increased
by twice the square of the median to it

This theorem is attributed to Appollonius of Perga (c. 225 b.c).

//. The difference between the squares of two sides of a tri-
angle is equal to twice the product of the third side and the
projection Qf the median upon it Digi,i,ed by Google

260

PLANE GEOMETRY

Given: AABC in which a>c, mb is the median to h and XD sp is its projeo-
p tion upon b.

Prove: I. a'+c' s2QY4-2»i^".

II. o«-c*s26p.
Outline of proof, which the student
is asked to write.
(1) Show that ^BDC is obtuse
and 2^BDA is acute by com-
paring A ABD and BDC.

(2) Use propositions 39a and 396 to give the values of a* and c*.

(3) Combine these values as indicated in I and II.
Consider this proposition when a=c.

Cor. 1. If m^ represents the length of the median to side b of
the triangle whose sides are a, b, c, then

Solve proposition 39c I for 7n^.

Theorem 39d. If h^ represents the
altitude upon side a of a triangle whose
sides are a, b, c, and s represents its
a + b + c,

semiperimeter, i.e., s=-

then

hMVs(s^a)(s-b){s-c).

Fig. 1.

Proop

(1) At least one of the angles B or C is acute.
Suppose C is acute (Figures 1 and 2) .

(2) Thenc«aa«+6»-2a-CD.

(3)...CD-«-!±^
(4) But ha^^b^-CDK

-.( —

+2ab+b^-c^

\/ c»,(aa-2a6+5a) \

2a /\ 2a

(a+b+c)(a+b'-c)(c+a -b)ic-a+b)
* 4a«

All authorities to be given
by the student.

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SIMILARITY

261

»J^\/(«+&+c)(o+6-c)(a+c-6)(6+c-a) 4.

If«

a+b+c,
2

then o+6H-c«2«

aH-6-cs2«-2c
a-fc-6s28-26
6-fc-as2«-2a
V'2«-2(s-c)2(« -6)2(s-o)

I $- -\/«(«— a){s— 6)(s--c)
2o ^

•a)(«— 6)(«— c).

Show that this proof holds for Figures 3 and 4, i.e., when
C is obtuse or right.

Cor. 1. If A stands for the area of a triangle
whose sides are a, b,c, a nd whose semiperime ter is s, then

il = V5(5-a)(5-&)(5-C)
By what must the value of ha be multiplied to obtain the value of Af
This formula is known as Heron of Alexandria's, to which atten-
tion was called on p. 98, Ex. 359 of the First Study.

The Principle of Continuity, By considering both positive and
negative properties of quantities, a theorem may frequently be
stated so as to include several theorems. For instance. Theorems
39, Cor. 2, 39a and 396 may be stated as a single theorem if we
take into consideration the direction of the projection of c upon b.

If AD, the projec-
tion of c upon 6 in
Fig. 1 be considered
positive, AD in Fig.
3 will be negative.
Therefore we may
say in general that
a-362+c2-26.pc6, for in Fig.
1 that means a^=b^+c^^2b
If (+AD), or a2s62+c«-26.pcft,
while in Fig. 2 it means a's
6«+c^ 26(0), or a^ = 6* + c*,
and in Fig. 3 it means a*s6*+c* -26(-DA), or a^sb^+c^+2b'p^.

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Fig. 8.

262 PLANE GEOMETRY

EXERCISES. SET LXXXIII. METRIC RELATIONS

Theorems 39a through 39d enable us to calculate the altitudes
and medians of a triangle in terms of its sides, and the length of the
projection of any side upon any other, as well as to determine
whether a triangle is acute, right or obtuse.

973. If in solving the identity a^sfe^+c* -26*pe6 for peb for any
particular values of a, b, and c, we find (a) pch=0, then '^A =90®,
(b) Pc6>0, then ^A <90^ (c) pcb <0, then <A>90®. Explain.

974. Give the formulas for b and c corresponding to those given
in propositions 39a and 39b for a.

976. From the three formulas obtained in Ex. 974 derive

formulas for Peb, Pab, Pea or for Pbe, Pbay Pac

976. Give a formula for mo, for mc, for A&, for Ac.

977. In a right or an obtuse triangle, the greatest side is opposite
the right or the obtuse angle. Hence if a is the greatest side of a
triangle, show that if a^>V'\'<? the triangle is acute, if €?=b^'\'i?
the triangle is right, and if a^<b^'\'C^ tl^e triangle is obtuse.

dt978. Show that pcb^c cos A and hence that the general
formula might be a^^b'^+c^ -26c cos A.

Theorem 39e. If similar polygons are constructed on the sides
of a right triangle, as homologous sides,the polygon on the hypot-
enuse is equal to the sum of the polygons on the other two sides.

If Pi, Pi, and Ps be similar polygons constructed on a, 6, and c respect-
ively, as homologous sides, when "^^C is a right angle in A ABC, 77-— ?>

Pz ' Pz

Give the complete proof.

EXERCISES. SET LXXXIII (concluded)
Numeric

979. The base of an isosceles triangle is 48 in. Find the altitude
if each arm equals 50 in.

980. Let ABC be a right triangle. The two sides about the right
angle C are respectively 455 and 1,092 feet. The hypotenuse
AB is divided into two sects AE and BE by the perpendicular
upon it from C. Compute the lengths of AE, BE, and CE,

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SIMILARITY 263

98L (a) If two sides of a triangle equal 15 and 26, respectively,
and the projection of 15 upon 25 equals 9, what is the value of the
third side?

(6) Is the triangle right, acute, or obtuse?

982. The altitude of a triangle is 20 in. A line parallel to the
base and 12 in. from the base cuts off a triangle that is what part
of the given triangle?

983. If the side of an equilateral triangle equals 10 in., what
is the length of the projection of one side upon another?

984. Find the projection of AB upon a line XF, if AB and XY
include an angle of 45®, and AB=^,

986. Find the side a of the square equal to an equilateral tri-
angle whose side is s. Solve the equation for s in terms of o.

986. Two sides of a triangle are 5 and 8, respectively, and include
an angle of 30°. Find the area.

987. Find the area of an equilateral triangle of which
(o) The side is 30,

(6) The altitude is 34,

(d) The altitude is h.

988. Find the area of a trapezoid, given:
(a) The median m, and altitude h]

(6) The median m, one leg I, and the angle between this and the
base 30°;

989. In a trapezoid, given the two bases a, 6, and the altitude h.
The legs are divided into three equal parts by lines parallel to the
bases. Find in terms of a, 6, and A, the areas of the three parts
into which the trapezoid is divided.

990. Find the side of a rhombus composed of two equilateral
triangles and equal to another rhombus whose diagonals are 12
and 18.

991. ABC is a triangle and AD the altitude upon BC. If
ilD = 13, and the length of the perpendiculars from D to AB and
AC are 5 and lOf , respectively, find the area of the triangle.

992. Find the area of a square in terms of (a) its perimeter p,
(6) its diagonal d.

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264 PLANE GEOMETRY

993. find the dimensions of a rectangle given:

(a) Its perimeter p and area a :

(b) Its length I and diagonal d;

994. Find the projection of AB upon XY, if AB=m, and the
two lines include an angle (o) of 60°, (b) of 30"*.

995. In triangle abc, a = 8, & = 15, and the angle opposite c equals
60^ Findc.

996. In triangle abc, a»3, b=5, and the angle opposite c equals
120°. Findc.

997. In triangle abc, a»7, b=8, and the angle opposite c equals
120°. Findc.

998. Two sid^ of a triangle are 20 and 30, respectively, and
include an angle ol 45°. Find the third side.

999. Two sides of a triangle are 16 and 12 in., respectively, and
include an angle of 60°. Find the third side.

1000. Find the area of a rectangle in terms of its length I, and
diagpnal d.

1001. In triangle abc, 11=20, b = 15, and c = 7. Find the projec-
tion of b upon c. Is the triangle obtuse or acute?

1002. In a quadrilateral ABCD, AB-= 10, BC = 17, CD = 13,
DA =20, and AC =21. Fmd the diagonal BD.

1003. Two sides of a triangle are 17 and 10; the altitude upon the
third side is 8. What is the length of the third side?

1004. The sides of a triangle are 7, 8, and 9, respectively. Find
the length of the median to 8.

1006. The sides of a triangle are 10, 5, and 9, respectively.
Find the length of the median to 9.

1006. The sides of a triangle are 22, 20, and 18, respectively.
Find the length of the median to 18.

1007. The sides of a triangle are 9, 10, and 17, respectively.
Find the three altitudes.

1008. The sides of a triangle are 11, 25, and 30, respectively.
Find the three altitudes.

1009. The sides of a triangle are 12, 14, and 15 respectively.
Find the three altitudes.

1010. (a) Find the altitude of an equilateral triangle with side «.

(b) Find the side of an equilateral triangle with altitude h. i

lOOgle

SIMILARITY 285

1011. Khd the area of a triangle whose sides are respectively
(a) 13, 14, 15, (b) 9, 10, 17, (c) 11, 25, 30.

dl012. The sides of a triangle are as 8 to 15 to 17. Find the
altitudes if the area is 480 sq. ft.

1013. Find one diagonal of a parallelogram, given the sides
a, 6, and the other diagonal g.

Construction

1014. Given any sect as unit, construct a sect which is y/2y
y/S, y/b units.

PA 1

1015. In a given sect AB find a point P such that (o) p^ = — -?=,

1016. Construct an equilateral triangle equal to (a) the siun of
two given equilateral triangles, (6) their diflference.

1017. Construct a polygon similar and equal to (a) the sum of
two given similar polygons, (6) their difference.

1018. Construct a square equal to the sum of 3, 4, 5 given
squares.

(For other construction problems based on this character, see

Chapter VII.)

Theoretic

1019. The median drawn from the extremities of the hypot-
enuse of the right triangle ABC are BE, CF; prove that 4tBW+
4CF^=5BC^.

dl020. In a certain triangle ABC, AC^ -W^^^AB^; show
that a perpendicular dropped from C upon AB will divide the
latter into sects which are to each other as 3 to 1.

1021. If ABC is a right triang le, C the vertex of the right angle,
D any point in AC, then BD^+AC^^AB^+DC\

1022. The sum of the squares of the four sides of a parallelogram
is equal to the sum of the squares of its diagonals.

^023. If Jn the parallelogram ABCD ^A=60^ IU*sJB*+
BC^+AB^BC.

1024. The siun of the squares of the medians of a triangle is
three-fourths the sum of the squares of its sides.

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286 PLANE GEOMETRY

dl025. The sum of the square on the difference of the legs of a
right triangle, and twice the rectangle whose sides are the legs of
the triangle, is equal to the square on the hypotenuse.

State this exercise in algebraic form and prove it.

1026. One-half the sum of the squares on the sum and difference
of the legs of a right triangle is equal to the square on the
hjrpotenuse .

State and prove this exercise algebraically.

dl027. Two similar parallelograms are to each other as the pro-
ctlicts of their diagonals.

1028. If in triangle ABC, AB^BC and altitudes AD and BE

intersect at 0, then 77)=t^-

1029. If in a triangle the ratio of the squares of two sides is
equal to the ratio of their projections upon the third side, the
triangle is a right triangle.

41030. The sum of the squares of the sides of a quadrilateral
is equal to the sum of the squares of the diagonals increased by
four times the square of the sect joining their mid-points.

dl031. If perpendiculars are drawn to the sides of a triangle
from any point within it, the sum of the squares of three alternate
sects cut oflf on the sides is equal to the sum of the squares of the
three remaining sects.

1032. If ABCD is a sect such that AB^BC^CD, and P is
any other point, prove that PA^+SPC^^PD^+SPB*.

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CHAPTER IV

LOCUS

Theorem 40. The locus of points equidistant from the ends of
a sect is the perpendicular bisector of the sect
Cor. 1. Two points equidistant from the ends of a sect fix its
perpendicular bisector.

Theorem 41* T/ie locus of points equidistant from the sides of
an angle is the bisector of the angle.
Cor. 1. The locus of points equidistant from two intersecting

Unes is a pair of lines bisecting the angles.
(Concurrent lines are those which pass through a common point.)

Theorem 41a. The bisectors of the angles of a triangle are
concurrent in a point equidistant from the sides of the triangle.

Given: AABC; ^BAZ^t^ZAC, ^ACY^^YCB, ^CBX^^XBA.

To prove: AZ^ CY, and BX are concurrent in a point equidistant from

the sides.
Suggestions for proof: If BX were parallel to CY what relation would exist
between ^YCB and ^CBXf
If O is a point in FC, how is it located with regard to BC and ACT
If O is a point in BX, how is it located with regard to BC and ABt
Why, then, must be on AZf

267

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PLANE GEOMETRY

Theorem 41b.
a Mangle are
vertices.

The perpendicular bisectors of the sides of
concurrent in a point equidistant from the

Given: AABC; MMi ±XmB, RRi
±1EC, PPi±CFS; AMs
MB,AR^RC,CP^PB.
To prove: MMi, PPi, RRi are
ooncurrent in a point equi-
distant from Af By and C.
Suggestions for proof: If MMi
were parallel to PPi what
relation would exist between
MMi and CB?
Therefore, what relation
would exist between CB and BAf
If is a point in MMi how is it located with regard to A and B?
If is a point in PPi how is it located with regard to C and B?
Why, then, is a point in RRi?

Theorem 41c.
Given: AABC;

The attitudes of a triangle are concurrent.

CH±

XffF, BT± CfA]
AL± ULB,
To prove: AL, BT, CH

are concurrent.
Analysis of proof: If AL,
BT, CH were the
perpendicular bisec-
tors of the sides of
AAiBiCi they would be concurrent.

If they are to be such, how should the sides
of AAiBiCi be drawn through A, B, and C to
make AL±BiCiy BT±AiCiy and CH±AiBi?

UACi^ABiy BCi^BAiy and CAi^CBiy then
this construction would make possible a proof.

Give a synthetic proof.*

Note. — ^Where an obtuse triangle is involved, show that no separate
proof is necessary.

* For notes on various tyi)es of proof see Chapter VI, p. 297.

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LOCUS

269

OMmdMiO^OY.

Theorem 41cf. Tfie medians of a triangle are concurrent in a
point of trisection of each.
Given: AABC;BM^MC,
CMt^MtA, AMi^
MiB; M in BC, M ,
in BA, Mi in AC.
To prove: AM, BMt, CMi
meet in a point O
such that ilOs2MO,
S0^2Mi6ySindW^
2M^.
Notes on proof: Why can-
not CMi be parallel
to AM?
Bisect AO at X and CO at Y.
How is XY related to ilC? (Consider AAOC.)
How is MM 1 related to AC? (Consider AABC.)
Therefore, how is MMi related to XY?
Prove AXOY^ AMOMi, and hen^XO
Therefore, C0=2M^ and AD=2M0,

Consider, now, medians AM and BMt. Can they be parallel, and, if
not, why would their point of intersection Oi coincide with 0?

EXERCISES. SET LXXXIV. LOCUS *

1033. Find the locus of the mid-points of sects drawn from a
common point to a given line.

1034. Find the locus of the points in which the sects mentioned
in Ex. 1033 are divided in the ratio 5 to 8.

1035. Find the locus of the points in which the sects mentioned
in Ex. 1033 are divided in the ratio of two given sects a and 6.

1036. Find the locus of the mid-points of sects connecting points
on two parallels.

1037. Lines are drawn parallel to one side of a triangle and are
terminated by the other two sides. What is the locus of their
mid-points?

* Various terms are used in stating locus exercises. We shall follow the
most usual interpretations, which are: (1) No locus exercise need be proved
unless a proof is definitely called for, indicated by "prove"; (2) an accurate
construction is called for when the terms "plot" or "construct" are used;
(3) the terms "describe" or "find" are used in calling for a statement of what
the locus is.

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270 PLANE GEOMETRY

1038. Parallel sects are drawn with their extremities in the sides
of an angle. Find the locus of their mid-points.

1039. What is the locus of the vertices of triangles having (a) a
given base and a given altitude? (6) a given base and a given area?

dl040. Find a point within a triangle such that the lines joining
it to the vertices shall divide the triangle into three equal parts.

lOU. If AB be a fixed sect, find the locus of a point which moves
so that its distance from the nearest point in AB is always equal
to a given sect c.

How does this locus differ from the one obtained if for the word
" sect " we substitute " line "?

1042. If PQRS be a rhombus, such that Q and S lie on two fixed
lines through P, find the locus of R.

1043. If PQRS be a parallelogram of constant area and given
base PS, find the loci of R and Q.

1044. If A be a fixed point, BC a fixed line, n any integral
number, P any point in BC, and Q a point in AP or PA produced
so that AQ^n-AP, find the locus of Q.

1046. Find the locus in the last exercise if AP=n'AQ.

1046. If in the APQR a sect QS be drawn to any point in the

base, find the locus of a point T on this sect such that the ratio ■-—

is constant.

Justify the two expressions " the locus of points " and " the
locus of a point."

1047. If from the intersection of the diagonals of a parallelo-
gram sects are drawn to the perimeter, find the locus of the point
in these sects such that the ratio of the parts into which the sect

is divided is (a) constant, (6) equal to a given ratio — ^ or (c) equal

to the ratio of two given sects a and 6.

1048. Given a square with side 3 in. Construct the locus of a
point P such that the distance from P to the nearest point of the
square is 1 in.

1049. Upon a given base is constructed a triangle, one of whose
base angles is double the other. The bisector of the larger base
angle meets the opposite side at the point P. Find the locus of P.

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LOCUS 271

dl050. What is the locus of points, the distances of which
from two intersecting lines are to each other as m to n?

dl061. Find the locus of points the sum of whose distances from
two given parallel lines is equal to a given length. Discuss all
possible cases.

dl062. Find the locus of points the difference of whose distances
from two given parallel lines is equal to a given length. Discuss.

dl063. Find the locus of points the sirni of whose distances from
two given intersecting lines is equal to a given length.

dl064. Find the locus of points the difference of whose distances
from two given intersecting lines is equal to a given length.

1066. The vertex -4 of a rectangle -AfiCD is fixed, and the direc-
tion of the sides AB and AD also are fixed. Plot the locus of the
vertex C if the area of the rectangle is constant.

dl066. Plot the locus of a point if the product of its distances
from two perpendicular lines is constant.

dl067. Plot the locus of a point P such that the sum of the
squares of its distances from two fixed points is constant.

dl068. Plot the locus of a point such that the difference of the
squares of its distances from two fixed points is constant.

dl069. Given the base of a triangle in magnitude and position
and the difference of the squares of the other two sides, plot the
locus of the vertex.

1060. Given a square ABCD. Let E be the mid-point of CD,
and draw BE. A line is drawn parallel to BE and cutting the
square. Let P be the mid-point of the sect of this line within the
square. Construct the locus of P as the line moves, always remain-
ing parallel to BE.

Other locus exercises will be found in the chapter on ''Circles,"
pp. 273, 275, 276, 283, 284, 288, as well as in the chapter on
" Methods of Attacking Problems," p. 306, et seq.

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CHAPTER V

THE CIRCLE
Theorem 42. Three paints not in a straight line fix a circle.

Theorem 43. In equal circles, equal central angles intercept
equal arcs,* and conversely.

Theorem 43a. In equal circles, the greater of two central angles
intercepts the greater arc, and conversely.

Suggestion : Lay off the smaller central angle on the greater to prove the direct.
What may be done in the case of the converse?

Theorem 44. In equal circles, equal arcs are subtended by
equal chords, and conversely.

Theorem 44a. In equal circles, unequal arcs are subtended by
chords unequal in the same order, and conversely.

Suggestions: If radii are drawn, what do we know of the triangles formed?
Then what method of proving sects unequal may be used in the proof of

the direct?
In the proof of the converse, what is the only method you are ready to

use in order to prove arcs unequal?
Write the proofs of both parts of this theorem.

Theorem 46. A diameter perpendicular to a chord bisects it
and its subtended arcs.
Cor. 1. A radius which bisects a chord is perpendicular to it
Cor. 2. The perpendicular bisector of a chord passes through
the center of the circle.

Theorem 46. In equal circles, equal chords are equidistant
from the center, and conversely.

Theorem 46a. In equal circles the distances of unequal chords
from the center are unequal in the opposite order, and conversely.

* Such arcs are actually congruent, but we are following custom in using
the word "equal."

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THE CIRCLE

273

Axioms of Inequal-
ity (continued). 5.
Squares of positive un-
equals are unequal in
the same order. Illus-
trate.

Given: 0C«0Ci; chord

ilB>chord DE; CT±I7W, UJ^^UXF.
Ptove: CYKCiX.

Proop of the Direct

Authorities left for the student to
Bupply.

But3B>I5B, .-. FB>X1;

/. YC^kZCTx, and /. TT7<X77;.

Can the ftame method be used for the proof of the converse?
Give the proof m full.

Note. — ^Why is it better to use a direct method rather than the method
of exclusion in the proof of the converse?

Theorem 47. A line perpendicular to a radius at its outer ex-
tremity is tangent to the circle.
Cor. 1. A tangent to a circle is perpendicular to the radius

drawn to the point of contact
Cor. 2. The perpendicular to a tangent at the point of contact

passes through the center of the circle.
Cor. 3. A radius perpendicular to a tangent passes through the

point of contact
Cor. 4. Only one tangent can he drawn to a circle at a given

point on the circle.
Theorem 48. Sects of tangents from the same point to a circle
are equal.

Theorem 49. The line of centers of two tangent circles passes
through their point of contact.

Theorem 49a. The line of centers of two intersecting circles
is the perpendicular bisector of their common chord.

What is the locus of points equally distant from the ends of a sect? Where,
then, do the centers of these circles lie?

IS Digitized by Google

274 PLANE GfeOMETRY

EXERCISES. SET LXXXV. THE STRAIGHT LINE AND THE

CIRCLE

1061. What methods can you now add to those known before
this chapter of showing:

(a) Sects equal? (6) Angles equal? (c) Sects unequal? (d)
Angles unequal? (e) Sects perpendicular?

1062. Can you now mention certain relations of a new kind of
element? If so, what are they?

Numeric

ID63. Two parallel chords of a circle are 4 and 8 units in length,
and their distance apart is 3 units. What is the radius?

1064. Two parallel chords of a circle are d and k in length, and
their distance apart is /. What is the radius?

1066. Find the length of a tangent from a point 15" from the
center of a circle whose radius is 5".

1066. Find the radius of a circle if the length of a tangent from
a point 23" from the center is 16".

1067. Find the length of the longest chord and of the shortest
chord that can be drawn through a point 1' from the center of a
circle whose radius is 20".

1068. The radius of a circle is 5". Through a point 3" from
the center a diameter is drawn, and also a chord perpendicular to
the diameter. Find the length of this chord, and the distance (to
two decimal places) from one end of the chord to the ends of the
diameter.

1069. The span (chord) of a bridge in the form of a circular arc
is 120', and the highest point of the arch is 15' above the piers.
Find the radius of the arc.

1070. The line of centers of two circles is 30. Find the length
of the common chord if the radii are 8 and 26 respectively.

1071. Two circles touch each other, and their centers are 8"
apart. The radius of one of the circles is 5". What is the radius
of the other? (Two solutions.)

1072. If the radii of two concentric circles are denoted by a and
6, respectively, find the radius of a third circle which shall touch
both given circles and contain the smaller.

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THE CIRCLE 275

Locus

1073. Find the locus of the center of a circle which has a given
radius and is tangent to a given circle.

1074. Find the locus of the extremity of a tangent of given length
drawn to a given circle.

1075. Two equal circles are tangent to each other externally.
Find the locus of the centers of all circles tangent to both.

1076. A sect so moves that it re-
mains parallel to a given line, and

so that one end lies on a given cir- y^ ^">^ '. ^,f^ N^

cle. Find the locus of the other end. I /\ / / \

Does the accompanying diagram
give the complete locus?

1077. What is the locus of the

mid-points of parallel chords of a circle? Prove the correctne^ of
your statement.

1078. From a point outside how many tangents are there to a

circle? Prove.

1079. Find the locus of the
mid-point of a sect that is drawn
from a given external point to a
given circle.

1080. A straight line 3 in. long
moves with its extremities on the

perimeter of a square whose sides are 4 in. long. Construct the
locus of the mid-point of the moving line.

1081. A circular basin 16 in. in diameter is full of water, and
upon the surface there floats a thin straight stick 1 ft. long. Shade
that region of the surface which is inaccessible to the mid-point
of the stick, and describe accurately its boundary.

1082. The image of a point in a mirror is apparently as far
behind the mirror as the point itself is in front. If a mirror revolves
about a vertical axis, what will be the locus of the apparent image
of a fixed point 1 ft. from the axis?

1083. In the rectangle ABCD the side AB is twice as long as the
side EC. A point E is taken on the side AB^ and a circle is drawn

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276 PLANE GEOMETRY

through the points C, D, and E. Plot the path of the center of the
circle as E moves from A to B.

1084. Find the locus of a point P such that the ratio of its dis-
tances from two .fixed points is equal to the constant ratio mton.

Construction *

1085. Find the center of a given circle.

1086. Inscribe a circle in q. given triangle.

1087. Circumscribe a circle about a given triangle.

1088. Escribe circles about a given triangle. (See p. 359.)

1089. Through a given point in a circle draw the shortest pos-
sible chord.

1090. Inscribe a circle in a given sector. VII.

1091. With its center in a given line construct a circle which
shall be:

(a) Tangent to another given line at a given point.

(6) Tangent to two other given lines.f

1092. Construct a circle of given radius r, which shall:

(a) Pass through a given point and be tangent to a given line;
(6) Pass through a given point and be tangent to a given circle;

(d) Be tangent to two given circles. VII.

1093. Construct a circle tangent to two given lines and having
its center on a given circle. VII.

1094. An equilateral triangle ABC is 2 in. on a side. Construct
a circle which shall be tangent to AB at the point A and shall pass
through the point C. VII.

1095. To a given circle draw a tangent that shall be parallel
to a given line.

1096. Draw two lines making an angle of 60**, and construct all
the circles of J^ in. radius that are tangent to both lines.

* While some more or less difficult construction problems have been inserted
at this point, they have been primarily inserted for the benefit of those pupils
who wish to test their power, and when found too difficult may well be omitted
until Chapter VII has been studied. Such problems will be followed by the
Roman number ** VII."

t See p. 311 for the section dealing with "The Discussion of a Problem."

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THE CIRCLE 277

dl097. Construct an equilateral triangle, ,'" ^]]^?f\

having given the radius of the circiunscribed / ,-' | \

circle. VII. /^/"^ / ! \

1098. Construct a circle, touching a given rc~ *- — \ }

circle at a given point, and touching a given \ ^^v^ j j

Une. VII. \ \iy

1099. In a given square inscribe four equal ^^ -'^'•

circles, so that each shall be tangent to two of the others, and also
tangent to two sides of the square.

dllOO. In a given square inscribe four equal circles, so that each
shall be tangent to two of the others, and also tangent to one and
only one side of the square. VII.

dllOl. In a given equilateral triangle inscribe three equal circles
tangent each to the other two, each circle being tangent to two
sides of the triangle.

1102. Draw a tangent to a given circle such that the sect inter-
cepted between the point of contact and a given line shall have a
given length. VII.

Theoreiic

1103. If two chords intersect and make equal angles with the
diameter through their point of intersection, they are equal.

1104. The area of a circimiscribed polygon is equal to half the
product of its perimeter by the radius of the inscribed circle.

1105. If two conmion external tangents or two conmion internal
tangents are drawn to two circles, the sects intercepted between the
points of contact are equal.

1106. If two circles are tangent externally, the conmion internal
tangent bisects the two conmion external tangents.

1107. A line tangent to two equal circles is either parallel to the
sect joining their centers or bisects it.

1108. A coin is placed on the table. How many coins of the same
denomination can be placed around it, each tangent to it and

tangent to two of the others?
Prove your answer.
1109. If through any point in the
j^.^^ /I ] convex arc included between two

tangents a third tangent is drawn,
a triangle will be formed, the peri-

278 PLANE GEOMETRY

meter of which is constant and equal to the sum of the two
tangents.

1110. If a triangle is inscribed in a triangle ABCy whose semi-
perimeter is s, the sects of its sides from the vertices to the points
of contact are equal to s -a, s -6, and s -c.

1111. The perimeter of an inscribed equilateral triangle is equal
to half the perimeter of the circumscribed equilateral triangle.

1112. The radius of the circle inscribed in an equilateral tri-
angle is equal to one-third of the altitude of the triangle.

dlllS. In a circumscribed quadrilateral the sirni of two opposite
sides is equal to the sum of the other two sides, and a circle can be
inscribed in a quadrilateral if the smn of two opposite sides is
equal to the smn of the other two sides.

dlll4. In what kinds of parallelograms can a circle be inscribed?
Prove.

1115. The diameter of the circle inscribed in a right triangle
is equal to the diflference between the smn of the legs and the
hypotenuse.

1116. All chords of a circle which touch an interior concentric
circle are equal, and are bisected at the points of contact.

Theorem 50. In equal circles central angles have the same
ratio as their intercepted arcs.

Cor. 1. A central angle is measured by its intercepted arc.

Theorem 51. Parallels intercept equal arcs on a circle.

Theorem 52. An inscribed angle, or one formed by a tangent
and a chord is measured by one-half its intercepted arc.

Theorem 52a. The mid-point of the hypotenuse of a right
triangle is equidistant from the three vertices.

What point in the circumscribed circle is the mid-point of the hypotenuse?

Theorem 53. An angle whose vertex is inside the circle is mea-
sured by half the sum of the arcs intercepted by it and its vertical.

Theorem 54. An angle whose vertex is outside the circle is
measured by half the difference of its intercepted arcs.

Theorem 54a. The opposite angles of a quadrilateral inscribed
in a circle are supplementary.

(Proof left to the pupil.)

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THE CIRCLE 279

Cor. 1. A quadrilateral is inscriptible if its opposite angles are
supplementary.

Suggestion: Show, by the method of exclusion, that the fourth vertex of the
quadrilateral lies on the circle passing through three of its vertices.

EXERCISES. SET LXXXVI. MEASUREMENT OF ANGLES

Numeric

1117. Find the value of an angle which (a) is inscribed, and in-
tercepts an arc of 160°, (6) is inscribed in a segment of 250°.

1118. If the tangents from a point to a circle make an angle of
60°, what are the values of the arcs they intercept? What if the
angle is a right angle?

1119. Find the angle whose sides are tangents drawn from a
point whose distance from the center of the circle is the diameter
of that circle

1120. An angle between two chords intersecting inside a circle
is 35°, its intercepted arc is 25° 18'; find the arc intercepted by its
vertical.

1121. A triangle is inscribed in a circle, and another triangle is
circumscribed by drawing tangents at the vertices of the inscribed
triangle. The angles of the inscribed triangle are 40°, 60°, and 80°.
Find all the other angles of the figure.

1122. The arcs subtended by three consecutive sides of a quad-
rilateral are 87°, 95°, 115°; find the angles of the quadrilateral;
the angles made by the intersection of the diagonals, and the
angles made by the opposite sides of the quadrilateral when
produced.

1123. Three consecutive angles of an inscribed quadrilateral
are 140° 30', 80° 30', and 39° 30'. Find the numbers of degrees in
the arcs subtended by the four sides if the arc intercepted by the
largest angle is divided into parts in the ratio of 4 to 5.

1124. Three consecutive angles of a circumscribed quadrilateral
are 85°, 122°, 111°. Find the number of degrees in each angle of the
inscribed quadrilateral made by joining the points of contact of the
sides of the circumscribed quadrilateral.

1125. The points of tangency of a quadrilateral, circiunscribed

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280 PLANE GEOMETRY

about a circle, divide the circiimference into arcs, which are to
each other as 4, 6, 10, and 16. Find the angles of the quadrilateral.
1126. If the sides AB and BC of an inscribed quadrilateral
ABCD subtend arcs of 60° and 130°, respectively, and the diagonals
form <^AED = 70'', find the number of degrees in (a) iCB, (6) ^,
(c) each angle of the quadrilateral.

1127. In this figure ^B=4r, ^A

=65°, and ^BCZ)= 97°. Find the

niunber of degrees in each of the other

1^ angles, and determine whether or not

CD is a diameter.

1128. In this figure <m=62° and

AD B ^n=28°. Find the number of de-

grees in each of the other angles, and deter-
mine whether or not AB is a diameter.

1129. At the vertices of an inscribed
quadrilateral tangents are drawn to the cir-
cle, forming a circumscribed quadrilateral.
The arcs subtended by the sides of the in-
scribed quadrilateral are in the ratio of 3 to 4 to 5 to 8.

(a) Find the angles of each quadrilateral.
(6) Find the angles between the diagonals of the inscribed
quadrilateral.

(d) Find the angles between the sides of the inscribed and those
of the circimiscribed quadrilateral.

1130. The vertices of a quadrilateral inscribed in a circle divide
the circumference into arcs which are to each other as 1, 2, 3, and 4.
Find the angles between the opposite sides of the quadrilateral.

1131. The sides of an inscribed quadrilateral subtend arcs in the
ratio (a) 1 to 2 to 3 to 4, (6) 3 to 5 to 7 to 9. How many degrees in
each angle of the quadrilaterals in (a) and (6)?

1132. The bases of an inscribed isosceles trapezoid subtend ares
of 100° and 120°. How many degrees in each angle of the trapezoid
(a) if the bases are on the same side of the center, (6) if they are
on opposite sides of the center?

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THE CIRCLE 281

Theoretic

1133. The angle formed by two tangents is equal to twice the
angle between the chord of contact * and the radius drawn to a
point of contact.

1134. If the tangents drawn from an exterior point to a circle
form an angle of 120°, the distance of the point from the center is
equal to the sum of the tangents.

1135. An isosceles trapezoid is inscriptible; that is, a circle can
be circumscribed about it.

1136. If in a circle two chords are drawn, and the mid-point of
the arc subtended by one choTd is joined to the extremities of the
other chord, the two triangles thus formed are mutually equi-
angular, and the quadrilateral thus formed is inscriptible.

1137. If A, By C, -4.1, Biy Ci are six points in a circumference,
such that AB is parallel to AiBi and AC is parallel to AiCi, then
BCi is parallel to BiC.

1138. Let A be any point of a diameter, B the extremity of a
radius perpendicular to the diameter, P the point in which BA
meets the circmnf erence, C the point in which the tangent through
P meets the diameter produced. Prove that AC=PC.

1139. If two circles touch internally, and the diameter of the
smaller is equal to the radius of the larger, the circumference of the
smaller bisects every chord of the larger which can be drawn
through the point of contact.

dll40. Two circles touch internally in the point P, and 4B is a
chord of the larger circle touching the smaller in the point C.
Prove that PC bisects the angle APB.

1141. If two circles intersect at the points A and 5, and through
A a variable secant be drawn cutting the circles in C and D, the
angle CBD is constant for all positions of the secant.

1148. If two circles are tangent externally, the corresponding
sects of two lines drawn through the point of contact and ter-
minated by the circles are proportional.

* By the chord of contact is meant the sect joining the points of contact
of a pair of tangents.

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282

PLANE GEOMETRY

1143. If two circles are tangent to each other and a sect be
drawn through the pomt of tangency, terminating in the circles,
the diameters from the extremities of this sect are parallel.

Case I. Circles tangent externally.
Case II. Circles tangent internally.

1144. If two circles are tangent to each other and a sect be
drawn through the point of tangency and terminating in the
circles, tangents at the extremities of this sect are parallel.

Case I, the circles tangent externally; and Case II, tangent
internally.

dll45. If two sects OA and OB, not in a str aight line, are divided
in C and D, respectively, so that OA-OC^OB^OD, then A, B, C, D
are concydic, that is, lie on the same circle.

dll46. The altitudes of a triangle bisect the angles of the tri-
angle determined by their feet, i.e., the angles of the pedal triangle.

dll47. The feet of the medians and the feet of the altitudes of a
triangle are concyclic.

Hint: Pass a circle through the feet of the medians, then prove that the
foot of any one of the three altitudes will lie on this circle.

1148. If two circles are tangent externally, and a secant is drawn
through the point of contact, the chords formed are proportional
to the radii. ^^

1149. If C is the mid-point of AB, and chord CD cuts chord AB

. J, CE CA

' CA CD
dllSO. If two circles are tangent externally, the common tangent
is a mean proportional between the diameters.

dll61. The
line joining the
extremities of
two parallel radii
.p of two circles
passes through
the direct center
of similitude if
the radii have the
same direction,
and through the inverse center if the radii have opposite directions.

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THE CIRCLE 283

dll62. Taking any point as a center of similitude of two circles,
the two radii of one of them, drawn to its points of intersection
with any other line passing through that center of similitude, are
parallel, respectively, to the two radii of the other, drawn to its
intersections with the same line.

Hint: Use an indirect proof depending upon Ex. 1151.

dll63. All secants, drawn through a direct center of similitude
P of two circles, cut the circles in points whose distances from P,
taken in order, form a proportion.

dll54. If in the last exercise, the line of centers cuts the circles
in points A, B, C, D, and any other secant through P cuts the
circle in points ilf , iV, fi, /S, prove that PN*PR is constant and equal
to PB'PC.

dll66. The common external tangents to two circles pass
through the direct center of similitude, and the conmion interior
tangents pass through the inverse center of similitude.

What method of drawing the common tangents to two circles
may be derived from this fact?

1166. ABC is an isosceles triangle inscribed in a circle, BD a

chord drawn from its vertex cutting the base in any point E.

^. W AB

"ove =====

15 BE

1157. If two circles are tangent internally, all chords of the
greater circle drawn from the point of contact are divided pro-
portionally by the circumference of the smaller.

1168. If two circles touch at Af , and through M three lines are
drawn meeting one circle in A, B, C, and the other in D, E, F,
respectively, tiie triangles ABC and DEF are similar.

Locus

1169. An angle of 60® moves so that both of its sides touch a
fixed circle of radius 5 ft. What is the locus of the vertex?

1160. Find the locus of the mid-point of a chord drawn through
a given point within a given circle.

1161. Through a point A on a circle chords are drawn. On each
one of these chords a point is taken one-third the distance from A
to the end of the chord. Find the locus of these points.

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284 PLANE GEOMETRY

1162. The locus of the vertex of a triangle, having a given base
and a given angle at the vertex, is the arc which forms, with the
base, a segment capable of containing the given angle.

1163. Find the locus of the points of contact of tangents drawn
from a given point to a given set of concentric circles.

1164. A variable chord passes, when prolonged, through a
fixed point outside a given circle. What is the locus of the mid-
point of the chord?

1166. Upon a sect AB a segment of a circle containing 240** is
constructed, and in the segment any chord CD subtending an arc
of 60° is drawn. Find the locus of the intersection of AC and BD,
and also of the intersection of AD and BC.

1166. The locus of the centers of circles inscribed in triangles
having a given base and a given angle at the vertex is the arc
which forms with the base a segment capable of containing a right
angle plus half the given angle at the vertex.

1167. The locus of the intersections of the altitudes of triangles
having a given base and a given angle at the vertex is the arc
forming with the base a segment capable of containing an angle
equal to the supplement of the given angle at the vertex.

dll68. Find the locus of a point from which two circles subtend*
the same angle.

1169. If A and B are two fixed points on a given circle, and P
and Q are the extremities of a variable diameter of the same
circle, find the locus of the point of intersection of the lines
AP and BQ.

dll70. The lines li and U meet at right angles in a point A. is
any fixed point on U.» Through draw a line meeting Zi in B.
P is a var3ring point on this line such that OB'OP is constant.
Plot the locus of P as the line swings about as a pivot.

Theorem 66. A tangent is the mean proportiondl between any
secant and its external sect, when drawn from the same points
to a circle.

* If tanoents from the same point to two circles form equal on^€t, the cireUa
are eaid to eybtend equal angles from thcU point.

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THE CIRCLE

286

Cor. 1. The product of a sec-
ant and its external sect
from a fixed point out-
side a circle is constant.
Hint: Draw tangent PT.
What is the constant in this corollary 7
State the corollary in another way.

Theorem 56a. If chords intersect inside a circle, the product
of their sects is constant

Prove by means of similar triangles.

Applying the principle of continuity, Theorem 55, Cor. 1, and
Theorem 55a can be stated as a single theorem. State them bo.

Theorem 66b. The square of the bi-
sector of an angle of a triangle is equal
to the product of the sides of this angle,
dirrtinished by the product of the sects
made by that bisector on the third side.

Given: AABC with bisector tb cutting AC at D

into sects q and r.
Prove: tb^^ac—qr.

Proop

Circumscribe OO about ABC, and
extend BD to E in QO and draw
chord CE.

Let DJ?a<.

(1) ABDA^ABCE

(2) .-. -^«^

(3) ,\ac^tb*+tb8

(4) But tbsmqr

(5) .\acmtb*+qf

(6) .\tb*mac-qr

Theorem 66c.

(1) Why?

(2) Why?

(3) Why?

(4) Why?

(5) Why?

(6) Why?

In any triangle the product
of two sides is equal to the product of the
diameter of the circumscribed circle and the
altitude on the third side.
Hint: Prove AABX^ AEBC,

Consider the special case where 4- B is a right
angle and evolve a formula for ^.

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286 PLANE GEOMETRY

Cor. 1. IfR denote the radius of the circle circumscribed about
a triangle whose sides are a, b, c,and semiperimeter s,
then

j^_ obc

Wsis -a){8 -b){s -c):
In the figure for Theorem 55c, ac^hj).
r^ ac T% oc

b =

.R=

But hb=^y/s{s '-d){a -6)(« -c)

ac obc

^\/s{8 -a)(s -6)(s -c) 4\/s(s -a){s -6)(s -c)
b

EXERCISES. SET LXXXVII. METRIC RELATIONS
Numeric

1171. A point P is 10 in. from the center of a circle whose radius
is 6 in. Find the length of the tangent from P to the circle.

1172. The length of a tangent from P to a circle is 7 in., and the
external sect of a secant is 4 in. Find the length of the whole
secant.

1173. A point P is 8 in. from the center of a circle whose radius
is 4. Any secant is drawn from P, cutting the circle. Find the

• product of the whole secant and its external sect.

1174. From the same point outside a circle two secants are
drawn. If one secant and its external sect are 24 and 15, respec-
tively, and the external sect of the other is 7, find that secant.

1176. Two chords intersect within a circle. The sects of one
are m and n and one sect of the other is p. Find the remaining sect.

1176. If a tangent and a secant drawn from the same point to a
circle measure 6 in. and 18 in., respectively, how long is the ex-
ternal sect of the secant?

1177. Two secants are drawn from a common point to a circle.
If their external sects are 12 and 9, and the internal sect of the
first is 8, what is the length of the second?

1178. The radius of a circle is 13 in. Through a point 5 in.
from the center a chord is drawn. What is the product of the two

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THE CIRCLE 287

sects of the chord? What is the length of the shortest chord that
can be drawn through that point?

1179. One sect of a chord through a point 3.5 units from the
center of a circle is 2 units in length. If the diameter of the circle
is 12 units, what is the length of the other sect of the chord?

1180. The radius of a circle is 2 units. If through a point P,
4 units from the center, secant PQR is drawn, and QR is oneimit,
what is the length of PQ?

1181. AABC is inscribed in a circle of radius 5 in. Find the
altitude to BC if AB is 4, and AC is 5 in.

1182. The sides of a triangle are 4, 13, and 15, respectively.
Find the radius of the circmnscribed circle.

1183. In AabCy a = 20, 6 = 15, and the projection of b upon c
(Pjc) is 9. Find the radius of the circumscribed circle.

1184. In Aabc, a = 9 and 6= 12. Find c if the diameter of the
circumscribed circle is 15.

1186. The sides of a triangle are 18, 9, and 21, respectively.
Find the angle bisector corresponding to 21.

1186. The sides of a triangle are 21, 14, and 25, respectively.
Find the angle bisector corresponding to 25.

1187. The sides of a triangle are 22, 11, and 21, respectively.
Find the angle bisector corresponding to 21.

1188. The sides of a triangle are 6, 3, and 7, respectively. Find
the angle bisector corresponding to 7.

1189. In a triangle the sides of which are 48, 36, and 50, where
do the bisectors of the angles intersect the eides? What are the
lengths of the angle bisectors?

1190. In each of the Exercises 1181 to 1189 what kind of triangle
is involved?

Construction

1191. Construct the mean proportional between two^veneects,
using in turn the methods suggested by the following propositions:

(a) 39 Cor. 1, (6) 39 Cor. 4, (c) 55.

1192.* Construct a square equal in area to that of a given: (a)
rectangle; (6) triangJe; (c) trapezoid.

* For discussion and illustrations of the type of analysis applicable see
pp. 318 to 321, and Problems 17, 19, 21, 22, 26, 27, Chapter VII.

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288 PLANE GEOMETRY

1193, Draw thro ugh a given external point P a secant PAB to a
given circle so that IW^PA^PB. VII.

dll94. Draw through one of the points of intersection of two
given intersecting circles a common secant of given length. VII.

dll95. From a point outside a circle draw a secant whose exter-
nal sect is equal to one-half the secant.

Locus

dll96. Given the fixed base of a triangle and the sum of the
squares of the other two sides, describe the locus of the vertex.

dll97. Repeat Ex. 1 196, given the difference of the squares of the
other two sides.

dll98. Through P any PMN is drawn, cutting a c ircle K in M
and N, and P moves so that the product of the sects PM PN has
the constant value k^. Find the locus of P.

Theoretic

1199. If chord AB bisects chord CD, either sect of chord CD
is a mean proportional between the sects of AB.

1200. If two circles intersect, their common chord produced
bisects the common tangents.

1201. In the diameter of a circle points A and B are taken
equally distant from the center, and joined to a point P in the
circumference. Prove that AP^+BP^ is constant for all posi-
tions of P.

1202. If a tangent is limited by two other parallel tangents to
the same circle, the radius of the circle is the mean proportional
between its sects.

1203. The tangents to two intersecting circles, drawn from any
point in their common chord produced, are equal.

dl204. The sum of the squares of the diagonals of a trapezoid is
equal to the sum of the squares of the legs plus twice the product
of the bases.

1206. In an inscribed quadrilateral the product of the diagonals
is equal to the sum of the products of the opposite sides. (Ptolemy's
Theorem.)

dl206. If the opposite sides of an inscribed hexagon intersect,
they determine three coUinear points. (" Mystic Hexagram,"
discovered by Pascal when he was 16 years of age.)

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/
/

THE CIRCLE 289

dl207. If a circumference intersects tha sides a, b, c, of a AABC
in the points Ai and At, Bi and Bi, Ci and Cj, respectively, then
ACiBAiCBiAC^BAiCBi , ,„ ,, ^, ,

C^-AlC'B^J-^-SC-^^l- (C^°* ' Theorem.)

Theorem 66. A circle may be circumscribed about, and in-
scribed within, any regular polygon
Cor. 1. An equilateral polygon inscribed in a circle is regular.
Cor. 2. An equiangular polygon circum" /\ A

scribed about a circle is regular. / \ / \

Cor. 3. The area of a regular polygon is a/- -ir^

equal to half the product of its apo-
them and perimeter.
Suggestion: What is the area of AAOJB? B^

Theorem 67. If a circle is divided into' any number of equal
arcs, the chords joining the successive points of division form a
regular inscribed polygon; and the tangents drawn at the points
of dimsion form a regular circumscribed polygon.

Cor. 1. Tangents to a circle at the vertices of a regular inscribed
polygon form a regular circumscribed polygon of the
same number of sides.

Cor. 2. Lines drawn from each vertex of a regular inscribed
polygon to the mid-points of the adjacent arcs sub-
tended by its sides form a regular inscribed polygon
of double the number of sides.

Cor. 3. Tangents at the mid-points of the arcs between con-
secutive points of contact of the sides of a regular
circumscribed polygon, form a regular circumscribed
tolygon of double the number of sides.

Cor. 4. The perimeter of a regular inscribed polygon is less
than that of a regular inscribed polygon of double
the number of sides; and the perimeter of a regular
circumscribed polygon is greater than that of a
regular circumscribed polygon of double the number
of sides.

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290

PLANE GEOMETRY

Cor. 6. Tangents to a circle at the mid-points of the arcs sub-
tended by the sides of a regular inscribed polygon,
form a regular circumscribed polygon, of which the
sides are parallel to those of the original polygon and
the vertices lie on the prolongations of the radii of the
one inscribed.

Suggestions: Show that AB and AiBi are
both perpendicular to OP and are,
therefore, parallel.

Show .'. that 2S^A = 2S^Ai, 2S^B =
4.JB1, ....

What kind of n-gon is then circum-
scribed?

BP-=BT and J5iPi=5iri and,
therefore, B and Bi lie on bisector
of ^POT,
^^-^ Radius OBi bisects iS^POT.

Theorem 68. A regular polygon the number of whose sides
is 3 •2" may be inscribed in a circle.

Theorem 69. If i„ represent the side of a regular inscribed
polygon of n sides and i2 „ the side of one of 2n sides and r the

radius of the circle, i2n=\'2r^ -rVij^ -i/.

Theorem 60. If i„ represent the side of a regular inscribed

polygon of n sides, c„ that of a regular circumscribed polygon of n

2ri„
sides, and r the radius of the circle, c„= / ^ .^ >

Theorem 61. The perimeters of regular polygons of the same
number of sides compare as their radii and also as their apothems.

Theorem 62. Circumferences have the same ratio as their
radii.

Cor. 1. The ratio of any circumference to its diameter is
constant.

Cor. 2. In any circle c=2irr.

Theorem 63. The value of w is approximately 3.14169.

Theorem 64. The area of a circle is equal to one-half the
product of its radius and its circumference.

Cor. 1. The area of a circle is equal to w times the square of
its radius.

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THE CIRCLE 291

Cor. 2. The areas of circles compare as the squares of their

radii.
Cor. 3. The area of a sector is equal to half the product of its

radius and its arc.

(Proof left to the student.)

Note. — Cor. 3 does not suggest the most convenient method of determining
the area of a sector. Suggest a more convenient one.

A segment of a drde is a portion of it bounded by an arc and its
subtending chord. Similar sectors and similar segments are those
of which the arcs contain the same number of degrees.

Cor. 4. Similar sectors and similar segments compare as the
squares of their radii.

Suggestions: How do circles
and Oi compare in area?
How, then, would like parts
of them compare?

Wliy are similar sectors
like parts?

Why is AAOBco
AAiOiBi? How, therefore,
do the triangles compare
in area?

Justify the following alge-
braic statements which form the basis of proof here:

sec. AOB _/R^\_ AAOB . sec.AQJg ^ sec.^iQiBi
sec.AiOiBi\r^)~AAiOiBi ''' AAOB AAiOiBi
. seg.^QB _ seg.AiQijgi seg. AOB _ AAOB _R*
" AAOB " AAiOiBi ^^ seg.AiOiBr AAiOiBi r^'

EXERCISES. SET LXXXVIII. MENSURA.TION OF THE CIRCLE

Num£ric

1208. Find

the circumferences of circles

with diameters as

follows:

(a) 9 in.

(g) 29 centimeters

(b) 12 in.

(d) 7.3 in. (/) 3i in.

(h) 47 millimeters

1209. Find

the diameters of circles with

circumferences as

follows:

(a) 15

(g) 3361.512 in.

Q>) TT*

(d) lira^ (/) 219.912 in.

(A) 3173.016 in.

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292 PLANE GEOMETRY

1210. Find the diameter of a carriage wheel that makes 264
revolutions in going half a mile.

1211. The diameter of a bicycle wheel is 28 in. How many
revolutions does the wheel make in going 10 mi.?

1212. Find the radii of circles with circumferences as follows:
(a) 7w (c) 15.708 in. (c) 18.8496 in. {g) 345.576 ft.
(6) 3^ (d) 21.9912 in. (/) 125.664 in. Qt) 3487.176 in.

1213. Find the radius of a circle whose circumference is m units.

1214. An arc of a certain circle is 100 ft. long and subtends an
angle of 25° at the center. Compute the radius of the circle correct
to one decimal place.

1216. The circiunference of a circle is 10. Find the circumfer-
ence of one having twice the area of the original.

1216. Find the central angle of a sector whose perimeter is equal
to the circiunference.

1217. Find the areas of circles with diameters as follows:
(a) 16afe (c) 2.5 ft. (e) 3f yd. {g) 3 ft. 2 m.
(6) 247r2 id) 7.3 in. (J) 4f yd. Qt) 4 ft. 1 in.

1218. Find the area of circles with radii as follows:

(a) bx (c) 27 ft. (c) 3^ in. {g) 2 ft 6 in.

(6) 27r {d) 4.8 ft. (/) 4f in. Qi) 7 ft. 9 in.

1219. Find the radii of circles with areas as follows:
(a) Tra^b* (c)x (e) 12.5664 {g) 78.54
(6) 47rmV (d) 2ir (/) 28.2744 (A) 113.0976

1220. Find the areas of circles with circiunferences as follows:
(a) 27r (c) ira (c) 18.8496 in. {g) 333.0096 in.
(6) 47r {d) Uwa^ (/) 329.868 in. (h) 364.4256 in.

1221. Find the area of a circle whose circumference is C.

1222. Find the area of a sector whose radius is 6 and whose
central angle is 40°.

1223. Find the area of a fan that opens out into a sector of 120°,
the radius of which is 9| in.

1224. The arc of a sector of a circle 2^ in. in diameter is 1-| in.
What is the area of the sector?

1226. Find the central angle of a sector whose area is equal to
the square of the radius.

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THE CIRCLE

293

1226. Find the circumference of a circle whose area is S.

1227. A circle has an area of 60 sq. in. Find the length of an
arc of 40°.

1228. Find the radius of a circle equivalent to a square the side
of which is 6.

1229. The circumferences of two concentric circles are 30 and 40,
respectively. Find the area bounded by the two circiunferences
by the shortest method you know.

1230. In an iron washer here shown, the dia-
meter of the hole is 1|- in., and the width of
the washer is -f in. Find the area of one face
of the washer.

1231. The area of a fan which opens out
into a sector of 111° is 96.866 sq. in. What is
the radius? (Use 7r= 3.1416. Why?)

1232. The radius of a circle is 10 ft. Two parallel chords are
drawn, each equal to the radius. Find that part of the area of the
circle lying between the parallel chords.

1233. A square is inscribed in a; circle of radius 10. Find the
area of the segment cut off by a side of the square.

1234. Find a semicircle equivalent to an
equilateral triangle whose side is 5.

1236. A kite is made as shown in the dia-
gram, the semicircle having a radius of 9 in.,
and the triangle a height of 25 in. Find the
area of the kite.

1236. Two circles are tangent internally, the
ratio of their radii being 2 to 3. Compare their
areas, and also the area left in the larger circle
with that of each of the circles.

1237. A reservoir constructed for irrigation
purposes sends out a stream of water through

a pipe 3 ft. in diameter. The pipe is 1000 ft. long. How many
times must it be filled if it is to discharge 10,000 acre-feet of
water? (An acre-foot of water is the amount required to cover
1 acre to a depth of 1 ft.)

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294 PLANE GEOMETRY

1238. Each side of a triangle is 2n centimeters, and about each
vertex as center, with radius of n centimeters, a circle is described.
Find the area bounded by the three arcs that lie outside the triangle,
and the area boimded by the three arcs that lie inside the triangle.

1239. From a point outside a circle whose radius is 10, two
tangents are drawn. Find the area bounded by the tangents
and the circumference, if they include an angle of 120°. Find
both results.

1240. Upon each side of a square as a diameter semicircles are
described inside the square. If a side of the square is s, find the
sum of the areas of the four leaves.

A 1241. Find the area bounded by three arcs each
of 60° and radius 5 if the convex sides of the arcs are
turned toward the area.
1242. Find the area bounded by three
arcs each of 60° and radius 5 if the
concave sides of the arcs are turned toward the
area.

dtl243. The flywheel of an engine is connected
by a belt with a smaller wheel driving the machinery of a mill. The
radius of the flywheel is 7 ft., and of the driving wheel is 21 in. (a)
How many revolutions does the smaller wheel make to one of the
larger wheel? (6) The distance between the centers is 10 ft. 6 in.
What is the length of the belt connecting the two wheels if it is
not crossed? (c) If it is crossed?

1244. Given a circle whose radius is 16, find the perimeter and
the area of the regular inscribed octagon.

^ 1246. The following is Ceradini's approximate

method of constructing a sect equal in length to a
circle: Draw diameter AB and tan-
gent BK at B. Draw OC making
<COB = 30°. Make CD =30B. Draw
AD J and prolong it, making ZB =
2AD. Then AE is the required sect.
Determine the accuracy of this con-
struction by computing the ratio of AE to AB.

Suggestion : Let r = radiuB. Compute AB and AE in terms of r, then divide.

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THE CIRCLE 296

1246. Another method of finding the approximate value of the
circle is as follows: Draw diameter CD. Make central angle AOB
=30°. jyvsmABLCD. Draw ^ ^^
CE tangent at C and equal
to ZCD. Draw BE. Then!
BE equals the circle. Deter-
mine the accuracy of this
construction.

1247. In making a drawing for an arch it is required to mark
off on a circle drawn with a radius of 5 in. an arc that shall be 8 in.
long. This is best done by finding the angle at the center. How
many degrees are there in this angle?

1248. The perimeter of the circimiscribed equilateral triangle is
double that of the similar inscribed triangle.

1249. Squares are inscribed in two circles of radii 2 in. and 6 in.,
respectively. Find the ratio of the areas of the squares, and also
of the perimeters.

1260. Squares are inscribed in two circles of radii 2 in. and 8 in.
respectively, and on their sides equilateral triangles are con-
structed. What is the ratio of the areas of these triangles?

1261. A log a foot in diameter is sawed so as to have the cross-
section the largest square possible. What is the area of this square?
What would be the area of the cross-section of the square beam cut
from a log of half this diameter?

1262. If r denotes the radius of a regular inscribed polygon, a
its apothem, s a side, A an angle, and C the angle at the center,
show that:

(a) In a regular inscribed triangle «=rV3, a^^r^ As 60°,
C=120°.

(6) In a regular inscribed quadrilateral ssrv2, a=^rV2,
As90°, Cs90°.

( d) In a r egular inscribed decagon «s^r(v5-l), as

\r\\0+2y/by A = 144°, C=36°.

li263. If r is the radius of a circle, a the apothem of a regular
inscribed n-gon, and i„ one of its sides, Hn a side of a regular inscribed

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296

PLANE GEOMETRY

2n-gon, Cn a side of a regular circumscribed n-gon; An the area of
a regular inscribed n-gon, and A^n that of a regular inscribed
2n-gon, fill out the accompanying table:

Given

Required

Given

Required

1. r^in

15. in,n»3

2. r,in

itn

16. f, n=3

3. r, tin

17. in,n-6

4. r, in

18. f,n=6

5. r, n-3

in, a

19. in, n = 12

6. r, n=6

in, a

20. f,n =20

7. f,n = 12

in, a

21. in, n = 12

8. r, n-4

in, a

22. f,n = 12

9. r, n«8

in, a

23. in,n»4

10. r, n = 10

in, a

24, f, n=4

11. r,n=5

in, a

25. in, n=8

12. int n, a

26. f, n=8

13. int n, r

27. f, n=5

14. tn, n, r

Ajn

28. r,n = 10

Theoretic
1264. The area of a regular inscribed hexagon is a mean propor-
tional between the areas of the inscribed and circumscribed equi-
lateral triangles.

dl265. An equilateral polygon circumscribed about a circle is
regular if the number of its sides is odd.

dl266. An equiangular polygon inscribed in a circle is regular if

the number of its sides is odd.

1257.. If C be a point in the
straight line AB, the three eemi-
circles, drawn respectively upon
sects ABy AC, and CB as dia-
1^ meters, bound an area equal to a
circle of which the diameter is
the perpendicular CD, D being in the largest semicircle.

1268. If upon three sides of a right tri-
angle semicircles be drawn as indicated in
the diagram, the area of the right triangle
is equal to the sum of the two crescent-
shaped areas, bounded by the semi-
circles. (Hippocrates' Theorem.)

1269. Give a simpler proof for Ex. 859. (b) Generalize this fact.

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CHAPTER VI

METHODS OF PROOF

There are two general methods of proving theorems, the direct
or synthetic, and the indirect method. Each of these methods of
proof may be in its natm-e geometric or algebraic.

Further, indirect proofs, whether geometric or algebraic, may
take different forms. Thus a theorem may be proved indirectly,
either by means of exclusion or redaction to an absurdity, or by
arudysis.

It is the object of this chapter to give an illustration of each
of these methods of proof, together with several exercises that will
be most naturally proved by that method.

A. THE DIRECT OR SYNTHETIC METHOD

In this method we employ either superposition or start with the
data, and combining them with known truths proceed step by
step until we arrive at the desired conclusion.
I. GEOMETRIC PROOF.*

Illustration: The bisector of the vertex
angle of an isosceles triangle bisects
the base.
Given: AABC, ABsBC, X in AC bo that

2^.ABX^2^.CBX.
Prove: AX^XC.

Proof
In AABX and ACBX.

(1) AB^BC, 25.ABZa4.CBX.

(2) BX^BX.

(3) /. AABX^ACBX,

(4) .\AX~CX.

(1) Data.

(3) Two sides and the included
angle determine a triangle.

(4) Horn, sides of cong. A are
equal.

♦ The method of superposition should be very rarely used. For illustra-
tions of it, recall the proofs of the fimdamental theorems in congruence of
triangles.

297

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PLANE GEOMETRY

EXERCISES. SET LXXXIX. SYNTHETIC METHODS OF PROOF
Give a sjnithetic proof of each of the following:

1260. The bisector of the vertex angle of an isosceles triangle
is perpendicular to the base.

1261. If the. perpendicular bisector of the base of a triangle
passes through the vertex, the triangle is isosceles.

1262. Any point in the bisector of the vertex angle of an isosceles
triangle is equidistant from the ends of the base.

n. ALGEBRAIC PROOF.

Illustration: If one leg of an isosceles triangle
is extended through the vertex by its own length,
the sect joining its end to the end of the base
is perpendicular to the base.
Given: ABDast. line, AB=BD=BC.
Prove: DC±AC,

Proof

(1) 2^A^-4.i>^-4.DCA»l8o^ .

(2) •/ AB^BCmdBC^BD.

(3) ^As^BCA, ^Ds^DCB.

(4) .'. substituting (3) in (1),
^BCA + 2S^DCB+ ^DCA s l80^

(6) But ^BCA-{-2S^DCB^:S^DCA.

(6)
(7)

(8)

(9)

24DCA

90^

^DCA is a rt. -4..
DC±AC,

(1) The sum of the 4. of a A is
a st. 2^.

(2) Data.

(3) Base^ofanisoscelesAareequal.

(4) Quantities may be substituted
for their equals in an equation.

(5) The whole equals the sum of all
its parts.

(6) See (4).

(7) Quotients of equals divided by
equals are equal.

(8) The niuneric measure of a rt. 4.
is 90^

(9) Bydef.of ±.

This method is especially adapted to the proof of numerical
relations between angles or sects.

The following procedure is generally used in applying the alge-
braic synthetic type of proof.

1. Observation of the numeric relations that immediately follow
from the data.

2. Statement of these relations in algebraic form — the equality
or the inequality.

3. Reduction of these algebraic relations by the help of axioms
until the desired conclusion is reached. GoOqIc

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METHODS OP PROOF 299

EXERCISES. SET LXXXIX (concluded)

Give an algebraic synthetic proof of each of the following:

1263. The bisectors of two supplementary adjacent angles are
perpendicular to each other. "^ ,

1264. If the bisectors of two adjacent angles are perpendicular
to each other, those angles are supplementary.

1266. If two sides of a triangle are imequal the angles opposite
them are unequal in the same order.

1266. The sum of the altitudes of a triangle is less than its
perimeter.

1267. The angle whose sides are th^ altitude from and the bisector
of an angle of a triangle is equal to one-half the difference bertween
the remaining angles of the triangle.

B. THE INDIRECT METHODS
I. GEOMETRIC; or n. ALGEBRAIC.

a. By the Method of Exclusion.

Two magnitudes of the same kind may bear one of three rela-
tions to each other. The first may be less than, equal to, or greater
than the second. If it can be shown that two of these relations
are false, the third is of necessity true. Similarly the position of a
point may be fixed by the method of exclusion.

lUvstration 1, Theorem 21c: If two angles of a triangle are

unequal, the sides opposite them are p

unequal in the same order.

Given: AABC in which ^A > 4C.

Prove: a>c, ^

Proof

a>c^ a^c, ora<c.

Suppose a<c.

Then 4il<4C.

But this contradicts the data.

.-. a<c. _

Suppose a^c.

Then ^As^C.

But th^ contradicts the data.

.'. a^ c.

.'. a>c.

Authorities left to the student.

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aqp

PLANE GEOMETRY

Illustration 2, Theorem 64a, Cor. 1:
A quadrilateral whose opposite angles
are supplementary is inscriptible.
-^^ a Given: Quadrilateral ABCD in which ^A

4-4-C = 180% 4B-f4i> = 180^
Prove: A, B, C, D ooncyclic.

Authorities left to the student.

Proof

(1) A circle may be passed through A, B,C.

(2) D lies outside, inside, or on this circle.

(3) Suppose D Ues inside QABC in the
position l>i.
Then ^.B* M^ and 4.Di«§(cS^+5>).

(4) /. 2i.B+}S^Di^yHj(TC+6SX+in')^
^(360•+:f?)>180^

(5) .*. D cannot lie inside the circle.

(6) Suppose D lies outside OABC in the
position Z>t.
Then ^B^-HiP^and^Da^-H iCST-fP).

(7) .•.25.B-f4.D,«H(A^+C^Si -^ =
^(360•-ZP)<180^

(8) .*. D cannot lie outside the circle.

(9) .*. D lies on the circle and ABCD is in-
sonptible.

EXERCISES. SET XC. PROOF BY THE METHOD OF EXCLUSION
Prove the following facts by means of the method of exclusion:

1268. Knowing (1) that equal chords subtend equal arcs, and
(2) that unequal chords subtend arcs unequal in the same order,
prove the converse of each of these facts by the method of exclusion.

1269. In a fashion similar to that used to prove Ex. 1268, show
(1) when as6, csd;

that if (2) when a>6, c>d; then the converse of each of these

(3) when a<6, c<d,
facts is true. (This is known as the Law of Converses,)
b. By Reduction to an Absurdity. (Reductio ad absurdum.)
This method is similar to that of exclusion in that it makes an
assiunption which results in a contradiction of the data, but
diflfers from it in that but one such assumption is made before a
final conclusion is reached. Briefly, in proving a proposition by
reduction to an absurdity, we do so simply by proving that the
theorem which contradicts the conclusion of the original* is false.
* Such a theorem is called the contradictory of the direct.

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METHODS OF PROOF

301

Illustration: If the median to
the base of a triangle meets it
obliquely, the remaining sides are
unequal.

Given: D in AC in AABC ; ADmDC,

\ADB7£ H^BDC.
Prove: AB^^BC, -^

^ Proof

(1) Suppose ABmBC, ' Authorities left to the student.
Then AABD^ABDC,

(2) .-. 4BDils4.CDB.

(3) But this contradicts the data and
/. the assumption is absurd.

(4) .'.ABy^BC,

EXERCISES. SETXCI. PROOF BY REDUCTION TO AN ABSURDITY
Prove the following exercises by the
method of reduction to an absurdity:

1270. Prove that if two angles of a tri-
angle are equal, the sides opposite them
are equal.

Hint: Suppose AB>BC and take AX ^BC.

1271. If upon a common base an isosceles
and a scalene triangle are constructed, the

line joining their vertices does not bisect the vertex angle of the
isosceles triangle.

Hint: Assume that it does bisect it.

1272. If perpendiculars are drawn to the sides of an acute angle
from a point inside the angle, they enclose an oblique angle.

1273. Prove that if two triangles resting on a common base
have a second pair of sides equal, and the third vertex of each
outside the other triangle, their third sides are unequal.

c. By the Method of Analysis,

The method of analysis, which is attributed to Plato, was
undoubtedly used by Euclid, but was probably emphasized by
the former.

The analysis of a theorem is a course of reasoning, whether con-
scious or unconscious, by means of which a proof is discovered

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302 PLANE GEOMETRY

It consists of discovering the immediate condition under which the
conclusion would be true, and continuing to do this with each
new condition until one known to be true is reached.

Analjrsis always takes the following form: Suppose we wish to
prove that AsB if C^D.

We start by saying: A sB if x^y.
But x^ y if wsn.

and ms n if C=D.

But CsD by data.

The proofs of theorems are usually put in the synthetic form,
but they are derived analytically and then rearranged by retracing
the steps taken.

The anal3rtic method might also be called the method of reduc-
tiorif or of successive substitutions.

If we wish to discover how to prove a theorem, we should alwajrs
use the anal3rtic method. It is much more likely to suggest those
helpful auxiliary Unes which are frequently needed before any
relation between the theorem to be proved and a known theorem
is apparent.

On the contrary, nothing in the S3aithetic method suggests the
use of suitable auxiliary lines, and though we may continue to
make deductions as they occur to us, we often waste time and
energy without getting any nearer the conclusion.

Illustration 1, Theorem 216; If two sides of a triangle are imequal,
the angles opposite them are unequal in the same order.
B/s^ Given: a>~c in AABC.

Prove: 2i.A>4.C.
Analysis: 4A > ^C if part of 4-^1 > 4.C.

Part of 4A > 4C if that part can
be made an exterior 2^ of a A in which
■^ 4 C is a non-adjacent interior ^ .
This cannot be done as the 2^ are now placed.

How, then, can we find an 2;. which is equal to part of 2(.A and placed
s desired?

Since c <a, an isos. A can be formed by laying off BX ==^cona.
Hence the auxiliary line AX is suggested, and }S^A > 2<^C if 2^.BAX > 21.C.
4jBAX > 4C if TS^BXA > 2^.0.
But 4BXA > 4C.
.*. We may reverse the steps and give a brief synthetic proof if desired.

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METHODS OF PROOF 303

The analytic method does not always lead at once to the shortest
proof of a theorem, though it is far more likely to do so than the
synthetic method. At times an analysis suggests several methods
of proof, and our selection will depend upon which seems the
shortest. Skill in selection can be acquired only by practice.

Ilhistration 2; If one median in a triangle is intersected by a
second, the sect between the vertex and the point of intersection
is double the other sect. »

Given: Medians CD and A^ intersecting y^ \

at O in AABC. y/^ \

Prove: aDs20S, and CO s20P. 9"^^ 'J^^^

Analysis: One sect may be proved double y^ ^ \

another by proving (1) half the longer j/^^^!,^ti:i^:^9^~::i>^^\^^

=the shorter, or (2) double the shorter ^'^""^^^ ^^"^^

=the longer. -^ ^

The first method suggests that we take F and G in AO and CO so that
AF=FO»and CGsGO._

Now AO ^20E and CO =20D if FO =0E and GO ^OD.
FO =0E and GO =0D if GD and FE are diagonals of a O-
FE and GD are diagonals of a O if DE=FG and DE || FG,
DE^FG and DE\\FG ii DE bisects AB and BC in AABC, and if
FG bisects AO and CO in A AOC, for then DE= }(AC) s FG and Z>jEf ||
AC II FG.
.• . a synthetic proof can now readily be given.

Success in this tjrpe of work very often depends on the selection
of suitable auxiliary Unes. Those which are most often of use are
discovered by

(a) joining two points,

(b) drawing a line parallel to a given line,

(d) bisecting given sects as in the last illustration,

(e) producing a sect by its own length, as might have been done
in the last illustration.

Since we have not proved, even if it be true, that the algebraic
processes employed in the proof of theorems in proportion are
reversible, only a synthetic proof is valid. Of course, to suggest
the synthetic line of argument it is desirable to give an analysis
first if needed.

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304

PLANE GEOMETRY

Illustration:

Given: -^s^.

Analysis

(1)

Operation performed.

ifo«6«+aM«=a«6«+6\;«

(2)

(1) .6H6«+d»)

(2) istrueifa«d«s6«c«

(3)

(2) -a«6»

(3) is true if ad^bc

(4)

+ V(3)

(4) Is true if ^a^

(5)

(4) ^bd

But ^ej by data.

.' . the following synthetic proof

a c

(1)

(1) Given.

(DM, .\ad^bc

(2)

(2) Products of equals multiplied
by equals are equal.

(2) », .-. ay»36«c«

(3)

(3) Like powers of equals are equal.

(3) +a«&«, /.a^d^H-a«6«s

(4) Sums of equals added to equals

6^»H-a26«

(4)

are equal.

(4).6«(6«+d«),.-.^4+^,

(5)

(5) Quotients of equals divided by
equals are equal.

EXERCISES. SET XCII. ANALYTIC METHOD OF PROOF

Give an analysis of each of the following exercises, and follow
it by a concise sjmthetic proof.

1274. Prove the second illustra-
tion under the analytic method by

(a) proving ADOE^AFOO or
ADOF ^ AGOE (p. 303) .

(6) Doubling DO and OE, Use
each of the three methods sug-
gested by the following figures.

FIQ. 2

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METHODS OF PROOF 305

1275. Prove Theorem 31, Cor. 4, analytically.

1276. Prove Theorem, 31 Cor. 5, analytically.

1277. If one acute angle of a right triangle is double the other,
the shorter leg is one-half the hypotenuse. Prove first by drawing
auxiUary lines outside the triangle, and then by drawing them
inside the triangle.

1278. A median of a triangle is less than half the sum of the
adjacent sides.

1279. Prove the theorem given under A II analjrtically (p. 298).

1280. Prove the theorem given under Bla analjrtically (p. 299).

1281. Prove Ex. 1270 analytically.

1282. If (a+6+c+d)(a-6-c+d)s(a-6+c-d)(a-b6-c-d)

prove that r= j.
a

1283. If r-s- then -5 — 5-^=-; — j^-j.

b d' a^-Sab c^-Scd

1284 If ^=5. prove that ^-^-^^-^ ^-^-^^.
1286. If U% then ^±=^^'1^^.

While practice alone can give skill in the proof of theorems, the
following suggestions may be of help to the student.

First Make as general a figure as possible.

If a fact is to be proved concerning triangles in general the
figure should be that of a scalene triangle, since many facts are
true of isosceles or equilateral that are not true of scalene triangles.
Again, if a fact is to be proved concerning quadrilaterals in general,
it might be misleading to draw a parallelogram or even a trapezoid.

Second. Always have clearly in mind what is given and what is to
be proved.

Third. If (he proof is not readily seen, resort to analysis.

Fomih. Give a proof by the method of redtiction to an absurdity
only as a last resort.

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CHAPTER VII

CONSTRUCTIONS. METHODS OF ATTACKING
PROBLEMS

The solutions of the following fourteen problems and the corol-
laries to them are typical of one class of solution of construction
problems, namely, those solutions which are at once found to rest
directly upon some known theorem, and in addition, at times, upon
some known construction.

Problem l.'^ Draw a perpendicular to a given line from a given
point (a) outside the line, (&) on the line.\

The constructions rest directly upon Theorem 40, Cor. 1, the
fact that two points equidistant from the ends of a sect determine
the perpendicular bisector of that sect.

.Problem 2. Bisect a given (a) sect, {b) angle, (c) arc.

The construction of (a) rests inmiediately upon Problem 1.

The construction of (6) rests inmiediately upon Theorem 5,
the fact that three sides determine a triangle.

The construction of (c) rests immediately upon Theorem 43,
the fact that equal central angles intercept equal arcs, and
Problem 2 (6).

Problem 3. Reproduce a given angle.

The construction rests directly upon Theorem 5.

Problem 4. Draw a line through a given point, and parallel to a
given line.

The construction may rest directly upon Theorem 11, the fact
that if when lines are cut by a transversal the alternate-interior
angles are equal, the lines thus cut are parallel, and Problem 3.

* The first thirteen problems of the syllabus were taken up as exercises
in the First Study, but are repeated here (with only suggestions for their con-
struction) as an integral part of a syllabus of constructions.

t According to Proclus, this problem was first investigated by (Enopides,
a Greek philosopher and mathematician of the 5th century B.C. Proclus
speaks of such a line as a "gnomon'' — ^the conmion name for the vertical piece
on a simdial.

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METHODS OF ATTACKING PROBLEMS 307

Problem 6. Construct a triangle, given any three independent
parts; (a) two angles and the included side, (b) two sides and the
included angle, (c) three sides, (d) the hypotenuse and a leg of a
right triangle.

(What can you say in case (c) if one side is equal to or greater
than the sum of the other two sides?)

Problem 6. Divide a sect into n equal parts.

The construction may rest upon Theorem 31, Cor. 3, the fact
that parallels which intercept equal sects on one transversal do
so on all transversals, and Problem 3. For further help see Ex. 429,
p. 116.

Problem 7. Find a common measure of two commensurable
sects.

Given: ZB and CD commensur- "^
able sects.

JL 2_^

Required : A common measure m. C- • ^ D

Solution: Lay off AFsCD on A5.

ThenABaCD-l-rB.

LayoffCZs2FBonCD.

LayoffFXs^onFB.

Lay off XB on ZD.

It is found to be contained exactly in ZD,
^ Then XB is the greatest common measure of AB and CD, and any

integral part oi XB is a common measure of them. Prove it.

Problem 8. Pass a circle through three non-collinear points.
The construction rests directly upon Theorem 416.
Cor. 1. Circumscribe a circle about a triangle.*

Problem 9. Divide a given sect into parts proportional to n
given sects.

The construction rests directly upon Theorem 31, Cor. 2.

Problem 10. Divide a given sect harmonically in the ratio of
two given sects.

Use (1) the method suggested in Problem 9, or (2) the method
suggested by Theorem 326, Cor. 1.

. * The center of a circle circumscribed about a polygon is called its circum-
center.

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308 PLANE GEOMETRY

Problem 11. Find a fourth proportional to three given sects.

The construction rests directly upon Theorem 31. The solution
is given, but the proof is left to the pupil.

„ ^ Given: Sects a, &, c.

Required: Sect x, so that ~ »=.
& d

av-''

Solution: Construct any angle EST,
\ OniSr, layoflf5F«a.

S>'" \ On SR, lay off SWmh.

/ \ On VT, lay off VQmc,

y<r \ DrawTF.

\ \ Draw 57 II YW and cutting SR

^ ^^ ^ ^ 5r~R Then FF is the required sect*.

Cor. 1. find a third proportional to two given sects.

What is the only modification needed in the construction of

Problem 11 in order to find x, so that -^ s — ?

b X

Problem 12. Upon a given sect as homologous to a designated

side of a given polygon construct another similar to the original

polygon.

The construction rests directly upon Theorem 366 and
Problem 11.

Problem 13. Construct a square equal to the sum of two or
more given squares.

Use the Pythagorean Theorem.

Cor. 1. Construct a square equal to the difference of two given
squares.

In this case the larger of the given squares will be the square
on which side of the right triangle?

The construction therefore rests upon Problem 5 (d)

Cor. 2. Construct a polygon similar to two given similar poly^
gons and equal to (a) their sum, (b) their difference.

How do the areas of similar polygons compare?

Problem 14. Inscribe in a circle, regular polygons the number
of whose sides is (a) 3-2", (6) 4-2".

(a) The construction rests directly upon Theorem 58 and the
construction of an equilateral triangle given its side (here the
radius of the circle) in order to obtain a central angle of 120®.

(b) What is the central angle in the case of the square?

METHODS OF ATTACKING PROBLEMS 309
THE SYNTHETIC METHOD OF ATTACKING A PROBLEM

The preceding type of construction problem is the simplest, and
the solution of one of that nature is usually so readily seen, that
without further explanation (except for one or two suggestions
appended to the first exercises) the pupil is asked to solve the fol-
lowing set of exercises.

EXERCISES. SET XCIII. SYNTHETIC SOLUTIONS

1286. Trisect a right angle. (We know that each angle of an
equilateral triangle is two-thirds of a right angle. What construc-
tion does this therefore suggest?)

1287. Divide an equilateral triangle into three congruent tri-
angles. (We know that the bisectors of the angles of a triangle
are concurrent, and that triangles are determined by two angles
and the included side.)

1288. Construct an equilateral triangle with a given sect as
altitude. (What fact about the altitude of an equilateral triangle
suggests the construction?)

1289. Construct a square having given its diagonal.

1290. Through two given points draw straight lines which shall
make an equilateral triangle with a given straight line.

1291. On a given sect construct a rhombus having each of one
pair of opposite angles double each of the other pair.

1292. On a given base construct a rectangle equal to (a) a given
square, (b) another given rectangle, (c) a given triangle, (d) a
given trapezoid.

1293. The sides of a polygon are 5, 7, 9, 11, 13. Construct one
similar to it having the ratio of similitude 3 to 5.

1294 Construct a polygon similar to the accompanying polygon,
having the ratio of simili-
tude equal to that of the two
given sects a and b.

1296. Construct a poly-
gon similar to two (or more)
given similar polygons and
equivalent to their sum (or difference);

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310 PLANE GEOMETRY

1296. Construct a circle equal to the sum of two given circles.

1297. Construct a circle equal to 2 o

the difference of two given circles. ^ a ^^

Problem 15. Construct a ^<;^ B^^^

triangle given two sides and ^ '' — ^^^7' \

the.angle opposite one of them. y^^ / \

Given: Sides a and c; ^A. ^^^^^ /^ ^\

Required: AABC. ^^^1 ^n/ _ >/

Construction: Construct4PilOs25.A. A 'C '^"^ZZZ.'^^ T

OnAQ, layoff ZBsc.

With B a£i center and a as radius, strike an arc cutting AP in C and Ci,
touching it in only C, or not at all, according as a> A*, a s^, or o<A6.
Then AABC and AABCi fulfill the required conditions.

Q Discussion: I. Conditions under which

^^ there are two solutions.

^ — —

B^'^ (1) 2^A acute, a <c, but>^, the

^^^\ perpendicular from B to XP,

o^^'' / I \ II. Conditions under which there is

^"^ ^1 S/ifcx * but one solution.

^^^^ . ^ I ! \ ^^ (1) i^A'^acute, as^ or c.

^ .^__L.^ j^ ^2) 2^^ acute, right, or obtuse,

and a>c.

.u lJ\ — _\

A y ^^ A ^^ A

in. Conditions under which K Vv a

there is no solution. } *^? \ ^^'^^J

(1) t^A right or obtuse ^| ^\/ ^\ /
and a<c, — j--^ ^^ ^^'^

(2) ^A acute and a <kb. A " A

THE DISCUSSION OP A PROBLEM

As in the case of Problem 15, many exercises in construction
call for a discussion, by reason of the fact that the number of solu-
tions of the problem varies under different conditions.

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METHODS OF ATTACKING PROBLEMS

311

III. ^A obtuse.

(1) a<c.

(2) a^c.

(3) a>c.

It will be noticed that the discussion of Problem 15 has been
arranged under three heads, based upon the number of possible
solutions. The discussion might have been arranged under entirely
di£Ferent headings. To show this, the pupil is asked to fill in the
discussion imder the following heads.
I. <A acute. II. <A right.

(1) a<A6. (1) a<c.

(2) asAft. (2) a^c.

(3) a>A6but <c. (3) a>c.

(4) a=c.

(5) a>c.
or,

I. a <c. II. a^c. III. a>c. (Fill in all the necessary subheads.)

Further illustration of what is meant by the discussion of a problem

in construction: To find a point X which shall be equidistant from

points Pi and Pj and at a given distance d from P|.

Given: Points Pi, Pj, Pi; sect d. d

Required: Point X.such that PiX =

PtX and PtX ^d.
Construction: (1) Draw PiPi and

bisect it at A.

(2) Erect ??JBxPiPi.

(3) With Pt as center and ct
as radius describe a circle cut-
ting ^R at Xf Xi.

Then X, Xi are the required points.

Proop

(1) PiXaPiX (1) The locus of points equidistant
and PiXi^PiXi from the ends of a sect is the perpen-
•.* X, Xi lie on QR. dicular bisector of that sect.

(2) '-' X and Xi are on GPii PsXa (2) Const., and the locus of points
PsXi b3. at a given distance from a given point

is a circle of which the center is the
given point, and the radius the given
distance.
Discussion: DrawK7^±"5S.

I. Two solutions: If 3>PlF, for then a portion of QR will be a chord
of the circle P|.

II. One solution: If J^ PtNj for then QR will be tangent to OPt at X.

III. No solution: If J<P|iV, for then all points in QR will he further
from the center Pi than the length of the radius.

>Pi

9. 14 ^rJ^Tr

'P.

^^Xi

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312 PLANE GEOMETRY

LOCI AND PROBLEMS SOLVED BY THE METHOD OF
THE INTERSECTION OF LOCI

In many problems we are asked to find the position of a point
which satisfies two given conditions. Each of the conditions
determines a locus on which the point lies, and the solution of
the problem is therefore the point or points common to both loci.

The last exercise illustrates this type of problem. The points
X and Xi fulfill two conditions. The first condition determined
the locus of points equidistant from Pi and P2, the second deter-
mined the locus of points at a given distance d from Ps. The solu-
tion of the problem was those points conunon to both loci.

EXERCISES. SET XCIV. INTERSECTION OF LOCI
Give a full discussion of each of the exercises in this set.
To find a point X such that:

1298. X shall be at the distance di from point Pi, and d^ from
point P2.

1299. X shall be equidistant from two parallel lines and at the
distance d from point P.

1300. X shall be equidistant from two intersecting lines, and
also equidistant from points P and Pi.

In line I find point X so that:

1301. X shall be at distance d from P.

1302. X shall be equidistant from P and Pi.

1303. X shall be at distance d from k,

1304. X shall be equidistant from two given parallel lines.
1306. Draw a circle of given radius to touch two given lines.

Draw a circle with a given radius:

1306. Passing through two given points.

1307. Passing through a given point and touching a given line.

1308. Passing through a given point and touching a given circle.

1309. Touching a given line and a given circle.

1310. Touching two given circles.

1311. Describe a circle touching two parallel lines and passing
through a given point.

1312. Construct a right triangle, having given the hypotenuse
and the altitude on the hjrpotenuse.

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METHODS OF ATTACKING PROBLEMS

313

1313. Construct a right triangle^ given sects of hypotenuse
made by the bisector of the right angle.

1314. Construct a triangle, given an altitude and the sects made
by the altitude upon the opposite side.

1316. Construct a right triangle, given the sects of the hypo-
tenuse made by the altitude upon the hypotenuse.

Problem 16. Through a given point draw a tangent to a given
circle, (a) when the

point is on the circle, — ^J^ ^

(b) when the point
is outside the circle.

GiYen: PL P (o) on QC,

(b) outside OC.
Required: TQ tangent to

OC.
Construction: For (a) For (6)

(1) Draw CF. (1) Draw sect CP,

(2) Draw FQ±UF. (2) On CP as diameter describeOO
Tlien PQ is the required tangent. cutting OCin Q and Qi.

(3) Draw PQ and FQi, each of
which is the required tangent.

Pboofs
For (6)
Draw T?Q and C^i.

(1) Then A PCQ and PCQi are
right A. Why?

(2) .-. PQ and PQi±CQ and CQi
respectively. Why?

(3) .'. PQ and PQi are tangent to
QC. Why?

Discussion: When the point is on the circle there is but one solution. Why?
When it is outside the circle there are two and only two possible solu-
tions. Why?

The solution of all construction problems calls for at least four
parts. Namely:

1st — ^The statement of what is given.

2nd — ^The statement of what is required.

3rd — ^The construction of what is required.

4th — ^The proof of the correctness of this construction.

For (a)

Since ]^.CPQ is a rt. 4., and TJP a
radius, PQ is tangent to QC. Why?

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314 PLANE GEOMETRY

We have already seen that many problems call for a discussion,
and in many of the problems in the syllabus an analysis will be
given.

Omitting data and what is requu-ed, one might say then that

the systematic solution of a problem consists of four parts : (1) The

analysis, (2) the construction, (3) the proof, and (4) the discussion.

2T Problem 17. Construct a

Q common (a) external and (b)
""■"^^-^^ internal tangent to two circles.
f^!\^^^'^'-'-^^^ P Y Given: 0« Cand
f — ^ H ■j^-:^'^ Ci of radii r and
J>'^"^^ Required: The
,^^^^^ ^\ common (a) ex-
_ temal tangents
T[ TN and TiNi,
and (6) internal tangents TN and TiN\,
Analysis of (a): Suppose one of the required tangents, TN, drawn.
Then if r T^n.l'NX will intersectTC^y at P.
Then ACTPcoACiiNTP.

But r and n are known.

.-. to find P divide CCi externally in the ratio of -*

/. to find tangent TN, draw PN tangent to OCi and show that if pro-
duced it will be tangent to OC

The completion of the problem is left to the student, who should discuss
the case where r^ri.
Analysis of (&): The analysis in this case is similar to that in case (a) except
that CCi will be divided internally in the ratio of r to n.
Construction and proof left to the student.

Problem 18. Inscribe a circle in a triangle.*

The center is determined by the intersection of what two loci?

Cor. 1. Find the centers of the escribed circles of a triangle.^

The excenters are determined by the intersections of what loci?

♦ The center of the circle inscribed in a polygon is called the incenter of the
polygon.

t The circles which are tangent to one side of a polygon and to the two consecu-
tive sides prodttced, are called the escribed circles, and their centers are called the
excenters of the polygon.

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METHODS OF ATTACKING PROBLEMS 315

Problem 19. Construct upon a given sect the segment of a
circle capable of containing a given angle.

GiYen: Sect AB; 4.C.

Required: Segment of OO resting on AB as chord such that the inscribed

Analysis: If the construction were complete ABC would be any one of a num-
ber of triangles of base AB and a vertex 4.C

The circle circiunscribed about any one AABC would give the required
locus.

.'. the following:
Construction: Const. ^.ZCFa^C.

With Bf any convenient pt. in CY as center, and radius = sect AB,
strike an arc cutting CX in A.
Circiunscribe a circle about AABC.
Then segment ABC is the required segment.
(Proof left to the student.)

Problem 20. Construct a mean proportional between two giveri
sects.
Given: Sects a and b, '

a c h

Required: Sect c, so that = ==* — ^— ^— _^^____

c
Hint: Can you word Proposition 39, Cor. 4, to relate to the sects of a diameter
of a circle?

Extreme and Mean Ratio. If sect AB is so divided by the pt. P

thai ====, the sect is said to be divided in extreme and mean

PB AB
ratio. That is, one part of the sect is a mean proportional between

the entire sect and the other part.

This division of a sect is known as the
"golden section." As in harmonic division, a sect may be divided
internally and externally into extreme and mean ratio.

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316

PLANE GEOMETRY

.^-.^v

Problem 21. Divide a given sect into mean and extreme ratio.*
GiTen: Sect «.

Required: Sects a and b so that (1) - e-^^ and (2) I=— fi;-
^ .a «— o s-j-o

Analysis: Suppose the construction completed.
Then(l) •«-•-*

a ,8

I and r

8— a b

8+b'

These suggest /

a tangent as the i^ §.

mean propor- "1 " """

tional between a ^

secant and its external sect or the leg of a rt. A as the mean proportional be-
tween the hypotenuse and its projection upon the hypotenuse. As the propor-
tions now stand there is but one known term.

But by composition we get (1) tit? a - and (2) -^m ?

This suggests 8 as the tangent to a circle, H-a as the entire secan t, an d a as
its external sect on the one hand, and "b as the entire secant and 6— « as its
external sect on the other hand.

The most natural secant to select is that which passes through the center.

Hence the following:
Construction: (1) Const. OO of diameter 8 tangent to XF. (XF=i) at Y,

(2) Draw secant XQOP.

(3) On XY lay off XQi^XQ, and on YX produced lay off XPi=XP.

(4) Then Qi and Pi are the required points. (Proof left to the student.)

Problem 22. Inscribe in a circle a regular decagon.

Given: OO.

Required: AB, the side of the regular in-
scribed decagon.

Analysis: If AB aside of reg. decagon,
4A0B=36^
.-. 2S^0AB^^0BA^ 180^36^ ^ ^^o

and if PB bisects 4 ABO, 4PBO » 36*-
40 and 4 APB = 180* - (36° +72*) - 72*.

.'. AAPB<oAABO and AOPB is

isosceles.

.•.4£-g. ButAB-PB-OP.

... 4^s^. ... to find OP-=AB divide OA, the radius of OO, into

OP OA
extreme and mean ratio, and use the mean as AB, (Proof left to student.)

♦ A painting is said to be most artistically arranged when its center of inter-
est is so placed that it divides the width of the picture into extreme and mean
ratio. If the student is especially interested in the topic, he will find references
in Chapter X.

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METHODS OF ATTACKING PROBLEMS

317

Cor. 1. Inscribe in a circle regular polygons the number of
whose sides is (1) 6-2", (2) 16*2".

(1) Left wholly to the student.

(2) Hint: A^^i-A- D
To transform a polygon means to

change Us shape biU not its
area.

Problem 23. Transform a

polygon into a triangle.

Given: Polygon AJ?CZ)J^/^.
Required: AA^ABCDEF.
Hint: With CA as base, what is

the locus of the vertices of triangles whose areas area AABC?

To eUminate side AB, by what particular

A shall we then replace AABC7
Similarly, what A is to replace ADEF7
Similarly, how dispose of side DH7

Problem 24. Construct the square
equal to a given (a) parallelogram, (&)
triangle, (c) polygon.

Problem 26. Construct a parallelogram equal to a given square
having given (a) the sum of base and altitude equal to a given sect,
(b) the difference of base and altitude equal to a given sect.
Given: Square S;

sect a.
Required: AO^S
with (a) base +
alt.Ba,(&)base
—alt. aso.

T^^

L.
P

(a) From Fig. 1 show that any parallelo-
gram with QR as base and PQ as altitude »
square S, and that it also fulfills the second /
condition. /

0- /

FXQ. 1

(6) From Fig. 2 show that any parallelo-
gram with QR as base and PQ as altitude »
square s, and that it also fulfills the second
condition.

This problem solves geometrically the
algebraic problems, given (o) X'\-y'=a
and xy^s, (6) x—y^a and xy=«, findx
and y.

\?/

/ I ia

FxQ. 2

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318 PLANE GEOMETRY

Problem 26. Construct a polygon similar to one and equal to
another of two given polygons.

Given: Polygons P and Q.
Required: Polygon R so

that iicoP and i2=Q.
Analysis: If R^P and x
homologous to s, then

If P and Q equal m* and n* respectively, and Q^R, then — r s — .
P m^ ,P 8^ "* ^

/. — s— , /, x is the fourth proportional to the side of a square =P, the

side of a square ^Q, and the side in P homologous to x.
(Construction and proof left to the student.)

Problem 27. Construct a square which shall have a given ratio

to a given square. p ^

Given: Square S of side a, ratio -. \
Required: Square R of side r so that ^v

Analysis: If a rectangle were to be found N

^, ^ rect. AE p .i ,. ,

so that o — -i the alt. 8 of

square S q'

square S would simply have to be divided so that — ■ — s 2.

Then it remains to convert rect. AE into a square.

(Construction and proof left to the student.)

Cor. 1.* Construct a polygon equal to any part of a given
polygon and similar to it.

(Construction left to the student.)

FORMAL ANALYSIS OF A PROBLEM

When problems .(as is usually the case) are of such nature that
the application of known theorems and problems to their solution
is not at once apparent, the only way to attack them is by a method

* This corollary is included in the syllabus because of its practical value in
drafting, since it is by this method that the draftsman finds his scale, in enlarg-
ing or reducing a diagram of any sort.

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METHODS OF ATTACKING PROBLEMS 319

resembling the analytic method of proof of a theorem. This
method is called the analysis of the problem.

In brief, the directions for following this method are:

1. Suppose the construction completed.

2. Draw a figure showing all the parts concerned. (Given parts
in heavy lines, and required parts in dotted lines.)

3. Study the relations of the parts, and try to find some relation
which will suggest a possible construction.

4.. If a first attempt fails, introduce new relations by means of
auxiliary lines, and continue the study of relations until a clue to
the correct construction is derived.

5. Look for that clue in a rigid part of the figure — usually a
triangle. * .

A few more illustrations to show just what is meant by these
directions will be helpful.

1. Construct a triangle, having given the base, a base angle, and
the sum of the remaining sides.

Suppose ABC to be the completed triangle.

In it we know AC^b, ^A, and c+a.

.'.produce AB to X so that BX=a,

Then ABCX is an isos. A.

But AAXC is determined (6,
^A, a+c).

But ^XCB^^X •.• BX^BC.

.'. AABC is determined.

S. Construct a triangle, having given one side, the median to thai

side, and the altitude to a second side,

GiTen: h, mb, he.

Analysis: Suppose ABC to be the required A.
Then the known parts are AC

.\ AEmEC^%CD^hcyBE^ ^c ^X / \

fli6, and 4 at Dsrt. 2<«. y / \^ \

ADAC is determined. Whv? y"" ^/ ^ \

ADAC is determined. Why?

Using this triangle as the basis of con- y/^ V \\

Btruction, what means have we of fixing the
vertex B?

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320

PLANE GEOMETRY

Conatrud a triangle having given its medians.

- Given;

Jfflo

1^=^

'^^

tnat ^^f ^Wir*

Required: AABC.

Analysis: Suppose ABC

to be the required A.

Evidently no 4.

within the A is fixed

by the medians.

But it is known
ik&tOC^}ime,OEm

^ yim,OAm%ma, and

that OE bisects AC.

.*. if OE is produced to X, so that XE=EO, a parallelogram is detei^
mined whose sides and one diagonal are known.
Qenoe the following:
Constfttction: (1) Trisect mat ^^t ^^*

(2) Construct ACXO with %me, ^nia, and ^mb as its sides.

(?) Complete the OOCXA.

(4) Produce EO by %mh, and vertex B is determined.

4. Construct a trapezoid given the bdses and the base angles.
Analysis: Suppose ABCD to be the com-
pleted trapezoid.
We know AB, DC, 4D, and 4.C.
Is, then, any part of the figure de-
termined?

K we draw BDi \\ AD, then ABDiD
will be a parallelogram and DDi ^AB,

^^.BDiC^^ADC and n^^UC-XS and ABDiC is thus determined.
(Construction, proof and discussion are left to the student.)
The preceding exercises are illustrative of analysis geometric in
form, but the anal3rsis of a problem in construction may also be
algebraic in form, as, for example, in Problems 21, 26, and 27.

Note 1. — ^These diagrams show the symbolism used in reference to tri-
angles and trapezoids throughout this text.

D c O

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METHODS OF ATTACKING PROBLEMS

321

NoTB 2. — Construction of trapezoids.

The relations given by the following constructions are often
oseful. Let ABCD be any trapezoid. Draw CF and BH WAD,
CO and AH \\ DB. Join H (the in-
tersection of AH and BH) with F,
C, 0. The figures ACGH, ADBH,
FCBH, are parallelograms. BF^a
-c, AG^AB+BG^a+c, ^ACG
B e, ^CAH s ^CGH s 180** - e,
^CFH^^CBHsa+p. ^FCB^^FHB

a80^-(a+/8).

EXERCISES. SETXCV. PROBLEMS CALLING FOR ANALYSIS

Construct a triangle, having given:
1316. a, b, mb. 1317. a, b, ht.

1319. a, m,

a, "^B.
ay '*«•

1318. a, hcf vftat
1321. 6, c, Aa.
1324. Aa, Ac, *C.

1320. ma, fc«, <B.
1322. a, Aa, A.. 1323. A«, <B, ^C.
1326. a, 6, <il + <B.

1326. The base, the altitude, and an angle at the base.

1327. The base, the altitude, and the angle at the vertex.
Construct an isosceles triangle, having given:

1328. The base and the angle at the vertex.

1329. The base and the radius of the circumscribed circle.

1330. The base and the radius
of the inscribed circle. -

1331. The perimeter and the
altitude.

Hi 3 5 B ^ Let ABC be the required A,

EF the given perimeter. The alti-
tude CD ^ passes through the mid-
point of EF, and the As EAC, BFC
are isosceles.

1332. Construct a triangle, having
given two angles and the sum of A*"
two sides.

Can the third ^ be found? Assume the problem solved. If
AX^AB+BC, what kind of triangle is ABXC? What does
<C£A equal? Is <X known? How can C be fixed?

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322

PLANE GEOMETRY

1333. Construct a triangle, having given a side, an adjacent
angle, and the difference of the other sides.

\IAB, <il, and AC -BC are known,
what points are determined? Then can
XB be drawn? What kind of triangle is
AXBC1 How can C be located?

1334. Construct an isosceles triangle,
having given the sum of the base and
'^ ""an arm, and a base angle.

1336. Construct a triangle, having given the base, the sum of
the other two sides, and the angle included by them.

1336. Construct a triangle given the mid-points of its sides.

1337. Draw between two ' sides of a c^
triangle a sect parallel to the third side,
and equal to a given sect.

If PQ = d, what does AR equal? How

will you reverse the reasoning? A B B

1338. Draw through a given point P between the sides of an

angle AOB a sect terminated by the
sides of the angle and bisected at P.
If PM=PN, and PR \\ AO, what
can you say about OR and RNt Can
you now reverse this? Similarly, if
PQ II BO, is Oe=QM?

1339. Given two perpendiculars,
AB and CD, intersecting in 0, and
a line intersecting these perpendiculars in E and F; construct
a square, one of whose angles shall coincide with one of the right
angles at 0, and the vertex of the opposite angle of the square
shall lie in EF. (Two solutions.)

1340. Draw from a given point P in the base AB of a triangle
ABC a line to AC produced, so that it may be bisected by BC.

Construct a rectangle, having given:

1341. One side and the diagonal.

1342. One side and the angle formed by the diagonals.

1343. The perimeter and the diagonal.

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METHODS OF ATTACKING PROBLEMS 323

Construct a parallelogram, having given:
1344. Two independent sides and one altitude
1346. Two independent sides and an angle.

1346. One side and the two diagonals.

1347. One side, one angle, and one diagonal.

1348. The diagonals and the angle formed by the diagonals
Construct a rhombus, having given:

1349. The two diagonals.

1360. The perimeter and one diagonal.
13R1. One angle and a diagonal.

1362. The altitude and the base.

1363. The altitude and one angle.

1364. Construct a square, having given the diagonal.
Construct a trapezoid, having given:

1366. The four sides.

1366. The bases, another side, and one base angle.

1367. The bases and the diagonals.

1368. One base, the diagonals, and the angle formed by the
diagonals.

1369. Inscribe a square in a given triangle.

1360. In a given triangle inscribe a rectangle similar to a given
rectangle. c K

Hint: Let ABC be the i)/' JN^^^--"""' j i 1

given triangle and R the /\ ^4---''' |N. | R

given rectangle. On the a X — ' ^ — ^ ' ^

altitude CH construct a '^ ^ ^ ^ ^ ^

rectangle CL similar to the given rectangle. The line AK will determine a

point E which will be one of the vertices of the required rectangle. Why?

1361 Inscribe a square in a semicircle.

1362. Divide a given triangle into two equal parts by a line
parallel to one of the sides.

1363. Bisect a triangle by a line parallel to a given line.

1364. Bisect a triangle by a line drawn perpendicular to the base.
Construct a circle which shall be tangent to a given:

1366. Line, and to a given circle at a given point.

1366. Circle, and to a given line at a given point.

1367. Transform a triangle ABC so that < A is not altered, and
the side opposite the angle A becomes parallel to a given line M^^

CHAPTER VIII

SUMMARIES AND APPLICATIONS

A. SYLLABUS OF THEOREMS

1. Vertical angles are equal.

2. Two sides and the included angle determine a triangle.

3. Two angles and the included side determine a triangle.

4. The bisector of the vertex angle of an isosceles triangle
divides it into two congruent triangles.

Cor. 1. The angles opposite the equal sides of an isosceles tri-
angle are equal.

Cor, 2. The bisector of the vertex angle of an isosceles triangle
bisects the base and is perpendicular to it.

Cor. 3. An equilateral triangle is equiangular.

Cor. 4. The bisectors of the angles of an equilateral triangle
bisect the opposite sides and are perpendicular to them.

Cor. 6. The bisectors of the angles of an equilateral triangle
are equal.

5. A triangle is determined by its sides.

5a. Only one perpendicular can be drawn through a given point
to a given line.

56. Two sects drawn from a point in a perpendicular to a given
line, cutting off on the line equal sects from the foot of the per-
pendicular, are equal and make equal angles with the perpendicular.

5c. The sum of two SQcts drawn from any point within a triangle to
the ends of one of its sides is less than the sum of its remaining sides.

5d. Of two sects drawn from a point in a perpendicular to a
given line and cutting off unequal sects from the foot of the per-
pendicular, the more remote is the greater, and conversely.

Cor. 1. All possible obliques from a point to a line are equal
in pairs, and each pair cuts off equal sects from the foot
of the perpendicular from that point to the line.

6. The perpendicular is the shortest sect from a point to a line.
6a. The shortest sect from a point to a line is perpendicular to it.

7. The hypotenuse and an adjacent angle determine a right
triangle.

8. The hypotenuse and another side determine a right triangle.

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SUMMARIES AND APPLICATIONS 325

9. lines perpendicular to the same line are parallel.

10. A line perpendicular to one of a series of parallels is perpen-
dicular to the others

11. If when lines are cut by a transversal the altemate-interioi
angles are equal, the lines thus cut are parallel.

Cor. !• If the alterncate-exterior angles or corresponding angles
are equal when lines are cut by a transversal, the lines
thus cut are parallel.

Cor. 2. If either the consecutive-interior angles or the con-
secutive-exterior angles are supplementary when lines
are cut by a transversal, the lines thus cut are parallel.

12. Parallels cut by a transversal form equal alternate-interior
angles.

Cor. 1. Parallels cut by a transversal form equal corresponding
angles and equal alternate-exterior angles.

Cor. 2. Parallels cut by a transversal form supplementary con-
secutive-interior angles and supplementary consecutive
exterior angles.

12a. Two angles whose sides are parallel each to each or per-
pendicular each to each, are either equal or supplementary.

13. The sum of the angles of a triangle is a straight angle.
Cor. 1. A triangle can have but one right or one obtuse angle.
Cor. 2. Triangles having two angles mutually equal are mutu-
ally equiangular.

Cor. 3. A triangle is determined by a side and any two homolo-
gous angles.

Cor. 4. An exterior angle of a triangle is equal to the sum of
the non-adjacent interior angles.

14. The sum of the angles of a polygon is equal to a straight
angle taken as many times less two as the polygon has sides.

Cor. 1. Each angle of an equiangular polygon of n sides equals

n— 2
the th part of a straight angle.

Cor. 2. The sum of the exterior angles of a polygon is two

straight angles.

Cor. 3. Each exterior angle of an equiangular polygon of n

2
sides is equal to the -th part of a straight angle.

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326 PLANE GEOMETRY

15. If two angles of a triangle are equal, the sides opposite them
are equal.

Cor. 1. Equiangular triangles are equilateral.

16. Either diagonal of a parallelogram bisects it.

Cor. 1. The parallel sides of a parallelogram are equal, and the

opposite angles are equal.
Cor. 2. Parallels are everywhere equidistant.

17. A quadrilateral whose opposite sides are equal is a parallelo-
gram.

18. A quadrilateral having a pair of sides both equal and parallel
is a parallelogram.

19. A parallelogram is determined by two adjacent sides and an
angle; or parallelograms are congruent if two adjacent sides and an
angle are equal each to each.

20. The diagonals of a parallelogram bisect each other.

21. A quadrilateral whose diagonals bisect each other is a
parallelogram.

21a. The difference. between any two sides of a triangle is less
than the third side.

216. If two sides of a triangle are unequal, the angles opposite
them are unequal in the same order.

21c. If two angles in a triangle are unequal, the sides opposite
them are unequal in the same order.

21d. If two triangles have two sides equal each to each, but the
included angles unequal, their third sides are unequal in the same
order as those angles.

21e. If two triangles have two sides equal each to each, but the
third sides unequal, the angles opposite those sides are unequal
in the same order.

21/. If one acute angle of a right triangle is double the other
the hypotenuse is double the shorter leg, and conversely.

22. Rectangles having a dimension of one equal to that of
another compare as their remaining dimensions.

23. Any two rectangles compare as the products of their
dimensions.

24. The area of a rectangle is equal to the product of its base
and altitude.

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SUMMARIES AND APPLICATIONS 327

26. The area of a parallelogram is equal to the product of its
base and altitude.
Cor. 1. Any two parallelograms compare as the products of

their bases and altitudes.
Cor. 2. Parallelograms having one dimension equal compare as

their remaining dimensions.
Cor. 3. Parallelograms having equal bases and equal altitudes

are equal.

26. The area of a triangle is equal to half the product of its
base and altitude.

Cor. 1. Any two triangles compare as the products of their

bases and altitudes.
Cor. 2. Triangles having one dimension equal compare as their

remaining dimensions.
Cor. 3. Triangles having equal bases and equal altitudes are

equal.
26a. The square on the hypotenuse of a right triangle equals
the sum of the squares on the two legs.

266. The areas of two triangles that have an angle of one equal
to an angle of the other, are to each other as the products of the
sides including those angles.

27. The area of a trapezoid is equal to hafi the product of its
altitude and the sum of its bases.

28. Any proportion may be transformed by alternation, i.e., the
first term is to the third as the second is to the fourth.

29. In any proportion the terms may be combined by addition
(usually called composition) ; i.e., the ratio of the sum of the first
and second terms to the second term (or first term) equals the ratio
of the sum of the third and fourth terms to the fourth term (or
the third term).

N.B. — "Addition'' and "sum'' are ijsed in the algebraic sense.

30. In a series of equal ratios, the ratio of the sum of any num-
ber of antecedents to the sum of their consequents equals the
ratio of any antecedent to its consequent.

31. A line parallel to one side of a triangle divides the other
sides proportionally.

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328 PLANE GEOMETRY

Cor. 1. One side of a triangle is to either of the sects cut off
by a line parallel to a second side as the third side is
to its homologous sect.

Cor. 2. Parallels cut oflf proportional sects on all transversals.

Cor. 3. Parallels which intercept equal sects on one trans-
versal do so on all transversals.

Cor. 4. A line which bisects one side of a triangle, and is parallel
to the second, bisects the third.

Cor. 6. A sect which bisects two sides of a triangle is parallel
to the third side and equal to half of it.

Cor. 6. The line (usually called median) joining the mid-points
of the non-parallel sides of a trapezoid is parallel to the
bases and equal to one-half their sum.

Cor. 7. The area of a trapezoid equals the product of its median
and altitude.

32. A line dividing two sides of a triangle proportionally is
parallel to the third side.

Cor. 1. A line dividing two sides of a triangle so that these
sides bear the same ratio to a pair of homologous sects,
is parallel to the third side.
32a. The bisector of an angle of a triangle divides the opposite
side into sects which are proportional to the adjacent sides.

326. The bisector of an exterior angle of a triangle divides the
opposite side externally into sects which are proportional to the
adjacent sides.
Cor. 1. The bisectors of an adjacent interior and exterior angle
of a triangle divide the opposite side harmonically.

33. The homologous angles of similar triangles are equal and
their homologous sides have a constant ratio.

34. Triangles are similar when two angles of one are equal each
to each to two angles of another.

34o. Triangles which have their sides parallel or perpendicular
each to each are similar.

35. Triangles which have two sides of one proportional to two
sides of another, and the included angles equal, are similar.

36. If the ratio of the sides of one triangle to those of another
is constant, the triangles are similar.

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SUMMARIES AND APPLICATIONS . 329

36a. The homologous angles of similar polygons are equal, and
their homologous sides have a constant ratio.
Cor. 1. The homologous diagonals drawn from a single vertex

of similar polygons divide the polygons into triangles

similar each to each.
366. Polygons are similar if their homologous angles are equal
and homologous sides proportional.
Cor. 1. If homologous diagonals drawn from a single vertex of

polygons divide them into triangles similar each to each,

the polygons are similar.

37. The perimeters of similar triangles are proportional to any
two homologous sides, or any two homologous altitudes.

Cor. 1. Homologous altitudes of similar triangles have the same

ratio as homologous sides.
Cor. 2. The perimeters of similar polygons have the same ratio

as any pair of homologous sides or diagonals.

38. The areas of similar triangles compare as the squares of any
two homologous sides.

Cor. 1. The areas of similar polygons compare as the squares

of any two homologous sides or diagonals.
Cor. 2. Homologous sides or diagonals of similar polygons have

the same ratio as the square roots of the areas.
38a. If two parallels are cut by concurrent transversals, the ratio
of homologous sects of the parallels is constant.

386. If the ratio of homologous sects of two parallels cut by
three or more transversals is constant, the transversals are either
parallel or concurrent.

39. The altitude upon the hjrpotenuse of a right triangle divides
it into triangles similar to each other and to the original.

Cor. 1. Each side of a right triangle is a mean proportional

between the hypotenuse gud its projection upon the

hypotenuse.
Cor. 2. The square of the hjrpotenuse of a right triangle is

equal to the sum of the squares of the other two sides.
Cor. 3. The altitude upon the hypotenuse of a right triangle

is a mean proportional between the sects it cuts off on

the hypotenuse.

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330 PLANE GEOMETRY

39a. In any triangle, the square of the side opposite an acute
angle is equal to the sum of the squares of the other two sides,
diminished by twice the product of one of those sides by the pro-
jection of the other upon it.

39&. In an obtuse triangle, the square of the side opposite the
obtuse angle is equal to the sum of the squares of the other two
sides, increased by twice the product of one of those sides by the
projection of the other upon it.

39c. I. The siun of the squares of two sides of a triangle is equal
to twice the square of half the third side, increased by twice the
square of the median to it.

II. The diflference of the squares of two sides of a triangle is
equal to twice the product of the third side and the projection of
the median upon it.

Cor. 1. If rria represents the length of the media n upon side a
of the triangle whose sides are a, 6, c, then ma^i\/2Q>^+c^) -a^.

39d. If Ao represents the altitude upon side a of a triangle whose
sides are a, 6, c, and s stands for its semiperimeter, i.e.,

ss — - — , then ha=-y/s{s -a)(s -6)(s -c).

^ Of

Cor. 1. If A stands for the area of a trian gle whose sides are
a, 6, c, and whose semiperimeter is s, then A = y/s(s-a) (s —6) (5— €).

39e. If similar polygons are constructed on the sides of a right
triangle, as homologous sides, the polygon on the hypotenuse is
equal to the siun of the polygons on the other two sides.

40. The locus of points equidistant from the ends of a sect is
the perpendicular bisector of the sect.

Cor. 1. Two points equidistant from the ends of a sect fix its
perpendicular bisector.

41. The locus of points equidistant from the sides of an angle
is the bisector of the angle. .

Cor. 1. The locus of a point equidistant from two intersecting
lines is a pair of lines bisecting the angles.

41a. The bisectors of the angles of a triangle are conciurent in a
point equidistant from the sides of the triangle.

416. The perpendicular bisectors of the sides of a triangle are
concurrent in a point equidistant from the vertices.

41c. The altitudes of a triangle are concurrent. ^ ,

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SUMMARIES AND APPLICATIONS 331 .

41d. The medians of a triangle are concurrent in a point of tri-
section of each.

42. Three points not in a straight line fix a circle.

43. In equal circles, equal central angles intercept equal arcs,
and conversely.

43a. In equal circles, the greater of two central angles intercepts
the greater arc, and conversely.

44. In equal circles, equal arcs are subtended by equal chords,
and conversely.

44a. In equal circles, unequal arcs are subtended by chords
unequal in the same order, and conversely.

45. A diameter perpendicular to a chord bisects it and its sub-
tended arcs.

Cor. 1. A radius which bisects a chord is perpendicular to it.
Cor. 2. The perpendicular bisector of a chord passes through
the center of the circle.

46. In equal circles, equal chords are equidistant from the center,
and conversely.

46a. In equal circles, the distances of unequal chords from the
center are unequal in the opposite order.

47. A Une perpendicular to a radius at its outer extremity is
tangent to the circle.

Cor. 1. A tangent to a circle is perpendicular to the radius

drawn to the point of contact.
Cor. 2. The perpendicular to a tangent at the point of contact

passes through the center of the circle.
Cor. 3. A radius perpendicular to a tangent passes through the

point of contact.
Cor. 4. Only one tangent can be drawn to a circle at a given

point on the circle.

48. Sects of tangents from the same point to a circle are equal.

49. The line of centers of two tangent circles passes through
their point of contact.

49a. The line of centers of two intersecting circles is the perpen-
dicular bisector of their conmion chord.

50. In equal circles central angles have the same ratio as their
intercepted arcs.

Cor. 1. A central angle is measured by its intercepted ace. ,

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382 PLANE GEOMETRY

51. ParaUels intercept equal arcs on a circle.

52. An inscribed angle, or one formed by a tangent and a chord
is measured by one-half its intercepted arc.

52a. The mid-point of the hypotenuse of a right triangle is
equidistant from the three vertices.

53. An angle whose vertex is inside the circle is measured by
half the siun of the arcs intercepted by it and its vertical.

54. An angle whose vertex is outside the circle is measured by
half the difference of its intercepted arcs.

54a. The opposite angles of a quadrilateral inscribed in a circle
are supplementary.
Cor. 1. A quadrilateral is inscriptible if its opposite angles are
supplementary.

55. A tangent is the mean proportional between any secant and
its external sect, when drawn from the same point to a circle.

Cor. 1. The product of a secant and its external sect from a
fixed point outside a circle is constant.

55a. If chords intersect inside a circle, the product bf their sects
is constant.

556. The square of the bisector of an angle of a triangle is equal
to the product of the sides of this angle, diminished by the product
of the sects made by the bisector on the third side.

55c. In any triangle the product of two sides is equal to the
product of the diameter of the circumscribed circle and the alti-
tude on the third side.

Cor. 1. If R denote the radius of the circle circumscribed about
a triangle whose sides are a, 6, c, and semiperimeter a,

then i2s —

Ws(s-aXs-b){s-c)

56. A circle may be circimiscribed about, and inscribed within
any regular polygon.

Cor. 1. An equilateral polygon inscribed in a circle is regular.
Cor. 2. An equiangular polygon circumscribed about a circle is

regular.
Cor. 3. The area of a regular polygon is equal to half the product

of its apothem and perimeter.

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SUMMARIES AND APPLICATIONS 333

57. If a circle is divided into any number of equal arcs, the chords
joining the successive points of division form a regular inscribed
polygon; and the tangents drawn at the points of division form a
regular circumscribed polygon.

Cor. 1. Tangents to a circle at the vertices of a regular inscribed
polygon form a regular circumscribed polygon of the
same number of sides.

Cor. 2 Lines drawn from each vertex of a regular inscribed
polygon to the mid-points of the adjacent arcs sub-
tended by its sides, form a regular inscribed polygon of
double the number of sides.

Cor. 3. Tangents at the mid-points of the arcs between conse-
cutive points of contact of the sides of a regular circum-
scribed polygon form a regular circumscribed polygon
of double the number of sides.

Cor. 4. The perimeter of a regular inscribed polygon is less
than that of a regular inscribed polygon of double the
number of sides, and the perimeter of a regular circum-
scribed polygon is greater than that of a regular cir-
cumscribed polygon of double the number of sides.

Cor. 6. Tangents to a circle at the mid-points of the arcs sub-
tended by the sides of a regular inscribed polygon
form a regular circumscribed polygon, of which the
sides are parallel to the sides of the original and the
vertices lie on the prolongations of the radii of the
inscribed polygon.

58. A regular polygon the number of whose sides is 3*2* may be
inscribed in a circle.

59. If in represent the side of a regular inscribed polygon of
n sides, and i2» the side ef one of 2n sides, and r the radius of the

circle, i2n= V2r2 -ry/^r^ -in^

60. If in represent the side of a regular inscribed polygon of n
sides, Cn that of a regular circumscribed polygon of n sides, and r

27*t

the radius of the circle, Cn^ . =^=»

61. The perimeters of regular polygons of the same number of
sides compare as their radii and also as their apothems.

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334 PLANE GEOMETRY

62. Circumferences have the same ratio as their radii.

Cor. 1. The ratio of any circle to its diameter is constant.
Cor. 2. In any circle c = 27rr.

63. The value of it is approximately 3.14159.

64. The area of a circle is equal to one-half the product of its
radius and its circumference.

Cor. L The area of a circle is equal to ir times the square

of its radius.
Cor. 2. The areas of circles compare as the squares of their

radii.
Cor. 3. The area of a sector is equal to half the product of

its radius and its arc.
Cor. 4. Similar sectors and similar segments compare as the

squares of their radii.

B. SYLLABUS OF CONSTRUCTIONS

Problem 1. Draw a perpendicular to a given line from a given
point (a) outside the line, (6) on the line.

Problem 2. Bisect a given (a) sect, (6) angle, (c) arc.

Problem 3. Reproduce a given angle.

Problem 4. Draw a line through a given point, and parallel to
-a given line.

Problem 6. Construct a triangle given any three independent
parts; (a) two angles and the included side, (6) two sides and the
included angle, (c) three sides, (d) the hypotenuse and a leg of a
right triangle.

Problem 6. Divide a sect into n equal parts.

Problem 7. Find a conunon measure of two commensurable
sects.

Problem 8. Pass a circle through three non-collinear points.
Cor. 1. Circumscribe a circle about a triangle.

Problem 9. Divide a given sect into parts proportional to n
given sects.

Problem 10. Divide a sect harmonically in the ratio of two
given sects.

Problem 11. Find a fourth proportional to three given sects.
Cor. 1. Find a third proportional to two given sects.

Problem 12. Upon a given sect as homologous to a designated

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SUMMARIES AND APPLICATIONS 335

side of a given polygon construct another similar to the original
polygon.

Problem 13. Construct a square equal to the sum of two or
more given squares.

Cor. 1. Construct a square equal to the difiFerence of two

given squares.
Cor. 2. Construct a polygon similar to two given similar poly-
gons and equal to (a) their sum, and (6) their difference.
Problem 14. Inscribe in a circle regular polygons the number
of whose sides is (a) 3-2'', (6) 4-2*'.

Problem 16. Construct a triangle, given two sides and the angle
opposite one of them.

Problem 16. Through a given point draw a tangent to a given
circle (a) when the point is on the circle, (b) when the point is
outside the circle.

Problem 17. Construct a common (a) external and (6) internal
tangent to two circles.
Problem 18. Inscribe a circle in a triangle.

Cor. 1. Find the centers of the escribed circles of a triangle.
Problem 19. Construct upon a given sect the segment of a
circle capable of containing a given angle.

Problem 20. Construct a mean proportional between two given
sects.
Problem 21. Divide a given sect into mean and extreme ratio.
Problem 22. Inscribe in a circle a regular decagon.
Cor. 1. Inscribe in a circle regular polygons the number of
whose sides is (a) 5-2'', (6) 15-2''.
Problem 23. Transform a polygon into a triangle.
Problem 24. Construct the square equal to a given (a) parallelo-
gram, (6) triangle, (c) polygon.

Problem 26. Construct a parallelogram equal to a given square,
having given (a) the sum of base and altitude equal to a given
sect, (6) the difiFerence of base and altitude equal to a given sect.

Problem 26. Construct a polygon similar to one and equal to
another of two given polygons.

Problem 27. Construct a square which shall have a given ratio
to a given square.

Cor. 1. Construct a polygon equal to any part of^a given
polygon and similar to it. Q'^'^^^ ^y V^OOgle

336

PLANE GEOMETRY

C. SUMMARY OF FORMULAS
I. SUM OF THE ANGLES OF A POLYGON,
a. Interior. S^n -2. {S stands for sum of angles in

n-2 {E stands for each angle of an
n ' equiangular 7^■gon, in st, ^:)

Equiangular. E^

b. Exterior

b.
c.
d.

S=2.

Equiangular. E^-.
. n

IL AREAS.

o. Parallelogram. A^bh.
h. Triangle A^^h.

c. Trapezoid. L A^ihQ)+bi).
2. A=hm.

UL METRIC RELATIONS IN TRIANGLES.

o. Bight triangle, c^^a^+b^. {^C=rt^.)

Acute triangle. c^=a^+b^ -2apb^. «C<r<.^.)
Obtuse triangle. c^^a^+b^+2api,a. {^C>rt.^.)
In any triangle.

1. a2+62=2(|)2+2mc2.

2. a^ -b^^ 2cpm,^

3. tnc^ W2(c^^+&^) -c^-

4. hc=^Vs(s -a){s -b)(s -c).

6. ab=heD. (Z) stands for diameter of circum-
scribed circle.)
6. te^^ab-pq, (pandgaresectsof cmadeby^)
7 £ = ^

8. A = Vs(s -d){$ -b){s -c).

IV. MENSURATION OF THE CIRCLE.

abc

h. Un

C. Cn

^y/8(8-a){s-'b){s-c)

sV2r^-rV4r^-in^.
_ 2rin

d. Cs27rr or C^tD.

e. irs3.14159

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SUMMARIES AND APPLICATIONS 337

V. KATIO Ain> PROPORTION.

sn: J. -^. then,

a. Product of means. . .

. admhe.

6. Alternation.

a h

e. Inversion.

a c'

d. Addition (Composition) — ^ = -t— or — r =» *

,^. . . V o— 6 c— d a— 6 c— d

(Division) —r- m —-J- or ■ .

^ b d a e

(Both) ^J-"-±-l

e. Series of equal ratios. Given ^ ■ ^ e — ■ . . ., then,

o-hc-hg-h- • • £
6-f-d-f-/+..."6'^- ••

VL DIVISION OF SECTS.

„ . AI AE A 7*~~S 'S

a. Hannomc. ^^sE' a f P E

b. Extreme and mean ratio.

AB AI .AB BE
AT^TB "^^ BE^JE

Vn. SIMILAR FIGURES.

P S (P and p stand for perimeter, S and a for
p 8 ' homologous sides.)

. A 5«

a «*

c. Circle. 1. - = -i.

2 £^^-^

c r d

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338 PLANE GEOMETRY

Vm. MEASUREMENT OP ANGLES.
a. C^a. (C stands for number of degrees in central angle.)

6. / 2c -. (/ stands for number of degrees in inscribed angle

■2*

or one formed by tangent and chord.)

TTSS— ;r-. (TF stands for number of degrees in angle with

vertex inside circle.)

I —8

d. Os-^. (0 stands for number of degrees in angle with

vertex outside circle.)
(o stands for number of degrees in intercepted arc.)
(v stands for number of degrees in arc intercepted by vertical
angle.)

(I stands for number of degrees in larger intercepted arc.)
(8 stands for number of degrees in smaller intercepted arc.)

IX. TANGENT AND SECANT,
a. T^^S'E. {T stands for length of sect of tangent from point
to circle.)
(8 stands for length of secant from same point

to circle.)
(E stands for length of external sect of the secant.)
6. pq&db. (pq stand for sects of a chord made by the inter-
section of a chord whose sects are a and 6.)

D. SUMMARY OF METHCjjDS OF PROOF
I. TRIANGLES CONGRUENT. \

Show that they have

(a) two sides and included angle respectively equal,
(6) a side and any two angles respectively equal,

(d) an isosceles triangle formed by the bisector of the vertex

angle,

(e) a parallelogram formed by the diagonal,
or, that they are right triangles, and have

(/) the hjrpotenuse and adjacent angle respectively equal,
(g) the hypotenuse and adjacent side respectively equal.

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SUMMARIES AND APPLICATIONS 339

EXERCISES. SET XCVI. CONGRUENCE OF TRIANGLES

1368. If a perpendicular is erected at any point on the bisector
of an angle and produced to cut the sides of the angle, two con-
gruent triangles are formed.

1369. In triangle ABC, BD (D in AC), the bisector of angle B
intersects AE (E in BC), a perpendicular to BD, in F. Prove that
triangles ABF and BFE are congruent.

1370. Triangles are congruent if two sides and the median to
one of these sides are equal respectively to two sides and the homol-
ogous median in the other. (Is this always true or is there any
exception?)

1371. Isosceles triangles are congruent if the base and a leg of
one are equal respectively to the base and a leg of the other.

1372. The three sects joining the mid-points of the sides of a
triangle divide the figure into four congruent triangles.

1373. If equal distances are laid off from the same vertex on
the l^s of an isosceles triangle, and these points of division are
joined with the opposite vertices, congruent triangles are formed.

1374. If two angles of a triangle are equal, the bisector of the
third angle divides the figure into two congruent triangles.

II. SECTS EQUAL.
Show that they are

(a) homologous parts of congruent polygons,
(6) sects from a point in a perpendicular cutting oflF equal
distances from its foot,

it by equal sects from the same point in the perpen-
dicular,

(d) bisectors of the angles of an equilateral triangle,

(e) sects of the base of an isosceles triangle formed by the

bisector of the vertex angle,
(/) sides opposite equal angles of a triangle,
(g) parallel sides of a parallelogram,

(h) sects of one diagonal of a parallelogram made by the other,
(t) sects on a transversal cut ofiF by parallels which cut off

eqiial sects on some other transversal,

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340 PLANE GEOMETRY

(j) chords in equal circles subtending equal arcs,

(k) chords in equal circles equally distant from the center,

(I) sects of tangents to a circle from the same point.

EXERCISES. SET XCVII. EQUALITY OF SECTS

1376. The bisectors of the base angles of an isosceles triangle
are equal.

1376. The sects joining the mid-points of the legs of an isosceles
triangle with the mid-point of the base are equal.

1377. The medians to the legs of an isosceles triangle are equal.

1378. If the base of an isosceles triangle is trisected, the sects
joining the points of division with the vertex are equal.

1379. If from any point in the circmnference of a circle two
chords are drawn making equal angles with the radius to the point,
these chords are equal.

1380. A circmnscribed parallelogram is equilateral.

m. ANGLES EQUAL.
Show that they are

(a) straight or right angles,

(6) supplements or complements of equal angles,

(d) homologous parts of congruent polygons,

(e) angles opposite the equal sides of a triangle,

(J) alternate-interior, alternate-exterior, or corresponding

angles of parallels,
(g) opposite angles of a parallelogram,
(A) homologous angles of similar polygons,
(i) measured by equal arcs,
or, that their sides are respectively
(j) perpendicular or parallel.

EXERCISES. SET XCVIII. EQUALITY OF ANGLES

1381. The tangent at the vertex of an inscribed equilateral
triangle forms equal angles with the adjacent sides.

1382. If in the base AC of an isosceles triangle ABC, two points
D and E are taken so that AE=AB and CD=BC, prove that
^DBE=^A,

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SUMMARIES AND APPLICATIONS 341

1383. If triangles are similar to the same triangle, they are
similar to each other.

1384. If through the vertices of an isosceles triangle lines are
drawn parallel to the opposite sides, they form an isosceles triangle.

1385. If the bisector of an exterior angle of a triangle is parallel
to the opposite side, the triangle is isosceles.

1386. Two circles intersect at the points A and B. Through
A a variable secant is drawn, cutting the
circles at Cand D. Prove that the angle
DBC is constant.

1387. Two circles touch each other inter-
nally at P. MN is a chord of the larger,
tangent to the smaller at C. Prove ^MPC

IV. LINES PARALLEL.
Show that they are

(a) perpendicular or parallel to the same line,
(6) opposite sides of a quadrilateral which can be shown to
be a parallelogram,

sides proportionally,

(d) side of a regular inscribed polygon and the side of a polygon

the sides of which are tangent to the circle at the mid-
points of the arcs subtended by the inscribed polygon,
or, that they have

(e) alternate-interior, alternate-exterior, or corresponding

angles equal,
(/) consecutive-interior or consecutive-exterior angles supple-
mentary.

EXERCISES. SET XCIX. PARALLELISM OF LINES

1388. If two sides of a triangle are produced their own lengths
through the common vertex, a line joining their ends is parallel
to the third side of the triangle.

1389. The line joining the feet of the perpendiculars dropped
from the extremities of the base of an isosceles triangle to the
opposite sides is paraUel to the base.

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342 PLANE GEOMETRY

1390. The tangents drawn through the extremities of a diameter
are parallel.

1391. If from one pair of opposite vertices of a parallelogram
lines are drawn bisecting the opposite sides respectively, the lines
are parallel.

1392. The bisectors of a pair of opposite angles of a parallelo-
gram are parallel.

1393. If two circles intersect and a sect be drawn through each
point of intersection terminated by the circumferences, the chords
which join the extremities of these sects are parallel.

1394. If through the point of contact of two tangent circles two
secants are drawn, the chords joining the points where the secants
cut the circle3 are parallel. Discuss both cases.

1396. If a straight line be drawn through the point of contact
of two tangent circles so as to form chords, the radii drawn to the
other extremities of these chords are parallel. (What two cases?)

1396. If a straight line be drawn through the point of contact
of two tangent circles so as to form chords, the tangents drawn at
the other extremities of the chords are parallel. (What two cases?)
V. LINES PERPEHDICULAR.

Show that

(a) two adjacent angles formed by them are equal,
(6) .one is the base and the other the bisector of the vertex
angle of an isosceles triangle,

(d) one is a tangent to a circle and the other a radius drawn

to the point of contact,

(e) one is a radius bisecting the other which is a chord of the

same circle,
(/) one is the shortest sect from a point to the other.
EXERCISES. SET C. PERPENDICULARITY OF LINES

1397. Every parallelogram inscribed in a circle is a rectangle.

1398. The bisectors of two interior angles on the same side of a
transversal to two parallels are perpendicular to each other.

1399. If either leg of an isosceles triangle be produced through
the vertex by its own length, and the extremity joined to the
extremity of the base, the joining line is perpendicular to the base.

SUMMARIES AND APPLICATIONS 343

1400. Two circles are tangent externally at A, and a common
external tangent touches them in B and C. Show that angle
BAC is a right angle.

1401. If two consecutive angles of a quadrilateral are right
angles, the bisectors of the other two angles are perpendicular.

1402. The line joining the center of a circle to the
mid-point of a chord is perpendicular to the chord.

1403. If the diagonals of a parallelogram are equ^l
the figure is a rectangle. ^,

1404. If the opposite sides of
an inscribed quadrilateral be produced to meet ^B'^
in A and F, the bisectors of the angles A and F
meet at right angles.

Hint: Ftoyq ^^.BGK-^^^CHK

VI. SECTS UNEQUAL.

Show that they are

(a) seQts from a point in a perpendicular to a line cutting off

imequal sects from its foot,
(6) sects on a line cut off by unequal sects drawn from a

point in the perpendicular to that line,

(d) third sides of two triangles which have the other two sides

respectively equal but the included angles unequal,

(e) chords of equal circles subtending unequal arcs,

(/) chords of equal circles unequally distant from the centers.

EXERCISES. SET CI. INEQUALITY OF SECTS

1406. A, B, Cj and D are points taken in succession on a semi-
circimiference and arc AC is greater than arc BD. Prove that
chord AB is greater than chord CD.

1406. Two chords drawn from a point in the circumference are
unequal if they make imequal angles with the radius drawn from
that point. Which of the chords is the greater?

1407. Two chords drawn through an interior point are imequal
if they make unequal angles with the radius drawn through that
point. Which is the greater one?

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344 PLANE GEOMETRY

1408. If two unequal chords be produced to meet, the secants
thus formed are unequal.

1409. In the equilateral triangle ABC, side BC is produced to
D, and DA is drawn. Prove that BD>AD,

1410. If the bisector of the right angle il of a right isosceles
triangle BAC cuts BC in D and is produced to X so that DK^AD,
then ifC<Aif; also than fiC.

Vn. ANGLES UNEQUAL.
Show that they are

(a) one exterior and one non-adjacent interior angle of a

triangle,
Q>) opposite the unequal sides of a triangle,

have two sides respectively equal but the third sides
imequal,

(d) measiu-ed by unequal arcs.

EXERCISES. SET CII. INEQUALITY OF ANGLES

1411. If A is the vertex of an isosceles triangle ABC and the leg
AC is produced to point D and DB drawn, prove that ^ABD>
<ADB.

1412. If the side BC of the square ABCD is produced to P and
P is joined with A, prove that ^APB< ^BAP.

1413. The angle formed by two tangents is equal to twice the
angle between the chord of contact and the radius drawn to a
point of contact.

1414. If the diagonals of quadrilateral ABCD bisect each other
at P, and side BC is longer than side AB, prove that ^BPC>
^BPA.

1416. In the quadrilateral MNRS, MS is the longest and
NR the shortest side. Prove that ^MNR>^MSR; also that
^NRS>^NMS.

1416. If in parallelogram ABCD, side BC>AB,ajid the diago-
nals intersect in P, prove that ^BPC>^BPA.
Vm. TRIANGLES SIMILAR.
Show that they have

(a) a center of similitude,

(6) two angles respectively equal.

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SUMMARIES AND APPLICATIONS 346

equal,
or, that

(d) their sides bear a constant ratio,

(e) they are triangles formed by homologous diagonals of

similar polygons,
(/) they are triangles formed by the altitude upon the hypo-
tenuse of a right triangle.

EXERCISES. SET CIII. SIMILARITY OF TRIANGLES

1417. If through the point of contact of two tangent circles
three secants are drawn, cutting the circumferences in A, B, C,
and ill, Bif Ci, respectively, then triangles ABC and AiBiCi are
similar.

1418. If from the point P outside a circle two secants are drawn
to meet the circumference in B and C, and D and E, respectively,
the triangles PBE and PCD are similar.

1419. If the bisector AD of ^A in the inscribed triangle ABC
is produced to meet the circumference in Ef then triangles ADD
and AEC are similar.

1420. If two chords, AB and CD, intersect in E, the triangles
AEC and BED are similar.

1421. If the altitudes AD and BE of triangle ABC intersect
at F, triangles AFE and BFD are similar.

14^. ilD and B£ are two altitudes of triangle il£C Prove that
triangles CDE and CBA are similar.

IZ. SECTS PROPORTIONAL.
Show that they are

(a) homologous parts of similar polygons,
(6) sects of sides of a triangle cut off by line parallel to third
side,

(d) two sides of a triangle and the sects of the third side made

by the bisector of the opposite (1) interior and (2)
exterior angle,

(e) the altitude upon the hypotenuse and the sects of the

hypotenuse made by it, or a leg, hypotenuse, and the
projection of the leg upon the hypotenuse.

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346 PLANE GEOMETRY

EXERCISES. SET CIV. PROPORTIONALITY OF SECTS

1423. Triangles are similar if an angle of one is equal to an angle
of another and the altitudes drawn from the vertices of the other
angles are proportional.

1424. If two circles are tangent externally, and through the
point of contact a secant is drawn, the chords formed are propor-
tional to the radii.

1426. If C is the mid-point of the arc AB, and a chord CD meets

the chord AB in E, then ^ = ^•

1426. The diagonals of any trapezoid divide each other in the
same ratio.

X. PRODUCTS OF SECTS EQUAL.

Show that the

(a) factors are sides of similar polygons,

(6) sects are sects of chords intersecting inside a circle,

the other are the entire secant from the same point and its

external sect.

EXERCISES. SET CV. EQUALITY OF PRODUCTS OF SECTS

1427. If from any point E in the chord AB the perpendicular

EC be drawn upon the

5^0 /^ rT^s^ diameter AD, then AC X

AD=ABXAE.

1428. If in the triangle
ABC the altitudes ADaM
gg meet in F, then BD X
DC^DFXAD.

1429. In the same diagram, BD XAC^BFXAU.

1430. If a chord be bisected by another, either sect of the first
is a mean proportional between the sects of the other.

The preceding summary is incomplete. The idea contained in
it may be developed by asking the pupil to make similar lists of
methods of establishing such geometric relations as the following:
inequality of arcs, conditions under which a quadrilateral is a
parallelogram, sums of sects unequal, differences of products
unequal, lines concurrent, points colhnear, points concyclic.

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SUMMARIES AND APPLICATIONS 347

EXERCISES. SET CVI. MISCELLANEOUS

1431. Draw, through a given point, a secant from which two
equal circles shall cut off equal chords.

1432. In a right isosceles triangle the hypotenuse of which is
10 in., find the length of the projection of either arm upon the
hypotenuse.

1433. Find the projection of one side of an equilateral triangle
upon another if each side is 6 in.

1434. If two sides of a triangle are 10 and 12, and their included
angle is 120°, what is the value of the third side?

1436. If two sides of a triangle are 12 and 16, and their included
angle is 45°, find the third side.

1436. Assuming the diameter of the earth to be 8000 mi., how
far can you see from the top of a mountain a mile high?

1437. Write the formula involving the median to 6, to c.

1438. If the sides of a triangle ABC are 5, 7, and 8, find the
lengths of the three medians.

1439. If the sides of a triangle are 12, 16, and 20, find the
median to side 20. How does it compare in length with the side
to which it is drawn? Why?

1440. In triangle ABC, a = 16, 6 = 22, and mc= 17. Find c.

1441. In a right triangle, right-angled at C, ^^=8^; what is c?
Find one pair of values for a and b that will satisfy the conditions
of the problem.

1442. If the sides of a triangle are 7, 8, and 10, is the angle
opposite 10 obtuse, right, or acute? Why?

1443. Draw the projections of the shortest side of a triangle
upon each of the other sides (a) in an acute triangle (6) in a right
triangle, (c) in an obtuse triangle. Draw the projections of the
longest side in each case.

1444. Two sides of a triangle are 8 and 12 in. and their included
angle is 60°. Find the projection of the shorter upon the longer.

1446. In Ex. 1444 find the projection of the shorter side upon the
longer if the included angle is 30°; 45°.

1446. Write the formula for the projection of a upon b.

1447. In triangle ABC, a = 15, 6 = 20, c = 25; find the projection
of 6 upon c. Is angle A acute, right, or obtuse?

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348 PLANE GEOMETRY

1M8. If the altitude upon the hypotenuse of a right triangle
divides the hypotenuse in extreme and mean ratio, the smaller
arm is equal to the non-adjacent sect of the hypotenuse.

1449. If two circles are tangent externally, their common ex-
ternal tangent is a mean proportional between their diameters.

1460. The sides of a triangle are 13, 17, 19. Find the lengths
of the sects into which the angle bisectors divide the opposite sides.

1461. The angles of a triangle are 30°, 60*", 90°. Find the lengths
of the sects into which the angle bisectors divide the opposite sides,
if the hypotenuse is 10.

1462. Find the sum of (a) the acute angles of a starred pentagon,
(6) all the interior angles.

1463. The difference of the squares of two sides of a triangle is
equal to the difference of the squares of the sects made by the
altitude upon the third side.

1464. The perpendicular erected at the mid-point of the base of
an isosceles triangle passes through the vertex and bisects the
angle at the vertex.

Construct a right triangle, having given:
1466. The hypotenuse and one side.

1466. One side and the altitude upon the hypotenuse.

1467. The median and the altitude upon the hypotenuse.

1468. The hypotenuse and the altitude upon the hypotenuse.

1469. The radius^ of the inscribed circle and one side.

1460. The radius of the inscribed circle and an acute angle.

1461. The hypotenuse and the difference between the arms.

1462. The hypotenuse and the sum of the arms.

1463. The sides of a triangle are 6, 7, and 8 ft. Find the areas
of the two parts into which the triangle is divided by the bisector
of the angle included by 6 and 7.

1464. The square constructed upon the siun of two sects is equal
to the sum of the squares constructed upon these two sects, in-
creased by twice the rectangle of the sects.

1466. The square constructed upon the difference of two sects
is equal to the sum of the squares constructed upon these sects,
diminished by twice their rectangle*

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SUMMARIES AND APPLICATIONS 349

1466. The difference between the squares constructed upon two
sects is equal to the rectangle of their sum and their difference.

1467. A straight rod moves so
that its ends constantly touch two
fixed rods perpendicular to each other.
Find the locus of its mid-point.

1468. If a quadrilateral has each
side tangent to a circle, the siun of one
pair of opposite sides equals the sum
of the other pair.

1469. Analysis of the regular inscribed hexagon — prove that:
(a) Three of the diagonals are diameters.

(6) The perimeter contains three pairs of parallel sides.

(d) Radii drawn to the alternate vertices divide the hexagon
into three congruent rhombuses.

(e) The diagonals joining the alternate vertices form an
equilateral triangle of which the area equals one-half that of the
hexagon.

(J) The diagonals joining the corresponding extremities of a
pair of parallel sides of tibe hexagon form with these sides a
rectangle.

1470. Inscribe in a given circle a regular polygon similar to
a given regular polygon.

1471. Construct the following angles: 60°, 30**, 72°, 18°, 24°, 42°.

1472. Construct AABC given ^B, b, and nib.

1473. Construct a triangle, having given the base, the altitude,
and the angle at the vertex.

1474. (a) Construct a square that shall be to a given triangle as
5 is to 4.

(6) Construct a square that shall be to a given triangle as m is
to n, when m and n are two given sects.

1476. The diagonals of a regular pentagon divide each other
into extreme and mean ratio.

1476. If the sect I is divided internally in extreme and mean
xatiO; and if ^ is its greater sect; what is the value of 8 in terms of 11

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350 PLANE GEOMETRY

1477. A sect 10 in. long is divided internally *in extreme and
mean ratio. Find the lengths of its sects.

1478. A sect 8 in. long is divided externally in extreme and mean
ratio. Find the length of its longer sect.

1479. Experience has shown that a book, photograph, or other
rectangular object is most pleasing to the eye when its length and
width are obtained by dividing the semiperimeter into extreme and
mean ratio. Find to the nearest integer the width of such a book
where its length is 8 in.

1480. If the bisector of an inscribed angle be produced until it
meets the circimiference, and through this point of intersection a
chord be drawn parallel to one side of the angle, it is equal to the
other side.

1481. If from the extremities of a diameter perpendiculars be
drawn upon any chord (produced if necessary), the feet of the
perpendicular are equidistant from the center.

1482. Find the locus of points, the distances of which from two

intersecting lines L and Li are as — .

The locus consists of two straight lines. Draw parallels to L and
Li, such that their distances from L and Li respectively shall be

as— ; these parallels will intersect in points belonging to the

required locus. Special case : -^ LLi = 60°, m = 2, n = 1 .

1483. Between the sides of a given angle a series of parallels
are drawn; find the locus of points which divide these parallels in

the ratio — . Special case : m = 4, n = 1.

1484. Construct a triangle equal to the sum of two given
triangles.

1486. Construct a triangle equal to the difference of two given
parallelograms.

1486. Construct a regular pentagon, given one of the diagonals.

1487. Prove that the following solution of the problem to divide
a given circle into any number of equal parts (say 3), by drawing
concentric circles, is correct.

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SUMMARIES AND APPLICATIONS 351

Trisect the radius at M and N. Draw semicircle on ilO as diam-
eter. Erect perpendiculars MB and
NC meeting semicircle at B and C.
With centers at and radii OB and
OCy draw circles.

14B8. Transform a rectangle into a
parallelogram, having given a diagonal.

1489. Transform a given triangle
into a right triangle containing a given
acute angle.

1490. The sides of a triangle are 8, 15, and 17. Find the radius
of the inscribed circle.

1491. The sum of the squares on the sects of two perpendicular
chords is equal to the square of the diameter of the circle.

If AB, CD are the chords, draw the diameter BE, and draw AC,
ED,BD. Prove that ilC=S2).

1492. Calculate the lengths of the common external and internal
tangents to two circles whose radii are 16 and 12 units respectively,
and whose line of centers is 40.

1493. Describe a circle whose circmnference is equal to the -
difiference of two circmnferences of given radii.

1494. Construct a circle equal to three-fifths of a given circle.

1495. Construct a circle equal to three times a given circle.
14%. Construct a semicircle equal to a given circle.

1497. Construct a circle equal to the area boimded by two con-
centric circumferences.

Required to circumscribe about a given circle the following
regular polygons: ^

1498. Triangle. 1499. Quadrilateral. 1600. Hexagon.
1501. Octagon. 1502. Pentagon. 1503. Decagon.

1504. Construct a square equal to a given (a) parallelogram,
(6) triangle, (c) polygon.

1505. Construct a triangle similar to a given triangle and equal
to another given triangle.

1506. Construct a polygon similar to a given polygon and having
a given ratio to it.

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PLANE GEOMETRY

1607. Construct x="

1608. Construct x^Va^+b^-c*.

1609. Construct the roots of x^+ax+b^O.

1610. Given the sects a, b, c, co nstruc t sect x if:

(a) x^V^ob. (6) x^Va^-bK (c) x^y/a^-bc.

1611. Transform a given triangle into an equilateral triangle.

1612. Transform a given parallelogram into an equilateral
triangle.

1613. Construct an equilateral triangle equal to one-half a given
square.

1614. Transform a triangle into a right isosceles triangle.

1616. Divide a given sect into two sects such that one is to the
given sect as y/2 is to -v/sT

1616. In a circle a line £F is drawn per-
pendicular to a diameter AB, and meeting
it in G. Through A any chord AD is drawn,
meeting EF in C. Prove that the product
AD XAC is constant, whatever the direc-
tion of AD.

Draw BD and compare the ▲ ACG^
ADB. Is the theorem also true when G
lies outside the circle?

1617. If two sects OA, OB, drawn through a point are divided
in C, D, respectively, fo that OA XOC
^OS XOD, a circle can be described
through the points A,B,C, D.

Show that A DAO, CBO are sim-
ilar, and the '^•DAC and CBD equal.
Therefore, if a segment is described
upon CD capable of containing the
^DAO, the arc of this segment will pass through B.

1618. Divide a quadrilateral into four equal parts by lines drawn
from a point in one of its sides.

1619. Draw a common secant to two given circles exterior to
each other, such that the intercepted chords shall have the given
lengths a and 6.

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353

1620. Divide quadrilateral ABCD
into three equal parts by straight
Unes passing through A.

(a) Transform ABCD into AilDJSf.

Divide AADE into the three equal
parts ADF, AFG, and AGE,

As the last two parts do not lie
entirely in the given quadrilateral
draw GH \\ CA.

Then AFG-=-AFCH, and AF and
AH are the required lin^r - - -

Or (6) Trisect DB. Draw AF,
AE, CFy and CE.

Then the broken lines AFC and
AEC divide the figure into three
equal parts. To transform these parts so as to fulfill the
conditions, draw FH and EK parallel to AC. AH and AK are
the required lines.

1621. If the diagonals of a trapezoid are equal, the trapezoid is
isosceles.

1622. If the altitude BD of AABC is intersected by another

altitude in G, and EH and HF^
are perp endicular bisectors of AC
and CB, prove BG^2HE.

1623. The line joining the point
^ ^i ^\^^^ of intersection of the altitudes of

-O l^ _v^ a triangle and the point of inter-
section of the three perpendicular
bisectors, cuts off one-thitd of the B
corresponding median.

1624. The points of intersection of
the altitudes, the medians, and the
perpendicular bisectors of a triangle
lie in a straight line.

1626. If, through the points of intersection of two circumfer-
ences, parallels be drawn terminated by the circumferences, they
are equal.

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364 PLANE GEOMETRY

1626. If from any point in the circumference of a circle chords
be drawn to the vertices of an inscribed equilateral triangle, the
longest chord equals the sum of the smaller chords.

1627. Triangles are similar if two sides and the radius of the
circumscribed circle of one are proportional to the homologous
parts of another.

1628. Construct a square that shall be to a given triangle as 5
is to 4.

1629. Construct a square that shall be to a given triangle as
m is to n, when m and n are two given sects.

1630. If in a circle a regular decagon and a regular pentagon be
inscribed, the side of the decagon increased by the radius is equal
to twicS the apothem of the pentagon.

1631. If from a point 0, OA, OB, OC, and OD are drawn so that
the ^AOB is equal to the ^BOC, and the ^BOD equal to a right
angle, any line intersecting OA, OB, OC, and OD is divided
harmonically.

1632. Inscribe in a given circle a regular polygon of n sides,
n being any whole number.

The following construction is found in most cases to be suffi-
ciently exact for practical purposes:

^ Divide the diameter AB into n

equal parts (in the figure n=7).

V Draw the radius CD J. 4-B, produce

>:>v CB to E, and CD to F, making BE

\ and DF each equal to one of the

^.^^'Yk parts of the diameter ; draw EF,

^ 1 I 1^ cutting the circle for the first time

in (?. Then the line GH joining G

and the third point of division of AB, counting from B will be

very nearly equal to one side of the inscribed polygon of n sides.

For n=3 and n=4, this construction is impossible; forn=5 it

is useless, on account of its inaccuracy; but for n>5 it gives a

very close approximation to the exact value of the side required.

Inscribe a hexagon, a heptagon, an octagon, a nonagon, and

a decagon.

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355

1633. Draw a line meeting the sides C-A, CB, of AABC in D,
E, respectively, so that:

(a) DE II AB, BE^ED, N.B. Bisect

<^' _ _

(6) DE II AB, DE=AD+BE.

(c) DE^CD^BE. (See figure.)
Analysis: Suppose the problem solved, and
draw BD. The A BDE and DCE are isosceles, ^
whence A-J^BE^:S^BDE^h7i^DEC^h:i^DCE, which is known,
mines the point D, and E is easily found, since DE^DC,

Examine this problem for the special cases
when <ACB=90°, and when ^ACE = \2{f.

1634. Draw through a given point P in the
arc subtended by a chord AB a chord which
shall be bisected by AB,
^C On radius OF take CD equal to CF. Draw
BE\\BA,

1636. Through a given point F inside a
given circle draw a chord AB so that the ratio
AF^jrn
BF^n

Draw OFC so that d7^=^. Draw

CA equal to the fourth
proportional to n, wi, and the radius of the circle.

1636. Draw through one of the
points of intersection of two circles a
secant so that the two chords that are
formed shall be in the feiven ratio
mton.

1637. Draw a common tangent to
two given circles:

(a) Non-intersecting circles
(1) unequal, (2) equal.

(b) Intersecting

(1) unequal, (2) equal.

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PLANE GEOMETRY

d'^e

1638. Find the shortest path from P to Pi which shall touch
a given line, if P and Pi are two given points on the same side
of the line.

1639. Given points P, Pi, P2; through P draw a line which shall
be equidistant from Pi and P2.

1640. Given P, Pi, P2; through P draw a line so that the dis-
tances from P to the feet of the perpendiculars dropped from Pi,
Pj to the line, shall be equal.

1641. Find the direction in which a billiard ball must be shot

from a given point on the table,
so as to strike another ball at a
given point after first striking
one side of the table. (The angle
at which the ball is reflected from
a side is equal to the angle at
which it meets the side, that is,

FiG.1. ^JS?CB=^ilCF.)

Suggestion: Construct BE i. that side of the table which the ball is to
strike, and make ED^BE.

1642. The same as the preceding problem, except that the
cue ball is to strike two sides of the

table before striking the other ball.
(Fig. 1.)

Suggestion: BiEisEiDuDiH^HFi,

1643. Solve Ex. 1542 if the cue ball
is to strike three sides before striking
the other ball, also if it is to strike all'
four sides. (Fig. 2.)

1644. A biUiard ball
is placed at a point P

on a billiard table. In what direction must it be
shot to return to the same point after hitting all
four sides?

Suggestion: (a) Show that the opposite sides of the quadrilateral along
which the ball travels are parallel, (b) If the ball is started parallel to a diag-
onal of the table, show that it will return to the starting point.

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SUMMARIES AND APPLICATIONS 357

1646. Show that in the preceding problem the length of the path
traveled by the ball is equal to the sum of q

the diagonals of the table.

1646. Show that in the diagram B C=U
and OB=ivi and 0C=i6where OC±AD, OD
=iOC, and BZ)=Z)C. J-^T O 'P

dl647* Describe a circle the ratio of whose area to that of a
given circle shall be equal to the given ratio m to n.

dl648* In an inscribed quadrilateral the product of the diagonals
is equal to the sum of the products of the opposite sides. (Ptolemy.)

dl649. Find a point such that the perpendiculars from it to
the sides of a given triangle shall be in the ratio p to g to r.

dl660. Find a point within a triangle such that the sects joining
the point with the vertices shall form three triangles, having the
ratio 3 to 4 to 5.

dl66L Given a circle and its center; find the side of an inscribed
square by means of the compasses alone. (" Napoleon's Problem.")

dl662. Bisect a trapezoid by a line parallel to the bases.

dl663. The' feet of the perpendiculars dropped upon the sides
of a triangle from any point in the circumference of the circum-
scribed circle are collinear. (" Simpson's Line.")

dl664. The points A,B,C, D^are collinear. Find the locus of a
point P from which the sects AB and CD subtend the same angle.

dl666. Transform a given triangle into one containing two given
angles.

dl666. Transform a given triangle into an isosceles triangle,
having a given vertex angle.

Hint: Construct a A similar to a given isosceles A and equal to a gi^en
At or transform a A into one containing two given 2(.8.

dl667. Construct an equilateral triangle that shall be to a given
rectangle as 4 is to 5.

dl668. In a given triangle, ABC, inscribe a parallelogram similar
to a given parallelogram, so that one side lies in AB, and the other
two vertices lie in BC and AC respectively.

dl669. Divide a pentagon into four equal parts by lines drawn
through one of its vertices.

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PLANE GEOMETRY

dl660. Right triangles are similar if the hypotenuse and an
*ann of one triangle are proportional to the hypotenuse and an arm

of another.

dl661. If from a point A two equal
tangents, AB and AC, are drawn to
two circles, and Oi, and AD is per-
pendicular to OOi, then 5S -Oi-D s

US' -0^.

dl662. Conversely, if, in the same

2 t ^-^t S

diagram, D is taken so that OD -OiD ^OB - OiC , then the
tangents drawn from any point in the perpendicular, AD, to the
circles are equal. ^

dl663. Divide a ^

trapezoid into two ^4;

similar trapezoids by ( "^^«^^«}. ^^jj^ p

a line parallel to the

dl664. Through P
secants are drawn to a circle 0; find the locus of points which

divide the entire secants in the ratio —

^ dl666. Through a fixed pomt

F a secant is drawn to a given
circle, and through its intersec-
tions A, B with the circumfer-
ence tangents are drawn inter-
secting in a point P. If the se-
cant revolves about F, find the
locus of P.

dl666. Find the locus of the

vertex of a triangle, having given the base and the ratio of the

other two sides.

CONSTRUCTIONS LEADING TO TEE PROBLEM OF APOLLONIUS

(200 B.C.)
dl667. Construct a circle which shall pass through two given
points and be tangent to a given line.

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SUMMARIES AND APPLICATIONS

359

dl668. Construct a circle which shall pass through two given
points and be tangent to a given circle.

dl669. Construct a circle which shall pass through a given point
and be tangent to two given lines.

dl670. Construct a circle which shall pass through a given point
and be tang^it to a given line and to a given circle.

dl671. Construct a circle which shall pass through a given point
and be tangent to two given circles.

dl672. Construct a circle which shall be tangent to three given
lines.

dl673. Construct a circle which shall be tangent to two given
lines and to a given circle.

dl674. Construct a circle which shall be tangent to a given line
and to two given circles.

dl676. Construct a circle which shall be tangent to three given
circles. (" Problem of ApoUonius.")

THE TRIANGLE AND NINE OF ITS CIRCLES

The three escribed circles are represented "by arcs; the other
six by centers. Prove the construction of the nine circles. Find
three others.

Circles.

Circumscribed,
Inscribed,
Nine points,
Pedal (3),
C is the cenfroid; and 0, the
crthocenter; 13=S0.

dl676. The circumscribed
circle bisects the straight lines
joining the center of the in-
scribed circle with the centers
of the escribed circles.

dl677. Each vertex of the tri-
angle is collinear with centers
of two of the escribed circles.

dl678. The center of the inscribed circle is collinear with the
center of any escribed circle and the opposite vertex.

Coiten.

Radii.

Perp. from

4,5,6

40,50,60

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PLANE GEOMETRY

dl679. Each center of the inscribed or the escribed circles is the
orthocenter of the triangle having the other three centers as its
vertices.

dl580. The four circles, each of which passes through three of
the centers of the escribed and inscribed circles, are equal.

dl681. The three circles, the circumference of each of which
passes through the extremities of any side of a triangle and the
orthocenter, equal one another.

Nineteen circles in all have been mentioned.

THE NHfE-POINTS CIRCLE
The orthocenter is the paint at which the aUitudes of a triangle

The centroid cf any triangle is the point at which the medians of
the triangle meet,

1682. The mid-points of the sides of a triangle are concyclic with
the feet of the perpendicular from the opposite vertices, and with
the mid-points of the sects joining the orthocenter with the vertices.
(Nine-points circle.)

1683. The center of the nine-points circle is the mid-point of
the sect joining the orthocenter and the center of the circumscribed
circle.

1684. The diameter of the nine-points circle is equal to the radius
of the circumscribed circle.

1686. The orthocenter and the centroid are coUinear with the
centers of the nine-points and the circiunscribed circles.
1686. The nine-points circle is tangent to the inscribed and
escribed circles of a triangle.

Give proofs different from those sug-
gested in the text for the following
theorems:

1687. Theorem 216. Suggestion: (Kg. 1).
Suggestion: (Figs. 2 and 3).

Fig. 1

1688. Theorem 21d.

FiQ. 2

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SUMMARIES AND APPLICATIONS

361

1689. Theorem 21c. Prove by the method of exclusion.

1690. Theorem 21/. Suggestion: (Fig. A).

1691. Theorem 26a. For suggestions see Heath's Mathematical
Monographs, Numbers 1 and 2.

Fig. a

1692. Theorem 266.

1693. Theorem 40.

1694. Theorem 41.
1696. Theorem 53.

Fig. B

Suggestion: (Fig. B).
Prove by means of direct and opposite.
Prove as in Ex. 1593.
Suggestion: (Fig. 1).

FiQ. 1 Fig. 2 Fio. 3

1696. Theorem 54. Suggestions: (Figs. 2 and 3)

1697. Theorem 55, Cor. 1. Suggestion: ,(Fig. 4).

1698. Use the following analysis to find
another solution for Problem 21.
Analysis: Suppose the construction completed.

Then-s
a

a
s—a
or a»s8(«— a)
or a*^8*—8a
or a'-ha«s8«

Completing the square, a*+a«-h( 2 y ^**'^V^/

FiQ. 4

which suggests that (fl+0 ^ ^^^ hypotenuse in which the legs are

8 and 2 respectively.

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CHAPTER IX

COLLEGE ENTRANCE EXAMINATIONS

THE UNIVERSITY OF CHICAGO

Examination for Admission, June, 1908

MATHEMATICS (2)— PLANE GEOMETRY

(time allowed — ONE HOUR AND THIRTY MINUTES.)

[In writing, use only one side of the paper, put your name in full -at the
top of each sheet, and number your work according to the numbers on the
printed paper.]

[When required, give all reasons in full, and work out proofs and problems
in detail.]

State:

(a) At what school you studied this subject.
(6) How many weeks.

(d) What textbook you used.

I. Given two circles of unequal radii and lying exterior to each
other; make the construction by which a straight line may be
drawn tangent to both circles and shall cross the Une joining
their centers.

II. Prove that the sum of the exterior angles of any convex
polygon, made by producing each of its sides in consecutive order,
is equal to four right angles.

III. In any triangle PQR perpendiculars are let fall to the oppo^
site sides from the vertices P and Q. Show that the lines joining
the feet of these perpendiculars to the middle point of the side
PQ are equal. [Draw two figures, in one of which the angle at Q
shall be obtuse, in the other acute.]

IV. Two fields are of similar shape, one having five times the
area of the other, (a) If they are both circles and the radius of
the first is 25 rods, find the radius of the second, (b) If they
are both equilateral triangles, and the side of the first is 25 rods,
find the side of th6 other.

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COLLEGE ENTRANCE EXAMINATIONS 363

Examination fob Admission, September, 1916

MATHEMATICS (2)— PLANE GEOMETRY

(time allowed — ONE HOUR AND FIFTEEN MINUTES.)

{Id. writing use only one side of the paper, put your name in full at
the top of each sheet, and number your work according to the numbers on the
printed paper.]

[When required, give all reasons in full, and work out proofs and problems
in detail.]

1. Prove that a straight line parallel to one side of a triangle
divides the other two sides proportionally.

2. Find the locus of all points such that the two lines joining
each to two fixed points always make a given angle with each other.
What does the locus become when the given angle is a right angle?

3. Prove that in a triangle with a given fixed base the median
from the vertex opposite this base is greater than, equal to, or less
than half the base according as the vertical angle is less than, equal
to, or greater than a right angle.

4. Construct a circle passing through a given point and tangent
to two given intersecting straight lines.

5. Two tangents to a circle intersect in a point which is 50 inches
from the center of the circle. The area of the four-sided figure
formed by the two tangents and the two radii drawn to the two
points of contact is 625 square inches. Find the length of the tan-
gents and the radius of the circle.

6. Show how to construct geometrically a square that shall
contain the same area as a given rectangle whose base is b and
whose altitude is a. Prove the result.

HARVARD UNIVERSITY
JUNE, 18(>4
(In solving problems use for w the approximate value S^-.)

1. Prove that any quadrilateral the opposite sides of which are
equal, is a parallelogram.
A certain parallelogram inscribed in a circumference has two

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3M PLANE GEOMETRY

sides 20 feet in length and two sides 15 feet in length; what are
the lengths of the diagonals?

2. Prove that if one acute angle of a triangle is double another,
the triangle can be divided into two isosceles triangles by a straight
line drawn through the vertex of the third angle.

Upon a given base is constructed a triangle one of the base
angles of which is double the other. The bisector of the larger
base angle meets the opposite side at the point P. Find the locus
of P.

3. Show how to find a mean proportional between two given
straight lines, but do not prove that your construction is correct.

Prove that if from a point, 0, in the base, SC, of a triangle, ABC,
straight lines be drawn parallel to the sides, AS, AC^ respectively,
so as to meet AC in M and AB in iV, the area of the triangle AMN
is a mean proportional between the areas of the triangles BNO
and CMO.

4. Assuming that the areas of two parallelograms which have
an angle and a side common and two other sides imequal, but
commensurable, are to each other as the unequal sides, prove that
the same proportion holds good when these sides have no common
measure.

5. Every cross-section of the train house of a railway station
has the form of a pointed arch made of two circular arcs the centres
of which are on the ground. The radius of each arc is equal to
the width of the building (210 feet) ; find the distance across the
building measured over the roof, and show that the area of the
cross-section is 3675 (4ir -3^ 3) square feet.

SEPTEMBER, 1894

One question may be omitted.

(In solving problems use Jar v the approximate value S^.)

1. Prove that any quadrilateral the diagonals of which bisect
each other is a parallelogram.

The diagonals of a parallelogram circumscribed about a circum-
ference are 60 inches and 80 inches long respectively. How long
are the sides?

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COLLEGE ENTRANCE EXAMINATIONS 365

2. Prove that the difference of the angles at the base of a tri-
angle is double the angle between a perpendicular to the base
and the bisector of the vertical angle.

The sum of the base angles of each of a number of triangles con-
structed on a given base 10 inches long is 150**. What is the locus
of the vertices of these triangles?

3. Show how to find a fourth proportional to three given lines,
but do not prove that your construction is correct.

One circle touches another internally at 0, and a chord ilB of
the larger circle touches the smaller one at C. Prove that AO
makes with the common tangent to the circles an angle equal to
ABO and that CO bisects the angle AOB. State without proof
some relation that exists between the lines AO^ CB, BO, and AC.

4. Assuming that the areas of two rectangles which have equal
altitudes are to each other as their bases when the latter are com-
mensurable, show that the same proportionality exists when the
bases have no common measure.

5. A kite-shaped racing track is formed by a circular arc and
two tangents at its extremities. The tangents meet at an angle
of 60°. The riders are to go round the track, one on a line close to
the inner edge, the other on a line everywhere 5^ feet outside the
first line. Show that the second rider is handicapped by about
22 feet.

JUNE, 1895

One question may be omitted.

(In solving problems use for ir the approximate value S^-.)

L Prove that if two straight lines -are so cut by a third that
corresponding alternate-interior angles are equal, the two lines
are parallel to each other.

2. Prove that an angle formed by two chords intersecting within
a circumference is measured by one-half the sum of the arcs inter-
cepted between its sides and between the sides of its vertical angle.

Two chords which intersect within a certain circimiference
divide the latter into parts the lengths of which, taken in order,
are as 1, 1, 2, and 5; what angles do the chords make with each
other?

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866 PLANE GEOMETRY

3. Through the point of contact of two circles which touch each
other externally, any straight line is drawn terminated by the
circumferences; show that the tangents at its extremities are paral-
lel to each other.

What is the locus of the point of contact of tangents drawn
from a fixed point to the different members of a system of concen-
tric circimiferences?

4. Prove that, if from a point without a circle a secant and a
tangent be drawn, the tangent is a mean proportional between
the whole secant and the part without the circle.

Show (without proving that your construction is correct) how
you would draw a tangent to a circmnference from a point
without it.

5. Prove that the area of any regular polygon of an even number
of sides (2n) inscribed in a circle is a mean proportional between
the areas of the inscribed and the circumscribed polygons of half

the number of sides. If n be indefinitely
increased, what limit or limits do these three
areas approach?

6. The perimeter of a certain church win-
dow is made up of three equal semi-circum-
ferences, the centres of which form the vertices
of an equilateral triangle which has sides 3^
feet long. Find the area of the window and the length of its
perimeter.

SEPTEMBER, 1895

One question may be omitted.

(In aolmng problems use for v the approadmate value 34-.)

1. Prove that every point in the bisector of an angle is equaUy
distant from the sides of the angle. State the converse of this
proposition. Is this converse true?

2. Prove that an angle formed by two secants intersecting with-
out a circumference is measured by half the difference of the arcs
which the sides of the angle intercept.

A certain pair of secant lines which intersect without a circle
divide the circumference into parts the lengths of which, taken in

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COLLEGE ENTRANCE EXAMINATIONS 367

order, are to one another as 1, 2, 3, and 4. What angles do the lines
make with each other?

3. Two given circles touch each other externally at the point P,
where they have the common tangent PC. They are also touched
by the line AB in the points A and B respectively. Show that the
circle described on AB as diameter has its centre on PC, and
touches at P the straight line which joins the centres of the two
given circles.

4. Show how to describe upon a given straight line a segment
which shall contain a given angle.

A and B are two fixed points on the circumference of a circle,
and PQ is any diameter. What is the locus of the intersection of
PA and QB?

5. C is any point on the straight portion, AB, of the boundary
of a semicircle. CD, drawn at right angles to AB, meets the cir-
cumference at D. DO is drawn to the centre, 0, of the circle, and
the perpendicular dropped from C upon OD meets OD at E Show
that DC is a mean proportional to AO and DE.

State the fundamental theorem in the method of limits as used
in Plane Geometry.

6. A horse is tethered to a hook on the inner side of a fenc^
which bounds a circular grass plot. His tether is so long that he
can just reach the centre of the plot. The_area of so much of the
plot as he can graze over is ^ (4ir -3\/3) square rods; find the
length of the tether and the circumference of the plot.

JUNE, 1896
One question may be omitted.

(In solving problems use for r the approximate value S^-.)

1. Prove that if two oblique lines drawn from a point to a straight
line meet this line at unequal distances from the foot of the per-
pendicular dropped upon it from the given point, the more remote
is the longer.

2. Prove that the distances of the point of intersection of any
two tangents to a circle from their points of contact are equal.

A straight line drawn through the centre of a certain circle and

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368 PLANE GEOMETRY

through an external point, P, cuts the circumference at points
distant 8 and 18 inches respectively from P. What is the length
of a tangent drawn from P to the circumference?

3. Given an arc of a circle, the chord subtended by the arc,
and the tangent to the arc at one extremity, show that the per-
pendiculars dropped from the middle point of the arc on the tangent
and chord, respectively, are equal.

One extremity of the base of a triangle is given and the centre
of the circumscribed circle. What is the locus of the middle point
of the base?

4. Prove that in any triangle the square of the side opposite
an acute angle is equal to the sum of the squares of the other two
sides diminished by twice the product of one of those sides and the
projection of the other upon that side.

Show very briefly how to construct a triangle having given the
base, the projections of the other sides on the base, and the pro-
jection of the base on one of these sides.

5. Show that the areas of similar triangles are to one another
as the areas of their inscribed circles.

The area of a certain triangle the altitude of which is V^i is
bisected by a line drawn parallel to the base. What is the distance
of this line from the vertex?

6. Two flower beds have equal perimeters. One of the beds is
circular and the other has the form of a regular hexagon. The
circular bed is closely surrounded by a walk 7 feet wide bounded
by a circumference concentric with the bed. The area of tha walk
is to that of the bed as 7 to 9. Find the diameter of the circular
bed and the area of the hexagonal bed.

SEPTEMBER, 1896

One question may be omitted

{In solving problems use for v the approximate value 3^.)

1. Prove that, if one of two convex broken lines which have
the same extremities envelops the other, the first is the longer.

2. Prove that, when two circumferences intersect each other,
the line which joins their centres bisects at right angles their
common chord.

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COLLEGE ENTRANCE EXAMINATIONS 369

The centres of two circles of radii 8 inches and 6 inches respec-
tively are 10 inches apart. Show that the common chord is 9.6
inches long.

3. Show that, if two parallel tangents to a circle are intercepted
by a third tangent, the part of the third tangent between the other
two subtends a right angle at the centre of the circle.

State briefly how you might find a fourth proportional to three
given straight lines.

4. Prove that in any obtuse-angled triangle the square of the
side opposite the obtuse angle is equal to the sum of the squares
of the other two sides increased by twice the product of one of these
sides and the projection of the other upon that side.

What is the locus of the vertices of all the triangles so constructed
on a given base that the radii of their circumscribed circles are all
equal to a given line?

5. The line passing through the centres of two circles which
touch each other externally at A^ meets a common tangent, which
touches the circles at B and C respectively, at the point 5. Show
that iSA is a mean proportional between SB
andSC.

6. The perimeter of a certain church win-
dow is made up of three equal circular arcs
the centres of which are the vertices of an
equilateral triangle. Each of the arcs subtends
an angle of 300° at its own centre. Find the
area of the window, assvuning the length of the perimeter to be

110 feet.

JUNE, 1906

(One Hour and a Half)
The University Provides a Syllabus

1. Prove that the sum of the three angles of any triangle is
equal to two right angles.

On a line AE choose a point J?, and construct an isosceles triangle
ABC with AB as base, the base angles being less than 45°. With
B as vertex construct an isosceles triangle BCD whose base CD
lies in AC produced. Show that the angle DBE is three times the
angle il.

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370 PLANE GEOMETRY

2. Prove that the bisector of an angle of a triangle divides the
opposite side into segments proportional to the sides of the angle.

The hypotenuse of a right triangle is 10 inches long and one of
the acute angles is 30^. Compute the lengths of the segments into
which the short side is divided by the bisector of the opposite angle.

3. A chord BC of a given circle is drawn, and a point A moves
on the longer arc BC, Draw the triangle ABC, and find the locus
of the centre of a circle inscribed in this triangle.

4. Three equal circular plates are so placed
that each touches the other two, and a string
is tied tightly around them. If the length
of the string is 10 feet, find the radius of the
circles correct to three significant fig\ires.

5. Let be the centre of a circle and P
any point outside. With P as centre and

radius PO draw the arc of a circle, cutting the given circle at A and
B, With A and B as centres and AB as radius draw arcs inter-
secting in Q. Prove:

(a) that the points 0, Q, P are in a straight line; and

(6) that OPX 00 = (0.1)2.

Siiggestion. — ^Join A and B with 0, P, and Q, and join Q with
OandP.

SEPTEMBER, 1907

(One Hour and a Half)

The University Provides a Syllabus

1. Prove that in an isosceles triangle the angles opposite the
equal sides are equal.

On BC, the longest side of a triangle ABC, points B' and C are
taken so that BB'=-.BA and CC'=^CA. Show that the angle
B'AC equals half the s\im of the angles B and C.

2. Prove that the area of a triangle is one-half the product of
the base and the altitude.

Show that if a point move about within a regular polygon, the
sum of the perpendiculars let fall upon the sides (or the sides
produced) will be condtant

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COLLEGE ENTRANCE EXAMINATIONS 371

3. A rod 8 feet long is free to move within a rectangle 8 feet
long and 6 feet wide. Describe accurately the boundary of
the region within which the middle point of the rod will always
be found.

4. A circle is described upon one side of an equilateral
triangle as diameter. Compute the area of the part of the
triangle which lies outside the circle, correct to one per cent, of its
value.

5. Two circles intersect at right angles. The radius of one of
them is of length a, and its centre is the point 0. Show that if
any line be drawn through cutting the second circle in the points
P and P', then

OPxOP'=^a^.

1908

(One Hour and a Half)
The University Provides a Syllabus

1. Prove that if in a quadrilateral a pair of opposite sides be
equal and parallel, the figure is a parallelogram.

In a certain quadrilateral, one diagonal, and a line connecting
the middle points of a pair of opposite sides bisect each other.
Prove that the quadrilateral is a parallelogram.

2. Prove that in any circle equal chords subtend equal arcs.

Show that the bisector of an angle of a triangle meets the per-
pendicular bisector of the opposite side on the circimiference of the
circumscribed circle.

3. The radii of two circles are 1 inch, and V 3 inches respectively,
and the distance between their centres is 2 inches. Compute their
common area to three significant figures.

4. Determine a point P without a given circle so that the sum of
the lengths of the tangents from P to the circle shall be equal to
the distance from P to the farthest point of the circle.

5. The image of a point in a mirror is, apparently, as far behind
the mirror as the point itself is in front. If a mirror revolve about
a v^tical axis, what will be the locus of the apparent image of a
fixed point one foot from the axis?

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372 PLANE GEOMETRY

1909

(Two Hours)
The University Provides a SyUatras

1. Show that if a triangle be equilateral, it is also equiangular.
Is this theorem true in the case of a quadrilateral? Give your

reason.

2. Prove that the square of the hypotenuse of a right triangle is
equal to the sum of the squares of the other two sides.

Deduce from this a proof that of two oblique lines drawn from
the same point of a perpendicular and cutting off unequal distances
from the foot of the perpendicular, the more remote is the greater.

3. Prove that if the products of the segments into which the
diagonals of a quadrilateral divide one another are equal, a circle
may be circumscribed about the quadrilateral.

4. A square 10 inches on a side is changed into a regular octagon
by cutting off the comers. Find the area of this octagon.

6. A wheel 40 inches in diameter has a flat place 5 inches long
on the rim. Describe carefully the locus of the centre as the wheel
rolls along the level.

1910

(Two Hours)
The University Provides a Syllabus

1. Prove that if the sides of one angle be perpendicular respectively
to those of another, the angles are either equal or supplementary.

2. Show how to inscribe a circle in a given triangle.

How many circles can be drawn to touch three given lines? Are
there any positions of these lines for which the number is less?

3. Define "incommensurable magnitudes." Give a careful proof
of some theorem where such magnitudes occur.

4. ABCD are the vertices in order of a quadrilateral which is
circumscribed to a circle whose centre is 0. Prove that /^AOB
and /^COD are supplementary.

5. Two radii of a circle OA and OB make a right angle. A second
circle is described upon AB as diameter. Prove that the area of the
crescent-shaped region outside of the first circle, but inside of the
second, is equal to that of the triangle AOB. (Hippocrates, fifth
century b.c.)

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COLLEGE ENTRANCE EXAMINATIONS 373

1915

(Two Hours)
The University Provides a Syllabus

1. Two straight lines are cut by a third, and the alternate
interior angles are equal. Prove that the two straight lines are
parallel.

Prove that the bisector of an exterior angle at the vertex of an
isosceles triangle is parallel to the base. !

2. Prove that an angle formed by a tangent to a circle and a
chord through the point of contact is measured by one-half of
the intercepted arc: !

Tangents are drawn at the extremities of a chord of a circle, and |
the perpendicular bisector of the chord is drawn. The point in
which this last line cuts the minor arc is connected with one end
of the chord. Show that this connecting line bisects the angle
between the chord and the tangent.

3. Two unequal circles are tangent to each other externally, and
their common tangent at the point of contact meets one of the other
common tangents at the point P. Lines are drawn from P to the
centres of the two circles. Prove that these
two lines are perpendicular.

4. Three equal circular plates of radius r
are so placed that each is tangent to the other
two. Find the length of the shortest string
that can be tied around the three circles, and
the area enclosed by this string.

5. Two circles ABC and ADE touch internally at i4, and through
A straight lines, ABD and ACE, are drawn to cut the circles.
Prove that ABDE^AD^BC.

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CHAPTER X

SUGGESTIONS

The aim of this chapter is to suggest appropriate material for
reading and discussion outside of class, and in mathematics clubs.
To this end certain books and topics are mentioned, but it is
understood that these lists are intended to be suggestive rather
than comprehensive, and include only subject matter adapted to
pupils at the stage of development represented by this text.

After topics in the second list, parenthetical reference is fre-
quently made to books from the first list, in which at least some-
thing about the topic may be found, and in the third list definite
references for a few topics are given as a guide to the beginner.
The practice of seeking such references without help is particularly
valuable as a training for college work; hence, this list includes
only a few topics, and is to serve for direction at the outset only.

A. LIST OF TOPICS SUITABLE FOR STUDENTS' DISCUSSION

GENERAL

Mathematical Games. (Ball.)

Card Tricks.

Problems on a Chess Board.
History and Elementary Idea of Calculus. (White.)
Some Applications of Mathematics to Astronomy. (Ball.)
Parcel Post Problems. (School Science and Mathematics.)
Mathematics of Common Things.

Optical Illusions. (Smith's Teaching of (Geometry.)
Navigation. (Richards.)
Instruments.

Historic. (School Science and Mathematics.)

Astrolabe, squadra, carpenter's level, baculus mensorus, sundial, etc.
Pantograph. (Philhps and Fisher's Elements of Geometry.)

Map-making.
Planimeter. (Scientific American.)
Symmetry. (White; Dobbs, Synmietry, Chapt. VI.)

In nature.

Design.
Maxima and Minima. (Texts on Plane Geometry.)

374

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SUGGESTIONS 375

Linkages. (White.)

Fourth Dimension. (Flatland.)

Angles of a General Polygon. (School Science and Mathematics.)

Study of Rupert's ** Famous Problems in Geometry." (Heath's Monographs.)

Fallacies. (Ball.)

Odd or Purely Mechanical Constructions. (Becker.)

ARITHMETIC

History of. (Ball, Fink, Cajori, Brooks.)

Primitive Numeration.

Systems of Notation.

P^blems in Number Systems Other Than Our Own.

Special Consideration of Duodecimal System.

Kinds of Number.

Fundamental Operations.

Russian Peasant Multiphcation.
Calculation. (Langley's Computation. Longmans, Green & Co.)

Short Cuts. (Jones; White.)

Calculating Machines.

Shde Rule. (The Teaching of Mathematics, by Schultze, Macmillan.)

Approximation. (White.)

Repeating Decimals. (White.)
Peculiarities and PossibiUties of Our Electoral System.
Number Curiosities.

9 and its properties. (White.)

Tricks. (Dudeney; Jones; White.)
Alligation. (The Mathematics Teacher.)

ALGEBRAIC

History of. (Ball, Fink, Cajori, etc.)

Stages — Rhetorical, Syncopated and SymboUc.

Symbols.
Fallacies. (White; Jones.)
Choice and Chance. (Text on higher algebra.)

GEOMETRIC

History. (Smith's Teaching of Geometry.)

More detailed study of some period other than is given in text.
Famous Problems. (Heath's Monographs.)
Trisection of an Angle.

Instruments.
Squaring of the Circle.
Pythagorean Proposition.

Various proofs and applications.

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376 PLANE GEOMETRY

B. TOPICS WITH DEFINITE REFERENCES

GEOMETRIC FALLACIES

Ri^t angle is obtuse. Ball's Recreations, p. 40.

Part of a sect equals' the sect. Ball's Recreations, p. 41.

Every triangle is isosceles. Ball's Recreations, p. 42.,

Misoellaneous fallacies. Ball's Recreations, pp. 43-46.

Hypotenuse equals sum of legs of triangle. Canterbury Puzzles, p. 26.

1«0. Wrinkles, p. 93.

Two perpendiculars from a point to a line. Wrinkles, p. 05.

NUMBER CURIOSITIES

Mystio Properties of Numbers.

Repeating products. White, p. 11 et seq.

Nine. White, p. 25.
Fallacies. Ball, p. 23 et acq.
Magic Squares. See Andrews's book and his references.

PYTHAGOREAN PROPOSITION

Various proofs.

Rupert's "Famous Geometrical Theorems and Problems."
Applications (curious).

Fly Problem. Jones p. 2, No. 10.

Historical Problems. See Beman and Smith's Academic Algebra, p. 153.

C. LIST OF BOOKS SUITABLE FOR STUDENTS' READING.

HISTORY

' Ajaman: Greek History from Thales to Euclid. Longmans.
Ball: History of Mathematics. Macmillan.

Primer of the History of Mathematics. Macmillan.
Brooks: Philosophy of Arithmetic. Normal Publishing Co.
Cajori: History of Mathematics. Macmillan.

History of Elementary Mathematics. Macmillan.
Conant: Nxunber Concept. Macmillan.
Fine: Number Systems of Algebra. Heath.
Fink: Brief History of Mathematics. Open Court Pub. Co.
Frankland: The Story of Euclid. Wessel and Co.
Gow: History of Greek Mathematics. Cambridge Press.
Klein: Famous Problems of Elementary Geometry. Ginn.
Heath: Diophantus of Alexandria. Cambridge Pl^ss.
Manning: Non-Euclidean Geometry. Ginn.
Miller, G. A.: Historical Introduction to Mathematical Literature.

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SUGGESTIONS 377

SiOTH, D. £. : Rara Arithmetica. Ginn.

Teaching of Geometry. Ginn.

Teaching of Elementary Mathematics. Macmillan.
Smith and Kabpinski: The Hindu-Arabic Numerals. Ginn.

History of Japanese Mathematics. Open Court Pub. Co.
Boybb: Histoire des Math^matiques. Paris.
Cai^tob: Vorlesungen Uber Geschichte der Mathematik. Leipzig.
Portraits of Mathematicians by Prof. D. E. Smith.

Portfolios. Open Court Pub. Co.
Lantern Slides by Prof. D. E. Smith, of Teachers' College, Columbia University.

RECREATIONS

Abbott: Flatland. Little, Brown.

Andbews: Magic Squares and Cubes. Open Court Pub. Co.
Anonymous: Flatland. Boston.
Ball: Mathematical Recreations. Macmillan.
Cavendish: Recreations with Magic Squares. London.
Dudenby: The Canterbury Puzzles. Dutton & Co.
Habpson: Paradoxes of Nature and Science. Dutton A Co.
Hatton: Recreations in Mathematics. London.
Hill: Geometry and Faith. Lee and Shepard.
Jones, S. L: Mathematical Wrinkles. Author, Gunther, Texas.
Kempe: How to Draw a Straight Line. Macmillan.
Latoon: On Common and Perfect Magic Squares. Cambridge.
de Mobgan: a Budget of Paradoxes. London.
Manning: Fourth Dimension Simply Explained. Munn.
Pebby: Spinning Tops. London.

Schofield: Another World. Swan, Sonnenschein; London.
Schubebt: Mathematical Recreations. Open Court Pub. Co.
Whife: Scrap Book of Elementary Mathematics. Open Court Pub. Co.
Ahbens: Mathematische Unterhaltungen und Spiele. Leipzig.
Lucas: Recreations Math^matiques. Paris.
L'Arithm^tique amusante. Paris.
Maupin: Opinions et curiosity touchant la math^matique. Paris.
Rebiebe: Math^matiques et Math^maticiens, Pense^s et Curiosity. Paris.

PRACTICAL

Bbeckenbidge, Mebsebeau and Moobe: Shop Problems in Mathematics.

Ginn.
Calvin, F. H.: Shop Calculations. McGraw Hill.
Castle: Manual of Practical Mathematics. Macmillan.
Cobb: Applied Mathematics. Ginn.
Cox: Manual of Slide Rule. Keuffel and Esser.
Dooley: Vocational Mathematics. Heath.
Hinds, Noble and Eldbeoge: How to Become Quick at Figures.

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878 PLANE GEOMETRY

Langlbt: Computation. Longmans.

Mabsh: Vocational Mathematics. Wiley and Sons.

Mitbrat: Practical Mathematics. Longmans.

Richabdb: Navigation and Nautical Astronomy. American Book Co.

Baxblbt: Practical Mathematics. Longmans.

GENERAL

Cabub: Foundations of Mathematics. Open Court Pub. Co.

Clifford: Common Sense of the Exact Sciences. Appleton.

Frankland: Theories of Parallelism. Cambridge Press.

Henrici: Congruent Figures. Longmans.

Lagrangb: Lectures on Elementary Mathematics. Open Court Pub. Co.

Row: Geometric Paper Fofding. Open Court Pub. Co.

Stkbs: Source Book crfjgroblems for Geometry. Allyn and Bacon.

VON H. Becker: Geometrisches Zeichnen. Sammlung Gdschen.

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INDEX

Abbreviations, xi
Abscissa, 149
Acute angle, 28

triangle, 37
Addition, in proportion, 107
Adjacent, 31
Ahmes, 5
Alternate, exterior, 53

interior, 53
Alternation, 107
Altitude, of parallelogram, 80

of rectangle, 78

of trapezoid, 82

of triangle, 81
Analysis,\297

.of problems, 319
Angle, acute, 28

adjacent, 31

bisection, 19

central, 161

central, of regular polygon, 181

complementary, 28

construction, 19

definition of, 27

exterior, of polygon, 60

inscribed, 174

measurement, 28, 31

obtuse, 28

of depression, 142

of elevation, 142

right, 28

straight, 28

subtended by a circle, 284

supplementary, 28

vertical, 32, 37
Angles, al^mate, 53

consecutive, 53

exterior, 53

interior, 53
Antecedent, 105

Antilogarithm, 95
Apollonius, problem, 358
Apothem, 186
Arc, 161

major, 161

minor, 161
Arch, Ogee, 170

Persian, 170
Archytas, 8
Area, 75

of circle, 180

of parallelogram, 81

of rectangle, 77

of trapezoid, 83

of triangle, 81
Axioms, 25, 49

of inequality, 227, 273

of variables, 187

Base, in logarithms, 86

of rectangle, 78

of trapezoid, 82

of triangle, 81
Bisection, of angle, 19

of line, 18
Bibliography, historical, 10
Boyle's Law, 109

Cam, 23
Cancellation, 74
Center, of circle, 160

of gravity, 72

of parallelograms, 69

of similitude, 119
Central angle, 161

of regular polygon, 181
Centroid, 359, 360
Ceradini's Method, 294
Characteristic, 92
Charles's Law, llOt

879

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380

INDEX

Chord, 162

common, 169
Chord of oontacti 281
Circle, definition of, 160

equation of, 155
Circumcenter, 307
Circumference, 160
Circumscribed circle, 180

n-gon, 180
Commensurable, 77
Common chord, 169

measure, how to find, 307
Complement, 28
Composition, 107
Concurrent, 2ff7
Congruence, definition of, 38
Consecutive, exterior, 53

interior, 53
Consequent, 105
Constant, 184
Contact, chord of, 281
Continuity, principle, 261
Contradictory, of the direct, 300
Converse, 56, 57

theorem, 155
Converses, law of, 300
Coordinate, 149
Coplanar, 51
Corollary, 30
Corresponding angles, 53
Cosine, 139
Curve, reversed, 169

Decagon, 63

construction of regular, 316
Definitions, 24
Denominator, 74
Depression, angle of, 142
Designs, 20, 21, 22
Diagonal, 62

scale, 127
Diameter, 160
Dimension of rectangle, 78
Direct method, 290

Direct theorem, 155

variation, 110
Discussion of solution, 311
Distance, 48

between parallels, 67
Dividing a sect, 117
Division, 108

external, 244

harmonic, 244

internal, 244
"Doubling the angle at the bow," 67

Elevation, angle of, 142
Elimination, 228
Equation, as locus, 152

of circle, 155
Equilateral triangle, 37
Escribed circle, 314
Euchd, 9
Eudoxus, 9
Excenter, 314
Exclusion, 228, 297
Exponents, fractional, 88

negative, 88

principles of, 87

zero, 88
Exterior angle, 63
External division, 244

tangent circles, 168
Extreme and mean ratio, 315
Extremes. 106

Foot of perpendicular to a line, 134
Formulas, summary, 336, et seq-
Fourth proportional, construction, 308
Fraction, 74

Geometry, derivation, 9
Golden Section, 315
Graph, 150

application of, 150

of equations, 151
Gravitation, Law of. 111, 112

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INDEX

381

Barmonic division, 244, 307

Heptagon, 63

Heron of Alexandria's formula, 98, 261

Hexagon, 63

Hexagram, Mystic, 288

Hippocrates' Theorem, 296

Homologous, definition, 42

Hypotenuse, 50

Hypothesis, 54

Hypsometer, 126

Incenter, 314

Inclined plane, law of, 109
Incommensurable, 77
Indirect method, 297
Inequality axioms, 227, 273
Initial side, 139
Inscribed angle, 174

circle, 180

n-gon, 180
Intercept, 162
Interest, compound, 98
Internal division, 244

tangent circle, 168
Interpolation, 94
Intersection of loci, 312
Inverse variation, 110
Inversion, 108
Isosceles triangle, 37

Joint variation. 111

Law of converses, 300
Limits, 184

inferior, 184

postulate of, 184

superior, 184
Line, 14

broken, 228

curved, 15

of centers, 168

straight, 15
Lines, parallel, 51
Locus, 152, 153; 269, 312

Logarithm, 86

tables, 103
Logarithms, common, 91

historical note, 90

theorems, 93

Major arc, 161

Mantissa, 92

Mean and extreme ratio, 315

proportion, 106

proportional, 106
construction, 315
Means, 106
Measure, 173
Median, 230 .

of trapezoid, 244
Method, direct, 297

indirect, 297

synthetic, 297
Methods of proof, summary, 338,

et aeq.
Minor arc, 161

Nonagon, 63
Numerator, 74

Obtuse angle, 28

triangle, 37
Octagon, 63
Opposite theorem, 155
Ordinate, 149
Origin, 149
Orthocenter, 359, 360

Pantograph, 131
Parallel, 51

postulate, 52
Parallelogram, 67

altitude of, 80

base of, 80

of forces, 71

ruler, 68, 72
Parallels, construction, 55
TT, 188

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382

INDEX

T, historical note, 188
Pentagon, 63

construction of regular, 317
Pentagram, 234
Penumbra, 179

Perpendicular, construction, 17, 18
Plane, 16
Plato, 8
Plumb-level, 52
Point, 14
Polygon circumscribed about a circle,

180
Polygon, convex, 63

definition, 37

inscribed in a circle, 180

regular, 63

similar, construction, 308
Pons asinorum, 46
Postulate, 26, 27

of angles, 31

of limits, 184

of parallels, 52

superposition, 38
Postulates of perpendiculars, 48, 161
Projection, 133

orthogonal, 135

point upon a line, 134

sect upon a line, 134
Projector, 139
Proportion, 106
Proportional compasses or dividers,

129
Protractor, 29
Pythagoras, 7
Pythagorean badge, 234

niunbers, 12

theorem, 135

Quadrant, 35
Quadrilateral, 63

Radian, 191
Radius, 160
of regular polygon, 186

Ratio, 74

extreme and mean, 315
Ratio of similitude, 119
Rectangle, 69

altitude, 78

area, 80

base, 78

dimensions, 78
''Reductio ad absurdum,'' 300
Reduction to an absurdity, 297
Regular polygon, 63
Resultant, 70
Rhomboid, 238
Right angle, 28

Scalene triangle, 37
Secant, 173
Sect, 15
Sector, 192

compasses, 129
Sectors, similar, 291
Segment, 291
Segments, similar, 291
Sextant, 62
Sinular, 119

figures, 119

polygon, construction, 308

sectors, 291

segments, 291
Similarity of sets of points, 119
SimiUtude, center of, 119

ratio of, 119
Simpson's Rule, 84
Sine, 139
SUde rule, 98
SoUd, 14

Specific gravity, 109
Speculmn, 125
Squadra, 135
Square, carpenters', 175
"Square of," 239
"Square on," 239

optical, 62

root scale, 138

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INDEX

383

Stax polygon, 234
Straight angle, 28
Subtraction, in proportion, 108
Summary of formulas, 336 et aeq.
of methods of proof, 338 et aeq.
Superposition, postulate, 38
Supplement, 28
Surface, 14
Symbols, xi
Synthetic method, 297

Tangent, 139, 178

circles, 168
external, 168
internal, 168

common, construction, 314

construction, 313

line, 165
Terminal side, 139
Terms of fraction, 74

of proportion, 106
Thales, 7
Theodolite, 35
Theorem, 36
Third proportional, construction,

308
Transforming a polygon, 317
Transversal, 53
Trapezoid, altitude, 82

area, 82

bases, 82

Trapezoid, definition, 82

Egyptian formula for area, 11

median, 244
Trapezoidal rule, 84
Trefoil, 170, 171
Triangle, 37

acute, 37

altitude, 81

area, 81

ba^, 81

equilateral, 37

isosceles, 37

medians of, 230

obtuse, 37

right, 37 •

scalene, 37
Trigonometric ratios, 139

Umbra, 179

Variable, 184

decreasing, 184

increasing, 184
Variation, 110

direct, 110

inverse, 110

joint, 111
Vertex of angle, 27

of polygon, 37
Vertical angle; 37

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