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Full text of "Plane geometry. I. Abridged and applied. II. College preparatory"

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PLANE 
GEOMETRY 

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




s^ \ VS<i'\^ 




GIFT or 




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PLANE GEOMETRY 



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Digitized by Google I 

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 


CO 


is similar to 


^ 


is congruent to 


» 


approaches as a limit 


OC 


varies as 


5L 


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 


II 


a parallel, or is parallel to 


J 


not a parallel, or is not parallel to 


ll< 


parallels 


± 


a perpendicular, or is perpendicu- 
lar to 


X. 


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 


;r 


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 

zi 

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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, 

3 

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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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6 



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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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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8 



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. 




i 



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 



13 



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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16 



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 



NT 



/ 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. 



^ 



M 



'^ 



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INTRODUCTION 



19 



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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20 



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 



21 





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 



29 




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 



31 



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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32 



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. 

(c) Draw two supplementary- 
adjacent angles. 

(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. 

(c) ^AOB+^BOC+^COD. 

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

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INTRODUCTION 



33 



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). 

(c) Look at this figure and state whether 

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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34 



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. 



B 




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. 




ZJ 



c 




E^ 



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



I 



L 



:z: 



X 



^ 



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 



35 



(c) What relation exists between a and j9? ' 

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

(c) What further relation do you notice that is based upon the 
relatons stated in (c) and (d)? 

(/) 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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36 



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 



cv 



/ 

/ 



/ 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? 

(c) Where did P fall and 
where did Q fall when you had 

^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 



39 



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 


7\ 


\'/ 


\ 


\ ^'^ /^N\ / 


7 


^ 


V 


IsVifi 13 /\ 

/\^^j /io\ 


6 




^ 




\ 


X 



* 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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40 



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. 





(c) W hat w ould happen if 
AiP and CiQ were produced? 

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

(e) What conclusion can 
you draw? 



?/ 



\« 



X 



X 



h 



M. 



(/) 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 



41 





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. 

(c) Are there any congruent triangles? If so can you tell why 
they are congruent? 

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

(c) Test to see if these facts are all true when the base b is 
greater than the equal sides. 

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 



43 



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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b 






c 


/ 


/^^. 


•V^ 






V 




^, 


s.$. 




/ 






"^^ 




/ 






V 


^ 


:/.-— 


_-h£. 


^. 




.-^.^^ 


\ 






^. 


^^-^ ' 


\ 






^•*'' 




\ 










a.N 




^••' 


-< 





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? 

(c) Use this fact to prove 
^XYZ^^XPZ. 

(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 



46 







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 

4 



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? 

(c) Is the direction of a second side fixed? Why? 

(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 



52 



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? 







E 




A 




X 


B 


C 


.__ 


D 






r-- 


^^-K 






\f 





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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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54 



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. 

(c) a and 6 with c 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. 

(c) If it rains, the ground is wet. 

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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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 

(c) What is the sum of the angles of the triangles thus formed? 

(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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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 


3 






4 






5 






6 






7 






8 






9 






10 







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? 

(c) What combinations of regular polygons have you seen 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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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- 



y^ 



• 



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 



69 



(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. 

(c) Could you read the resultant to any degree of accuracy by any 
other method? 

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)» 

(c) (o-x)^ _ / xV 
a2 "" V^ aj 

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 



79 



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. 





B. 



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 



81 




H 



M 




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. 

(c) Will the altitudes of a triangle always fall within the triangle? 

(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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82 



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 



88 




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 



I 



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^ 






X 



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 


B 


C 


D 


E 


F 


(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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84 



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) . 

(c) By Simpson's Rule find the area of the section of the vessel 
in Ex. 270 (d). 

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 

(c) logiil331=3 (J) logio.l=-l (i) logio500= 2.6990 



86 



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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? 


1 


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. 



E 

ax' 



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90 



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 


1 


0.00 


64 


1.50 


2 


0.25 


256 


2.00 


4 


0.60 


1024 


2.50 


8 


0.75 


4096 


3.00 


16 


1.00 


65536, etc. 


4.00 


32 


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? 

(c) log 4.37 is .64048? 

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.) 

(c) As a result of (a) and (b) only a table of mantissas need be given. 

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. 

(c) The logarithm of the n^ power of a number equals n times the 
logarithm of that number. 

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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94 



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 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


43 


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 



95 



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*. 



N 





1 


2 


3 


4 


6 


6 


7 


8 


9 


43 


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 

(c) Compuiation by logarithms. 

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 

A 

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? 

(c) 5% interest compounded 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? 

(c) 5% 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. 

(c) Of $1500 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? 

(c) A knot is the length of one degree measured along the equator. 
How many miles in a knot? 

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 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 


0414 


0453. 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


22 


3424 


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3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


35 


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5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


38 


5798 


5809 


5821 


5832 


5843 


5855 


6866 


5877 


5888 


5899 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


49 


6902 


6911 


6920 


6928 


•6937 


6946 


6955 


6964 


6972 


6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


61 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 



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104 



PLANE GEOMETRY 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 


7482 


7490 


7497 


7505 


7613 


7520 


7528 


7536 


7543 


7551 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


69 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


05 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 


8513 


8519 


8525 


8531 


8537 


8643 


8649 


8555 


8561 


8567 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 J 


8745 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 




9489 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9860 


9864 


9859 


9863 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 



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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) ^ « 


d 


or ^ = 


d^c 
d 






or (2) « « 


c 


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.? 

(c) The specific gravity of sea water is 1 .024. What is the weight 
of 5 gallons? 

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 

(c) How long will a body fall before acquiring a velocity of 
520 feet per second? 

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.? 

(c) How far will a body fall in 12 seconds? 

(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? 

(c) What is the pressure on the same house when the wind is 
blowing 60 miles per 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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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. 


V 


D 


(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 



EX 



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:='- 



2 



-, 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? 



P 




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: 




AD 


DB 


AE 


EC 


AB 


AC 


24 


28 


18 








14 


56 




42 








112 


12 


418 






27 




18 


342 







433. Divide a sect 11 units long into parts proportional to 3, 
5, 7, and 9. 

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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? 



We 

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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SIMILARITY 



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. 




-~-~^^ 



(c) Circles. (d) General Curvilinear 

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 

On 



. 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. 



In 





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. 




y 



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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124 



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. 

(c) Under what conditions, if ever, will the image be 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 




y 



^^ 



^--1 ^ 



E^ 



J LB 



A 



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 

(c) Can you find a sect equal to f 

(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 















~~ 


■~~ 


"^ 


— 




~~ 


~~ 


■^ 


















































I 


^ 
























TT* 


/ 


IP 
























/ 




V 






















/ 


f 


























/ 


























/ 
























•h 


y 










Hi 


















^ 










M 
















/ 
























A 


/ 














Tf 


















































_j 






-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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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 

(c) How may such an instrument be used to divide a sect? 

(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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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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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 



(c) Make a pantograph. A crude one can be made of stiff card- 
board and brass brads. 

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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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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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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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: 



f 



F 



-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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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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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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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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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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PLANE GEOMETRY 




IniJtial Side 




JP 

C Thus (1) the sine of angle ABC is=^- 

BP 

■DJ 

(2) The cosine of angle ABC is ==• 

BP 

JP 

-^ (3) The tangent of angle ABC is =• 

BJ 

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 






300 


Rectangle 


Base 18 


300 






300 


Rhombus 


Angle 60° 


300 






300 


Square 




300 






300 



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C 



^ 





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. 

(c) Which of these results is more accurate? Why? 

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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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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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 


I. 




Deg. 


Sine 


Cosine 


Tangent 


Deg. 


Sine 


Cosine 


Tangent 


Deg. 


Sine 


Cosine 


Tangent 


1 


.0175 


.9998 


.0176 


31 


.5150 


.8572 


.6009 


61 


.8746 


.4848 


1.8040 


2 


.0349 


.9994 


.0349 


32 


.5299 


.8480 


.6249 


62 


.8829 


.4695 


1.8807 


3 


.0523 


.9986 


.0524 


33 


.5446 


.8387 


.6494 


63 


.8910 


.4540 


1.9626 


4 


.0698 


.9976 


.0699 


34 


.5592 


.8290 


.6745 


64 


.8988 


.4384 


2.0503 


5 


.0872 


.9962 


.0875 


35 


.5736 


.8192 


.7002 


65 


.9063 


.4226 


2.1445 


6 


.1045 


.9945 


.1051 


36 


.6878 


.8090 


.7265 


66 


.9135 


.4067 


2-2460 


7 


.1219 


.9925 


.1228 


37 


.6018 


.7986 


.7536 


67 


.9205 


.3907 


2.3559 


8 


.1392 


.9903 


.1405 


38 


.6157 


.7880 


.7813 


68 


.9272 


.3746 


2.4751 


9 


.1564 


.9877 


.1584 


39 


.6293 


.7771 


.8098 


69 


.9336 


.3584 


2.6051 


10 


.1736 


.9848 


.1763 


40 


.6428 


.7660 


.8391 


70 


.9397 


.3420 


2.7475 


11 


.1908 


.9816 


.1944 


41 


.6561 


.7547 


.8693 


71 


.9455 


.3256 


2.9042 


12 


.2079 


.9781 


.2126 


42 


.6691 


.7431 


.9004 


72 


.9511 


.3090 


3.0777 


13 


.2250 


.9744 


.2309 


43 


.6820 


.7314 


.9325 


73 


.9563 


.2924 


3.2709 


14 


.2419 


.9703 


.2493 


44 


.6947 


.7193 


.9657 


74 


.9613 


.2756 


3.4874 


15 


.2588 


.9659 


.2679 


45 


.7071 


.7071 


1.0000 


75 


.9659 


.2588 


3.7321 


16 


.2756 


.9613 


.2867 


46 


.7193 


.6947 


1.0355 


76 


.9703 


.2419 


4.0108 


17 


.2924 


.9563 


.3057 


47 


.7314 


.6820 


1.0724 


77 


.9744 


.2250 


4.3315 


18 


.3090 


.9511 


.3249 


48 


.7431 


.6691 


1.1106 


78 


.9781 


.2079 


4.7046 


19 


.3256 


.9455 


.3443 


49 


.7547 


.6561 


1.1504 


79 


.9816 


.1908 


5.1446 


20 


.3420 


.9397 


.3640 


50 


.7660 


.6428 


1.1918 


80 


.9848 


.1736 


5.6713 


21 


.3584 


.9336 


.3839 


51 


.7771 


.6293 


1.2349 


81 


.9877 


.1564 


6.3138 


22 


.3746 


.9272 


.4040 


52 


.7880 


.6157 


1.2799 


82 


.9903 


.1392 


7.1154 


23 


.3907 


.9205 


.4245 


53 


.7986 


.6018 


1.3270 


83 


.9925 


.1219 


8.1443 


24 


.4067 


.9135 


.4452 


54 


.8090 


.5878 


1.3764 


84 


.9945 


.1045 


9.5144 


25 


.4226 


.9063 


.4663 


55 


.8192 


.5736 


1.4281 


85 


.9962 


.0872 


11.4301 


26 


.4384 


.8988 


.4877 


56 


.8290 


.5592 


1.4826 


86 


.9976 


.0698 


14.3006 


27 


.4540 


.8910 


.5095 


57 


.8387 


.5446 


1.5399 


87 


.9986 


.0523 


19.0811 


28 


.4695 


.8829 


.5317 


58 


.8480 


.5299 


1.6003 


88 


.9994 


.0349 


28.6363 


29 


.4848 


.8746 


.5543 


59 


.8572 


.5150 


1.6643 


89 


.9998 


.0175 


57.2900 


30 


.6000 


.8660 


.5774 


60 


.8660 


.5000 


1.7321 


90 


1.0000 


.0000 


00 



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. 

10 




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146 



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. 

(c) Find the area of the field. 

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? 

(c) —6 imits from the x-axis? (d) 17 units from the y-axis? 

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? 

(c) x+y^lO? (/) x=-3y1 (i) x^+y^=25? 

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. 

(c) Three times as far from the i/-axis as from the x-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. 

(c) If two lines meet at right angles they are perpendicular to 
each other. 

(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? 

(c) Of which of them are the converse facts false — or at least 
not necessarily 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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p 



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? 

(c) Where would all points that are both 2' from the floor and 
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. 



J 




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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THE CIRCLE 



163 



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. 


















/ 

/ 


A 


7 


Y 


\ 



V- — ^^' 


v 




\l \t 


A 




/ ^ 


A 


-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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164 



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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THE CIRCLE 



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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166 




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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168 



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? 

(c) Is this fact true of the centers of two cog-wheels when they 

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. 

(c) Construct the equilateral arches ADE and DBF and the 
curved triangle EFC. 

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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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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(c) Prove that C, ff, and B are coUinear. 

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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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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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? 

B 




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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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 








r 


// 
1/ 


Y^-J5^ 


\, 


\ \ \ / 


\\ 


1 


\ N ' / / 


\ 


\\ 


/ / 1 \ \ 


_> 


\\ 


A5iJ^A\ 


// 


\\ 


1 1 


A 


SC 1 ^ 


k 


/h 


^^st>>^ ^^^ 


\ 


/ 


6 r"-^^^I^--<-^ 


^ 



®®f^ 


i^;^ 






v$7^ 


'TTT^ 


M^ 


(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: 



41 


AE 


Bb 


EB 


AND 


35^ 


40° 




80° 




48° 


50° 






216° 


40^* 




60° 


60° 




60° 




54° 




190° 






45° 


90° 
108° 


180° 




34° 


164° 




12 



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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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PLANE GEOMETRY 




(c) What may be said of the umbra if the luminous body and 

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. 



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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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. 

(c) The length of a side of the inscribed regular 16-sided polygon. 

(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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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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Proof 
Inscribe in each circle a reg- 
ular polygon of n sides, and let 
p and pi be their perimeters. 



(1) Then !^ . 
Pi 


T 

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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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 



D 



E 



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.? 

(c) In which case is the increase proportionally larger? 

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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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 






300 


Circle 


300 






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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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. 






13 



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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". 





1 




21ft. 












1 




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PLANE GEOMETRY 




H 



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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THE CIRCLE 199 

(o) Are all equilateral polygons equiangular? 
(6) Are the diagonals of a parallelograni equal? 

(c) A diagonal of a parallelogram divides the figure into two 
congruent triangles. Is this proposition conversely true? 

(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? 

(c) Construct a quadrilateral 
similar to ilBCZ), but 60% greater 
in area. 

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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THE CIRCLE 201 

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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202 



PLANE GEOMETRY 



762. Fill out the blank spaces in the table by referring to the 
diagram. (Fig. 1.) 



43 


41 


i^4 


45 


Sa 


5i 


Xnd 


A^ 


Se 


AG 


FH 


^1 


35° 




41 


40° 








80** 




30^ 




43 




43 


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: 


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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Fig. 2. 



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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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THE CIRCLE 



209 



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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PLANE GEOMETRY 




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? 

(c) Is arc MDN tangent to arc AM and arc BN at M and N 
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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THE CIRCLE 



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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PLANE GEOMETRY 



111! 




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REVIEW OF FIRST STUDY 



221 



f 



a 1 2 1^- 

a a 5 '"• 

S I is 
^ ? ? 



A 



¥: 



a ^ 

+ 



i 5. 

•^ Hi 



rlfl 

-I- 

Sll — I 
a 



lO 



.■0 



I 

I 



J 
d 



H 



&- 



e 

L^D 



'I 

Si 






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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 



.s 






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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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. 

I 

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. 

I 



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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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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 

2 

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? 

(c) the sum of the interior angles exceeds the simi of the exterior 
angles by 4 rt. ^•? 3 st. <«? 9 st. ^«? 

(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? 

(c) How, if at all, does the 

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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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? 




c: 



/ 

/ 
/ 


\ 


/ 
/ 


' Vo 


/ 

/ 
/ 
/ 

f 


^\ 



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. 

(c) Area 756, sides in the ratio -J. 

(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. 

(c) If a is increased by m, and h diminished by n. 

(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 


CP 


AP 




whence 


AP- 


-CP 


AP 


AZ 


-Vz- 


AZ 


or 


AC 


AP 

= — 





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 

8 

— ^<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. 

(c) a, and the ratios of a to 6, and & 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) 



2o 

I $- -\/«(«— a){s— 6)(s--c) 
2o ^ 

a 



•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, 

(c) The side is a, 

(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°; 

(c) Bases &i and In, and legs each I. 

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; 

(c) Its diagonal d and the ratio of its length to its width r. 

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 

QT 

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? 

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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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R) 



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 

(c) Tangent at a given point to a given circle. VII. 

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; 

(c) Be tangent to a given line and 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. 

(c) Find the angles between the opposite sides of the inscribed 
quadrilateral produced to intersect. 

(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 

_2 

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. 



19 



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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. 


(c) 5.9 in. (e) 2| ft. 


(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 


(c) 27rr (e) 188.496 in. 


(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°. 

(c) In a regular inscribed hexagon e^Vy as^VS, As 120°, 
C-60°. 

( 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 


a 


15. in,n»3 


An 


2. r,in 


itn 


16. f, n=3 


An 


3. r, tin 


in 


17. in,n-6 


An 


4. r, in 


en 


18. f,n=6 


An 


5. r, n-3 


in, a 


19. in, n = 12 


An 


6. r, n=6 


in, a 


20. f,n =20 


An 


7. f,n = 12 


in, a 


21. in, n = 12 


An 


8. r, n-4 


in, a 


22. f,n = 12 


An 


9. r, n«8 


in, a 


23. in,n»4 


An 


10. r, n = 10 


in, a 


24, f, n=4 


An 


11. r,n=5 


in, a 


25. in, n=8 


An 


12. int n, a 


An 


26. f, n=8 


An 


13. int n, r 


An 


27. f, n=5 


An 


14. tn, n, r 


Ajn 


28. r,n = 10 


An 



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.DiǤ(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, 

(c) drawing a line perpendicular 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. 



X 



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. 



^ 



p. 



^^Xi 



R 



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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^^ 



I 

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- / 



a 

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. 



\ 



\?/ 



/ 



A, 

/ I ia 

I 

/ 

/ 



Q 



s 

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^=^ 






'^^ 



/E 



^A 



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 




y 



\ 




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 



Ri 



h. Un 



C. Cn 



^y/8(8-a){s-'b){s-c) 

sV2r^-rV4r^-in^. 
_ 2rin 



d. Cs27rr or C^tD. 



e. irs3.14159 



Cr 



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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 



2 



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, 

(c) three sides respectively equal, 
or, that they are parts of 

(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, 

(c) sects of a line cut oflf from the foot of a perpendicular to 

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, 

(c) vertical 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, 

(c) one side of a triangle and a sect cutting the other two 

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, 

(c) one of them is perpendicular to a line parallel to the other, 

(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, 

(c) sides of a triangle opposite unequal angles, 

(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, 

(c) the angles opposite the third sides of two triangles which 

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 

(c) two pairs of sides proportional and the included angles 

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, 

(c) sects of transversals cut off by parallels, 

(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, 

(c) factors of one product are the tangent and the factors of 

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. 

(c) Any diagonal which is a diameter divides the hexagon into 
two isosceles trapezoids. 

(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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382 



PLANE GEOMETRY 



1607. Construct x=" 



ab 



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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SUMMARIES AND APPLICATIONS 



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. 





23 



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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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SUMMARIES AND APPLICATIONS 



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. 

(c) Tangent (internally and externally) 

(1) unequal, (2) equal. 




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356 



PLANE GEOMETRY 









\ 




d'^e 


fi 








? 


\ 




A 


fi 


> 



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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368 



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 — 

n 

^ 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. 


1 


lA 


2 


Perp. from 


3 


34 


4,5,6 


40,50,60 




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360 



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. 

(c) How many recitations per weelc. 

(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 

Digitized by VjOOQIC 



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 



Digitized by VjOOQIC 



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 



Digitized by VjOOQIC 



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 



Digitized by VjOOQIC 



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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Digitized by VjOOQIC 



Digitized by VjOOQIC 



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