iiir'^ I SYKES C©MStOCK GIFT OF Publisher EDUCATION DEPT. PLANE GEOMETRY PLANE GEOMETRY By MABEL SYKES Instructor in Mathematics, Bowen High School. Chicago Author of "A Source Book of Problems for Geometry" and CLARENCE E. COMSTOCK Professor of Mathematics, Bradley Polytechnic Institute RAND M9NALLY & COMPANY CHICAGO NEW YORK 3^ c^ Copyright, 1918, by { ,♦ | r Rand M?Nally St Company Edition of 1922 Kf c-22 THE CONTENTS The Preface ix Chapter I. Introductory p^^g Points and Straight Lines 1 Circles 5 Angles 6 Summary and Supplementary Exercises 15 Chapter II. Congruent Triangles Introductory Definitions 20 Tests for Equal Angles and Equal Segments 21 Application of Congruent Triangles to Constructions .... 30 Nature of Theorems and Proofs 35 Miscellaneous Theorems and Exercises 37 Chapter III. Parallels, Perpendiculars, Angles, Angle-Sums Introductory 48 Parallels . 50 Angles Made by Parallels and Transversals 54 Angles in Triangles 59 Angles in Polygons 63 Miscellaneous Theorems 66 Supplementary Exercises 69 Chapter IV. Quadrilaterals Symmetry 75 Parallelograms 79 Special Quadrilaterals 84 Parallels and Segments on Transversals 88 Supplementary Exercises 94 Chapter V. Inequalities Assumptions for Combining Inequalities 99 Fundamental Tests of Inequality 100 Tests for Unequal Sides and Angles in One Triangle .... 102 Tests for Unequal Sides and Angles in Two Triangles . . . 105 Supplementary Exercises 106 V 571779 vi THE CONTENTS Chapter VI. Circles and Related Lines Introductory 108 Related Arcs, Chords, and Central Angles 110 Chords in General Ill Tangents 114 Two Circles and Related Lines 117 Supplementary Exercises 120 Chapter VII. Circles and Related Angles Relation between Central Angles and Their Arcs 123 Relation between Inscribed Angles and Their Arcs . . . . 126 Relation between Angles Formed by Tangents and Chords and Their Arcs 135 Summary and vSupplementary Exercises 137 Chapter VIII. Loci General Considerations 143 Loci of Points 146 Determination of Points by the Intersection of Loci .... 152 Loci of Centers of Circles 153 Supplementary Exercises 155 Chapter IX. Ratio and Proportion Measurement of Segments 161 Ratios 163 Theory of Proportion 165 Ratios of Segments Made by Parallels ........ 168 Similar Triangles : 176 Important Special Cases 182 Applications of Equal Ratios 191 Summary and Supplementary Exercises 197 Chapter X. Area and Equivalence Introductory 208 Measurement of Polygons 211 Equivalent Polygons 220 Summary and Supplementary Exercises 228 Chapter XL Similarity Introductory 240 Tests for Similar Polygons 240 Properties of Similar Polygons 244 Summary and Supplementary Exercises 248 THE CONTENTS vii Chapter XII. Regular Polygons Definition 252 Construction of Regular Polygons 252 Properties of Regular Polygons 259 Similar Regular Polygons 262 Summary and Supplementary Exercises 264 Chapter XIII. Measurement of the Circle The Circumference of the Circle 269 Areas of Circles, Sectors, and Segments 275 Ratios and Circles 276 Summary and Supplementary Exercises 277 Chapter XIV. Maxima and Minima Introductory 284 Triangles 284 Polygons in General 287 Regular Polygons 290 Notes on Arithmetic and Algebra 292 Tables 298 Outline Summary 301 Index 309 THE PREFACE This book is written with the firm conviction that it is possible to give to high-school young people a more sys- tematic training in the science of geometry than is furnished by any textbook on the market to-day. In this connection the two main features of the book should be noted : 1. The analytical method of attack is employed throughout. Analyses of proofs serve several purposes. When it has been found by actual classroom experience that any particu- lar proof is so difficult that the pupil cannot reasonably be expected to think it out for himself, the analysis gives him at the outset the gist of the argument, calls his attention to the method of proof employed, and gives him some idea of how the proof may have been originally invented. In such cases not only is the analysis given, but as much of the proof as experience has found necessary. In this con- nection the treatment of the proofs of Theorems 3, 4, 9, 14, 64, 77, 113, and 120 may be noted and compared with the usual treatment of these same proofs. More important, however, is the fact that the pupil can, with proper training, invent many of his own proofs. While nothing can do away entirely with the element of inspiration in getting originals, the fact remains that the method of analysis is the method by which every trained mind attacks difficulties. The teacher should see to it that the pupil is continually asking himself the necessary questions, that he sees clearly how each step follows from the preceding, and that the results are set down in the orderly form here em- ployed. The statements given in the analyses may seem formal, but clearness and definiteness are essential. No statement should be permitted which does not clearly indi- cate that the pupil sees all the essential steps of the argu- ment. The proof should in every case be obtained by working backward from the analysis. X THE PREFACE 2. 'J he work is so arranged as to throw emphasis on the important theorems and methods. Without emphasis effective analysis is impossible. It is said to be a fundamental characteristic of the mind that any lasting impression of a vast field requires distinctions in emphasis. Moreover, it is just here that much of our geometry teaching has failed. Attention is called to the following points in arrangement and presentation: a) The division into chapters is based on the important general ideas in geometry, such as congruence, loci, ratio, area, equivalence, and similarity. If the work is so pre- sented that each chapter is made to serve the special, definite purpose intended, many of the details in both analysis and proof for later theorems may be left to the pupil. Other- wise such details should be given. The purpose of chapter ii, for example, is to train pupils in the use of congruent tri- angles. If this purpose has been accomplished, the pupil can work out for himself the details for such congruent triangle work as that used in the proofs of Theorems 33, 36, and 37. Similarly, it is only on the assumption that chapter vii and §§209-212 have served their purpose that the analysis and proof for Theorems 103, 104, and 105 may be safely left to the pupil. b) The purpose of each chapter is carefully worked out in the order and grouping of the theorems and exercises. In the minds of the pupils the importance of a theorem depends solely upon the frequency with which it is used. To this end the dependence of the minor theorems upon the more fundamental ones is made evident, and all exercises given in connection with the various theorems are intended to illustrate the use of those special theorems. On pages 50-54, for example, Theorems 10, 11, and 12 depend directly on Theorem 9, while §67 consists entirely of exercises in which it is required to prove two lines parallel. The pur- pose of chapter ix is to train pupils in the use of equal ratios. THE PREFACE > xi The outline of this chapter and the arrangement and group- ing of the exercises in it should be noted. c) Theorems and problems whose interest is largely theo- retical and historical or which have no important place in the plan of the work are inserted in the supplementary exer- cises at the end of the chapters. These are marked with a dagger (f) and can be given when it is desired to extend the course or to prepare for the examinations of the College Entrance Examination Board. Inasmuch as the traditional order is the most convenient, it has been preserved except where emphasis required a change*. This accounts for the separation of the work on ratio, proportion, and similarity. This subject involves two points. Pupils should not only be familiar with the tests for similar figures and the properties of similar figures, but should be able to prove ratios equal. The work is therefore divided into two chapters. The chapter on similarity is given after the chapter on area and equivalence to permit of grouping together all theorems involving properties of similar figures. Attention is also called to certain minor features: The introduction is natural and interesting, concise and to the point. The nature of the exercises in this first chapter should be noted, as well as the preliminary use of paper folding in construction work. The algebraic form of statement for theorems and exer- cises is extensively used. See especially chapters ix, x, and xi. The formal theory of limits is omitted. The idea of a limit is presented informally by exercises, but proofs that are either lacking in rigor or are too difficult for the pupil are omitted entirely. The treatment of the measurement of the circle will be found satisfactory and comprehensive. There is a great variety of exercises with concrete setting. These include exercises taken from surveying, physics, architecture, and industrial design. For illustrations of xii THE PREFACE exercises in surveying, see pages 27, 46, 72, 191, and 219; for exercises from physics, see page 106; for exercises from architecture, see pages 157, 158, 204, 206, 265, and 280; for exercises from industrial design, see pages 69, 71, 97, 230, 231, and 235. Illustrations of similar problems will be found on pages 141, 142, 214, 266, and 279. The Notes on Arithmetic and Algebra and the Outline Summary preceding the Index will be found convenient for reference. Attention is also called to the unusually complete character of the index. Thanks are due Professor G. A. Miller of the University of Illinois for his criticisms of the historical notes. M. S. C. E. C. Chicago, Illinois May, iQi8 PLANE GEOMETRY CHAPTER I Introductory 1. Geometry treats of points, lines, surfaces, and solids. Plane geometry deals with lines and points on a plane surface. A plane surface is a flat surface, like a plain. In fact, years ago the name was "plain geometry," that is, the geometry of the plain. In the early part of the seventeenth century the spelling was changed from "plain geometry" to "plane geometry." The word geometry comes from two Greek words meaning * ' the earth ' ' and ' ' to measure. ' ' From earliest times the results of geometry have been used for practical purposes, such as building and surveying. The work of the early Egyptians furnishes an illustration. POINTS AND STRAIGHT LINES CONCRETE REPRESENTATION 2. We represent a point on paper by a dot made with a sharp-pointed pencil and designate it by a capital letter placed near the dot (see point A, Fig. 1). ^^ We represent a straight line on paper by a ** mark made with a sharp-pointed pencil and a ^^^- ^ straight ruler and designate it by a small letter placed on the mark (see line n, Fig. 1). A straight line resembles a tightly stretched string, such as a thread by which a weight is suspended. The word line comes from the Latin word linea meaning "a linen thread." The word straight comes from the Anglo-Saxon verb meaning "to stretch." 'jE ; : ; :_ , ". p.L^m GEOMETRY LOCATION OF STRAIGHT LINES 3. Ex. 1. How many straight lines can be drawn through (1) one given point? (2) any two given points? (3) any three given points? (4) any four given points? Illustrate answers by figures. If two points are given, the sljraight line passing through these points is said to be located definitely. When two points are given, as points A and ^ ^ B (Fig. 2), it is often best to designate the Fig. 2 line passing through them as the line AB, rather than by a small letter. This method of designating the line locates it definitely with respect to the two given points A and B. We shall assume as apparent that Only one straight line can pass through two given points. Ex. 2. Designate each of the points in the figures that you drew for Ex. 1 by a capital letter. Read each of the lines by naming two of its points. LOCATION OF POINTS 4, Ex. 1. How many points are there in a straight line? Ex. 2. What is the greatest number of points in which two straight lines can intersect? Ex. 3. Draw two straight lines with no possible intersections. Ex. 4. What is the greatest number of points in which three straight lines can intersect? Ex. 5. Draw three straight lines with (1) no possible inter- sections; (2) only one possible intersection; (3) only two possible intersections; (4) three intersections. Ex. 6. Name the greatest number of points in which four straight lines can intersect. Draw a figiire to illustrate your answer. If two given straight lines intersect, the point of inter- section is said to be located definitely. When two inter- secting straight lines are given, such as lines a and 6, it is often best to design 2'*:e the point of intersection as the point INTRODUCTORY 3 ab rather than by one capital letter. In Fig. 3 the two straight lines a and b were first given; their intersection then locates point 0, which may be called the point ab. This method of ^^ o designating the point locates it definitely ^^^^ ^""^ with respect to the two given Hnes. Fig. 3 We shall assume as apparent that Two different straight lines can intersect in only one point. Ex. 7. Designate each of the lines in the figures you drew for Exs. 5 and 6 by a small letter. Read each of the points whose loca- tion is determined by these lines; show how each point is located. STRAIGHT-LINE SEGMENTS 6. The i)ortion of a straight line terminated by two given points of the line is called a straight-line segment. Here- after the single word segment will be used to indicate a straight-line segment. If two segments are parts of the same straight line, they are said to be collinear. 6. We transfer segments from one Hne to another by the use of the compasses, the dividers, or a strip of paper. To transfer a given segment. Open the di- viders to the segment required by laying them on the given segment. Without changing the adjustment place one leg on point A and mark off the segment AB on the line c (Fig. 4). Ex. 1. Find a segment which is the sum of two given seg- ments. 4 PLANE GEOMETRY Ex. 2. Find a segment which is the difference between two given segments. Can any segment be subtracted from any other segment? Ex. 3. Show how to multiply a given segment by 3 ; by 5 ; by w. Note. The problem of dividing a segment by any given number is not as simple as the three foregoing problems. It will be studied later. In §9 we learn how to divide a given segment by 2. 7. Two segments are said to be congruent if they can be placed upon each other so as to fit exactly. To make two segments coincide, their extremities must be made to coincide. We shall assume as apparent that Only one segment can be drawn joining two given points. If the end points of a segment are known, the segment is located definitely. 8. A segment has a definite length. The length of a segment may be obtained by the successive application of some standard unit. The length of a segment joining'two given points is often called the distance between the two given points. Segments that have the same length are called equal segments. Equal segments are congruent and congruent segments are equal. 9. To find the mid-point of a given segment by paper folding. Fold the paper so that point A falls on point B. Crease the paper ^ ^ sharply. Hold it to the light and see Fig. 5 that one part of the segment falls exactly upon the other part. The crease marks the mid-point of the segment. We shall assume as apparent that A given segment has only one mid-point. Exercise. Divide a given segment into four eqval parts. 1 INTRODUCTORY 5 RAYS 10. A portion of a line which starts at a given point and extends indefinitely in a given direction is called a ray. The point is called the origin ^ ° of the ray (see ray a, origin O, Fig. 6). ^^^- ^ Two rays that have a common origin and extend in the same direction are said to be coincident. If they have the same origin and extend in opposite directions, they are collinear. A number of rays from the same origin form a pencil of rays. Make a drawing showing a pencil of rays. 11. If the origin and any other point of a ray are known, the ray is located definitely. We shall assume as apparent that Only one ray can be drawn having a given origin and passing through a second given point. CIRCLES 12. A closed curved Une every point of which is equally distant from a given point in the same plane is called a circle. The given point is called the center of the circle. In Fig. 7, O is the center of the circle. A segment drawn from the center to the circle is called a radius. A segment drawn through the center and terminating in the circle is called a diameter. In Fig. 7, OC, OEy and OD are radii and DE is a diameter of circle 0, ^^^- '^ A segment joining two points on a circle is called a chord of the circle. In Fig. 7, XY and DE are chords of circle 0. A part of a circle is called an arc. In Fig. 7 the part of the circle between points A and B is an arc of circle 0. Note. Arc comes from the Latin, and means "a bow"; chord, from the Greek, and means "the string of a musical instrument." 6 PLANE GEOMETRY If two circles can be made to coincide, they are said to be congruent. Circles with equal radii are congruent. The truth of this fact may be shown thus: Draw two different circles with equal radii. Cut them out. Put the same pin through the center of each circle. We shall assume as apparent also that Congruent circles have equal radii. Note. A circle is usually drawn with the compasses. In place of compasses a pencil and a thread or a piece of chalk and a string can be made to answer the purpose. ANGLES DEFINITIONS 13. A figure formed by two rays which have the same origin is called an angle. The rays are called the sides or arms of the angle. The origin of the rays is the vertex of the angle. An angle may be designated in various ways. Fig. 8 In Fig. 8 we have the following angles, reading from left to right: LA, Zab, /.a, Z2, ABAC. Notice that in the last case the letter at the vertex of the angle is read between the other two letters. An angle may be considered as formed by the rotation of a ray about its origin. The size of the angle de- pends upon the amount of rotation. In Fig. 9 which is the larger angle? Why? INTRODUCTORY 7 ADJACENT ANGLES 14. Angles that have a common vertex and a common side which separates the angles are called adjacent angles. Make a drawing to illustrate this definition. Ex. 1. How many angles are formed when a ray starts from a point in a given straight line? Illustrate your answer by a drawing and designate the angles in as many ways as possible. Ex. 2. How many angles are formed when two straight lines intersect? Illustrate your answer by a drawing and designate the angles in as many ways as possible. Ex. 3. In the drawings that you made for Exs. 1 and 2, how many pairs of adjacent angles can you find? Can you find angles that are not adjacent? CONGRUENT ANGLES 16. Two angles are said to be congruent if they can be so placed that their sides are coincident. To construct an angle that shall be congruent to a given angle. Method I. By means of tracing paper. The details of this method are left to the pupil. Method II. By means of an angle-carrier. Fasten two pieces of cardboard or very stiff paper together as shown in Fig. 10. Open the instrument until the edges a and b coincide with the sides of the given angle. Without changing the adjustment of the angle-carrier transfer i^^io. lu the instrument to the desired position and draw the angle formed. 16. Two angles may be added by placing them adjacent to each other. The angle formed by the two exterior arms is the sum of the two adjacent angles. In Fig. 11, Z3 is the sum of Z 1 and Z 2. Z 3 - Z 1 + Z ?.. 8 PLANE GEOMETRY A smaller angle may be subtracted from a larger angle by placing the smaller inside the larger so that they have a common vertex and a common side. The remaining part of the larger angle is the difference between the two angles. In Fig. 12, Z3 is the difference between Zl and Z2. Z3=Z1-Z2. Fig. 12 Ex. 1. Draw two angles that are not congruent and construct an angle equal to their sum. Ex. 2. Draw two angles that are not congruent and construct the angle equal to the difference between the larger and the smaller angle. Ex. 3. Construct an angle that is twice as large as a given angle. Ex. 4. At point on line AB draw a ray that shall make with the line AB an angle congruent to Z 1 . Can this line have more than one position? (Fig. 13.) o Fig. i; 17. To construct the bisector of a ^ given angle. To construct the bisector of ZBAC. By paper folding. Fold the paper on a line extending through the vertex A (Fig. 14) so that the ray AB will be coincident with the ray AC. Crease the paper sharply. The crease bisects ZA. Why? We shall assume as apparent that Only one ray can be drawn bisecting a given angle. Ex. 1. Divide a given angle into four congruent parts. Ex. 2. Draw two angles that are not congruent. Construct an angle that is one-half the sum of the two angles just drawn. Ex. 3. Draw two angles that are not congruent. Construct an angle that is one-half the difference obtained by subtracting the smaller angle from the larger. Fig. 14 INTRODUCTORY 9 RIGHT ANGLES AND PERPENDICULARS 18. When a ray starts from a point in a straight line and forms two congruent angles, the angles are called right angles, and the ray is said to be perpen- dicular to the line. In Fig. 15, Z 1 is congruent to Z 2, Z 1 and Z 2 are called right angles, and BC is said to be per- pendicular to Z) A . ^'''' ^^ In Fig. 15 the ray BA may be supposed to rotate about point B as origin. One complete rotation would carry it back to the position BA. A right angle is obtained by one- quarter of a complete rotation. The ray OA (Fig. 16) may start from the position OA and rotate until it ex- tends in a direction exactly opposite to its original position. It has made one- half of a complete rotation. The figure AOAf, is called a straight angle. The angle formed by one complete rotation is called a perigon. An angle less than a right angle is called an acute angle. An angle greater than a right angle and less than a straight angle is called an obtuse angle. 19. To construct a perpendicular to a line from a given point in the line. To construct a perpendicular to AB from point O. By paper folding. Fold the paper through point so that the ray OA is coincident with the ray OB. Crease the paper sharply. The crease is _L AB. Why? (Fig. 17.) Fig. 17 Note. In ordinary practice a perpendicular to a line from a point in the line is usually drawn by means of a rectangular card or a drafts- man's triangle. Place one edge of the card on line ^45 with the corner of the card at point O and draw a line along the other edge of the card. 10 PLANE GEOMETRY We shall assume as apparent that Only one perpendicular can be drawn to a line from a point in the line. 20. To construct a perpendicular to a line from a point not in the line. To construct a perpendicular to line / from point O. By paper folding. Fold the paper through point O (Fig. 18) so that line / will fall upon itself. .^ Crease the paper sharply. The crease is the perpendicular to line / from point j O. Why? Fig. 18 Ex. 1. Show how to construct a perpendicular to a line from a point not in the line by means of a rectangular card or a draftsman's triangle. We shall assume as apparent that Only one perpendicular can be drawn to a line from a point not in the line. Ex. 2, How many ±s can be drawn to a given line? How may one of these be located definitely? 21. To construct the perpendicular bisector of a given segment. To construct the perpendicular bisector of AB, By paper folding. Fold the paper so that point A falls on point B. Crease the paper sharply. The crease is the per- j ~ B pendicular bisector of AB. Why? fig. 19 (Fig. 19.) 22. We shall assume as evident that : I. All straight angles are congruent. II. All right angles are congruent. Ex. 1. Show how to bisect a straight angle. Ex. 2. Construct by paper folding an angle that is one-fourth of a straight angle. INTRODUCTORY 11 MEASUREMENT OF ANGLES 23. An angle is measured by the successive application to it of some other angle considered as a unit. Sometimes angles are measured by comparing them with a right angle: Thus an angle may be H of a right angle; H oi Si right angle; ^ of a right angle. For most purposes, however, the right angle is too large a unit. A good practical unit for measuring angles is -g-J^ of a perigon and is called a degree (°). Ex. 1. How many degrees in a straight angle? in a right angle? Ex. 2. How many degrees in /^, H, Ks, ^3'^4, ^^ of a perigon? Ex. 3. How many degrees in }^i, %, H, H, H oi a. right angle Each degree may be divided into 60 congruent parts called minutes (')• Each minute may be divided into 60 congruent parts called seconds ("). The number that tells how many times the unit angle is contained in a given angle is the measure or the measure number of the given angle. Angles that have the same measure number are said to be equal. Since congruent angles can be so placed that their sides take the same direction, congruent angles repre- sent the same amount of rotation and have the same measure number. Also angles that have the same measure number represent the same amount of rotation and can be made to coincide. In other words, equal angles are congruent and congruent angles are equal. Ex. 4. The sum of three angles is 360°. The second is three times the first, and the third is four times the first. Find the number of degrees in each. Can you construct a figure by paper folding to illustrate your answer? 12 PLANE GEOMETRY IMPORTANT SPECIALLY RELATED ANGLES 24. Two angles are called complementary if their sum is an angle of 90°. Each of two complementary angles is called the complement of the other. Ex. 1. Find the complements of the following angles: a. 60° d. 62° 27' g, 33° b. 44° e. 59° 18' h. 42° 2' c. 39° 10' /. 25° 20' i. ic° Note: Many of the exercises that follow can and should be solved by an algebraic equation. Ex. 2. If the complement of an angle is K of the angle, find the number of degrees in the angle and its complement. Ex. 3. If an angle is H of its complement, find the number of degrees in the angle and its complement. Ex. 4. Draw any acute angle. Construct the angle which is the complement of the first angle. Ex. 5. Draw two equal acute angles. Draw the complement of each of these angles. Cut out these complements and place one upon the other. Are they congruent? Ex. 6. How are the complements in Ex. 5 obtained by sub- tracting equal angles from equal angles? The fact illustrated in Exs. 5 and 6 will be assumed as evident. It may be stated as follows: Complements of equal angles are equal. 26. Two angles are said to be supplementary if their sum is an angle of 180°. Each of two supplementary angles is called the supplement of the other. Ex. 1. Find the supplement of each of the following angles: a. 75° d. 59° 22' g. 90° 21' b. 18° 25' e. 63° 18' h. 16° 18' c. 11° 65' /. 12° 30' i. x° Ex. 2. How many ( degrees in an angle that is H of its supple- ment? Ex. 3. An angle is ; H of its supplement. Find the number of degrees in the angle and its supplement. INTRODUCTORY 13 Ex. 4. Draw any angle. Draw the angle which is the supple- ment of the first angle. Ex. 5. Draw two equal angles. Draw the supplements of each of these angles. Cut out these supplements and place one upon the other. Are they congruent? Ex. 6. How are the supplements in Ex. 5 obtained by sub- tracting equal angles from equal angles? The fact illustrated in Exs. 5 and 6 will be assumed as evident. It may be stated as follows: Supplements of equal angles are equal. 26. Certain important facts concerning sums of angles are illustrated in the next three exercises. It is evident that the sum of all the parts of an angle is equal to the whole angle. Show which is the whole angle and what are its parts in each of these three exercises. Ex. 1. If in Fig. 20 the ray OB starts from point in the line CA, how many degrees are there in the sum oi A\ and 2? Ex. 2. In Fig. 21, if AOB is a straight line, what is the sum of Z1+Z2+Z3+Z4? Ex. 3. In Fig. 22 what is the sum of Z1+Z2+Z3+Z4+Z5? Fig. 22 The facts illustrated in Exs. 1-3 will be assumed as evi- dent. They may be stated as follows: I. If a ray starts from a point in a straight line, the sum of the two adjacent angles formed on one side of the line is 180°, or a straight angle. Such angles are called supplementary adjacent angles. 14 PLANE GEOMETRY 11. The sum of the adjacent angles on one side of a straight line formed by any number of rays having a com- mon origin on the line is 180°, or a straight angle. III. The sum of the adjacent angles formed by a number of rays from the same origin is 360°, or a perigon. Ex.4. If Zl+/2+Z3 = 180°, and Z2 is twice Zl,and Z3 is three times Z2, find the number of degrees in each angle. Ex. 5. If four equal angles form a perigon, how many degrees in each? Ex. 6. If six equal angles form a perigon, how many degrees in each? Ex. 7. If Zl+Z2+Z3+Z4 = 360°, and Z4 is twice Z3, Z3 twice Z2, and Z 2 twice Zl, how many degrees in each? 27. Statement I in the preceding section indicates a very close relation between supplementary adjacent angles and a straight angle. The next two exercises illustrate another phase of this relation. Ex. 1. Copy two right angles from cards or draftsman's triangles (Fig. 23). Place them so that point falls on point 0', side OA along side O'A', and OB opposite O'B'. What kind of an angle is A BOB'} Ex. 2. Draw any two equal angles, A 1 and 2 (Fig. 24). Construct Z3, the supple- ment of Z2. How are Zl and 3 related? Why? Cut out the three angles. Place Z 1 in the position occupied by Z2. What kind of an angle is /.B'OC} Fig 24 The fact illustrated in Exs. 1 and 2 will be assumed as evident. It may be stated as follows: If two supplementary angles are adjacent, their exterior sides are collinear. , INTRODUCTORY 15 28. Two an^'lcs arc called vertical or opposite if the sides of one are prolongations of the sides of the other. If two lines intersect, two pairs of vertical angles are formed . Thus in Fig. 25 lines h and k intersect, forming vertical angles, Z 1 and Z 3, also Z 2 and Z 4. Ex. 1. Draw two intersecting lines (Fig. 25). Show that A2 and 4 are the supplements of the same angle. Show that A 1 and 3 are the supple- ments of the same angle. Ex. 2. In Fv^. 25 why is Zl=Z3and Z2= Z4? In general, why are vertical angles equal? Fig. 25 The fact illustrated in Exs. 1 and 2 will be assurned as evident. It may be stated as follows: If two straight lines intersect, the opposite or vertical angles are equal. SUMMARY AND SUPPLEMENTARY EXERCISES SUMMARY OF GEOMETRICAL ASSUMPTIONS 29. The foregoing definitions and exercises justify the following assumptions: A. Concerning the definite location of points: As. 1. Two different straight lines can intersect in only one point, or two intersecting straight lines locate a point (§4). As. 2. A segment can have only one mid-point (§9\ B. Concerning the definite location of segments, rays, and straight lines: As. 3. Only one segment can be drawn between two points, or a segment is located definitely if its extremities are given (§7). As. 4. Only one ray can be drawn having a given origin and passing through a second given point (§ 11). As. 5. Only one ray can be drawn bisecting a given angle (§17). 16 PLANE GEOMETRY As. 6. Only one straight line can pass through two given points (§3). As. 7. Only one perpendicular can be drawn to a line fjcom a given point in the line (§19). As. 8. Only one perpendicular can be drawTi to a line from a given point not in the line (§20). C. Concerning circles: As. 9. Circles with equal radii are congruent (§12). As. 10. Congruent circles have equal radii (§12). D. Concerning equal angles: As. 11. All straight angles are equal (§22). As. 12. All right angles are equal (§22). As. 13. Complements of equal angles are equal (§24). As. 14. Supplements of equal angles are equal (§25). As. 15. Vertical angles are equal (§28). E. Concerning angle-sums: As. 16. If a ray starts from a point in a straight line, the sum of the two adjacent angles formed on one side of the line is 180°, or a straight angle (§26). As. 17. The sum of the adjacent angles on one side of a straight line formed by any number of rays having a common origin on the line is 180°, or a straight angle (§26). As. 18. The sum of the adjacent angles formed by a number of rays from the same origin is 360°, or a perigon (§26). F. Concerning straight angles: As. 19. If two supplementary angles are adjacent, their exterior sides are coUinear (§27). GENERAL ASSUMPTIONS 30. The following assumptions are also true: As. 20. If equal segments (or angles) are added to equal segments (or angles), the results are equal segments (or angles). INTRODUCTORY 17 As. 21. If equal segments (or angles) are subtracted from equal segments (or angles), the results are equal seg- ments (or angles). As. 22. If equal segments (or angles) are multiplied by the same number, the results are equal segments (or angles) . As. 23. If equal segments (or angles) are divided by the same number, the results are equal segments (or angles) . As. 24. Segments (or angles) that are equal to the same segment (or angle) are equal. As. 25. Equal segments (or angles) may be substi- tuted for equal segments (or angles). Note. Hereafter when these assumptions are used they should be quoted in the form in which they apply; for example: Equal segments may be substituted for equal segments, or equal angles may be sub- stituted for equal angles, as the case may require. MISCELLANEOUS EXERCISES 31. After solving each of the following exercises state clearly and in full the definitions or assumptions which it is intended to illustrate. 1. One of two supplementary angles is 40° 15' larger than the other. Find the angles. 2. Find an angle whose complement is }4 as large as the angle itself. 3. Show that if two lines intersect so that one angle is a right angle all the angles are right angles. 4. Find an angle whose complement is }4 of its supplement. Can you construct this angle by paper folding? 5. What angles do the hands of a clock make at 5 o'clock? at 10 o'clock? 6. If, in Fig. 26,Z2= Z6, show that fl. Z 3=Z 6 (/. Z 1=Z 8 ^>. Z 2=Z 7 d. Z 4=Z 5 c. Z 1=Z 5 /. Z 4=Z 8 18 PLANE GEOMETRY 7. Which of the following directions locate the required line or rays definitely? Why? a. Construct a perpendicular to line AB. b. Construct a perpendicular to line AB from point O outside of AB. c. Connect points C and D. d. Bisect A A. e. Draw a line through point A. J. From point O outside of line A B draw a ray cutting line AB. g. Draw a ray which shall make a given angle with a given line. h. Draw a perpendicular bisector to a given segment. i. Draw a segment bisecting a given segment. 8. Two rays start from point on the same side of line AB. Of the three angles formed the second is 3 times the first and the third is 5 times the first. Find the angles. 9. Find an angle whose complement is 25° smaller than the angle itself. 10. Can you find an angle whose complement is % of its sup- plement? 11. Draw four straight lines so that there will be: a. No possible intersections; h. Only one possible intersec- tion; c. Only three possible intersections; d. Four possible intersections; e. Five possible intersections; /. Six possible intersections. 12. Can you draw four straight lines so that there can be only two possible intersections? 13. How many degrees in the supplement of /.x \i its comple- ment is 23°? 14. In Fig. 27 show that Z H- Z44-Z5+ZG- 2rt. ^. What other combinations of angles in Fig. 27 will be equal to two right angles? Fig. 37 INTRODUCTORY 19 15. Draw two complementary adjacent angles and the bisec- tor of each. How many degrees in the angle made by the bisectors? Why? 16. Draw two supplementary adjacent angles and the bisec- tor of each. How many degrees in the angle made by the bisectors? Why? 17. Draw AABC (Fig. 28) so that Z1=Z2. Show that Z 3 = Z 4. 18. How many degrees in the complement of Z6 if a the supplement of Z ft is 110°? Fig. 28 32. ABBREVIATIONS AND SYMBOLS USED Fig. figure Ex exercise Exs exercises Def definition As assumption Ass assumptions Th theorem Ths theorems Cor corollary alt. int alternate interior comp complementary sup supplementary adj Adjacent § section .* therefore ° degree ' minute " second cm centimeter mm millimeter _ (equal to (is equal to ^ (congruent to (is congruent to c^ (similar to I is similar to > (greater than (is greater than <^ (less than (is less than Z , A angle, angles rt. Z right angle rt. A right angles (perpendicular to -^ -^us perpendicular ' to II (parallel to (is parallel to A, A triangle, triangles rt. A right triangle n , CS . . . /rectangle, rec- ' ( tangles D , [s] square, squares EJ, CsJ. . . (parallelogram (parallelograms y — ^ /-g-\ (trapezoid, trap- ' ( ezoids O, (D circle, circles .15 arc^B AB chord ^5 5 area per perimrjter CHAPTER II Congruent Triangles INTRODUCTORY DEFINITIONS 33. Any two figures that can be made to coincide are called congruent figures. In congruent figures corresponding sides or angles are sides or angles that coincide or that can be made to coincide. We shall add the following to the list of assumptions: As. 26. Any figure can be moved about in space with- out changing either its size or its shape. As. 27. Figures congruent to the same figure are con- gruent to each other. 34. A figure formed of three segments joined end to end consecutively is called a triangle. Such a figure has three sides, three angles, and three vertices. Unless it is other- wise stated, a triangle should be drawn with its sides of different lengths (Fig. 29). A triangle that has at least two sides equal is called an isosceles triangle. The angle included by the equal sides is called the vertex angle, and its vertex, the vertex of the triangle. The third side is called the base of the triangle. A triangle with all its sides equal is called an equilateral triangle. 20 CONGRUENT TRIANGLES 21 Fig. 3] TESTS FOR EQUAL ANGLES AND EQUAL SEGMENTS TESTS I AND II FOR CONGRUENT TRIANGLES 35. Ex. 1. Draw a triangle with two sides equal to the seg- ments a and b (Fig. 30). Draw this triangle with a soft « pencil on paper that is not j, too hesLvy. Compare your Fig, 30 figure with your neighbor's by placing one paper upon the other and holding them to the light. Ex. 2. Draw a tri- angle with two sides equal to the segments a and b and the included angle equal to ZC (Fig. 31). Compare your figure with your neighbor's as ex- plained in Ex. 1. Ex. 3. Draw a triangle with two angles equal to A A and B and the included side equal to segment c (Fig. 32). Compare your triangle with ^^ your neighbor's by placing \ one paper upon the other and holding them to the light. Fig~~32 Ex. 4. Draw any triangle and letter it ABC. Draw another triangle and letter it XYZ, but make two sides and the included angle of AXYZ equal to two sides and the included angle of AABC. Compare the two triangles. c Note. In Ex. 4 and in all exercises requiring the construction of figures the pupil should not only make the drawing but should tell how he did a- it. Sentences should be short and exact. For example (Fig. 33): (1) Make ZX=/.A. (2) Make XY = AB. Etc. ^ Fig. 33 22 PLANE GEOMETRY Theorkm 1.* If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent in all corre- sponding parts and are called congruent triangles. Fig. 34 Given AABC and AC = DF. ADEF, ZA=ZD, AB = DE, and To prove AABC and ADEF congruent in all correspond ing parts. Proof: STATEMENTS 1. Place AABC on ADEF so that yl Swill iallonDE,A on D, B on E, and C and F on the same side of DE, 2. Segment AC will fall along the line of DF. 3. Point C will fall on point F. 4. BC will coincide exactly with EF. REASONS 1. This can be done because AB = DE. 2. ZA= ZD. 5. AABC ^ ADEF. Note: An angle of a triangle is two adjacent sides. ♦The formal proof of Th. 1 may be teacher desires. 3. AC = DF. 4. If the extremities of two segments coin- cide, the segments will coincide exactly. 5. Two triangles that coincide exactly are congiTient. said to be included between its jjostponed imtil such time as the CONGRUENT TRIANGLES 23 36. Theorem 2.* If two angles and the included side of one triangle are equal to two angles and the included side of another triangle, the triangles are congruent in all corresponding parts and are called congruent triangles. Fig. 35 Given AABC and ADEF, /.A= Z.D, An= Z.E, and AB = DE. To prove A ABC and ADEF congruent in all correspond- ing parts. Proof: STATEMENTS 1. Place A.4^Con ADEF so that AB falls on DE, A on D, B on E, and C and F on the same side of DE. 2. AC will fall along the line oiDF. 3. BC will fall along the line of£F. 4. C \^all fall on the line of DF and also on the line of EF, 5. .*. C will fall on point F. REASONS 1. This can be done because ? 2. Why? 3. Why? 4. Why? 6. .*. AABC ^ ADEF. 5. Two lines can inter- sect at only one point, f). Why? Note. A side of a triangle is said to be included between the two angles adjacent to i1. ♦The formal proof of Th. 2 may be postponed until such time teacher desires. the 24 PLANE GEOMETRY * Exercise. Draw any triangle and letter it ABC. Draw another triangle and letter it XYZ, but make two angles and the included side of AXYZ equal to two angles and the included side oi AABC. In how many ways can you construct a triangle congruent to a given triangle? 37. Ex. 1. Construct an isosceles triangle with base 3 cm. and each of the equal sides 5 cm. (Fig. 36) . Note. Notice the wording for the directions for this -*^ ^s drawing. Fig. 36 I. Draw segment AB=S cm. II, With A as a, center and 5 cm. as a radius make an arc above AB, Etc. Ex. 2. Construct an equilateral triangle with each side 4 cm. Theorem 3. The angles opposite the equal sides of an isosceles triangle are equal. Fig. 37 Given the isosceles A ABC, AC = BC, To prove AA^ LB. Analysts and construction: I. To prove ZA= ZB, prove A A and B correspond- ing angles of congruent triangles. II. To obtain the triangles, bisect ZC, continue the bisector CO to meet the base at 0. III. .-. to prove ZA=ZB, prove AAOCmABOC. IV. To prove AAOC ^ ABOC, prove two sides and the included angle CONGRUENT TRIANGLES 25 Proof: STATEMENTS SEASONS .a. AC = BC. I. a. • Let the r»uDi^ h. Z1=Z2. c. CO = CO. d, :.AACO^ABOC. h. c. d. \ give the rea- sons for each statement. . ZA=ZB. II C or re s p ending angles of congruent triangles. Cor. An equilateral triangle has all of its angles equal; that is, it is equiangular. 38. The method used in Th. 3 is very important. One way to prove two segments or two angles equal is to find two triangles that contain these segments or these angles and to prove these triangles congruent. In general we say that Corresponding sides of congruent triangles are equal. Corresponding angles of congruent triangles are equal. In the following exercises use the analysis and proof of Th. 3 as a model. To make an analysis, it is necessary to ask one's self a series of questions. Start with what is to be proved, and by means of these questions unravel the proof step by step. In Th. 3, for example, you should begin by asking yourself, What are the tests for equal angles? The answer to this question gives the first step in the analysis and suggests the construction given in step II. You should then ask your- self, What are the tests for congruent triangles? The answer to this question gives step IV of the analysis. This process should be continued for each theorem or exercise until the entire proof is clear. The proof is the analysis done back- ward, with all authorities stated in full. 26 PLANE GEOMETRY Ex. 1. Construct Fig. 38. Make AD and BC perpendicular to AB. Find 0, the mid-point of AB. Make AD = BC. Join DO and CO. Prove DO = CO. How might AAOD be made to coincide with ABOC? Ex. 2. Construct Fig. 39. Make BC perpen- dicular to AO. Make Z 1 = Z2. Prove AB = AC and BX = XC. How might AAXB be made to coincide with AAXC? Ex. 3. Construct Fig. 40. Draw XY, the per- pendicular bisector of segment AB. Join F, any point in XY, to A and to 5. Prove PA=PB. Ex. 4. Construct Fig. 41. ^i5 is any seg- ment. Make Z1=Z2 and Z3= Z4. Which A sides can be proved equal? Prove them equal. How might A^^A' be made to coincide with AABY? Fig. 40 Ex. 5. Construct Fig. 42. angle. Draw BO bisecting Z B. Connect A, any point in OB, Prove ^A = CA. Draw ABC, any Make AB = BC. with A and C. Ex. 6. Construct Fig. 43. ABC is an isosceles triangle. is the mid-point of AB. AX = BY. Prove OX = OY. Ex. 7. Construct Fig. 44. Draw any seg- ment AB. Find O, its mid-point. Draw rays h and k so that Z 1 = Z 2. Through O draw any line that will intersect h and k. Call the ])oints of intersection D and C respectively. Prove that AD = BC and that DO = CO. How might ABOC be made to coincide with AAOD ? Fig. 42 43 Fig. 44 CONGRUENT TRIANGLES 27 Ex. 8. Show that congruent triangles may be used to find the distance across a pond as follows (Fig. 45) : First set up a stake at any convenient point, as 0, from which points A and B are both visible. Set up a stake at Z) in a straight line with O and A so that OD = 0A. In the same way place stake C so qz that CO = OB. Measure CD. Fig. 46 Ex. 9. Explain from Fig. 46 how to find 3^^^5^-^ the distance across a stream. Notice that the ^Sa^g^rrSv required distance AB is perpendicular to a line \J along the bank of the stream. ^ Fig. 46 Ex. 10. Show that the distance across a stream may be measured as follows (Fig. 47): A pole is set c up at A with a stick fastened at the top so that ■^'''''^Pv Z DC A may be altered by moving the stick CD. / A person stands with the eye at C and ^7 T^^^^^^^^"^ moves the stick CD until he just sees point B. Fig. 47 The pole is then turned around. Standing with eye at C and looking along CD' he locates point B'. He then measures AB' . Could you do this using the visor of your cap rather than a pole and a stick? Is it necessary to turn through 180°? This device is said to be an old one.* Note. Theorems 1 and 2 were probably known to Thales. Thales is regarded as the founder of one of the earliest Greek schools of mathe- matics, about 600 B.C. It is said that Th. 2 was used in those days to determine the distance of a ship at sea. It is not known how this was done; but a tower or a cliff might have been used as the base of a triangle of which the ship formed the vertex. The base angles of this triangle could be found by observation. Show from Ex. 10 how the solution to the problem could be completed. Ex. 11. In Fig. 48, AB and CD are two b intersecting lines. AO = OB and CO = OD. ^\''y;^:^C\ Prove that AC = BD. Join CB and AD and ^^^ ° prove CB = AD. ^^°- ^^ ♦See W. E. Stark, "Measuring Instruments of Long Ago," School Science and Mathematics, 1910, 28 PLANE GEOMETRY Ex. 12. Construct Fig. 49. Draw AB any segment, struct h and k perpendicular to ^j5 at A and B respectively. Find 0, the mid-point oiAB. Make Z 1 = Z 2 and extend the sides imtil they inter- sect h and k. Call the points of intersection D and C respectively. Prove that AD = BC and DO = OC. Ex. 13. Construct Fig. 50. ABC is an isosceles triangle. O is the mid-point oi AB. Zl= Z2. Prove AX = BY and OX=-OY. Con- TEST III FOR CONGRUENT TRIANGLES 39. Ex. 1 . Construct two triangles ABC and X YZ so that AB = XF, BC=YZj and CA=ZX. Draw these triangles with a soft pencil on two pieces of fairly thin paper. Compare them by placing one paper on the other and holding the papers to the light. Before these triangles can be proved congruent what must you know? Ex. 2. Construct a triangle having its sides equal to three given segments. Theorem 4. If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent. B V Fig. 51 Given AABC and ADEF, AB = DE, AC = DF, and CB=^FE. To prove AABC m ADEF. CONGRUENT TRIANGLES 29 Analysis and construction: I. To prove AABC ^ADEF, prove AC = /.F. II. " " ZC=ZF, place AABC so that AB, the longest side of AABC, coincides with DE, the longest side of ADEF, and point C is opposite point F. Join CF and prove that ZC^and ZF are made by adding the base angles of two isos- celes triangles. Proof: STATEMENTS REASONS I. a. EF = EC. I. a. Given. b. AEFC is isosceles. b. A triangle that has two sides equal is isosceles. c. Z2 = Z4. c. The base angles of an isosceles A are equal. II. a. DF = DC. II. a. Why? b. ADFC is isosceles. b. Why? c. Z1 = Z3. c. Why? III. .'.ZC=ZF. III. Equal angles added to equal angles give equal angles. IV. ADEF ^ ADEC. IV. Why? V. ADEF m AABC. V. Triangles congruent to the same triangle are congruent to each other. Ex. 3. Prove Th. 4 by proving that ZA = ZD; with Z C and Z F obtuse angles. Ex. 4. Construct Fig. 52. ABC is an isosceles tri-^ — ^ ^ angle. O is the mid-point of ^ B. Prove Z1=Z2. Fig. 52 30 PLANE GEOMETRY APPLICATION OF CONGRUENT TRIANGLES TO CONSTRUCTIONS CONSTRUCTION OF ANGLES 40. Problem 1. At a given point in a given line to con- struct an angle equal to a given angle. B X Fig. 53 Given line /, point X in line /, and Z BAG. To construct at point X in line / an angle equal to Z BAG, Analysis and directions: I. In order to construct an angle WXO equal to /.A, construct two congruent triangles that shall contain A A andX II. With A as a center and any radius cut the sides of ZA at D and E. III. With X as a center and the same radius draw an arc cutting Hne / at W. IV. With 1^ as a center and DE as a radius cut the last arc at O. V. Join X and O. Proof: STATEMENTS REASONS I. a. AE = XO. b. AD = XW. Let the pupil give all c. DE = WO. reasons in full. d. .'.ADAE^AWXO. II. .-. ,CA==ZX. Exercise. How is it possible to construct the ZX in more than one position on the line I and still have it fulfill the require- ments? CONGRUENT TRIANGLES 31 DIVISION OF ANGLES 41. Problem 2. To construct the bisector of a given angle. Given ZBAC. To construct the bisector of ZBAC. Analysis and directions: I. In order to construct the bisector of Z.A, construct a line through point A so that Z 1 equals Z 2. II. In order to construct Zl = Z2, construct congruent triangles that contain A 1 and 2. III. With i4 as a center and any convenient radius draw an arc cutting the sides of /.A Sit E and D. IV. With £ as a center and any convenient radius draw an arc. V. With Z) as a center and the same radius cut this last arc at O. VI. Join A and 0. Let the pupil give the proof. Ex. 1. Show that a carpenter's steel square may be used to bisect an angle as follows: Mark o off equal distances OA and OB on the sides of the angle. Place the square as shown in Fig. 55. Mark point D. Join D and O. Fig. 55 Ex. 2. Construct with ruler and compasses the figures for Exs. 4 and 5, §38. 32 PLANE GEOMETRY Ex. 3. Draw two vertical angles and bisect each. Show by As. 19 that the two bisectors are collinear. Ex. 4. Construct one-half the supplement of any given angle. Ex. 5. Construct a triangle congruent to a given triangle. In how many ways can this be done? Ex. 6. Draw any triangle and bisect each angle. Bisect each angle of an equilateral triangle. 42. Note. Trisection of Angles. By the method given in §41 we can bisect any angle that we choose. The trisection of an angle is a much more difficult problem. We shall find a little later how to trisect a right angle. In elementary geometry we confine ourselves to the circle and straight line and use no instruments except the com- passes and the straightedge. A straightedge is a ruler that is not graduated to any scale. It has been shown that angles generally cannot be trisected by the use of these instruments only. The Greeks learned How to trisect any angle, and since their time many ways of trisecting angles have been found. These methods, however, have always required other curves than the circle and other instruments than the compasses and straightedge. Draftsmen's methods for trisecting angles are approximations. A number of different instruments for trisecting angles have been made. Fig. 56 shows one such instrument. What segments are made equal? Show that Fig. 54 is used twice in Fig. 56. Why is the angle trisected? How could an instrument for bisecting angles be made? Such an instrument is sometimes used by carpenters for cutting and fitting moldings. CONGRUENT TRIANGLES 33 CONSTRUCTION OF PERPENDICULARS 43. Problem 3. To construct a perpendicular to a line from a given point in the line. / / t- V \ \ \. Fig. 57 Given line / and point in line /. To construct a perpendicular to line / from point 0. Analysis and directions: I. In order to construct OX ± / at 0, construct Z1=Z2. II. In order to construct Zl= Z2, construct two con- gruent triangles. III. With as center and any convenient radius draw an arc cutting line / at Y and Z. IV. With Y as center and a longer radius draw an arc. V. With Z as center and the same radius cut the last arc at X. VI. Join X and 0. Let the pupil give the proof (see §18 for the proof to I). Ex. 1. Construct at point in a given line / an angle of 45°; of 135°; of 22>^°; of 157K°. Ex. 2. Divide a given angle into four equal parts. Ex. 3. Construct Fig. 39 with ruler and compasses. Ex. 4. Construct the complement of any acute angle. Can the complement have more than one position? Ex. 5. Construct two complementary adjacent angles. Con- struct the bisector of each. How many degrees in the angle made by the bisectors? 34 PLANE GEOMETRY 44. Problem 4. To construct a perpendicular to a line from a point not in the line. q \ 4\ r\^ Fig. 58 Given line / and point not in line /. To construct a perpendicular to / from 0. Directions: I. With O as center and any radius draw an arc cut- ting / at y and Z. 11. With Y as center and a radius greater than half YZ draw an arc opposite O. III. With Z as center and the same radius cut this arc at X. IV. Join and X. Analysis: I. To prove OX ± /, prove Z 1 = Z 2. II. " " Z1=Z2, join OY and OZ and prove AYOAUAZOA. III. To prove /\YOA ^ AZOA, prove Z3= Z4. IV. '* " Z3=Z4, ..... Let the pupil complete the analysis and give the proof. Note. The proof for Prob. 4 will be much simpler after intersecting circles are studied. The reasons for the directions are then apparent. Ex. 1. Draw any triangle and construct the perpendicular from each vertex to the opposite side. ^ Ex. 2. Make the drawing called for in Ex. 1 for an equilateral triangle and also for a triangle containing an obtuse angle. CONGRUENT TRIANGLES 35 46. Pkoblkm T). To construct the perpendicular bisector of a given segment. 01 B i I I Fig. 59 Given the segment AB. To construct the perpendicular l:)isector of AB. Directions: ^ I. With i4 as a center and any radius greater than H AB construct arcs above and below AB. II. With B as a center and the same radius intersect these arcs at C and D. III. Join C and D. Let the pupil give the analysis and the proof. Note. Like the proof for Prob. 4, the proof for Prob. 5 is simpler after one studies intersecting circles. Ex. 1. In the solution for Prob. 5, which radii must be made equal? Which radii need not be made equal? Draw the figure and give the directions, analysis, and proof for this problem, varying the radii as much as possible. Ex. 2. Divide a given segment into four equal parts. NATURE OF THEOREMS AND PROOFS 46. The statements of geometrical facts in this and the preceding chapter have received three different narnes: assumptions, theorems, and corollaries. A theorem is a statement of a fact that is to be proved true. An assumption is a statement of a fact that is taken as true without proof. Its ^mth is taken for granted. 36 PLANE GEOMETRY Many of the assumptions in chapter i were obtained by observation. Many were shown to be true informally. The proof of a theorem shows that if certain facts are known to be true a certain other fact must be true. The facts that are known or given constitute the hypothesis. The fact to be proved is called the conclusion. In Theorem 3 the hypothesis is: The triangle has two equal sides. The conclusion is: The angles opposite these sides are equal. In Theorem 4 the hypothesis is : Three sides of one triangle are equal to three sides of another triangle. The conclusion is: The triangles are congruent. The hypothesis and conclusion of a theorem may be found by studying the grammatical construction of the statement of the theorem. If the sentence contains a clause beginning with "if," this clause is the hypothesis. If there is no such clause, the complete subject is the hypothesis and the complete predicate the conclusion. In each case the proof consists in showing that the con- clusion must follow from the hypothesis. The analysis shows how the proof has been or may be thought out. The proof is the analysis worked backward and is set down in what is called the synthetic form. The synthetic method is the opposite of the analytical method. The special method used for the proofs of Ths. 1 and 2 is called proof by superposition. This word is derived from two Latin words. What is its literal translation? A theorem that follows easily from another theorem is called a corollary of that theorem. Proofs are required for all theorems and exercises that follow unless some specific statement is made to the con- trary. In all proofs a warrant must be given for each step. Some of the exercises have been called problems. Any geometrical problem calls for the construction of some CONGRUENT TRIANGLES 37 geometrical figure to fulfill certain stated requirements. Unless it is otherwise stated, these constructions must be performed with compasses and straightedge only. In all problems it is necessary to prove that the completed figure fulfills the stated requirements. MISCELLANEOUS THEOREMS AND EXERCISES 47. OUTLINE REVIEW A. We have the following methods for proving two segments or two angles equal : I. In the case of segments look for a. Sums, differences, equal multiples, or equal parts of equal segments. b. Sides of isosceles or of equilateral triangles. c. Corresponding sides of congruent triangles. 11. In the case of angles look for a. Sums, differences, equal multiples, or equal parts of equal angles. b. Right angles. c. Supplements of equal angles. d. Complements of equal angles. e. Vertical angles. /. Corresponding angles of congruent triangles. g. Base angles of an isosceles triangle. B . Either segments or angles may be proved equal by prov- ing them corresponding parts of congruent tri- angles. To prove the triangles congruent compare I. Two sides and the included angle of one with two sides and the included angle of the other. II. Two angles and the included side of one with two angles and the included side of the other. III. Three sides of one with three sides of the other. 38 PLANE GEOMETRY 48. Theorem 5. If a perpendicular be erected to a straight line, oblique segments drawn from the same point in the perpendicular cutting the straight line at equal dis- tances from the foot of the perpendicular are equal. Hypothesis: Line XY l.iQ line I and the oblique segments CA and CB are drawn from Cso that AO = OB. Conclusion: CA=CB. Analysis: I. To prove CA=CB, prove AAOC ^ ABOC. IL " ** AAOC ^ ABOC, compare The proof is left to the pupil. Exercise. How could you make the congruent triangles in Fig. 60 coincide? 49. In the following exercises segments and angles are to be proved equal. The analysis given for Th. 3 may now be shortened as shown under Th. 5. Note. The figures for all exercises and theorems should be con- structed according to the hypothesis with ruler and compasses. Ex. 1. Construct an isosceles triangle ABC. Let CO, the bisector of Z C, meet the base A B at 0. Prove that the seg- ments joining with the mid-points of ylC and BC are equal. Ex. 2. Two isosceles triangles stand on opposite sides of the same base. Prove that the segment joining the vertices bisects both vertex angles. Ex. -3. Investigate the case, Ex. 2, in which the two isosceles triangles stand on the same side of the same base. CONGRUENT TRIANGLES 39 50. Theorem 6. The bisector of the vertex angle of an isosceles triangle is the perpendicular bisector of the base. AADC is isosceles, AC = BC, and CO CO is a ± bisector of AB. Hypothesis: bisects Z C. Conclusion: Analysis: I. To prove CO a _L bisector oi AB, prove II. prove Z2= Z3. ^AO==BO. Z2= Z3 AO = BO Let the pupil complete the analysis and give the proof. 61. Theorem 7. The segment which joins the vertex of an isosceles triangle with the mid-point of the base bisects the vertex angle and is perpendicular to the base (Fig. 61). Hypothesis: A ABC is isosceles, AC = BC, and CO joins the vertex C with 0, the mid-point of base AB. f 1. CO bisects ZC. I 2. CO±AB. Analysis: I. To prove that CO bisects ZC, prove Z5= Z6. 11. '• " '• CO±AB, prove Z2= Z3. "^- lz2=Z3 I'P^^^^ Let the pupil complete the analysis and give the proof. C Otic lus ion 40 PLANE GEOMETRY Fig. 62 EXERCISES INVOLVING CONGRUENT TRIANGLES 62. In the exercises that follow, special care is needed in making the analysis. Study carefully the analyses given for Ths. 6 and 7 and for Ex. 1 below. As compared with the analyses for Ths. 3 and 5, one or more extra steps are required. The pupil should carefully ask himself the proper questions at each step and should be perfectly sure that he understands how each step follows from the pre- ceding one. 1. The segments joining the mid-points of the sides of an isosceles triangle form an isosceles triangle. Analysis: I. To prove AXYZ isosceles, prove XZ=XY. IL " " ZZ=Z7, prove Let the pupil complete the analysis and give the proof. 2. How would Ex. 1 have read if A ABC (Fig. 62) had been an equilateral triangle? Give proof. 3. If the segment which joins the vertex of a triangle with the mid-point of the base is perpendicular to the base, the triangle is isosceles. 4. In Fig. 63, CX is a perpendicular bisector of AB. C moves upward along the ray CX, how will the sides of A ACS change? What are the upper and lower limits to these sides? No proof is needed. . 5. If the segment which bisects the vertex angle of a triangle is perpendicular to the base, the tri- angle is isosceles. 6. In Fig. 64, CX bisects ZACB. If the segment AB moves to the right . and to the left along the ray CX but is always perpendicular to CX, how will the C-* sides of ACBA change? What is the lower limit to the length of these sides? .^j^ What is the upper limit? No proof is Fig. 64 needed. CONGRUENT TRIANGLES 41 7. Given the equilateral AABC with equal segments AX, BY, and CZ measured off on the sides as shown in Fig. 65, prove that AXYZ is equilateral. 8. Given the isosceles AABC with the mid-point of the base AB. The equal sides CA and CB are extended beyond the base so that AE = BF. Prove that OE = OF. Fig. 65 9. Frames of various shapes can be made by fastening together pieces of wood or iron as shown in Fig. 66. Could you change the shape of any of these figures by pressing upon opposite sides or vertices? Which ones could be so changed? Fig. 66 10. How are roof trusses, elevated train structures, bridges, and the like made rigid? 11. A segment drawn from the center of a circle to the mid- point of a chord is perpendicular to the chord. Is there any special position of the chord for which the exercise and its proof have no meaning? 12. Show how the instrument shown in Fig. 67 may be used as a leveling device. The pieces are so fastened together that AB = AC. AD is a plumb line. How must the framework be placed if BC is level? This instrtmient is said to be very ancient.* Fig. 67 Suggestion. Notice that a level line is one that is perpendicular to a plumb line. 13. Draw any triangle ABC. Find X, the mid-point of AB. Join CX and extend so that XY = CX. Prove AC = BY. •See D. E. Smith. The Teaching of Geometry, p. 178. 42 PLANE GEOMETRY Fig. 68 14. Fig. 68 shows a common form of truss. The king-rod CO joins C with the mid-point of AB in the isosceles AABC. The braces OE and OF join O ^^ with E and F, the mid-points of ^C and CB respectively. Prove that OE = OF and that CO is perpendicular to AB. The segment joining any vertex of a triangle with the mid-point of the opposite side is called a median of the triangle. In Ex. 14 the king-rod is the median of the isosceles triangle that forms the truss. 15. In Fig. 09, AD and BC are ± ^^ at yl and B respectively. If AD = BC, prove that AC = BD. 16. Given ABC, an isosceles triangle with the equal sides extended through the vertex to X and Y. AX and BY are ± AB at points A and B. Prove that AX = BY (Fig. 70). Fig. 70 17. Given the isosceles AABC with the equal sides CA and CB extended beyond the base so that AX = BY, prove that AY = BX. 18. In Fig. 71, ABC is an isosceles triangle with AC = CB. If is the mid-point of AB, and AC and BC are extended so that CE = CD, prove that OE^OD. 19. If any point in the bisector of the vertex angle of an isosceles triangle be joined to the extremities of the base AB, AAOB is isosceles. 20. The segments drawn from the extremities of the base of an isosceles triangle to the mid-points of the opposite sides are equal. 21. In Fig. 72, ABC is an isosceles triangle. CX and CY are drawn so that Z1=Z2. Prove that AXCY is isosceles. CONGRUENT TRIANGLES 43 22. In Fig. 73, if ABC is given an isosceles triangle and CX and CY are drawn so that Z 1 = Z 2, prove that £ ACXY is isosceles. 23. The three medians of an equilateral triangle are equal. '^ '* ^ ^ Fig. 73 24. The segments that bisect the base angles of an isosceles triangle and are terminated by the equal sides are equal. 25. Draw several figures for Ex. 24. Let the diff'erent triangles have the same base but legs of different lengths. Are there any limits to the angle that the bisector of one of the base angles makes with the base? No proof is needed. 26. The segments that bisect the angles of an equilateral triangle and are terminated by the opposite sides are equal. 27. If from the ends of the base of an isosceles triangle seg- ments are drawn making equal angles with the base and termi- nated by the opposite sides, these segments are equal. REVIEW DIAGRAMS 53. Review the proofs of Ths. 3 and 4 by means of the following scheme and diagram: Theorem 4 If three sides of one triangle arc equal to three sides of another, the triangles are congruent. 1 1 \ 1 1 Hyp. Def. iaoBcelea Th.3 Aa.go Th.l triangle | Fig. 74 The theorem to be proved is written down in full with a horizontal line below it (Fig. 74). Below the various ver- tical lines shown are written the references to the authorities on which the proof depends. Some of these references will be theorems. These theorems (2 and 3 above) should then be proved and the references to the authorities used written in as for Th. 4. This process should be con- tinued until only definitions and assumptions remain. 44 PLANE GEOMETRY EXERCISES THAT MAY INVOLVE MORE THAN ONE PAIR OF CONGRUENT TRIANGLES 54. Note. When exercises are selected for review from this or from the next section it is often well to require the proofs for all theorems used in proving the exercises assigned. L Any point in the median to the base of an isosceles triangle is equally distant from the extremities of the base. 2. Are there any special positions of the arbitrary point mentioned in Ex. 1 for which the proof given does not hold? Q 3. In Fig. 75, CA = CB and CD = CE. If AD and BE intersect at O, prove that CO bisects ZC. 4. In Fig. 76, CO is the median to the base AB of the isosceles AABC. If CX = CYy prove that CO bisects ZXOY. 5. Investigate the case, Ex. 4, in which X ^ and Y are on BC and AC extended. ^^^- ^^ 6. In Fig. 77, AC=-AD, BC = BD, AX=^AY. Prove that BX = BY. 7. Investigate the case, Ex. 6, in which X and Y are on AC and AD extended. Fig 77 8. In Fig. 78, CO ± line h, Rays are drawn from point C so that Z 1 = Z 2, intersecting line h at points A and B. A and B are joined with any point in OC extended. Prove that AABD is isosceles. 9. Investigate the case, Ex. 8, in which D is in CO extended. 10. In Fig. 79, CA = CB and AD = BD. If X is any point in DC extended, prove that AX = BX. 11. Investigate the case, Ex. 10, in which X is between C and D. Fig. 79 CONGRUENT TRIANGLES 45 Fig. 80 12. In Fig. 80, CD is any segment. Zl= Z2 and Z3= Z4. Prove that CD is a perpendicular bisector of AB. 13. If, in Fig. 80, A ABC is constructed isosceles and Z>, any point in the bisector of Z.C, is joined to A and B, DA and DB make equal angles with AB. 14. Construct the figures for Exs. 12 and 13 with C and D on the same side oi AB. 15. In Fig. 81, ^ABC is isosceles. CO is the median to the base AB. If Z 1 = Z 2, prove that 1 CX = CV and that OX = OY. 16. In Fig. 82, AC = BD and AD = BC. Prove that AAOB is isosceles. Fig. 82 17. In Fig. 83, A ABC is equilateral, CE = BD = AF. Prove that AXZY is equilateral. 18. In Fig. 83, if A ABC is equilateral and Z1=Z2=Z3, prove that AXYZ is equi- lateral. Fig. 83 19. Prove that the base angles of an isosceles triangle are equal by means of the construction shown in Fig. 84. A ABC is the isosceles A. CX = CY. Analysis: To prove Z1=Z2, prove that /.YAC= /.CBX and that^ /.YAB^ AABX. Note. This method of proving the theorem is given in Euclid's Elements. In the thirteenth century the students of Oxford, England, nicknamed this theorem "elefuga" — that is, "the flight of the wretched" — because most of them found it so difficult that few cared to study the subject further. Two or three hundred years later 46 PLANE GEOMETRY the students called it "pons asinorum," or "the bridge of asses." Very little is known concerning Euclid himself. He was a Greek who lived and taught in Alexandria about 300 B.C. Some of the earlier students of geometry kept the results of their studies secret. Euclid's Elements was a great advance lipon the work of his predecessors, both in arrangement and in rigor. It is said that Ptolemy once asked him "if there was in geometry any shorter way than that of the Elements,'' and he answered that "there was no royal road to geometry." Another story told of him is that some one who had begun to read geometry with EucHd asked, when he had learned the first theorem, "But what shall I get by learning these things?" Euclid called his slave and said, "Give him threepence since he must make gain out of what he learns." 20. Show that the distance between two inaccessible as A and B on opposite sides of a stream, may be found as (see Fig. 85): Set a stake at C, sighting it in line with AB. BC may be any convenient distance. Take D any point from which A, B, and C are visible. Sight E in line with D and C, making ED = DC. Sight F in line with D and B, making FD = DB. Sight G so that it will be in line with F and E and also ^^^ in line with D and A. What line should be measured the distance .45? points, follows . 85 to find 21. Suppose (Fig. 86) that P represents a fence post on one side of a stream and line / a fence on the other side of the stream. A fence is to be built in line with P and perpendicular to fence /. Show that the following method may be used: Fig. 86 Let P be the point and / the line. At any point A in the line / construct a perpendicular to line / and set stakes making AB = AC. On line / sight D in line with P and B, and E in line with C and P. Then sight F in line with D and C and at the same time in line with E and B. PF is 1. I and S is the point where the ± cuts the line /. CONGRUENT TRIANGLES 47 EXERCISES INVOLVING PROPERTIES OF AND TESTS FOR CONGRUENT TRIANGLES AND CONSTRUCTION OF TRIANGLES 66. 1. Corresponding medians of congruent triangles are equah Note. Two proofs may be given: one by means of congruent tri- angles and one by superposition. 2. Two triangles are congruent if two sides and the median to one of these sides are ecjual respectively to two sides and the corresponding median of the other. 3. Construct A ABC so that AB = 7 cm., AC =10 cm., and the median to AC = 4: cm. Suggestion. Draw any triangle ABC and the median to the side A C. Write the numbers 7, 10, and 4 on the proi)er segments. Look at the figure until it is evident to you wliich segments must be constructed first in order that each may have the required lenjiith. t4. Construct a triangle, given two sides and an angle opposite one of these sides. Suggestion. Let a and h be the given sides and ZA he opposite side a. Show how Fig. 87 is constructed. DescTil)e changes in the data given that will alter the results, using the following outline: I. Let ZA be an acute angle. II. Let ZA be a. right angle. III. Let ZA be an obtuse angle. In each case start with side a longer than ^^'" side b, then suppose side a to decrease gradually and note results. Note. This problem is used in trigonometry and surveying. 5. Construct an isosceles triangle so that one leg shall be 9.4 cm. and the median to that leg shall make with that leg an angle of 22K°. 6. Construct A.-li5C so that AB = H.6 cm., 5C = 6.5 cm., and the median to AB makes with AB an angle of 45°. 7. The bisectors of corresponding angles of congruent triangles are equal. 8. Constmct an isosceles triangle so that one base angle shall be H of a right angle and one leg 5 cm. CHAPTER III Parallels, Perpendiculars, Angles, Angle-Sums INTRODUCTORY PRELIMINARY THEOREM: TEST FOR UNEQUAL ANGLES 56. If one side of a triangle is extended, an angle is formed which is called an exterior angle of the tri- angle. Thus in Fig. 88, AB is extended. Z 1 is an exterior angle of AABC, The interior angle numbered 2 is adjacent to Z 1 ; Z 3 and Z 4 are the non-adjacent interior angles. Exercise. How many exterior angles has a triangle? 57. We may add the following to the list of general assumptions : As. 28. If one angle or segment is greater than a second and the second is equal to or greater than a third, then the first is greater than the third. As. 29. The whole is greater than any of its parts. 58. Theorem 8. An exterior angle of a triangle is greater than either of the non-adjacent interior angles. Hypothesis: In AABC the side AB is extended, forming the exterior Z. Conclusion: Zl > ZCor /.BAC. 48 PARALLELS AND ANGLES 49 Fig. 90 Analysis and construction: A. I. To prove Zl > ZC, prove part of Zl= ZC II. " " part of Z1=ZC, bisect CB at D, join AD, and extend, making DE = AD. Join EB and prove Z 2 = ZC. III. To prove Z2= ZC, prove ADBE ^ AADC. B. I. " " Zl > ZBAC, extend CB and prove Z5 = Zl and Z5 > ZBAC. c Let the pupil give the proof. Ex. 1. How many illustrations can you find in Fig. 90 of an exterior angle of a triangle? Show how Th. 8 applies in each case. Ex. 2. In Fig. 90, is any point inside AABC. Prove that ZAOB>ZC. TRANSVERSALS AND ANGLES 59. When two straight Hnes are crossed by a third straight line, various angles are formed which have special names. Thus, in Fig. 91: Zc, Zd, Zw, and Zx are interior angles. Za, Zb, Zjy and Zz are exterior angles. Zc and Zx, also Zd and Zw, are alternate interior angles. Za and Zz, also Zh and Zy, are alternate exterior angles. Zaand Zw, Z6and Zx, Zc and Z;y, also Zd and Zz, are corresponding angles. Exercise. In Fig. 92, name 8 pairs of alternate interior angles, 16 pairs of corre- sponding angles, and 8 pairs of alternate ex- terior angles. How many pairs of supple- mentary adjacent angles are there? Fig. 92 Fig. 91 50 PLANE GEOMETRY PARALLELS 60. Lines in the same plane that do not intersect however far we may follow them are called parallel lines. Two arbitrary straight lines in the same plane will ordinarily intersect and determine a point. Two parallel lines do not determine a point. This definition is the fundamental test for parallels. Five other tests for parallels are con- tained in the group that follows. Th. 9 is the fundamental theorem of the group. Exercise. Find in the room in which you are sitting illustra- tions of parallel lines and of lines which do not intersect and yet are not parallel. FUNDAMENTAL THEOREM: TEST FOR PARALLELS 61. Exercise. Construct two straight Hnes cut by a third straight line so that the alternate interior angles are equal. Theorem 9. If two straight lines in the same plane are cut by a third straight line so that the alternate interior angles are equal, the two straight lines are parallel. Fig. 93 Hypothesis: Lines a and b are cut by Hne c and Z 1 = Z 2. Conclusion: Line a \\ line 6. Analysis: I. To prove line a \\ line b, show that Hne a and Hne b cannot meet either on the right or on the left. IL To prove that line a cannot meet line b, show that if line a met line b, an exterior angle of a triangle would be equal to an opposite interior angle. PARALLELS AND ANGLES 51 Proof: STATEMENTS REASONS I. a. Line a might meet L a. Supposition, line h on the right. 6. Z 1 would be greater 6. Why ? than Z2. c. Zl = Z2. c. Given. d. :. line a does not d. Supposition a meet line b on the leads to a contra- right, diction. IL a. Line a might meet II. a. Supposition, line b on the left. Let the pupil complete the proof. 62. The ordinary direct synthetic proof is explained in §46 and has been used for theorems and exercises in chapter ii. The proof used for Th. 9 is an indirect proof. In technical terms it is called proof by reductio ad absurdum. This is a Latin phrase. What is its hteral translation? For such proofs we must : 1. Determine all the possible cases obtained by contra- dicting the given conclusion. 2. Then, since either the conclusion or one of the con- tradictory statements must be true, we must eliminate all but one of these by proving them absurd. Proofs of this character are very common, not only in mathematics, but in all argument. Their validity depends among other things upon the presentation of all possibilities. The proofs for many of the theorems and exercises that follow are clearer if expressed in algebraic notation. In some cases the solution of an equation is necessary. In other cases the use of algebraic manipulations and iden- tities are required without the solution of an equation. Be definite and accurate. 52 PLANE GEOMETRY DEPENDENT TESTS FOR PARALLELS 63. Theorem 10. If two straight lines in the same plane are cut by a third straight line so that one pair of correspond- ing angles are equal, the two straight lines are parallel. Fig. 94 Hypothesis: Lines a and b are cut by line n so that Z1 = Z2. Conclusion: Line a \\ line b. Analysis: I. To prove a \\ b, prove Z 2 = Z 3. IL '* '' Z 2 = Z 3, compare Z 2 and Z 3 with Z 1. Let the pupil give the proof. Ex. 1. Prove Th. 10 by proving that Z4=Z5. Use sup- plements of equal angles. Theorem 11. If two straight lines in the same plane are cut by a third straight line so that the interior angles on the same side of the transversal are supplements, the two straight lines are parallel. Hypothesis: Lines a and b are cut by line n so that Z24-Z4 = 2rt. Z. Conclusion: Line a \\ line b. Analysis (see Fig. 94) : L To prove a \\ b, prove Z 2 = Z 3. II. " " Z 2 = Z 3, show that Z 2 and Z 3 are each supplements of Z 4. Let the pupil give the proof. PARALLELS AND ANGLES 53 Theorem 12. Two straight lines in the same plane per- pendicular to the same straight line are paralleL Suggestion. Prove the alternate interior angles equal. Ex. 2. Would Ths. 9, 10, 11, and 12 be true if the phrase in the same plane were omitted? CONSTRUCTION OF PARALLELS 64. Problem 6. To draw a straight line through a given point parallel to a given straight line. Show that the problem may be solved by using Th. 9, Th. 10, or Th. 12. Make the construction by each method and prove it. How many lines may be drawn fulfilling the requirements ? Exercise. Show how to solve Prob. 6 by paper folding, or with a ruler and a card, or two draftsman's triangles. FUNDAMENTAL ASSUMPTION REGARDING PARALLELS 65. As. 30. Only one line can be drawn through a given point parallel to a given line. DEPENDENT TEST FOR PARALLELS 66. Theorem 13. Two lines parallel to a third line are parallel to each other. Suggestion. Prove by the indirect method. EXERCISES INVOLVING TESTS FOR PARALLELS 67. To prove two lines parallel, we must prove one of the following: 1. The alternate interior angles are equal; 2. The corresponding angles aYe equal; 3. The. interior angles on the same side of the trans- versal are supplements; 4. They are perpendicular to the same line; 5. They are parallel to the same line. 54 PLANE GEOMETRY Ex. 1. If two straight lines in the same plane are cut by a third straight line so that the alternate exterior angles are equal, the two straight lines are parallel. Ex. 2. If, in Fig. 95, lines h and k are cut by line n so that Za4-Zft = 2 vt.A, prove line k \\ line h. Ex. 3. If any two segments bisect each other, the segments joining the extremities are parallel. Ex. 4. In Fig. 96, ABCD is a four-sided figure with x = y and z = 'w. Prove x \\ y and w II z. ^ ^ ^ Fig. 96 ANGLES MADE BY PARALLELS AND TRANSVERSALS FUNDAMENTAL THEOREM: TEST FOR EQUAL ANGLES 68. Theorem 14. If two parallel lines are cut by a third straight line, the alternate interior angles are equaL u aA h A / Fig. D w 95 / / Fig. 97 Hypothesis: Line h || line k and the transversal line n cuts lines h and k. Conclusion: Z 1 = Z 2. Analysis and construction: I. To prove Zl= Z2, construct AB so that /.BAC = Z2 and prove Z.BAC=- Zl. IL To prove Z.BAC= Z.\, prove BA coincides with line h. III. To prove line BA coincides with line /t, show that BA II line k. and line h || line k. PARALLELS AND ANGLES 55 ProoJ: STATEMENTS REASONS ^I. a. Line h |1 line k. L a. Given b. Line AB \\ line k. h. Why? c. Line AB coincides with c. Only one line can line h. be drawn through a given point par- allel to a given line. IL Z 1 coincides with LB AC. IL Why? in. Z1=Z2. III. Z 1 coincides with an angle that is equal to Z2. DEPENDENT THEOREMS CONCERNING ANGLES MADE BY PARALLELS AND TRANSVERSALS 69. Theorem 15. If two parallel lines are cut by a third straight line, the corresponding angles are equal. Fig. 98 Hypothesis: Line k || line k, and line n cuts lines h and k. Conclusion: Z 1 = Z 2. Analysis: To prove Z 1 = Z2, compare Z 1 with Z3 and Z2 with Z3. Let the pupil give the proof. Theorem 16. If two parallel lines are cut by a third straight line, the interior angles on the same side of the transversal are supplements of each other. Suggestion. To prove that Z2 is the supplement of Z4, prove that Z2 is equal to an angle that is the supplement of Z4. 56 PLANE GEOMETRY APPLICATION OF PARALLELS TO TEST FOR PERPENDICULARS 70. Theorem 17. A line which is perpendicular to one of two parallels is perpendicular to the other. k n h 2 Fig. 99 Hypothesis: Line h \\ line k and line n J_ line h. Conclusion: Line n ± line k. Analysis: I. To prove line n ± line k, prove Z 1 = a rt. Z . II. " " Zl = art.Z, compare Zland Z2. The proof is left to the pupil. Ex. 1. Prove Th. 13 by constructing a line perpendicular to one of the given lines and proving it perpendicular to the others. Use Ths. 17 and 12. Ex. 2. Prove Th. 13 by drawing a transversal and using Th. 9. 71. If Ths. 9 and 14 are compared, it will be seen that they are related to each other in a peculiar way. In Th. 14 two straight lines are given parallel to prove that the alternate interior angles are equal. In Th. 9 thfe alter- nate interior angles are given equal to prove that the two lines are parallel. Th. 14 is Th. 9 turned around. If two theorems are so related that the hypothesis and conclusion of one are respectively the conclusion and the hypothesis of the other, each theorem is called the converse of the other. If a theorem is proved true, its converse may or may not be true. The truth of the converse must not be assumed; it must be proved. Ex. 1. State and investigate the truth of the converse of the theorem: If two triangles are congruent, the angles of one are equal respectively to the angles of the other. Ex. 2. State the converse of Th. 15 and of Th. 16. PARALLELS AND ANGLES 57 EXERCISES INVOLVING ANGLES MADE BY PARALLELS J2. L If, in Fig. 100, h \\ k and Z 6 = 27° 30', find the number of degrees in each angle formed. 2. If two parallel lines are cut by a third h a/i straight line, the alternate exterior angles are equal. 3. If, in Fig. 100, line h \\ line k, prove ^^^ ^^^ that /.a-\- /.x = 2 rt.A. What other angles in the figure can be proved supplementary in the same way? 4. If, in Fig. 101, Z6 = 44°, find the number of degrees in each angle of the figure. 5. If, in Fig. 101, Ax- Zy = 33°, find the number of degrees in each angle of the figure. 6. If, in Fig. 101, ZA is ^^ of Z.y, find Fig. 101 the number of degrees in each angle of the figure. 7. In Fig. 102, /.BAC is any angle. is any point on the ray AB. If the point X starts at point A and moves along the ray AC indefinitely, what are the limiting values of the Z XOA and of the Z BOX ? Investigate the case in which the point X moves from A along AD. Fig. 102 MISCELLANEOUS EXERCISES 73. In several theorems which we have proved, certain lines were constructed and used which were not given in the hypothesis. Such lines are called construction lines. Their use is not only permissible but often necessary. Occasionally they may be more or less arbitrary. When such lines are located defiftitely care should be taken that no facts are assumed which require proof. In general, two points or one point and a direction locate a line. For methods of locating points, lines, rays, or segments see Ass. 1-8. 1. What theorems have we had that were proved by the aid of construction lines? Tell how the line was located in each case. 58 PLANE GEOMETRY A line, ray, or segment that is located definitely is often called a fixed line, ray, or segment. ' The use of construction lines in the following exercises should be carefully noted : 2. Lines which are perpendicular to parallel lines are parallel (Fig. 103). t3. If two angles have their sides parallel right side to right side and left side to left side, the ~ I angles are equal (Fig. 104). Fig. 103 Note. The right side of an angle is the side on the right as one stands in the angle and faces out. t4. If two angles have their sides «// / parallel right side to left side and left / ^ — - — / / side to right side, the angles are supple- F' ' ^' / mentary. Fig. 104 5. The bisectors of a pair of alternate interior angles of parallel lines are parallel. 6. If through the vertices of an isosceles triangle lines are drawn parallel to the opposite sides, a triangle is formed which has two angles equal. 7. How would Ex. 6 read if the given triangle had all of its angles equal? If it had none of its angles equal? Give proof. If, in Fig. 105, line h \\ line k and is an arbi- ±. trary point between the parallels, prove Zb= Za -\- Zc. Fig. 105 9. If a four-sided figure has both pairs of opposite sides parallel, the opposite angles are equal. 10. If a four-sided figure has both pairs of opposite sides parallel and one angle a right angle, all of its angles are right angles. 11. A ray parallel to the base of an isosceles triangle through the vertex bisects the exterior angle at the vertex. 12. What would be true in Ex. 11 if the ray parallel to the base of the isosceles triangle cuts the sides of the triangle or the sides extended? Give proof. PARALLELS AND ANGLES 59 13. If a segment between two parallel lines is bisected, any other segment between the parallels and through the point of bisection is also bisected by this point. 14. In Fig. 106, AW\\BZ, BX = AY, and BZ = AW. Prove that XZllYW. 15. In Fig. 106, if AW \\ BZ, XZ\\ YW, and BX = AY, prove that BZ = AW. Fig. 100 ANGLES IN TRIANGLES FUNDAMENTAL THEOREM 74. Theorem 18. The sum of the interior angles of triangle is two right angles. Fig. 107 Hypothesis: A ABC is any triangle. Conclusion: Zl+Z2+Z3 = 2rt.^. Analysis and construction: I. To prove Zl + Z2 + Z3 = 2 rt.Z, compare Z 1 -h Z 2 + Z 3 with angles whose sum is 2 rt. Z . II. Construct XY through C parallel to AB and com- pare Z 1, 2, and 3 with A 4, 5, and 3 respectively. Let the pupil give the proof. Note. Th. 18 is one of the most famous theorems of geometry. It is supposed that the ancient Greeks knew that it was true for equi- lateral and for isosceles triangles before they l<:ncw that it was true for all triangles. The proof given above is supposed to be that of Pythagoras (about 500 B.C.) and may be one of the earliest proofs for this theorem. Ex. 1. Can you verify Th. 18 by tearing off the corners made by Al, 2, and 3 in Fig. 107 and rearranging them? 60 PLANE GEOMETRY Ex. 2. Are there any other ways of putting in construction lines for Th. 18 so that Z1 + Z2 + Z3 may be compared with angles whose sum is two right angles? Give proof. (See Fig. 108.) Fig. 108 Ex. 3. Find the number of degrees in the third angle of a triangle if the other two angles are: a. 40°, 56° d, 59° 25', 58° 42' b. 29°, 58° 10' e. 72° 16', 68° 42' c. 38° 40', 72° 18' /. 58° 18', 79° 53' COROLLARIES: VALUES AND COMPARISONS OF INTERIOR ANGLES OF TRIANGLES 75. Cor. I. Each angle of an equilateral triangle is 60°. Ex. 1. Construct angles of 30°,- 15°, 75°, 7° 30', 67° 30', 165°, 150°. Cor. II. If two angles of one triangle are equal respec- tively to two angles of a second triangle, the third angles are equal. Fig. 109 Hypothesis: A ABC and AXYZ have ZA= ZX and ZB=ZY. Conclusion: ZC= ZZ. Analysis: To prove ZC= ZZ, Prove- ZA-\-ZB-}-ZC=ZX-\-ZY-}-ZZ, ZA+ZB==ZX-{-ZY. Ex. 2. Segments drawn from an arbitrary point in the base of an isosceles triangle perpendicular to the opposite sides make equal angles with the base. PARALLELS AND ANGLES 61 Cor. IIL The acute angles of a right triangle are com- plements of each other. Ex. 3. The vertex angle of an isosceles triangle is 42°. The perpendiculars are drawn from the ends of the base to the opposite sides. Find the number of degrees in the angles that these per- pendiculars make with the base. THE EXTERIOR ANGLE OF A TRIANGLE 76. Theorem 19. The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles. Fig. 110 Hypothesis: In AABC the side AB is extended, forming the exterior Z XBC. Conclusion: Z XBC= Z 1 + Z 2. Analysis and constrtcction: L To prove ZXBC= Z 1 + Z2, divide ZXBC into two parts and compare these parts with Z 1 and Z 2 respectively. II. .*. draw BY through B\\AC and compare Z 4 with Z 2 and Z5 with Zl. Let the pupil give the proof. Ex. 1. An exterior angle of a triangle is 145°, one of the opposite interior angles is 62°. Find the number of degrees in each of the angles of the triangle. Ex. 2. If the exterior angle at the vertex of an isosceles triangle is 128°, find the number of degrees in each of the angles of the triangle. Ex. 3. Find the sum of the exterior angles formed when the hypotenuse of a right triangle is extended in each direction, 62 PLANE GEOMETRY EXERCISES INVOLVING THE ANGLES OF A TRIANGLE 77. 1. If one angle of a triangle is double the second, and the third is double the first, find each angle of the triangle. 2. How many degrees in each angle of an isoscples triangle if the vertex angle is (1) three times the sum of the base angles? (2) ^2 the sum of the base angles? (3) \i the sum of the base angles? (4) equal to the sum of the base angles? 3. A triangle can have only one right angle or one obtuse angle. May a triangle have a right angle and an obtuse angle? 4. If the vertex angle of an isosceles triangle is 42°, find the number of degrees in the angles at the intersection of the bisectors of the base angles. Find also the number of degrees in the angles at the intersection of the bisectors of the exterior angles at the base of the triangle. 5. In Fig. Ill, J\,ABC is isosceles. AO and OB bisect A A and B respectively. If the legs CA and CB are made to increase and to decrease in length, what is the upper limit to AAOB'i What is the lower limit? ^i^- ^ 6. If the vertex angle of an isosceles triangle is 62°, find the number of degrees that the bisector of one base angle makes with the opposite side. 7. If the vertex angle of an isosceles triangle is 70°, find the number of degrees in the angles at the intersection of the perpen- diculars drawn to the equal sides from the extremities of the base. 8. In Fig. 112, l\ABC is isosceles. AX and BY are per- pendicular to BC and AC respectively. If the legs c AC and BC are made to increase and to decrease in length, what is the upper limit to ZAOB ? What is its lower limit? 9. If, in AABC, ZA is 4 times ZC and ZB is 3 ^ig. 112 times Z C, find the number of degrees in each of the angles of the triangle. 10. Find the number of degrees in each of the angles of an isosceles triangle if each base angle is (1) >! the vertex angle; (2) 3 times the vertex angle. PARALLELS AND ANGLES 63 ANGLES IN POLYGONS 78. If several segments are joined end to end and the free end of the last is joined to the free end of the first, the figure formed is called a polygon. The segments are called the sides of the polygon; the common end points of the segments are called the vertices. Segments joining any two non-consecutive vertices are called diagonals. The sum of the sides is called the perimeter of the polygon. A polygon is said to be convex if no side can be extended so as to enter the polygon. Otherwise it is said to be concave and has one or more re-entrant angles. Fig. 113 In Fig. 113 polygon 1 is convex; polygon 2 is concave with one re-entrant angle; polygon 3 is concave with two re- entrant angles; polygon 4 is a cross polygon. Hereafter unless otherwise stated a convex polygon is intended. Polygons are named according to the number of sides: a polygon of 3 sides is called a triangle ; a polygon of 4 sides is called a quadrilateral; a polygon of 5 sides is called a pentagon ; a polygon of 6 sides is called a hexagon; a polygon of 7 sides is called a heptagon; a polygon of 8 sides is called an octagon ; a polygon of 10 sides is called a decagon; a polygon of 12 sides is called a duodecagon; a polygon of 15 sides is called a pentadecagon. Polygons are sometimes called by their English instead of by their Latin or Greek names, thus: 4-side, 7-side, 8-side, 9-side, etc. 64 PLANE GEOMETRY A polygon with all of its sides and all of its angles equal is a regular polygon. THE SUM OF THE ANGLES OF ANY POLYGON 79. Ex. L Find the sum of the four angles of a quadrilateral. Ex. 2. Find the sum of the five angles of a pentagon (Fig. 114). Analysis: To find the value of ZA + ZB+ ZC+ ^D-\- ZE, i divide the polygon into triangles whose vertices are the , //' |\ vertices of the polygon and find the sum of all the / j \ angles of all the triangles. .*. draw the diagonals from / ! >^^ one vertex and multiply the sum of the angles in one ^^^p triangle by the number of triangles. pj^ j j^ Ex. 3. Find the sum of the angles of a hexagon; of an octagon; of a decagon. Theorem 20. The sum of the iriterior angles of a poly- gon of n sides is 2(n-2) right angles. Let the pupil give the analysis. Proof: STATEMENTS 1. The diagonals divide the polygon into tri- angles. 2. The sum of the angles of each triangle is . 3. The sum of the angles of the triangles is . Ex. 4. By substituting the proper number for n in the formula given in Th. 20 find the sum of the angles of a hexagon; of an octagon; of a 15-side; of a 16-side; of a 20-side; of a 24-side; of a 32-side. Fig. 115 Ex. 5. Prove Th. 20 by means of the constructions shown in Fig. 115. PARALLELS AND ANGLES 65 Ex. 6. How many degrees in each angle of a regular octagon? of a regular pentagon? of a regular decagon? of a regular 12-side? of a regular 16-side? of a regular 20-side? of a regular 24-side? of a regular n-side? Ex. 7. Is it possible to have a regular polygon each of whose angles is 108°? 150°? 144°? 128°? 160°? If such polygons are possible, how many sides would there be in each case? Ex. 8. Find the sum of the interior angles of a peAtagon that has one re-entrant angle; of an octagon with two re-entrant angles. Ex. 9. How many regular triangles can be placed adjacent with their vertices at the same point? Will the space about the point be filled exactly? Why? Ex. 10. Can regular hexagons be placed with their vertices at the same point and the space be filled exactly? Why? Can regular quadrilaterals be so placed? regular pentagons? regular octagons? Why? THE SUM OF THE EXTERIOR ANGLES OF ANY POLYGON 80. Ex. 1. Find the sum of the exterior angles of a pentagon. Analysis: To find the sum of the exterior angles of ABCDE, subtract the sum of the interior angles from the sum of the interior and exterior angles (Fig. 116). ^^ -_ Let the pupil give the proof. Fig. 116 Ex. 2. Find the sum of the exterior angles of an octagon; of a decagon; of a 12-side. Theorem 21. The sum of the exterior angles of a poly- gon of n sides is four right angles. Let the pupil give the analysis and the proof. Ex. 3. Is it possible to have a regular polygon each of whose exterior angles is 24°? 36°? 40°? If so, how many sides would these polygons have? 66 PLANE GEOMETRY MISCELLANEOUS THEOREMS TEST I FOR CONGRUENT RIGHT TRIANGLES 81. A triangle that contains a right angle is called a right triangle. The side opposite the right angle of a right triangle is called the hypotenuse. The perpendicular sides are called the legs of the right triangle. Ex. 1. Construct a right triangle with the hypotenuse equal to a given segment and one acute angle equal to a given angle. Theorem 22. Two right triangles are congruent if the hypotenuse and an acute angle of one are equal to the hypotenuse and an acute angle of the other. Fig. 117 Hypothesis: In AABC and AXYZ, AC = XZ, AA = Z X, and Z B and Z Y are rt. A . Conclusion: AABC m AXYZ. Analysis: To prove AABC ^ AXYZ, prove ZC= ZZ. Let the pupil give the proof. Use Cor. II, §75. Ex. 2. Perpendiculars dropped from the mid-points of the equal sides of an isosceles triangle to the base are equal. Ex. 3. Perpendiculars from the mid-point of the base of an isosceles triangle to the legs are equal. Ex. 4. If, in Fig. 118, CO is the perpendicular bisector of AB, what are the limiting values of the length of the perpendicular from O to the segment BX as X moves along the ray OC? PARALLELS AND ANGLES 67 TEST II FOR CONGRUENT RIGHT TRIANGLES 82. Ex. 1. Construct a right triangle with the hypotenuse and one side equal respectively to given segments. Theorem 23. Two right triangles are congruent if the hypotenuse and a side of one are equal to the hypotenuse and a side of the other. Fig. 119 Hypothesis: In the A ABC and XYZ, AC = XZ, BC = YZ, and Z B and ZY are rt. A . . Conclusion: AABC ^ AXYZ. Analysis and construction: I. Toprove AA^C ^ AXYZ, prove ZA= ZX, II. To prove ZA = ZX, place AABC so that BC falls on YZ, B on Y, C on Z, and A opposite X, and prove XYAZ an isosceles triangle. III. To prove XYAZ a triangle, prove XYA a straight line (As. 19). Let the pupil give the proof. For I use Th. 22. Cor. If a perpendicular is erected to a straight line, equal segments drawn from the same point in the per- pendicular cut off equal distances from the foot of the perpendicular. Ex. 2. A line from the center of a circle perpendicular to a chord of the circle bisects the chord. Ex. 3. Construct an isosceles triangle, given one leg and the perpendicular from the vertex to the base. Ex. 4. Perpendiculars drawn from an arbitrary point in the bisector of an angle to the sides of the angle are equal. 68 PLANE GEOMETRY TESTS FOR ISOSCELES TRIANGLES 83. Theorem 24. If two angles of. a triangle are equal, the triangle is isosceles. The analysis and the proof are left to the pupil. We now have two tests for isosceles triangles. To prove a triangle isosceles, prove that L Two sides are equal, or II. Two angles are equal. Of these two tests the first is derived from the definition of an isosceles triangle and is therefore the fundamental one. Ex. L The bisectors of the base angles of an isosceles triangle form a second isosceles triangle. Ex. 2. The bisectors of the exterior angles at the base of an isosceles triangle form a second isosceles triangle. Ex. 3. Construct an isosceles triangle whose base shall be equal to a given segment and whose vertex angle shall be equal to a given angle. Ex. 4. In Fig. 120, ABC is an isosceles triangle with CA = CB. If CD — CE and AD and BE intersect a^ at 0, prove that AAOB is isosceles. Fig. 120 Ex. 5. In Fig. 121, AABC is isosceles. AX = BY and CZ = CW. Extend XZ and YW to meet at / c^ 0. Prove that AXOY is isosceles. zy^w AX Y B Ex. 6. If, in Fig. 121, AXOY is constructed Fig. 121 isosceles, XA = YB and XZ=YW, prove that AABC is isosceles. PROPERTY OF ISOSCELES TRIANGLES 84. Theorem 25. A segment from the vertex of an isos- celes triangle perpendicular to the base bisects the base and the vertex angle. The analysis and the proof are left to the pupil. For other properties of isosceles triangles, see Th. 6 and Th. 7. PARALLELS AND ANGLES 69 Fig. 122 SUPPLEMENTARY EXERCISES EXERCISES INVOLVING ANGLES OF POLYGONS 85. 1. In Fig. 122, ABCDE is a convex polygon of five sides, all of whose angles are obtuse. The sides of the polygon are extended until they intersect, form- ing the star polygon shown. Find the sum of the angles in the points of the star. What would be the sum of the angles in the points of the star if the polygon had six sides? If it had eight sides? 2. Suppose the convex polygons used in Ex. 1 were regular, how many degrees would there be in the angle at any point of the star polygons formed? 3. Fig. 123 shows a regular triangle with each side divided into three equal parts. If the points are joined as indicated, prove that DEFGHK is a regular hexagon. 4. Fig. 124 shows a kite formed of two regular triangles with the same base AC. If the sides are bisected and the points joined as shown, prove that AEFCGH is a regular hexagon. 5. Fig. 125 shows a regular hexagon ABCDEF, If alternate sides are extended in both directions as indicated, prove that a regular triangle XYZ is formed. Note. Tiled and mosaic floors are often made of equilateral triangles so colored that various patterns are formed. Fig. 126 XJLXJL ^ Fig. 126 represents three of these designs containing the figures used in the preceding exercises. The use of equilateral triangles, squares, and hexagons for tiles probably dates back to the ancient Egyptians. 70 PLANE GEOMETRY MISCELLANEOUS EXERCISES 86. Note. Be prepared to prove the theorems on which each of the following exercises depends. A perpendicular from any vertex of a triangle to the opposite side is called an altitude of the triangle. 1, Make review diagrams for Ths. 11, 15, 19, and 25. 2. If ABC is an isosceles triangle and if the perpendiculars A Y and BX drawn from the extremities of the base AB\.o the equal sides BC and ^C intersect at 0, t\AOB is isosceles. 3. In Fig. 127, ^5 is any segment, Zl= Z2, and AD = BC. From D and C perpendiculars are drawn to AB. What segments and angles are Fig. 127 equal? Why? 4. Investigate the case, Ex. 3, in which Zl and Z2 are obtuse angles. Give proof. 5. If the ray which is drawn through the vertex of a triangle parallel to the base bisects the exterior angle at the vertex, the triangle is isosceles. 6. A segment drawn from an arbitrary point in the bisector of an angle to one side of the angle and parallel to the other side forms with the bisector and the side to which it is drawn an isosceles triangle. 7. In Fig. 128, BD\\ XY. AB bisects ZX^IC and AD bisects Z CA F. Prove that BC = CD. x_ 8. Construct an isosceles triangle with the vertex angle Vs of a right angle and the alti- b c" tude 4.7 cm. ^^^- ^28 9. In Fig. 129, ABC is an isosceles triangle with D any point in ^C extended. From D a per- pendicular is drawn to AB cutting CB at E. Prove that CDE is an isosceles triangle. 10. Investigate the case, Ex. 9, in which the point F is on J 5 extended. PARALLELS AND ANGLES 71 U. In Fig. 130, ABC is an isosceles right triangle with ZC a right angle. CD is perpendicular from Cto AB. Construct XY=YC and YZ= YC. Note. The patterns used in applied design are usually made by repeating at regular intervals some very simple figure called the unit. Figs. 131 and 132 show two parquet floor patterns. Show how Fig. 130 in whole or in part is used in each of these designs. The possi- bilities of Fig. 130 as a design unit may be discovered by making four or eight copies of the figure, coloring the spaces to suit one's fancy, cutting the figures out, and fitting them together in various ways. ^^ m m Fig. 131 Fig. 132 ABC is an 12. Construct the fan truss shown in Fig. 133. isosceles triangle. AD = DC = CE = EB. AF = FD = DG = GC. BK = KE = EH = HC. <^^ Compare LEDC and LDAC. A l> ^- 13. Construct an isosceles triangle, given p^^, jgg the perimeter equal to a given segment and From a roof truss dcsiKn one base angle equal to a given angle. Suggestion. Study Fig. 133. CD + CE+DE is to be equal to the given segment AB, and Z.EDC is to be equal to the given angle. 14. Construct a triangle, given the perimeter and the two base angles. 15. Show that a sailor may calculate his dis- tance from a lighthouse as follows: In Fig. 134 C represents the lighthouse and BA the direction in which the boat is moving. With the proper instru- ments he may take the angle CBA. Then, if he notices the moment when the angle ( Z CAD) which the lighthouse makes with his course is twice Z B, he can find the length of CA, for he can obtain the length oi AB from his log. (From the Final Report of the National Committee of Fifteen.) Fig. 134 72 PLANE GEOMETRY 16. In surveying, it is often necessary to run a line, as CD (Fig. 135), so that it will be in the same straight line with LA but on the othfer side of some obstacle to vision, such as a house ^^^ or wood. Show that the following method Fig. 135 will give the desired result: Lay off the Z a so as to clear the obstacle. Take AB, a. convenient distance. Lay off Zb = 2Za. Make BC=BA, Lay off Zc= Za. ^ 17. Perpendiculars from the ends of the base of an isosceles triangle to the opposite sides are equal. 18. In any AABC perpendiculars from points A and B to the median to the side AB are equal. For what special case of the arbitrary AABC is the proof of this exercise meaningless? 19. If two parallel lines are cut by a third straight line, the bisectors of the interior angles on the same side of the transversal are perpendicular to each other. 20. If two parallel lines are cut by a third straight line, the bisectors of the four interior angles form a quadrilateral with four right angles. 21. If a perpendicular be drawn from the vertex of the right angle of a right triangle to the hypotenuse, the two triangles formed are mutually equiangular. 22. Construct an isosceles triangle with altitude 4.3 cm. and the base angles each H of a right angle. 23. Corresponding altitudes of congruent triangles are equal (Fig. 136). 24. Two triangles are congruent if two ^ g, sides and the altitude to the third side of one are equal respectively to two sides and the altitude to the third side of the other (Fig. 136). B D Fig. 136 25. Prove Ex. 24 when /.B and Z £ (Fig. 136) are obtuse. 26. Construct AABC so that AC = 10 cm., BC = 7 cm., and the altitude on AB = 6 cm. PARALLELS AND ANGLES 73 27. Two triangles are congruent if two sides and the altitude on one of these sides in one triangle are equal respectively to two sides and the corresponding altitude of the other. 28. Construct A ABC so that AB = 7 cm., AC^b cm., and the perpendicular from C to ^45 = 3 cm. 29. The line bisecting the exterior angle at the vertex of an isosceles triangle is parallel to the base. 30. The angle made by the bisectors of the base angles of an isosceles triangle is equal to the exterior angle at the base of the triangle. 3L If ABD is an isosceles triangle with DA and DB equal sides, and if i4D is extended through D to point C until DC = DA, prove that BC is perpendicular to AB. 32. If an exterior angle of a triangle is bisected and also one of the interior non-adjacent angles, the angle, made by the two bisectors is ^ the other interior non-adjacent angle. 33. Construct through a given point a ray that shall make a given angle with a given line. Is there more than one solution for this problem? 34. If two medians of a triangle are extended beyond their bases and segments are taken on the extended lines equal to the corresponding medians, the points thus found and the other vertex of the triangle are on a straight line that is parallel to the opposite side of the triangle. 35. In Fig. 137, ABC is an isosceles triangle with ZA twice ZC. Find the number of degrees in each angle shown in the figure ii AX bisects ZA. Note. Fig. 137 cannot be constructed at present without a protractor. Fig. 137 36. In Fig. 138, ABC is an isosceles triangle with ZB = 54°. AX±CB from A . Find the number of degrees in Z 1 and Z 2. 37. Construct AABC with ZA=30°, ZB = 45°, and the perpendicular from C to AB 6.4 cm. Pig. 138 74 PLANE GEOMETRY t38. In each of the figures shown in Fig. 139, lines a and a' are perpendicular to each other, also lines b and b'. Prove that Zab= Za'b'. Suggestion. Apply Cor. Ill, §75. Fig. 139 139. Investigate the truth of the statement that if two angles have the sides of one perpendicular respectively to the sides of the other the angles are equal. 40. In Fig. 140, A BCD is a four-sided figure with its sides equal and its angles right angles. AE = BF = BG = CH = etc. EX and HY are parallel to ^C; XF and MW are parallel to DB. NW, KZ, GY, and LZ are similarly drawn. Prove that ANOE, EOF, and EXF are isosceles. How many isosceles triangles does the figure contain? What triangles in the figure are congruent? If H and E are joined, prove EH || AC. 41. Find the relation between the base angles of two isosceles triangles if the vertex angles are supplementary. 42. If a segment meets the sides of an isosceles triangle at equal distances from the vertex, it is parallel to the base. 43. If the sides of two angles are parallel right side to right side and left side to left side, the bisectors of the angles are parallel. / 44. Find the number of degrees in the angles at the intersec- tion of the bisectors of the acute angles of a right triangle. 45. Prove that a convex polygon cannot have more than three obtuse exterior angles or more than three acute interior angles. CHAPTER IV Quadrilaterals SYMMETRY 87. We have seen in chapter ii that under certain cir- cumstances two figures can be placed one upon the other so as to coincide exactly. Moreover, if we wish to prove two segments or two angles equal, we often look for two triangles that contain these segments or angles and try to prove these triangles congruent. In many cases we can make one of these triangles coincide with the other by folding the figure along some line in the figure, or by rotating part of the figure about some one point. No. 1 Fig. 141 shows four figures of this kind which we have had. If Nos. 1 and 2 are folded along the line AB, the two parts will coincide. If Nos. 3 and 4 are rotated about point O through 180°, each figure will coincide with its original impression. Figures or parts of figures that can be made to coincide in either of these ways are said to be symmetric. Points, lines, segments, or angles that coincide under these circumstances are said to be symmetric to each other. Exercise. Can you find any other figures in chapter ii or chapter iii which are symmetric? Can you find any figures in these chapters which are not symmetric? 75 76 PLANE GEOMETRY DEFINITIONS OF AXIAL SYMMETRY 88. A figure is said to be symmetric with respect to a line as an axis if one part coincides with the remainder when it is folded on that line as an axis (Fig. 142). Two figures are said to be symmetric with respect to a line as an axis if one figure coincides with the other when the plane in which it lies is folded on that line as an axis > (Fig. 143). Such a figure or such figures are said to have axial symmetry. Fig. 143 THEOREMS AND EXERCISES INVOLVING AXIAL SYMMETRY 89. Theorem 26. The bisector of the vertex angle of an isosceles triangle is an axis of symmetry of the triangle. Analysis: ^ To prove that CD is an axis of symmetry of A ABC (Fig. 144), prove that AACD will coin- ^ ^ ^ cide with ABCD if AABC is folded on CD as Fig. 144 an axis. Ex. 1. The end points of a segment are symmetric with •respect to the perpendicular bisector of that segment as an axis. Ex. 2. What axes of symmetry have two parallel lines? Ex. 3. Show how to place two congruent triangles so that they are symmetric with respect to a side of one as an axis. 90.- Theorem 27. Two polygons are symmetric with respect to an axis if the vertices of one are symmetric to the corresponding vertices of the other. Ex. 1. How would Th. 27 read if the two polygons were parts of the same polygon? Ex. 2. Show how to construct a pentagon symmetric to a given pentagon with a given line as axis, QUADRILATERALS 77 DEFINITIONS OF CENTRAL SYMMETRY 91. A figure is said to be symmetric with respect to a point as a center if one part of the figure coin- cides with the remainder when it is rotated through an angle of 180° about the point as a fcenter (Fig. 145). Fig. 145 Two figures are said to be symmetric with *?=^^^ respect to a point as a center if one figure ^^^ •/ coincides with the other when it is rotated /;\ through an angle of 180° about the point as ^\|^::i^ a center (Fig. 146). Fig, ^q Such a figure or such figures are said to have central symmetry. These definitions give us the following test for central symmetry: A figure is symmetric with respect to a point as a center if for every point in it there is a cor- responding point so situated that the two points are sjrmmetric with respect to the center. Can you state a similar test for axial symmetry? THEOREMS AND EXERCISES INVOLVING CENTRAL SYMMETRY 92. Ex. 1. The center of symmetry of two points is the mid- point of the segment joining the two points. Ex. 2. Two vertical angles are symmetric with respect to their vertex as a center. Ex. 3. Find a center of symmetry of two parallel lines. How many such centers are possible? Theorem 28. Two polygons are symmetric with respect to a center if the vertices of one are symmetric to the cor- responding vertices of the other. Ex. 4. How would Th. 28 read if the two polygons were halves of the same polygon? Ex. 5. Which letters of the alphabet have central symmetry? Which ones are symmetric with respect to an axis? 78 PLANE GEOMETRY RELATION BETWEEN AXIAL AND CENTRAL SYMMETRY 93. Theorem 29. Any figure that has two axes of sym- metry at right angles to each other has the intersection of the axes as a center of symmetry. v> Fig. 147 Hypothesis: P is a point on any figiire which is symmetric with respect to xx' and yy' as axes. Axis xx' _L axis yy' . Conclusion: 0, the intersection of xx' and yy' , is the center of symmetry of the figure. Analysis: I. To prove the figure s)mimetric with respect to point O, prove that for every point in the figure there exists a point symmetric to it with respect to as a center. II. .*. let P be any point in the figure and P' be sym- metric to P with respect to yy' and P" be symmetric to P' with respect to xx' and prove O the mid-point of PP". III. To prove O the mid-point of PP", join PO, P'O, P"0, and prove PO = P'V and POP" a straight line. IV. To prove PO = P"0, prove them both equal to P'O. V. To prove that POP" is a straight line, prove that Zi-1-Z2+Z3+Z4 = 2rt. A. VI. To prove that Z 1+ Z2+ Z3+ Z4 = 2 rt. A, prove Z1=Z2, Z3=Z4, and Z2-}-Z3 = l rt. Z. Let the pupil give the proof in full. QUADRILATERALS 79 Ex. L How many axes of symmetry has an equilateral tri- angle? Has it a center of symmetry? Ex. 2. Prove that a quadrilateral with its four sides equal has a center of symmetry. Note. Symmetric figures are much used in ornament. Designers make constant use of the idea of symmetry. Illustrations may bo found in ornamental windows, wall paper, etc. Symmetry also occurs in nature, as in snow crystals, flowers, etc. Fig. 148 shows three sym- metric encaustic tile designs. Let the pupil find other illustrations. How might kaleidoscopes and mirrors be used by designers? K^SZrasa ^V/?f^1 Fig. 148 PARALLELOGRAMS DEFINITIONS 94. Ex. 1. Construct ZABC = (jO°. Make ^5 = 5.3 cm. and BC = G.7 cm. From A construct AD parallel to ^C. From C construct CD parallel to AB. Ex. 2. Construct a quadrilateral with its opposite sides parallel. Make one angle 45° and the sides that include the angle 3.8 cm. and 5.9 cm. respectively. A quadrilateral with each side parallel to its opposite is called a parallelogram. In Fig. 149, X and z, also w and y, are called opposite sides ; w and x are called consecutive sides; A A and C are called opposite angles; A A and D are called consecutive Fig. 149 angles; AC and BD are called diagonals. While any side may be considered as the base, x and z are usually called the bases. 80 PLANE GEOMETRY PROPERTIES OF PARALLELOGRAMS 95. The fundamental characteristic of parallelograms is stated in the definition, namely: The opposite sides are parallel. The next three theorems depend directly upon this fact. Theorem 30. Each diagonal of a parallelogram divides it into two congruent triangles. Theorem 31. The opposite sides of a parallelogram are equal. Theorem 32. The opposite angles of a parallelogram are equal. Ex. 1. Two consecutive angles of a parallelogram are supple- mentary. Ex. 2. The sum of the angles of a parallelogram is four right angles. 96. The diagonals of a parallelogram differ from the diagonals of other four-sided figures in important respects. Note the exercise on p. 81. We shall assume that the diagonals of convex quadri- laterals intersect. Theorem 33. The diagonals of a parallelogram bisect each other. Analysts: I. To prove that the diagonals bisect each other, prove that AO = OC and that BO = OD. IL To prove I ^^^^^, prove ADOC ^ AAOB . QUADRILATERALS 81 Exercise. Fig. 151 shows a parallelogram, a convex quadri- lateral, and a concave quadrilateral. In V ~~7 / 1 i -i which cases do the diagonals intersect? In \\ ^^ 1 / / which case do they bisect each other? v ^^~^ Fig. 151 97. Theorem 34. The intersection of the diagonals of a parallelogram is the center of symmetry of the parallelogram. CONGRUENCE OF PARALLELOGRAMS 98. The OR I'M 35. Two parallelograms are congruent if two sides and the included angle of one are equal to two sides and the included angle of the other. Fig. 152 Hypothesis: In [s] ABCD and A'B'CD\ w = w', x = x\ and ZA= ZA'. Conclusion: /Z7 ABCD U EJ A 'B'CD'. Analysis: To prove CJ ABCD ^ OJ A'B'CD\ prove that they will fit when superposed. Proof: STATEMENTS I. Place 'eJABCD upon EJ A'B'C'D' so that the sides of ZA will fall on the sides of Z.A\ x lying along x' and w along w' . II. Point B will fall upon point B'. III. Point D will fall upon point D'. IV. 1. ZD and /.D' are supplements of ^ ^4 and A\ respectively. .'. ZD= AD'. 2. z will fall along the line of z\ V Let the pupil give all reasons and complete the proof. 82 PLANE GEOMETRY TESTS FOR PARALLELOGRAMS 99. The definition of a parallelogram is the fundamental test for parallelograms. Other tests are dependent primarily upon the fundamental one. 100. Exercise. Construct a quadrilateral A BCD so that the opposite sides AD and BC are parallel and equal. Theorem 36. If a quadrilateral has one side equal and parallel to its opposite, it is a parallelogram. Fig. 153 Hypothesis: In the quadrilateral ABCD, x = z and x\\z. Conclusion: ABCD is a ZZ7. Analysis and construction: I. To prove ABCD a /Z7, prove w \\ y. II. *' " w\\y, draw AC and prove Z 1 = Z 2. III. " " Z1=Z2, prove Let the pupil complete the analysis and give the proof. 101. Exercise. Construct a quadrilateral ABCD so that the opposite sides, AD and BC, also the sides AB and DC, are equal. Theorem 37. If a quadrilateral has each side equal to its opposite, it is a parallelogram. Fig. 154 Hypothesis: In the quadrilateral ABCD, x = z and w = y. Conclusion: ABCD is a O. The analysis and the proof are left to the pupil. QUADRILATERALS 83 102. Thp:orkm 3S. If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. EXERCISES INVOLVING TESTS FOR PARALLELOGRAMS 103. To prove that any given quadrilateral is a parallelo- gram, prove that it has (1) Each side parallel to its opposite, or (2) Each side equal to its opposite, or (3) One side equal and parallel to its opposite, or (4) The diagonals bisecting each other. 1. Construct a parallelogram, given two sides and the included angle. In how many ways is this possible? 2. If two sides of a quadrilateral are parallel and two opposite angles are equal, the figure is a parallelogram. 3. If A BCD is a parallelogram and E and F are respectively the mid-points of the opposite sides AB and CD, prove that AECF is a parallelogram. 4. Given EJ ABCD, with points E and F on the diagonal .4C so that AE = CF, prove Fig. 155 that BFDE is a parallelogram (Fig. 155). „ 5. Investigate Ex. 4 if points E and F are on the diagonal extended. 6. Given HJABCD, AE = CG, AH = CF, prove that EFGH is a parallelogram (Fig. 156). Fig. 156 7. Construct a parallelogram, given the sides and one diagonal. THEOREMS AND EXERCISES INVOLVING TESTS FOR EQUAL AND PARALLEL SEGMENTS 104. Ths. 36 and 37, together with the definition of a par- allelogram, give an additional test for equal and for parallel segments : To prove two segments equal or parallel, find a quadri- lateral of which these segments are opposite sides and prove that the quadrilateral is a parallelogram. 84 PLANE GEOMETRY Theorem 39. Segments of parallels intercepted between parallel lines are equal. Theorem 40. Segments of perpendiculars intercepted between parallel lines are equal. Ex. 1. Construct through a given point a line that shall be parallel to a given line. Ex. 2. If A BCD is a parallelogram and E and F are respec- tively the mid-points of the opposite sides A B and CD, prove that AF\\CE. Ex. 3. Given CJABCD with E, F, G, and H the mid-points of the sides, and the points joined as indicated in the figure; prove that AXCY is a parallelogram (Fig. 157). Ex. 4. In what other way may the points ^ f^ b in Fig. 157 be joined so as to form a parallelo- ftg. 157 gram? Ex. 5. Construct a line that shall be intercepted by the sides of a given triangle and be parallel to the base of the triangle and shall have a given length. SPECIAL QUADRILATERALS 105. A four-sided figure is called a quadrilateral. Unless otherwise stated, a quadrilateral should be drawn with no sides equal and no sides parallel. The following special kinds of quadrilaterals are important : A quadrilateral formed by two isosceles triangles on oppo- site sides of the same base is called a kite (Fig. 158, No. 1). A quadrilateral with but one pair of parallel sides is called a trapezoid (Fig. 158, No. 2). If a trapezoid has its non-parallel sides equal, it is called an isosceles trapezoid (Fig. 158, No. 3). The non-parallel sides of a trapezoid are sometimes called the legs of the trapezoid. The parallel sides of a trapezoid are called the bases. QUADRILATERALS 85 The perpendicular distance between the bases of a trape- zoid is called the altitude of the trapezoid. In Fig. 158, No. 2, b and b' are the bases and a is the altitude. A quadrilateral with two pairs of parallel sides has been defined as a parallelogram. The perpendicular distance between the bases of a paral- lelogram is called the altitude of the parallelogram. A par- allelogram has two altitudes, since each pair of parallel sides may be 'Considered as bases. In Fig. 158, No. 4, b may be considered as a base with a as the ' corresponding altitude, or b' may be considered as a base with a' as the cor- responding altitude. The following special kinds of parallelograms are of con- siderable importance and of widespread occurrence: A parallelogram with one right angle is called a rectangle (Fig. 158, No. 5). A parallelogram with two consecutive sides equal is called a rhombus (Fig. 158, No. 6). A rectangle with two consecutive sides equal is called a square (Fig. 158, No. 7). The segment joining the mid-points of two opposite sides of a quadrilateral is called a median of the quadrilateral (segment XY, Fig. 158, No. 8). The definition of any particular figure is the fimdamental test for the determination of that figure. 7 86 PLANE GEOMETRY KITES 106. Ex. 1. One diagonal of a kite is an axis of symmetry. Ex. 2. The axis of symmetry of a kite bisects the angles through whose vertex it passes. Theorem 41. The diagonals of a kite are perpendicular to each other, and the one which is an axis of symmetry bisects the other. ISOSCELES TRAPEZOIDS 107. Ex. 1. The base angles of an isosceles trapezoid are equal and the diagonals are equal. Ex. 2. The segment joining the mid-points of the parallel sides of an isosceles trapezoid is an axis of symmetry. - Ex. 3. If the base angles of a trapezoid are equal, the trape- zoid is isosceles. Ex. 4. Construct an isosceles trapezoid, given the two bases and the altitude. Ex. 5. Construct an isosceles trapezoid, given two consecutive sides and the included angle. RECTANGLES 108. Ex. 1. Construct a rectangle, given two adjacent sides. Theorem 42. All the angles of a rectangle are right angles. Ex. 2. What properties has a rectangle by virtue of the fact that it is a parallelogram? Ex. 3. The diagonals of a rectangle are equal. Ex. 4. If the diagonals of a parallelogram are ^^°- ^^^ equal, the parallelogram is a rectangle (Fig. 159). Analysis: I. To ipToveOJABCD a n , prove ZA=aTt.Z. II. •' *' Z^=art. Z, prove Z.4 = Z5. III. " •' Z^ = ZB, prove Let the pupil complete the analysis and give the proof. Ex. 5. Construct a rectangle, given one side and one diagonal. QUADRILATERALS 87 Ex. 6. Construct a rectangle, given one diagonal and the angle between the diagonals. Ex. 7. The medians of a rectangle bisect each other at right angles. Ex. 8. Make a list of all of the properties of the rectangle. Which of these properties are special properties of the rectangle? RHOMBUSES 109. Ex. 1. Construct a rhombus, given one side and one angle. Theorem 43. All the sides of a rhombus are equal. Theorem 44. The diagonals of a rhombus are perpen- dicular to each other and bisect the angles through which they pass. Ex. 2. The diagonals of a rhombus are axes of symmetry. Ex. 3. If the diagonals of a quadrilateral bisect each other at right angles, the figure is a rhombus. Ex. 4. The altitudes of a rhombus are equal. Ex. 5. Construct a rhombus, given one side and one diagonal. Ex. 6. Construct a rhombus, given one side and the altitude. When is this problem impossible? Ex. 7. What properties has a rhombus by virtue of the fact that it is a parallelogram? Make a list of all the properties of the rhombus. Which of these are special properties of the rhombus? SQUARES 110. Ex. 1. Construct a square, given one side. Ex. 2. Show that a square may be classified as a special kind of a rectangle, or rhombus. From these facts make a list of all the properties of the square. Ex. 3. Each diagonal and each median of a square is an axis of symmetry. Ex. 4. If the diagonals of a quadrilateral are equal and bisect each other at right angles, the figure is a square. Ex. 5. Construct a square, given one diagonal. 88 PLANE GEOMETRY PARALLELS AND SEGMENTS ON TRANSVERSALS TEST FOR EQUAL SEGMENTS 111, Theorem 45. If a series of parallels cuts off equal segments on one transversal, it cuts off equal segments on all transversals. Fig. 160 Hypothesis: li,\\l2,\\h,\\U, cut the transversal h so that segments A, b, and c are equal, and cut off the segments X, y, and z on transversal k. Conclusion : x = y = z. Analysis: I. To prove x = y = z, draw MN, PQ, and RS \\ h and prove AMNP ^ APQR ^ ARST. II. To prove AMNP m APQR \ ^^7= Z2= Z3 mARST, prove] ^^^^g^^g'. Let the pupil complete the analysis and give the proof. Ex. 1. Prove Th. 45 by drawing the construction lines from the points of division on h instead of from those on k. Problem 7. To divide a given segment into any num- ber of equal parts. Solution: Let AB he the given segment. From point A draw a ray making any convenient angle with AB. From point A lay off equal divisions on this ray. The number of divisions must be the same as the number of parts into which ^5 is to be divided. Join the last point of division with B. Draw parallels from the other points of division. Let the pupil draw the figure and give the proof. QUADRILATERALS Ex. 2. A line may be divided into any number of equal parts (for example, 5) by the construction shown in Fig. 161. Give the complete directions and the proof. Note. A sheet of ruled paper may be used to divide a segment into a given number of equal parts. Number the lines ps in Fig. 162. If the segment is to be divided into 7 equal parts, put the ends of the segment on lines and 7. Why? Ex. 3. Show how a carpenter's steel square may be used to divide a board into strips of equal width (Fig. 163). Note. By the width of the board is meant the ^ perpendicular distance between the sides. Fig. 163 RELATED THEOREMS CONCERNING TRIANGLES 112. Theorem 46. A segment parallel to the base of a triangle and bisecting one side is equal to half the base. \ \ '\ \ . Jvr Fig. 161 •'//' A t \ 2 \ :i \ 4 \ ,5 \ « \ 7 \ s Ji Fig. 162 > /y\,:i <- % // \\« ^ I ^ W.-/ ?* ;^ N\i. K. Fig. 164 Hypothesis: In AABC, XY bisects AC and || AB. Conclusion: XY = }/2 AB. Analysis and construction: I. To prove XY = }/2 AB, prove XY is equal to a seg- ment that is }/2 AB. II. .-. construct YZ from Y\\AC and prove XY=AZ Sind XY = ZB. Let the pupil give the proof. 90 PLANE GEOMETRY 113. Theorem 47. A segment parallel to the base of a triangle and bisecting one side bisects the other side also. Suggestion. Draw a line through the vertex of the triangle parallel to the base and apply Th. 45. 114. Theorem 48. A segment bisecting two sides of a triangle is parallel to the third side. Fig. 165 Hypothesis: In A ABC, XY joins X and Y, the mid- points oi AC and BC respectively. Conclusion: XY\\AB. Analysis and construction: I. To prove XY\\AB, prove that XY coincides with a segment that is || to AB. 11. /. draw XZ\\AB from X and prove that XZ coin- cides with XY, III. To prove that XZ coincides with XY, show that XZ and XY both pass through X and Y. Proof : STATEMENTS I. a. XY passes through X and Y. b. XZ passes through X. c. XZ passes through Y. 11. XY and XZ coincide. III. XYWAB, Let the pupil give reasons. Apply Th. 47 in Ic. For II see As. 6. Exercise. A segment bisecting two sides of a triangle is equal to half the base. QUADRILATERALS 91 115. If three or more lines pass through a common point, they are said to be concurrent. Theorem 49. The medians of a triangle are concurrent in a point that is two-thirds the distance from each vertex to the mid-point of the opposite side. Hypothesis: ABC is any triangle. Conclusion: (1) The medians are concurrent. (2) The point of intersection is two-thirds the distance from each vertex to the mid-point of the opposite side. Analysis and construction for (1) : I. To prove that the medians are concurrent, let any two medians, as AF and BG, meet at O. Join CO and extend to E and prove that AE = EB. II. To prove that AE = EB, prove that AB may be the diagonal of a £17 . III. .-. extend CE to H so that OH = CO, join HB and HA. Prove AH BO a O. IV. .-. prove HB \\ OF (part of AF) and AH \\ OG (part oiBG). The proof is left to the pupil. Analysis for (2) : Prove CO = % CE. Exercise. Segments drawn from one vertex of a parallelogram to the mid-points t/ of the opposite sides trisect the diagonal which they intersect (Fig. 167). Fig. 167 Suggestion. Draw diagonal AC and apply Th. 49. 92 PLANE GEOMETRY 116. Theorem 50. The median from the vertex of the right angle of a right triangle to the hypotenuse is one-half the h5rpotenuse. Hypothesis: In the rt. A ABC, AX is the median from the /.A to the hypotenuse CB. Conclusion: AX = }/2 CB. Analysis: I. To prove that AX = y2 CB, prove AX- X5 isosceles. II. " " AX-XB isosceles, from X construct XO \\ AC and prove XO a ± bisector of AB. Let the pupil give the proof. 117. The following exercises are applications of Ths. 46-50. Ex. 1. The segments joining the mid-points of the sides of a triangle divide it into four congruent triangles. Ex. 2. Construct a triangle, given the mid-points of its sides. Ex. 3. Perpendiculars from the 'mid-points of two sides of a triangle to the third side are equal. How might this exercise be proved if the given triangle were isosceles? Ex. 4. If D is any point in the side ^C of a AABC, the seg- ments joining the mid-points oi AD, DC, CB, and AB form a parallelogram. Ex. 5. Through a given point within an angle draw a seg- ment terminated by the sides of the angle and bisected by the given point. Ex. 6. Divide a right triangle into two isosceles triangles. Ex. 7. Would Ths. 46, 47, and 48 be true for parallelograms as well as for triangles? Give proofs. QUADRILATERALS 93 TRAPEZOIDS 118. Theorem 51. The segment joining the mid-points of the non-parallel sides of a trapezoid is parallel to the bases. .L \ Fig. 169 Hypothesis: In C^ABCD, XY joins X and Y, the mid- points of the non-parallel sides AD and BC respectively. Conclusion: XY \\ AB and CD. Analysis and construction: I. To prove XY \\ AB and therefore || DC, prove that XY coincides with a segment that is || AB. II.. .*. draw XZ \\ AB from X and prove that XZ coin- cides with XY, III. To prove that XZ coincides with XY, show. . . . . Let the pupil complete the analysis and give the proof. 119. Theorem 52. The segment joining the mid-points of the non-parallel sides of a trapezoid is equal to one-half the sum of the bases. ~vO Fig. 170 Analysis: To prove A'F = K> (AB+DC), draw D^ inter- secting XY at and prove XO = y^ AB and OY = }i DC. Ex. 1. Construct a trapezoid so that AB-Q cm., BC = 3.2 cm., CD=4: cm., AD = 2.S cm. Ex. 2. If two trapezoids have the four sides of one equal respectively to the four sides of the other, the angles of one are equal respectively to the corresponding angles of the other. 94 PLANE GEOMETRY SUPPLEMENTARY EXERCISES EXERCISES INVOLVING PARALLELOGRAMS 120. Note. Be prepared to prove the theorems on which any of the following exercises depend. 1. Make a review diagram for Th. 36. 2. Any segment drawn through the intersection of the diago- nals of a parallelogram, and terminated by the sides of the paral- lelogram, is bisected by the point of intersection of the diagonals. 3. Perpendiculars drawn to a diagonal of a parallelogram from the opposite vertices are equal. 4. The bisectors of two consecutive angles of a parallelogram are perpendicular to each other. 5. The bisectors of two opposite angles of a parallelogram are parallel. 6. The bisectors of the angles of a parallelogram form a rectangle. 7. The median to one pair of opposite sides of a parallelogram is equal and parallel to the other two sides. 8. The medians of a parallelogram bisect each other. H 9. In Fig. 171, ABCD is a parallelogram. The sides are extended through A and C so that CE = AG and CF = AH. Prove that EFGH is a parallelogram. 10. In Fig. 172, ABCD is a parallelo- gram. CD and AB are extended so that DE = BF, DF cuts CB at Y and BE cuts DA at X. Prove DX = BY. Fig. 172 11. If, in Fig. 172, DX is made equal to BY, and BX and DY are drawn and extended to meet CD at E and AB at F respectively, prove DE = BF. 12. In Fig. 173, ABCD is a parallelogram. ^" DX and BY are perpendicular to ^C from D and B respectively. DY and BX are joined. Prove DXBY a parallelogram. Fig. 173 QUADRILATERALS 95 AE = CG and D 13. In Fig. 174, A BCD is a parallelogram. AH = CF. Prove A£0^ ^ AGOF. 14. Investigate the case, Ex. 13, in which __^ E, Hj F, and G are on the side of the paral- pic. 174 lelogram extended. From a Roman floor design 15. In Fig. 175, OXZY is any parallelogram, OC is any line through O. YA , XB, and ZC are drawn from F, X, and Z respectively perpendicu- lar to OC. Prove OC = OA + OB. 16. Fig. 176 shows two forms of parallel rulers often used in mechanical drawing for construct- ing parallel lines. Show how each is constructed and upoii what theorems in geometry the con- struction depends. Which one of these is some- times used in folding gates? 17. If the sides of ADEF are parallel respec- tively to the sides of AABC and pass through the vertices of AABC, prove that FE = 2AB, and that BF and AC bisect each other (Fig. 177). Fig. 177 18. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. 19. The sum of the perpendiculars drawn from an arbitrary point in the base of an isosceles triangle to the equal sides is equal to the perpendicular from one end of the base to the opposite side. 20. Investigate the case, Ex. 19, in which the arbitrary point is in the base extended. 21. The sum of the perpendiculars drawn from an arbitrary point within an equilateral triangle to the sides is equal to the altitude of the triangle. 22. Investigate the case, Ex. 21, in which the arbitrary point is outside of the triangle. 96 PLANE GEOMETRY EXERCISES INVOLVING SPECIAL QUADRILATERALS 121. 1. Are the diagonals of a parallelogram perpendicular to each other? of a rhombus? of a square? of a trapezoid? of a kite? of a rectangle? , 2. Name the quadrilaterals in which the diagonals are equal to each other. 3. Name the quadrilaterals in which the angles are bisected by the diagonals. 4. What quadrilaterals have one axis of symmetry? What quadrilaterals have two axes of symmetry? Have any quadri- laterals more than two axes of symmetry? Name the axis or the axes in each case. 5. What quadrilaterals have a center of symmetry? Name the center of symmetry in each case. 6. The figure formed by joining the mid-points of the sides of a rectangle is a rhombus. 7. The figure formed by joining the mid-points of the sides of a rhombus is a rectangle. 8. The figure formed by joining the mid-points of the sides of a square is a square. 9. In Fig. 178, A BCD is a square with the equal distances AE, BF, BG, CH, etc., measured on the sides in each direction from the vertices; prove that EHKN and FGLM are rectangles and that XYZW ^' is a square. ^ £ JTB Fig. 178 10. Investigate the case, Ex. 9, in which A BCD is a rhombus or a rectangle instead of a square. 11. In Fig. 179, ABDC is a rectangle, AB and CD are divided into the same number of equal parts and the points joined as indicated. Prove that the figures formed are rhombuses. Would it be possible to construct the figure so that squares are formed instead of rhombuses? C K L M y A E F G H Fig. 179 QUADRILATERALS 97 12. In Fig. 180, A BCD is a square with its diagonals A C and BD. AE = BF=CG = DH. GW and YE are parallel to AC, and FZ and HX are parallel to DB. Prove that WXYZ is a square. 13. Given A BCD a square with i4 C a diagonal. U AX = CY, prove that 5FZ)X is a rhombus (Fig. 181). Is this exercise true if A BCD is a rhombus? If A BCD is any parallelogram? Fig. 180a Parquet floor design Fig. 181 Fig. 181a A parquet floor design 14. Investigate the case, Ex. 13, hi which X and Y are on AC extended. 15. In Fig. 182 given the DABCD with £, F, G, and H the mid-points of the sides, and the p_ points joined as indicated. Prove that WXYZ is a square. Analysis: To prove WXYZ a D, prove it a a with WX = XY and Zl=a rt. Z. Fig. 182 Fig. 182 a Parquet floor design ■P z 16. In Fig. 183, A BCD is a square with AX = BY CZ = DW. Prove that PFXFZ is a square. ^ z B Fig. 183 From a Roman floor design 17. In Fig. 184, ABC is an isosceles triangle with ZC = art.Z,C0±^5 from C. AX=XY=YB. XW and YZ are ± AB from X and Y. Prove that PFXFZ is a square. Fig. 184 18. Construct a square that shall have its vertices on the sides of a rhombus. 98 PLANE GEOMETRY EXERCISES INVOLVING PARALLELS AND TRANSVERSALS 122. Note. Be prepared to prove the theorems on which any of the following exercises depend. 1. Make review diagrams for Ths. 45, 49, and 50. 2. Name two important special cases of Th. 45. 3. Prove the converse to each of the theorems called for in Ex.2. 4. The segments which join the mid-points of the sides of a quadrilateral taken in order form a '[/-/ ^f parallelogram (Fig.. 185). 5. Prove Ex. 4 for a concave quadrilateral J^ ^ and for a cross quadrilateral. Fig. 185 6. The medians of a quadrilateral bisect each other. 7. The segments which join the mid-points of two opposite sides of a quadrilateral to the mid-points of the diagonals form a parallelogram. 8. If from two opposite vertices of a parallelogram segments are drawij to the mid-points of the opposite sides, these segments trisect the diagonal joining the other two vertices. 9. It is said that an Arab, about 900 a.d., trisected a given segment as in Fig. 186. Show that this is the same construction as that given in Ex. 8. 10. A segment joining the mid-points of the non- parallel sides of a trapezoid bisects both diagonals. p^,^ jgg^ 11. The segment joining the mid-points of the diagonals of a trapezoid is parallel to the bases and equal to >^ their difference. 12. If one leg ot a trapezoid is perpendicular to the bases, the segments joining the mid-point of the other leg to the extremities of the first leg are equal. 13. If the mid-points of the legs of an isosceles triangle are joined to the mid-point of the base, the figure formed is a rhombus. 14. Given A BCD a trapezoid with the base AB twice the base CD. The diagonals AC and BD intersect at 0. Prove that ^O is twice OC. Suggestion. Join the mid-points of .40 and OB with D and C. CHAPTER V Inequalities ASSUMPTIONS FOR COMBINING INEQUALITIES 123. To As. 28 and As. 29 in chapter iii, which were assumptions of inequality, we must now add the following: As. 31. If equal segments (or angles) are added to unequal segments (or angles), the resulting segments (or angles) are unequal in the same order. As. 32. If equal segments (or angles) are subtracted from unequal segments (or angles), the resulting segments (or angles) are unequal in the same order. As. 33. If unequal segments (or angles) are added to unequal segments (or angles) , the greater to the greater and the lesser to the lesser, the resulting segments (or angles) are unequal in the same order. As. 34. If unequal segments (or angles) are subtracted from equal segments (or angles), the resulting segments (or angles) are unequal in the opposite order. As. 35. If unequal segments (or angles) are mtdtiplied by the same number, the resulting segments (or angles) are unequal in the same order. As. 36. If unequal segments (or angles) are divided by the same number, the resulting segments (or angles) are unequal in the same order. Exercise. Illustrate each of the foregoing assumptions, using numbers to represent the lengths of the segments or the number of degrees in the angles. 99 100 PLANE GEOMETRY FUNDAMENTAL TESTS OF INEQUALITY 124. The fundamental test of inequality is As. 29, given in §57, the whole is greater than any of its parts. TEST FOR UNEQUAL ANGLES 125. The fundamental theorem for unequal angles is Th. 8, an exterior angle of a triangle is greater than either of the non-adjacent interior angles. TESTS FOR UNEQUAL SEGMENTS 126. The two following tests for unequal segments are important. As. 37 may be called the fundamental test for unequal segments. It is known to every one who goes 'cross lots rather than around the corner. As. 37. The sum of two sides of a triangle is greater than the third. As. 38. The difference between two sides of a triangle is less than the third side. Ex. 1. How many triangles can be formed with the sides 7, 11, 19, and 15 cm.? Ex. 2. If two sides of a triangle are 8 cm. and 15 cm., what are the upper and lower limits of the third side? Ex. 3. If D is an arbitrary point in the side BC of AABC, AB+BC+AC > 2AD. Ex. 4. The sum of the diagonals of any quadrilateral is greater than the sum of either pair of opposite sides. Ex. 5. The perimeter of* a quadrilateral is greater than the sum of its diagonals. Ex. 6. If fiom a point within a triangle segments are drawn to the vertices of the triangle, the sum of these segments is greater than >^ the perimeter of the triangle. / / Ex. 7. The median to one side of a triangle is less J'' than }4 the sum of the other two sides (Fig. 187). Fig. 187 INEQUALITVES." 101 127. Theorem 53. If from a point within a triangle seg- ments are drawn to the extremities of one side, their sum is less than the sum of the other two sides of the triangle. Pig. 188 Hypothesis: Ift AABC, segments DA and DB are drawn from point D to the extremities of AB. Conclusion: AD+DB < AC-\-CB. Analysis and construction: To prove AD-\-DB< AC-\-CB, extend AD to meet BC at E and prove DB< DE-\-EB and AD-\-DEZC. Ex. 3. If O is any point in the side AC of AABC, OC+OB /.A. Conclusion: AC > CB. Analysis and construction: I. To prove AC > CB, construct a triangle with BC for one side such that the sum of the other two sides shall equal AC. II. /. construct OB from point B so that OB = OA, that is, so that Z 2 = Z 1, and compare AC with OC-^OB and OC+OB with CB. The proof is left to the pupil. 129. Theorem 55. If one side of a triangle is greater than a second, the angle opposite the greater side is greater than the angle opposite the lesser side. Fig. 190 Hypothesis: In AABC, AOCB. Conclusion : AB> /.A. INEQUALITIES 103 Analysis and construction: I. To prove ZB>ZA, compare ZB with an angle that is greater than /.A. II. .*. on AC take CD = CB, draw BD, and compare Z 1 with ZB and ZA. Proof: ^ STATEMENTS I. a. Zl= Z2. b. ZB >Z2. c. :.ZB>Z\. 11. Zl>ZA. III. .-. ZB> ZA. Let the pupil give the reasons. Ex. 1. Review Ths. 3 and 24 concerning the equal sides and angles of a triangle. Ex. 2. Prove Th. 55 by an indirect proof. Ex. 3. Prove Th. 55 by the following construction: ' Bisect Z C, continue the bisector to meet ABaX E. On CA take CD = CB. Join E and D. Ex. 4. In Fig. 191, XO is the perpen- dicular bisector of AB. ZW \\ AB. If point C moves to the right and to the left along ZW, how will the relative lengths of CA and CB change? What will be the limits of Z CAB ? How will the relative sizes ol AA and B . ^ Fig. 19i change? Ex. 5. The angles at the extremities of the greater side of a triangle are acute. Ex. 6. If, in a ^ BCD, BC < CD, the diagonal DB does not bisect Z Z), but Z 2 < Z 1 (Fig. 192). Ex. 7. State and investigate the converse of ^^ Ex. 6. Prove your conclusion. Fig. 192 Ex. 8. An angle of a triangle is right, acute, or obtuse accord- ing as the median from its vertex is equal to, greater than, or less than half the side that it bisects. 104 . PLANE GEOMETRY DISTANCES AND OBLIQUE SEGMENTS 130. Theorem 56. The perpendicular is the shortest segment from a point to a straight line. The proof is left to the pupil. Ex. 1. The altitude to one side of a triangle is less than half the sum of the other two sides. Ex. 2. The sum of the three altitudes of a triangle is less than the perimeter. The distance between a point and a line is defined as the length of the perpendicular from the point to the line. Theorem 57. If from a point in a perpendicular to a straight line two oblique segments are drawn cutting the straight line at unequal distances from the foot of the per- pendicular, the more remote is the greater. 0' o Fig. 193 Hypothesis: AO _L line / and the oblique segments AB and AC are so drawn that OB>OC. Conclusion: AB>AC. Analysis: I. To prove AB > AC, prove Z 2 > Z 1. II. *• " Z2>Z1, " Z 2 an obtuse angle. III. " " Z2 an obtuse angle, prove Z2 > Z4. Theorem 58. If from a point in a perpendicular to a straight line two unequal oblique segments are drawn, the greater cuts the straight line at the greater distance from the foot of the perpendicular. Suggestion. Use an indirect proof. INEQUALITIES 105 TESTS FOR UNEQUAL SIDES AND ANGLES IN TWO TRIANGLES 131. Theorem 59. If two triangles have two sides of one equal to two sides of the other, but the included angle of one greater than the included angle of the other, the third side of the first is greater than the third side of the second. Fig. 194 Hypothesis: In ^ABC sxid A' B'C\AB = A' B\BC = B'C\ and ZB> ZB\ Conclusion: AC>A'C. Analysis and construction: I. To prove AC>A'C\ compare AC with a segment that is greater than A'C II. .-. place AA'B'C on £\ABC so that A'B' coin- cides with AB, A' on A, and B' on B. Bisect LC'BC\ let the bisector meet AC at X\ join XC . Compare AXA-XC with AC and AC, III. To compare AX-\-XC' with AC, prove XC = XC. Let the pupil complete the analysis and give the proof. i Discussion. In constructing Fig. 194 according to state- ment II of the analysis it may happen that point C will fall on the Hne AC or within the triangle as well as in the posi- tion shown in the figure, li C falls within the triangle, the proof is the same as for the case given above. If C falls on AC, the theorem is evident without proof. Exercise. If, in EJABCD, ZA AB. 3. In quadrilateral A BCD, AD=BC and ZD> jLC, Prove LB > ZA. 4. CD is the median to side AB in AABC, and ZB > ZA. If any point X in CD is joined to A and B, prove ZXBA > ZXAB. 5. In Fig. 195 prove that ^X+^^<^I^+^I^. ^^' is bisected at right angles by ZY. J5^' is a straight line. Y is any point in ZY. Prove also that Zr=Z2. Pig. 196 Note. Ex. 5 illustrates some important facts from physics. In Fig. 196, XY represents a plane mirror with a candle C in front of it. E represents the eye. The reflection C of the candle seems to be as far behind the mirror as C is in front of it. The light from C strikes the mirror at M and is turned back to the eye at E so that the distance CME is the least possible. The light appears to come directly from C. Also Zl= Z2. Similar relations hold when an elastic object strikes a surface and rebounds freely. »c' Fig. 1! INEQUALITIES 107 in ■fB 6. Show how to find the path of a billiard ball which is at A (Fig. 197) and which is struck so as to rebound from the side of the table XY and strike ball B. Fig. 197 7. Show the path of a billiard ball which is struck so as to rebound from each side of the table and return to its original position. (From O. Henrici, Elementary Geometry, Congruent Figures.) 8. Prove Th. 54 by the construction shown in Fig. 198. A nalysis: I. To prove i4C>CS, construct a triangle with AC for one side such that the difference between the other two sides will be BC. II. .'. construct Z1=Z2 and compare 9. Prove Th. 55 by the construction shown Fig. 199. Analysis: I. To prove Z4 > Zl, compare Z4 with an angle that is greater than Zl. II. .'. construct ^O so that Z2= Z3 and compare . . 10. If two opposite sides of a quadrilateral are equal but the diagonals are unequal, the angles which are opposite the longer diagonal are respectively greater than the angles which are opposite the shorter diagonal. 11. If two sides of a triangle are unequal, the median drawn to the shorter side is longer than the median drawn to the longer side (Fig. 200). Analysis: I. To prove CX > AY, prove CO > AO. (See Th. 49.) II. To prove CO > AO, draw the median BZ and compare ACZO and AAZO. III. To compare ACZO and AAZO, prove Zl>' Z2. IV. To prove Zl >Z2, compare ACZB and AAZB, CHAPTER VI Circles and Related Lines INTRODUCTORY DEFINITIONS 134. We have already defined circle, radius, diameter, chord, and arc. (See §12.) Two circles or two arcs that can be made to coincide are called congruent circles or arcs. In succeeding chapters we shall consider two ways of measuring an arc, namely: by its length and by the number of degrees that it contains. Each method gives a numerical measure for the arc, but the measures and the methods are different. Before two arcs can be made to coincide they must have not only the same measure but the same radius. Congruent arcs will have, the same measure whichever method is used in measuring them and will be called equal arcs. The chord joining the ends of an arc is called the chord of the arc. Every arc has one and only one chord. Every chord, however, has two arcs. If the chord is a diameter, its two arcs are congruent and are called semicircles. If the chord is not a diameter, its two arcs are unequal. The larger arc of a chord is called the major arc and the smaller arc is called the minor arc. An angle with its vertex at the center of a circle is called a central angle. The sides of the angle cut off two arcs on a circle. The minor arc cut off by the sides of a central angle is said to be the arc intercepted by the angle, 108 CIRCLES AND RELATED LINES 109 ASSUMPTIONS CONCERNING CIRCLES 136. In §29 we have: As. 9. Circles with equal radii are congruent. As. 10. Congruent circles have equal radii. To these two assumptions we shall add the following: As. 39. The diameter of a circle is twice its radius. As. 40. A circle is located definitely if its center and its radius are known. As. 41. If a line passes through a point within a circle, the Hne and the circle intersect in two and only two points. As. 42. Every diameter bisects the circle. As. 43. A circle is symmetric with respect to any diame- ter as an axis and with respect to its center as a center. As. 44. Between the same two points on a circle there is one and only one minor arc of the circle, provided these points are not the ends of a diameter. As. 45. A segment joining a point within a circle and the center is shorter than the radius. As. 46. If a segment that has one end at the center of a circle is shorter than the radius, it lies wholly within the circle. As. 47. A segment joining a point without a circle and the center is longer than the radius. As. 48. If a segment that has one end at the center of a circle is longer than the radius, it extends without the circle and cuts the circle but once. As. 49. In the same circle or in congiiient circles equal central angles intercept equal minor arcs. As. 50. In the same circle or in congruent circles equal minor arcs intercept equal central angles. Note. Ass. 49 and 50 should be verified. Draw the figures on fairfy thin paper, place the centers together, and hold to the light. Figures may be drawn to illustrate Ass. 39-48. The assumptions in § 30 are true fpr ^rcs of the same circle. 110 PLANE GEOMETRY RELATED ARCS, CHORDS, AND CENTRAL ANGLES 136. Ass. 49 and 50 are closely related to the two follow- ing theorems and should be learned with them. Theorem 61. In the same circle or in congruent circles A. Equal chords intercept equal central angles. B. Equal central angles intercept equal chords. Fig. 201 Suggestion. Prove by congruent triangles. Theorem 62. In the same circle or in congruent circles A. Equal chords have equal minor arcs. B. Equal minor arcs have equal chords. Analysts A: To prove BC = ZT, prove /.A= ZX. Use Th. 61A and As. 49. Analysis B: To prove BC = ZY, prove /.A= AX. Use As. 50 and Th. 61 B. Ex. 1. If two circles are not congruent, can an arc of one be congruent to an arc of the other? Ex. 2. For what special cases do the proofs of Ths. 61 and 62 have no meaning? Ex. 3. Show how to bisect a given arc. « Ex., 4. A ray from the center of a circle through the mid-point of a chord is perpendicular to the chord and bisects the arc of the chord. CIRCLES AND RELATED LINES 111 CHORDS IN GENERAL FUNDAMENTAL THEOREM 137. Theorem 63. A radius perpendicular to a chord bisects the chord and its arc. • Fig. 202 Hypothesis: The OO is any circle; AB is any chord of OO, and the radius OC ± AB. Conclusion: CO bisects the AB and the AB. The analysis and the proof are left to the pupil. Ex. 1. Investigate the case in which AB (Fig. 202) is a diameter. Ex. 2. Construct through a given point within a circle a chord that shall be bisected at the given point. Ex. 3. If a diameter is perpendicular to a chord, the quadri- lateral formed by joining the extremities of the chord to the extremities of the diameter is a kite. Ex. 4. If two circles have the same center and a line intersects both of them, the segments intercepted between the circles are equal. Suggestion. Draw a radius perpendicular to the given line and sub- tract equal segments from equal segments. Ex. 5. A segment from the center of a circle to the mid-point of an arc is a perpendicular bisector of the chord of the arc. Ex. 6. If, in a circle whose center is 0, B is the mid-point of ACf perpendiculars from B to AO and CO are equal. Ex. 7. If two circles intersect, the segment that joins the centers bisects the common chord at right angles. Suggestion^ Show that a kite is formed by the radii drawn to the points of intersection. 112 PLANE GEOMETRY TEST FOR DIAMETERS 138. Theorem 64. The perpendicular bisector of a chord passes through the center of the circle. 11. IIL Fig. 203 Hypothesis: OO is any circle; AB is any chord of OO; Xy is the _L bisector of AB. Conclusion: XY passes through point O. Analysis and construction: I. To prove that XY passes through point 0, prove that XY coincides with a line that passes through O. .*. construct OZ from J_ AB and prove that XY and OZ coincide. To prove that XY and OZ coincide, show that they are both ± bisectors of AB. Proof: STATEMENTS I. a. Xy is the _L bisector of A6. 6. OZ _L AB. c. OZ bisects AB. II. XY coincides with OZ. III. XY passes through point O. CONSTRUCTION AND DEFINITE LOCATION OF CIRCLES 139. Ex. 1. To find the center for a given arc. Suggestion. Find the intersection of two diameters. Ex. 2. To construct a circle that shall pass through the three vertices of a triangle. Suggestion. The sides of the triangle will be chords of the circle. REASONS I. a. Given. h. Construction. c. Why? II. (See As. 7.) IIL Why? CIRCLES AND RELATED LINES 113 Theorem 65. One and only one circle can be drawn through three non-collinear points. The proof is left to the pupil. We have, therefore, two methods of locating circles definitely : a. If the center and radius are known, the circle is located definitely (As. 40). 6. If a circle passes through three given non-collinear points, it is located definitely (Th. 65). TEST FOR EQUAL CHORDS 140. Theorem 66. If in the same circle or in congruent circles perpendiculars from the center to two chords are equal, the chords are equal. Fig. 204 Hypothesis: OA = eX, AD ±BC from A, XZ ±YW iromX, SindAD = XZ. Conclusion : BC=YW, Analysis and construction: To prove BC—YW, prove CD=wz, CD=y2 CB, wz=y2 yw. Let the pupil complete the analysis and give the prooL Exercise. State all tests for equal chords. EQUAL DISTANCES 141. Theorem 67. In the same circle or in congruent circles perpendiculars from the center to two equal chords are equal. (See Fig. 204.) Let the pupil give the analysis and the proof. 114 PLANE GEOMETRY TANGENTS TESTS FOR TANGENTS 142. A line that touches a circle at one point but does not cut it is called a tangent to the circle. This definition is the fundamental test for tangents. The point at which the tangent touches the circle is called the point of contact or the point of tangency of the tangent. 143. Theorem 68. A line which is perpendicular to a radius at its outer extremity is a tangent to the circle. Fig. 205 Hypothesis: OO is any circle; radius OA is any radius; line AB ± OA at A. Conclusion: AB is tangent to OO at A. Analysis and construction: i. To prove that AB is tangent to OO at A, prove that all points in AB except A lie outside the circle. II. To prove that any point in AB other than A, such as - M, lies outside the circle, join O and M and prove OM > OA. Proof: STATEMENTS I. OM > OA. II. .'. M lies outside OO. III. All points except A lie outside OO. IV. .-. AB is tangent to OO. IV. Whyi REASONS I. Th. 56 (quote in full). II. As. 48. III. Since M is any point in AB other than A. CIRCLES AND RELATED LINES 115 144. Problem 8. To construct a tangent to a circle at a given point on the circle. Ex. 1. Construct a circle of given radius tangent to a given line at a given point. Ex. 2. Show that only one tangent can be drawn to a circle at a given point on the circle. TEST FOR PERPENDICULARS 145. Theorem 69. A tangent to a circle is perpendicular to the radius drawn to the point of contact. Fig. 206 Hypothesis: OO is any circle, AB is tangent to OO at A . OA is the radius drawn to the point of contact A . Conclusion: AO ± AB. Analysis and construction: L' To prove AO ± AB, show that the supposition that AB is not J_ AO'^leads to an absurdity. II. If AO is not J_ AB, suppose some other line, as DO^ is ± AB, and show that the supposition- that DO is ± AB contradicts the hypothesis. Ex. 1. Two tangents at the ends of a diameter are parallel. Ex. 2. A diameter bisects all chords that are parallel to the tangents at its extremities. LENGTHS OF TANGENTS; TEST FOR EQUAL SEGMENTS IIG. Theorem 70. If two tangents meet at a point with- out a circle, the distances from the intersection to the points of tangency are equal. The analysis and the proof are left to the pupil. 116 PLANE GEOMETRY TEST FOR DIAMETERS 147. Theorem 71. A perpendicular to a tangent at the point of contact passes through the center of the circle. Fig. 207 Hypothesis: In QO, AB is tangent to OO at A and AC ± AB Sit A. Conclusion: AC passes through 0. Analysis and construction: I. To prove that AC passes through 0, prove that AC coincides with a Hne that does pass through 0. 11. .*. connect and A and prove that AO and AC coincide. III. To prove that AO and AC coincide, show that they are both ± to AB at A. (See As. 7.) The proof is left to the pupil. EXERCISES INVOLVING TANGENTS 148. 1. Construct a line that shall be tangent to a given circle and parallel to a given line. 2. Construct a line that shall be tangent to a given circle and perpendicular to a given line. 3. Construct a line that shall be tangent to a given circle and make a given angle with a given line. * 4. A tangent to a circle at the mid-point of an arc is parallel to the chord of the arc. 5. If two circles have the same center, a chord of the larger which is a tangent of the inner is b^isected at the point of tangency. CIRCLES AND RELATED LINES 117 6. If two circles have the same center, chords of the larger which are tangents of the inner are equal. 7. If two tangents to a circle meet at a point without the circle, the bisector of the angle between the tangents passes through the center of the circle. 8. Fig. 208 shows an instrument that may be used to find the center of a metal disk. How is it made and how would it be used? TWO CIRCLES AND RELATED LINES DEFINITIONS @©6)999 No. 1 So. 2 yo. 3 So. 4 So. 5 No. 6 Fig. 209 149. Fig. 209 shows the six possible relations of two circles. Two circles are said to be concentric if they have the same center (Fig. 209, No. 1). Two circles are said to be tangent if they have but one common point. They may be tangent internally as in Fig. 209, No. 3, or tangent externally as in Fig. 209, No. 5. This definition is the fundamental test for tangent circles. The line passing through the centers of two circles is called the line of centers of the two circles. The segment through the points of intersection of two circles is called the common chord of the two circles. 150. Since any diameter of a circle is an axis of symmetry of that circle, we will assume As. 51. Th^ line of centers of two circles is an axis of sjnnmetry of the two circles. Exercise. When have two circles a center of symmetry? 9 lis . PLANE GE:OMETRY INTERSECTING CIRCLES 151. Theorem 72. If two circles intersect in one point not on the line of centers, they intersect in two points. Fig, 210 Hypothesis: (DA and B intersect at point P not on the line of centers AB. Conclusion: (DA and B intersect in a second point. Analysis: To prove that (DA and B intersect at a second point, prove that there is on each circle one point P' which is symmetric to point P. Cor. If two circles intersect, the points of intersection are symmetric points. We will assume as evident that As. 52. Two circles cannot intersect at more than two points. Theorem 73. If any two circles intersect, the line of centers is the perpendicular bisector of the common chord. Suggestion. Prove by folding the figure on the axis of symmetry. Ex. 1. Use Th. 73 to construct a ± bisector to a given seg- ment and to construct a _L to a line from a point not in the line. Ex. 2. If two equal circles intersect, the common chord sub- tends equal central angles in the two circles. Ex. 3. In Fig. 211, ®0 and 0' are two equal (D intersecting at A and B. The line of cf centers meets the ©0 at C and QO' at D. Prove ZACB=ZADB, CIRCLES AND RELATED LINES 119 152. Theorem 74. If two congruent circles intersect, the common chord is an axis of symmetry of the figure. Fig. 212 Suggestion. Fold the figure on PP' as an axis. Point A will fall on point B. The circles will coincide. Cor. If two congruent circles intersect, the segment joining the centers and the common chord are perpendicular bisectors of each other. Exercise. Solve Ex. 1, §151, by Th. 74 Cor. TANGENT CIRCLES 153. Theorem 75. If two circles meet at a point on their line of centers, the circles are tangent. Fig. 213 Suggestion. Use an indirect proof. Suppose that they have a second point in common. See Th. 72 and Th. 65, §139. CoR. I. If the segment joining the centers of two circles is equal to the sum of the radii, the circles are tangent externally. CoR. IL If the segment joining the centers of two circles is equal to the difference between the radii* the circles are tangent internally. 120 PLANE GEOMETRY Theorem 76. If two circles are tangent, the point of contact is on the line of centers. Suggestion. Use an indirect proof. Ex. 1. In Fig. 214, ®A and B are tangent at point 0. Show that they have a common tangent, OX, at point 0. Construct OX. r^ Ex. 2. In Fig. 214, XY and XZ are tan- gent to d) A and B respectively from any point in the common tangent OX. Prove Fig. 214 that XY = XZ if Y and Z are the points of contact. Ex. 3. In Fig. 215, ^ and B are two equal circles tangent at C. XY is the common tangent at the point C. Prove that any point on XF may be the center of a circle tangent to ©A and B. Y Fig. 215 Ex. 4. In Fig. 216, A and B are two equal non-intersecting circles. XF is a ± bisector of the line of centers AB. Prove that any point on XF may be the center of a circle tangent to® A and B. Ex. 5. Fig. 216 Investigate the case, Ex. 4, if the circles intersect. Ex. 6. Show how to construct circles that will be tangent to each of two concentric circles. Ex. 7. With three given segments as radii construct three circles each tangent to the other two. SUPPLEMENTARY EXERCISES EXERCISES INVOLVING INSCRIBED AND CIRCUMSCRIBED POLYGONS 154. A polygon is said to be inscribed in a circle if its vertices are on the circle and its sides are chords of the circle. In this case the circle is said to be circumscribed about the polygon. A polygon is said to be circumscribed about a circle if its sides are tangent to the circle. In this case the circle is said to be inscribed in the polygon. CIRCLES AND RELATED LINES 121 1. Inscribe an equilateral octagon in a given circle. Suggestion. It is necessary to divide the pcrigon at the center of the circle into 8 equal parts. 2. Inscribe an equilateral hexagon in a given circle. Suggestion. One-sixth of the perigon is 60°. How are angles of 60° constructed? 3. Inscribe an equilateral polygon of 12 sides in a given circle; also one of 16 sides. 4. Prove that the polygons constructed in Exs. 1, 2, and 3 are regular. 5. If a regular pentagon is inscribed in a circle, its diagonals are equal. Note. The pupil cannot construct a regular pentagon at present without a protractor. 6. Any of the longer diagonals of a regular hexagon is a diameter of the circumscribed circle. 7. How many sets of equal diagonals has a regular octagon? Are any of them diameters of the circle? 8. AB and CD are two diameters of a circle perpendicular to each other. Prove that tangents at their extremities form a square. 9. The sum of two opposite sides of a circumscribed quadri- lateral is equal to the sum of the two remaining sides. 10. AX and BX are tangents to ©0 And meet at point X without OO. AO and BO are radii drawn to the points of contact A and B. Prove that Z0-\- ZX = 2 rt. A ; that OX bisects ZO and ZX and is a perpendicular bisector of the chord AB. 11. To circumscribe about a given circle a triangle whose angles shall be equal to three given angles. Analysis and directions: I. To circumscribe a triangle about a circle, construct the sides tangent to the circle. II. To construct ZA equal to one of the given angles, construct Z YOX equal to the supplement of Z^ (Fig. 217). Let the pupil complete the directions, construct the figure, and give the proof. 122 PLANE GEOMETRY MISCELLANEOUS EXERCISES 155. Note. Be prepared to prove the theorems on which any of the following exercises depend'. 1. If two lines intersect at a point within a circle and make equal angles with the segment joining the point of intersection and the center of the circle, the chords cut off are equal. 2. Investigate the case, Ex. 1, {a) when the point of intersection is on the circle; (b) when the point of intersection is without the circle. 3. If two equal chords intersect within a circle, they make equal angles with the segment joining the point of intersection and the center of the circle. 4. Investigate the case, Ex. 3, (a) when the equal chords inter- sect on the circle; (b) when the equal chords are segments of lines that intersect without the circle. 5. A perpendicular bisector of a chord bisects all chords parallel to the given chord. 6. A line joining the mid-points of two parallel chords passes through the center of the circle. 7. If a tangent and^a chord are parallel, they cut off equal arcs. 8. Two parallel chords in a circle cut off equal arcs. Suggestion. Draw a diameter perpendicular to one of the chords and use the As. : Equal arcs subtracted from equal arcs give equal arcs. 9. A chord through the point of tangency of tangent circles subtends equal central angles in the two circles. 10. A line through the center of a circle is cut by two parallel tangents. Prove that the segments cut from this line between the tangents and the circle are equal. 11. In Fig. 218, AB is any chord and CD is a diameter intersecting the chord. DE and CF are ± AB from D and C respectively. Prove that AE = BF. 12. In Fig. 219, ^^ is any chord in O 0. CD is a diameter drawn to the mid-point of arc AB. Prove /1=Z2. CHAPTER VII Circles and Related Angles RELATION BETWEEN CENTRAL ANGLES AND THEIR ARCS UNITS FOR MEASUREMENT *OF ARCS 156. There are two ways by which an arc is measured: by its length and by the number of degrees it contains. In this chapter we consider the measure of an arc in degrees. The number of degrees in an arc is closely related to the number of degrees in certain angles. The degree of angle is Meo of a complete rotation. Since in the same circle equal central angles intercept equal arcs, Heo of a complete rotation about any point, as O, will inter- cept Keo of any circle drawn with O as center. The arc intercepted by the unit angle of one degree at the center of a circle is taken as a unit for measuring the arcs of that circle. The unit arc is therefore Meo of the circle and is called a degree of arc. Smaller units are obtained by using smaller divisions of the dngle. An arc of one minute (') corresponds to a central angle of one minute. An arc of one second (") corresponds to a central angle of one second. Ex. 1. Are two angles of the same number of degrees con- gruent? Are two arcs of the same number of degrees always congruent? Illustrate by a figure. Ex. 2. If a wheel makes 250 revolutions a minute, through how many degrees does it revolve in one second? Ex. 3. If a a A BCD is inscribed in a circle, how many degrees in the arc AB? 123 124 PLANE GEOMETRY FUNDAMENTAL RELATION 167. It follows that a central angle of 30° intercepts an arc of 30°; a central angle of 42° 7' 15" intercepts an arc of 42° 7' 15". We shall accordingly assume that if the measure of a central angle is any number whatsoever, the measure of the intercepted arc is expressed by the same number. We have, therefore, As. 53. The measure of a central angle and its inter- cepted arc are expressed by the same number, or a central angle is measured by its intercepted arc. 158. A protractor is an instrument for measuring angles. It usually consists of a semicircle or circle divided into unit arcs and is used like any other scale for measuring. Fig. 220 shows one form of protractor. Ex. 1, To measure an angle with a protractor, Place the protractor with the center C on the vertex of the angle and the line of zeros CO along a side of the angle. Read off the number of degrees on the scale as indicated by the other side of the angle. Ex. 2. Show how to construct with the protractor an angle that shall have a given number of degrees. Ex. 3. Construct with the protractor angles of 54°, 72°, 125°, 40°, 18°. Draw a number of angles; measure each with protractor. Ex. 4. With a protractor divide a given circle into 10 equal parts; into 9 equal parts; into 15 equal parts. By this means inscribe in a given circle equilateral polygons of 10, 9, and 15 sides. Prove that these polygons are regular. CIRCLES AND RELATED ANGLES 125 INEQUALITIES IN CIRCLES 169. The following are the fundamental assumptions concerning inequalities in circles: As. 54. In the same circle or in congruent circles, if two central angles are unequal, the minor arc subtended by the greater angle is greater than the minor arc subtended by the lesser angle. As. 55. In the same circle or in congruent circles, if two minor arcs are unequal, the angle subtended by the greater arc is greater than the angle subtended by the lesser arc. fEx. 1. In the same circle or in congruent circles, if two chords are unequal, the central angle subtended by the greater chord is greater than the central angle subtended by the lesser chord. Suggestion. Draw radii to the ends of the chords and apply Th. 60. fEx. 2. Investigate the converse of Ex. 1 and prove your conclusion. fEx. 3. In the same circle or in congruent circles, if two minor arcs are unequal, the chord subtended by the greater arc is greater than the chord subtended by the lesser arc. Suggestion. Apply As. 55 and the preceding exercise. fEx. 4. Investigate the converse of Ex. 3. Prove your con- clusions. fEx. 5. In the same or in congruent circles the greater of two unequal chords is at the less distance from the center (Fig. 221). A nalysis: I. To prove perpendicular OY < perpendicular OX, prove OF < a part of OX. 11. .*. place /1 5 with point A on point C, and so that B lies between D and C, prove that OZ cuts ^~^ DC, and that OF < a part of OZ. . Fig. 221 III. To place AB in position required in II, prove AB < DC. fEx. 6. Prove the converse of Ex. 5 by an indirect proof. Ex. 7. Construct in a given circle the shortest chord that shall pass through a given point. 126 PLANE GEOMETRY RELATION BETWEEN INSCRIBED ANGLES AND THEIR ARCS THE MEASURE OF THE INSCRIBED ANGLES 160. An angle is said to be inscribed in a circle if its vertex is on the circle and its sides are chords of the circle. The arc cut off between the sides of an inscribed angle is called its intercepted arc. Exercise. Inscribe in a given circle angles of 44°, 72°, 105°, etc. Find the measure of the intercepted arcs. Use a protractor. 161. Theorem 77. An inscribed angle is measured by one-half its intercepted arc. Fig. 222 Hypothesis: ZCAB is inscribed in OO. Conclusion: Z CAB is measured by }4 EC. Case A. When the center of the circle is on one side of the angle. Analysis and construction: I. To prove that Z.A is measured by ^i BC, compare ZA with an angle whose measure is known. II. .*. connect C and and compare ZA and Zx. III. Tocompare ZAa.nd Z:^;, compare Z A -\- ZC with Zx^ Proof: STATEMENTS I. 1. ZA+ZC= Zx. 2. ZA=ZC. 3. :.2ZA=Zx. ZA Zx. CIRCLES AND RELATED ANGLES 127 II. Zx is measured by BC. ,'. ZA is measured by J^ BC. Let the pupil give all reasons. Case B. When the center of the circle is within the angle. Analysis: I. To prove that ZA is measured by >2 BC, compare Zi4 with angles whose measures are known. II. .% draw the diameter through point A and compare ZA with Zx and Zy. Proof: STATEMENTS REASONS 1. Zx is measured by 3^ 5X. L Why? 2. Z;v is measured by }4 CX. 2. Why? 3. .*. ZA is measured by }4 3. If equal numbers are BC. added to equal num- bers, etc. Case C. When the center of the circle is without the angle. The analysis and the proof are left to the pupil. Ex. 1. If two angles of an inscribed triangle are 70° and 50*, find the number of degrees in the arcs subtended by each side. Ex. 2. The arcs subtended by the sides of an inscribed triangle are in the ratio of 1:2:3. Find the num- f e ^ bar of degrees in each angle of the triangle. ^^^r^fT^^^c Ex. 3. In Fig. 223, the semicircle is divided into / \ 5 equal parts by^the points C, D, F, and G. E is the ^^^' ^23 mid-point of FD. Find the nurfiber of degrees in ^dSign^°° each angle of the figure. _Ex.^. In Fig. 224, i^ = 120° and j5c=100°. AD = BC. Find the number of degrees in each angle of the figure. Fig. 224 128 PLANE GEOMETRY COROLLARIES: TESTS FOR EQUAL ANGLES, RIGHT ANGLES AND SUPPLEMENTARY ANGLES 162. Cor. I. Inscribed angles measured by the same or by equal arcs are equal, and, conversely, arcs that measure equal inscribed angles are equal. Ex. 1, If the vertices of a square A BCD lie on a circle and E is any point in the arc AB, pt-ove that CE and DE trisect the ZAEB. An angle is said to be inscribed in an arc if its vertex lies on the arc affti its sides pass through the extremities of the chord of the arc. In this case the arc is said to contain the angle. Cor. II. An angle inscribed in a semicircle is a right angle. For summary of tests for perpendiculars and right angles, see p. 301. Ex. 2. The angle between the segments joining a point with- in a circle to the ends of a diameter is an obtuse angle. Ex. 3. The angle between the segments joining a point with- out a circle to the ends of a diameter is an acute angle. Ex. 4. If the radius of one circle is the diameter of a second, any chord of the larger drawn from the point of contact is bisected by the smaller circle. ^ Construct a right triangle, given Ex. 5. The hypotenuse and one leg. Ex. 6. The hypotenuse and an acute angle. CoR. III. Inscribed angles are supplementary if the sum of their intercepted arcs is 360°. Ex. 7. The opposite angles of an inscribed quadrilateral are supplementary. Ex. 8. If a triangle is inscribed in a circle, the sum of the angles inscribed in the arcs subtended by the sides is 4 right angles. CIRCLES AxND RELATED ANGLES 129 EXERCISES INVOLVING CENTRAL AND INSCRIBED ANGLES 163. The foregoing theorem and its corollaries give us L A new method for proving angles equal. XL A new method for proving arcs equal. III. A new method for proving angles supplementary. IV. A new method for determining and constructing right angles. 1. Find the arc which measures ZACD in Fig. 225. 2. If from point P on a circle PQ and a b\ diameter PR are drawn, a radius parallel to PQ bisects QR. Fig. 225 3. A circle is circumscribed about A ABC and P is the mid- point of AB. Prove that ZABP is K /C. 4. Construct a number of angles inscribed in the same arc and the bisector of each angle. What theorem can be inferred from the drawing? Prove it. 5. Show that the center of a given circle may be found with a carpenter's steel square as shown in Fig. 226. Give reasons (Th. 50). 6. ABC is a triangle inscribed in a circle whose center is 0. OD is perpendicular to AC. Prove that ZCOD=ZB. 7. Show that the center of a given arc may be found by using two carpenter's steel squares as shown in Fig. 227. Give reasons. 8. In Fig. 228, ©0 and 0' are two equal ^" circles intersecting at A and B. C and D are any points on CD and 0' respectively. Prove Z BCA = ZADB. Fig. 228 9. A central angle is twice an inscribed angle that intercepts the same arc. 10. Inscribe in a given circle a triangle whose angles are equal to the angles of a given triangle. Suggestion. Use Ex. 9. 130 PLANE GEOMETRY ANGLES FORiVLED BY INTERSECTING CHORDS 164. Theorem 78. An angle formed by two chords intersecting within a circle is measured by one-half the sum of the intercepted arcs. Fig. 229 Hypothesis: The chords AC and BD intersect within OO. Conclusion: Zl is measured by '^ {AD-{-BC). Analysis and construction: I. To prove that Z 1 is measured by }4 {AD -{-EC), com- pare Z 1 with angles whose measures are known. il. .*. connect A and B and prove Zl= Z.A-\- ZB. Let the pupil give the proof. _Ex. l._ If, in Fig. 230, ^ = 55°, ^'S=140°; and BC=% AD, find the number of degrees in each angle of the figurCo Ex. 2. Discuss the special case of Th. 78 in which the given chords intersect at the center of the circle. • Ex. 3. In Fig. 231, P is the mid-point of CD A PA and PB are any two chords from P. Prove that Zl=Zl'and Z2=Z2'. p,^ 231 Ex. 4. Find the sum of each pair of opposite arcs into which two perpendicular chords divide a circle. Ex. 5. If, in Fig. 230, AC had been a diameter, Z 1 had been 72°, and DC had been h AB, what would have been the number of degrees in each of the other angles of the figure? CIRCLES AND RELATED ANGLES 131 ANGLES FORMED BY SECANTS INTERSECTING WITHOUT THE CIRCLE 165. A line of indefinite length. which cuts a circle in two places is called a secant. Theorem 79. An angle formed by two secants inter- secting without a circle is measured by one-half the differ- ence of the intercepted arcs. Fig. 232 Hypothesis: The secants BA and BC intersect without oo. Conclusion: Z.B is measured by ^4 {AC—XY). Analysis and construction: I. To prove /LB measured by >^ {AC-XY), compare ' AB with angles whose measures are known. II. .*. connect C and X and prove AB= Z.AXC— AC. III. .-. prove AAXC = ZB + ZC. Let the pupil give the proof. Ex. 1. In Fig. 233, secants a and b meet without OO. If secant b moves until it comes into the positions b' and b'\ how are 41, 2, and 3 measured? Give the theorem that applies in each case. Ex . 2. If, in Fig. 234, Z C = 25° and XV = H A B, find the number of degrees in Z XA Y and in ZBYA. Ex. 3. If, in Fig. 234, ZC = 24°, XY^VsAB,^ and AV is 3. diameter, find the number of degrees in each angle of the figure. Fig. 233 Fig. 234 132 PLANE GEOMETRY ARCS FORMED BY PARALLEL CHORDS: TEST FOR EQUAL ARCS 166. Theorem 80. Parallel chords intercept equal arc: on a circle. Fig. 235 Hypothesis: OO is any circle and AB || DC. Conclusion: AD = BC. Analysis and construction: I. To prove AD = BC, prove that they are intercepted by equal inscribed angles. II. .*. join AC Let the pupil complete the analysis and give the proof. _Ex. 1. If, in Fig. 235, AD = 32° and minor DC is H of minor AB, find the number of degrees in each arc. Ex. 2. If on a circle AC = BD, prove that AD is either par- allel with or equal to BC. Ex. 3. If the vertices of a trapezoid lie on a circle, its diago- nals are equal and its base angles are equal. Ex. 4. Prove Th. 79 by comparing ZB with one inscribed angle. Suggestion. From D construct a line parallel toAB and compare ZB with ZFDC (Fig. 236). Fig. 236 Ex. 5. Prove Th. 78 by comparing Zl with one inscribed angle. Suggestion. From A construct a line || DB. CIRCLES AXD RELATED ANGLES 133 CONSTRUCTION OF PERPENDICULARS 167. Probleim 9. To construct a perpendicular to a line from a point in the line. /!->. ^-'' Fig. 237 Analysis and constniction: I. In order to construct a perpendicular to line I at A, construct a right angle at A . II. To construct a right angle at A, inscribe in a semi- circle an angle with its vertex at A. III. To construct a semicircle through A, take any point O as center and segment OA as radius Let the pupil complete the directions and give the proof. 168. Problem 10. To construct a perpendicular to a line from a point not in the line. ^^ r / / 1 1 ^^ \ / ^^ L )^ — I Fig. 238 Analysis and construction: I. In order to construct a perpendicular to line / through point A, construct a right angle with its vertex on / and with one side passing through point A. *. construct a circle through A cutting Hne /. *. use any oblique segment through A to line / as a diameter. II III Let 'the pupil complete the directions and give the proof. 10 134 PLANE GEOMETRY CONSTRUCTION OF TANGENTS 169. Problem 11. To construct a tangent to a circle from a given point without the circle. Fig. 239 Given QO and point A without the circle. To construct sl tangent to OO from point A. Analysis and construction: I. In order to construct a tangent to OO from point Ay construct a right angle whose vertex is on OO and whose sides pass through points O and A. II. /. join OA and construct a circle on OA as diameter, cutting OO at X and Y. Join AX and AY. Let the pupil give the proof. Ex. 1. Let P represent any point within OO (Fig. 240). Suppose P to move away from the center along the ray OX. How many tangents can be drawn to OO through point P when P is within the circle? when P is on the circle? when P is outside the pj^,^ 240 circle? How are these tangents constructed in each case? Locate various positions of P outside the circle and sketch in the tangents from P at each position of P. As P moves away from the circle how does the angle between the tangents change? How do the points of contact move? What are the limiting positions of these points? How does the length of the segment between P and the point of contact change? What are the limiting lengths of this segment? Ex. 2. Circumscribe an isosceles triangle about a circle, given the base of the triangle. Is the problem always possible? CIRCLES AND RELATED ANGLES 135 RELATION BETWEEN ANGLES FORMED BY TANGENTS AND CHORDS, AND THEIR ARCS MEASUREMENT OF THE ANGLE 170. Theorem 8L An angle formed by a tangent and a chord is measured by one-half its intercepted arc. Fig. 241 Hypothesis: AB is tangent to QO oX A and ^C is drawn from A , the point of contact of the tangent. Conclusion: Z 1 is measiired by ^ AC. Analysis and construction: I. To prove that Zl is measured by >^ AC, compare Z 1 with angles whose measures are known. II. .*. draw diameter AD and compare Zl wdth Z3 and Z2. III. To find the measure of Z3, prove Z3 = 90° and ACD = \m\ The proof is left to the pupil. Ex. 1. In Fig. 241, prove that ABAC Is measured by K ADC. Ex.2. Prove Th. 81 by comparing A CAB with one inscribed angle. Suggestion. Draw CD. Ex.3. Prove Th. 81 by comparing A CAB with a central angle (Fig. 242). Fig. 242 136 PLANE GEOMETRY ARCS FORMED BY PARALLEL CHORD AND TANGENT 171. Theorem 82. If a chord and a tangent are par- allel, they cut off equal arcs. The analysis and the proof are left to the pupil. Exercise. The chord of an arc is parallel to the tangent drawn to the mid-point of the arc. ANGLES FORMED BY SECANTS AND TANGENTS 172. Theorem 83. An angle formed by a secant and a tangent is measured by one-half the difference of the inter- cepted arcs. Theorem 84. An angle formed by two tangents is measured by one-half the difference of the intercepted arcs. Let the pupil review Ths. 78 and 79. Ex. 1. In Fig. 243, if AXC is 220°, find the number of degrees in ZB. Ex. 2. In Fig. 243, if Z 5 is 68°, find the number of degrees in AXC and A YC. Ex. 3. Two tangents are perpendicular to each other. How many degrees in the arcs formed by the points of tangency? Ex. 4. If line BC (Fig. 243) moves so that it remains parallel to its original position, how will ZB he measured when BC passes through point A? Ex. 5. In Fig. 244 let AB and CD represent any two chords intersecting within O 0. How is Z 1 measured? How will the measure of Z 1 change if CD revolves about point C in either Fig. 244 direction? Various positions of CD are shown by the dotted lines. How would Zl be measured in each case? Discuss the vase in which CD becomes parallel to AB. CIRCLES AND RELATED ANGLES 137 SUMMARY AND SUPPLEMENTARY EXERCISES 173. SUMMARY OF IMPORTANT POINTS IN CHAPTER VII I. Tests for measurement of angles. a. A central angle is m'easured by, etc. (§157). b. An inscribed angle is measured by, etc. (§161). c. An angle formed by a tangent and a chord, etc. (§170). d. An angle formed by two chords intersecting within, etc. (§164). e. An angle formed by two secants intersecting without, etc. (§165). /. An angle formed by a secant and a tangent (§172). g. An angle formed by two tangents, etc. (§172). II. General test. Angles and arcs are both measured in degrees. It is evi- dent, therefore, that in the same circle or in congruent circles : Angles are equal if they are measured by equal arcs. Arcs are equal if they measure equal angles. For summary of test of equal arcs, see p. 303. EXERCISES CONCERNING TANGENTS TO CIRCLES 174. 1. Prove that a tangent can be drawn to a circle from a given point without the circle by the following method (Fig. 24 r>) : Let yl be the given point. Join 0/1. With ,-- — ^-* as center and OA as radius draw circle ABC. / ^ — ^ '> Let OA cut the given circle at D. At D draw ;' [ o^^l.^P^^ BC tangent to the given circle and ciitting the \ V ^V>4/' outer circle at B and C. Draw OB and OC cut- '\^ ^^'l^' ting the given circle at E and F. AE and AF "*""!' i. X XI. • -1 Fig. 245 are tangent to the given circle. Suggestion. Prove Z1=Z2 = 1 rt. Z, by comparing AOEA with AODB. Note. The construction given in Ex. 1 is similar to Euclid's. 138 PLANE GEOMETRY t2. From the following figures (Fig. 246) give the analysis, directions, and proof for constructing the common tangents -V— , Fig. 246 to two given circles. Note the application of this to belts over pulleys. • . 3. A line that is tangent to each of two equal circles is parallel to the segment joining their centers or it bisects this segment. 4. Two common interior tangents to two circles are equal. 5. Is the same true of the exterior tangents? 6. How many common tangents can be drawn to two circles if they are in each of the possible positions shown in Fig. 209? Show how to construct the tangents in each case. 7. An angle between two tangents to a circle is double the angle between the chord joining the point of contact and the radius drawn to one point of contact. ^ 8. Circumscribe an isosceles triangle about a circle, given the altitude of the triangle. Is the problem always possible? 9. Circumscribe about a given circle a right triangle, given one leg. Is the problem always possible? MISCELLANEOUS EXERCISES 175. Note. Be prepared to prove the theorems on which any of the following exercises depend. 1. Make review diagrams for Ths. 82 and 83. 2. In Fig. 247 the sides of AXFZ are tan- gent to the circle at the vertices of the inscribed triangle ABC. If Zl=43° and Z3 = 62°, find^ ^ the number of degrees in each angle of the figure. Fig. 247 CIRCLES AND RELATED ANGLES 139 3. If, in Fig. 248, ZCAB = 5S°, find the number of degrees in ZD, ZE, and ZC^F if BF is given tangent to the circle. 4. A chord forms equal angles with the tan- gents at its extremities. Discuss the special case in which the chord is a diameter of the circle. 5. Prove Ths. 83 and 84 bv a method similar ^ to that suggested for Th. 79 in Ex. 4, § 166. 6. In Fig. 249, AC is tangent to OO at ^. Prove that Z CAB is equal to K ^ 0. 7. A tangent at the vertex of an inscribed angle forms equal angles with the sides of the given angle if these sides are equal. 8. If an isosceles triangle is inscribed in a circle, the tan- gent at the vertex makes equal angles with the legs and is parallel to the base. 9. In Fig. 250, AB equals CD and the chords are produced to intersect at P. Prove that the segment PA equals the segment PD. 10. If two chords intersect in a circle and a seg- ment of one is equal to a segment of the other, the chords are equal. Fig. 250 11. State and investigate the converse of Ex. 10. Prove your conclusions. 12. P and Q are the points of intersection of two arbitrary circles. PA and PB are the diameters through P. Prove that AB passes through Q. 13. A circle constructed on side ^B of A ABC as diameter passes through the feet of the perpen- ^ diculars from A and B to the sides BC and AC re- spectively (Fig. 251). See Th. 50. Fig. 25; 140 PLANE GEOMETRY 14. Circles constructed on two sides of a triangle as diameters will intersect on the third side. 15. A circle constructed on one leg of an isosceles triangle as a diameter passes through the mid-point of the base. 16. If semicircles are constructed on the sides of an equilateral triangle as diameters, they will intersect at the mid-points of the opposite sides at-^—^ - (Fig. 252). FiG.'252 Church window detail 17. The ray which bisects the angle formed by a tangent and a chord bisects also the intercepted arc. 18. If a tangent is drawn to a circle at the extremity of a chord, the mid-point of the intercepted arc is equally distant from the chord and the tangent. 19. If the extremities of any two diameters in a circle are joined in order, the figure formed is a rectangle. 20. If, in Fig._^53, (S)_0 and O' are tan- gent at X and AB and CD pass through the point of contact, prove that ^C is parallel to BD. Fig. 253 21. Prove Ex. 20 if the circles are tangent internally. 22. If, in Fig^ 254, (D and 0' are tan- gent at X and A B passes through point X, prove that tangents at A and B are parallel. Fig. 254 23. Prove Ex. 22 if the circles are tangent internally. 24. In Fig- 255, © O and 0' are tangent at X and AB passes through X. Prove that diameters from A and B are parallel. Fig. 255 25. Prove Ex. 24 if the circles are tangent internally. CIRCLES AND RELATED ANGLES 141 Fig. 256 Fig. 257 26. In Fig. 256, CD and X are tangent at C. AB is one of the common exterior tangents with A and B the points of contact. Prove that ZACB is a right angle. 27. If two equal circles intersect and a segment is drawn through either point of inter- section terminating in the circumferences, the segments joining the extremities with the other point of inter- section of the circles are equal. 28. In Fig. 257, AC and BD are tangent to OO at opposite ends of the diameter AB. CD is an arbitrary tangent intersecting AC and BD at C and D respectively. Prove that /.COD is a right angle. Suggestion. If X is the point of contact of tangent CD, draw OX, OC, and OD and prove Z2-f Z3 = }/i the straight angle. .'. prove Z 1 = Z 2 and Z 3 = Z 4. 29. Given a circle (Fig. 258) divided into eight equal parts and the points joined as indicated, prove a. AK = KB = BYr b. /.A = ZB=ZC. c. AK = KP. d. OPQRSTU Visa regulsLT «f octagon. e. WXYZ is a square. 30. In Fig. 258 find the num- ber of degrees in ZA, ZAKB, ZHVB, ZBVC. 31. Construct a figure similar to Fig. 258 by dividing the circle into sixteen equal parts and joining every seventh point of division, or by dividing the circle into twelve equal parts and joining every fifth point, and find the number of degrees in the angles formed. Note. Stars similar to the above, though often more complicated, are common in cut-glass designs. 142 PLANE GEOMETRY 32. Two circles intersect at the points A and B. Through A a variable secant is drawn cutting the circles in C and D. Prove that the angle CBD is constant for all positions of the secant. — College Entrance Examination Board, Plane Geometry Examina- tion, 1908. 33. In Fig. 259 the circle is divided into eight equal parts. Find the center of AOC so that it shall pass through the three points A, 0, and C. Prove that AOC is a semicircle. 34. In Fig. 259 prove that L is the mid-point of OLA and OLB. Prove that OL and OM are equal; also AL, LB, and BM. ^ ^,,, ' ' Fig. 259 35. Do you know any practical uses of rosettes similar to the above? 36. A quadrilateral A BCD is inscribed in a circle. At the points of division tangents are drawn forming a circumscribed quadrilateral. A^ = o5°, BC=UO°, CS = 35°. Find the number of degrees in each angle of the inscribed quadrilateral, in each angle of the circumscribed quadrilateral, and in the angles formed by the diagonals of each quadrilateral. 37. Fig. 260 shows a simple turnout used on street railroad tracks. The rails AE and CF are arcs of circles tangent to rails AB and CD at A and C respectively. Rail AE crosses rail CD at H. Prove that Zl made by rail CD and. the tangent to ^E at H is equal to Z2. O is the center for AE and CF. ^ 38. Fig. 261 shows two opposite turnouts from the end of a straight track. The curved rails are tangent to the straight ones at C and D respectively. Prove that the angle between the tangents at F is equal to ZX + ZY ii X and Y |' are the centers for the arcs. Fig. 261 Fig. 260 CHAPTER VIII Loci GENERAL CONSIDERATIONS DEFINITIONS 176. The exercises below illustrate the following defini- tions : A point which moves so as to fulfill some given require- ment is called a variable point. The path of a point which moves so as to fulfill some given requirement is called a locus. A line or group of lines is called a locus if they contain all points which fulfill some given requirement and contain no other points. Ex. 1. Find a point which is 2 in. from a fixed point O. Is there more than one such point? If you consider such a point as so moving that it shall always, remain 2 in. from 0, what will be its path? Is this path located definitely? How? Ex. 2. If A and B are two fixed points, find a point which is as far from A as from B. How many such points can you find? What is the path of a point moving so that it is always equally distant from two fixed points? Is this path a fixed path? TO FIND THE LOCUS 177. In some cases the line or set of lines that make up the locus can be found directly from a knowledge of the geometry involved. In other cases the locus may be found by locating several positions of the moving point. If positions enough are located, the locus may often be inferred from them. This is a method of discovery that is common to all scientific inquiry. The correctness of the inference 143 144 , PLANE GEOMETRY is then determined by a careful demonstration, the nature of which will be discussed in the next section. Note. One or more of the following exercises may be introduced here if the teacher desires. In many cases the locus cannot be found unless a very large number of points are located. Some of these exercises lead to other kinds of lines than the straight line and the circle, lines that are studied in more advanced courses in geometry. In these latter exercises the locating of the points must be by experi- ment, as the pupil knows no construction by which he can find them. Such exercises are valuable in that they force home to the pupil the possibility and the occasional necessity of finding loci by experiment and in that they set him to thinking on a subject entirely new to him. No proofs should be required. EXERCISES IN FINDING LOCI Find the following loci by experiment. 1. The ray a starts from point P on O X as origin. Find the locus of the mid-points of the chords cut from this ray by the circle as the ray moves about point P (Fig. 262). Fig. 262 2. Find the locus called for in Ex. 1 if the origin of a is without OX. 3. A ladder stands upright against a wall. Find the locus of a point on the middle round if the foot of the ladder is pulled out until the ladder is flat on the ground. 4. OX and OY are two lines at right angles to each other. Find the locus of a point which is twice as far from OX as from OY. 5. In Fig. 263, ABCD is a parallelogram and / any line. Imagine line / to move so as always to remain parallel to its original position. Find the locus of the mid-points of the segments cut from / by the sides of the parallelogram. Fig. 263 6. Find the locus of a point }4 the way up the ladder mentioned in Ex. 3. 7. Find the locus of centers of circles tangent to each of two given circles when (1) the one circle is within the other, (2) the circles intersect, (3) one circle is without the other, (4) the circles are tangent internally, (5) the circles are tangent externally. LOCI 145 8. Find the locus of centers of circles which pass through a given point and are tangent to a given line when (1) the point is on the given line, (2) the point is not on the given Hne. 9. Find the locus of centers of circles which are tangent to a given circle and to a given line when (1) the line intersects the circle, (2) the line is tangent to the circle, (3) the line is wholly without the circle. What is the locus of: 10. A point on the knob of a swinging door? 11. A point on the side of a spinning top? 12. The hub of a wheel of a moving bicycle? 13. A point on the tire of a moving wheel?- 14. A point on the rim of a plate moving about another plate? Note. No drawings are required for Exs. 10-12. COMPLETE PROOFS FOR LOCI 178. After the locus has been found by some method, it is necessary to give a complete formal proof that the line, or set of lines, found really constitutes the locus. It is necessary to show, therefore, that every point on the line or set of lines is a possible position of the variable point and that every possible position of the variable point is on this line or set of lines. If these two facts can be proved, it is evident that the line or set of lines constitutes the whole locus and nothing but the locus. To put it more formally: In order to prove that a line or set of lines is the locus of a point which moves so as to fulfill certain requirements, prove I. Every point in the line or set of lines fulfills the requirements, and II. Every point which fulfills the requirements is on the line or set of lines. Note. It is immaterial in which order I and II are proved. 146 PLANE GEOMETRY LOGI OF POINTS THE BISECTOR OF AN ANGLE 179. Theorem 85. The bisector of an angle is the locus of points equally distant from the sides of the angle. Fig. 264 Hypothesis: AX bisects Z CAB. Conclusion: ^X is the locus of points equally distant from AB and AC; that is, ^ I. Every point in AX is equally distant from AB and AC. IL Every point that is equally distant from AB and AC lies in AX. The analysis and the proof for I are left to the pupil. Analysis and constriction for II: 1. Let Q be a point equally distant from AB and AC. 2. To prove that Q lies in AX, join Q and A and prove that QA and AX coincide. 3. To prove that QA and AX coincide, show that both bisect /.A. Let the pupil complete the analysis and give the proof. Ex. 1. Can you determine a point equally distant from the sides of an angle without constructing the bisector of the angle? Ex. 2. What is the locus of a point equally distant from two intersecting lines? LOCI 147 THE PERPENDICULAR BISECTOR OF A SEGMENT 180. Theorem 8G. The perpendicular bisector of a seg- ment is the locus of a point equally distant from the ends of the segment. Hypothesis: OP is the perpendicular bisector of the segment AB. Conclusion: OP is the locus of a point equally distant from A and B ; that is, I. Every point in OP is equally distant from A and B, II. Every point that is equally distant from A and B lies in OP. The analysis and the proof for I are left to the pupil. Analysis and construction Jor II: 1. Let be a point such that QA=QB. 2. To prove that Q lies in OP, join Q and 0, the mid-point of AB, and prove that QO and OP coincide. 3. To prove that QO and OP coincide, show that both QO and OP are perpendicular bisectors oi AB. Let the pupil complete the analysis «and give the proof. 181. Cor. If two points are each equally distant from the extremities of a segment, the line passing through these points is the perpendicular bisector of the segment. Suggestion. Show that the two given points must both lie on the perpendicular bisector. 148 PLANE GEOMETRY Prove the following by Th. 86 and Cor. Ex. 1. The diagonals of a rhombus or of a square bisect each other at right angles. Ex. 2. One diagonal of a kite bisects the other at right angles. Ex. 3. If two circles intersect, the line of centers is a perpen- dicular bisector of the common chord. Ex. 4. If two equal circles intersect, the segment joining the centers and the common chord bisect each other at right angles. Ex. 5. The perpendicular bisector of a chord passes through the center of the circle. Ex. 6. A radius to the mid-point of an arc is the perpen- dicular bisector of the chord of the arc. Ex. 7. Two tangents to a circle where the center is O inter- sect at point X. Prove that OX is the perpendicular bisector of the chord joining the points of contact of the two tangents. OTHER SIMPLE LOCI 183. Find the loci called for in the following exercises; give complete proof for each as for Ths. 85 and 86. Find the locus of a point which is Ex. 1. At a given distance from a given point. Ex. 2. At a given distance from a given straight line. Ex. 3. Equally distant from two given parallels. CONCURRENT LINES 183. The two following exercises are preliminary and may be quoted as theorems in proving Ths. 87 and 89. Ex. 1. Lines that are perpendicular to intersecting straight lines will intersect. Suggestion. Use an indirect proof. AD and DB are the given intersecting lines. Join A and B. Show that. ii AC and BC were parallel, Z1+ Z2 would equal two right angles (Fig. 266). Ex. 2. The bisectors of any two angles of a Fig. 266 triangle will intersect. LOCI 149 184. Theorem 87. The perpendicular bisectors of the sides of a triangle are concurrent at a point which is equally distant from the vertices. Fig. 267 Hypothesis: AABC is any triangle. Lines x, y, and z are the perpendicular bisectors of AB, EC, and AC respec- tively. Conclusion: x, y, and z are concurrent at a point equally distant from the vertices. Analysis: I. To prove x, y, and z concurrent, prove a. That x and y intersect at some point, as 0. b. That the intersection of x and y is on z. IL To prove that the intersection of x and y is on z, prove that is equally distant from A and C. III. .'. prove equally distant from A and B, also from B and C. Proof: STATEMENTS I. X and y intersect at some point, as at 0. IL a. X is locus of points equally distant from A and B. b. .'. O is equally distant from A and B. III. a. y is locus of points equally distant from B and C. b. .: is equally distant from B and C. IV. .'. O is equally distant from A and C. ' V. a. z is locus of points equally distant from A and C. b. .'. O is on z. Let the pupil give the reasons. 11 150 PLANE GEOMETRY Theorem 88. The altitudes of a triangle are concurrent. Fig. 268 Hypothesis: AABC is any triangle, x, y, and « are the altitudes to sides AB, EC, and CA respectively. Conclusion: x, y, and z are concurrent. Analysis: To prove x, y, and z conctirrent, construct through A, B, and C lines parallel to the opposite sides of AABC and prove that x, y, and z are the perpendicular bisectors of the sides of the triangle so formed. Let the pupil put in the construction, complete the analysis, and give the proof. 185. Theorem 89. The bisectors of the angles of a tri- angle are concurrent at a point equally distant from the sides of the triangle. Hypothesis: AABC is any triangle, x, y, and z bisect AA, B, and C respectively. Conclusion: x, y, and z are concurrent at a point equally distant from the sides. LOCI 161 Analysts: I. To prove x, y, and z concurrent, prove a. That x and y intersect at some point, as at O. b. That the intersection of x and y is on z. 11. To prove that the intersection of x and y is on z, prove that O is equally distant from AC and BC. III. .*. prove that O is equally distant from AC and AB, also from AB and ^C 186. Problem 12. To circumscribe a circle about a tri- angle. Note. This problem has been discussed elsewhere (§139). The discussion here should be based upon Th. 87. Problem 13. To inscribe a circle in a given triangle. Fig. 270 Given AABC. To inscribe a circle in AABC. Analysis and constmction: I. To construct a circle inscribed in AABC, construct a circle tangent to the sides of AABC. II. /. find a point, such as O, so situated that perpen- diculars from O to the sides, such as OE, OF, and OG, are radii. III. .'. construct IV. To prove the perpendiculars radii, prove them equal. Let the pupil give the directions and the proof. Exercise. Construct a circle tangent to one side of a triangle and to two sides extended. A circle tangent to one side of a triangle and to two sides extended is said to be escribed to the triangle. 152 PLANE GEOMETRY DETERMINATION OF POINTS BY THE INTERSECTION OF LOCI 187. Give analysis, directions, proof, and discussion for the following exercises. Ex. L Find in a given line a point that is equally distant from two given points. A nalysis: I. Let / represent the given line and A and B the given points. II. To find a point in Hne / equally distant from A and B, find the intersection of line / with the locus of points equally distant from A and B. Let the pupil give the directions and the proof. Discussion: How many points can be found as required? Are there any special positions of the given line or the given points that will alter the results? Give all reasons. Find in a given line a point that is Ex. 2. At a given distance from a given point. Ex. 3. Equally distant from two given parallel lines. Ex. 4. At a given distance from a second given line. Ex. 5. Equally distant from two given intersecting lines. In the preceding five problems the required point must not only be in a given line, but must also fulfill a second requirement which calls for the construction of a locus. In the problems that follow, the construction of two loci are necessary in order to determine the point. The analysis should state clearly what loci are needed. We have seen that the intersection of two straight lines locates a point and that the intersection of a straight line and a circle locates two points. As some loci consist of two straight lines, it is often possible to find more than one point that fulfills the requirements. The pupil should begin by draw- ing a figure to illustrate the maximum number of points possible, and after giving the analysis with the directions for the construction and the proof he should discuss and draw figures for all special cases. LOCI 153 Ex. 6. Find a point which is at a given distance from a given point C and at the same time equally distant from two given points A and B. A nalysis: I. To find all points that are at a given distance from point C, construct the locus of II. To find all points equally distant from A and B, construct the locus of III. .'. construct Let the pupil give definite directions for the construction, also the proof and the discussion. Ex. 7. Points may be found which will fulfill any two of the following requirements: a. Be at a given distance (1>2 in.) frbm a given point. h. Be at a given distance (1>^ in.) from a given line. c. Be equally distant from two given points. d. Be equally distant from two given parallel lines. e. Be equally distant from two given intersecting lines. State and solve problems made by combining into pairs the requirements given above. LOCI OF CENTERS OF CIRCLES DETERMINATION OF THE LOCI 188. In the determination of the loci of centers of circles, we may think of a circle as changing. Fig. 271 shows a number of circles of the same radius tan- gent to line I. A great many more such (^^TT^T^^fHT^'T^ circles might be drawn. We might, however, consider these as various posi- Fig. 27] tions of one circle which rolls or slides along the line. In the same way in Fig. 272 a number of circles are shown tangent to AB and AC\ but we might consider them as various positions of one- circle which expands and contracts as it rolls Fig. 272 or slides between AB and AC. 154 ' PLANE GEOMETRY Find the locus called for in each of the following exercises and give complete proof for each. Ex. 1. ^Find the locus of the centers of circles which are tangent to the sides of an angle. I. To find the locus, sketch in a number of circles tangent to the sides of the given angle Z A . Let AX be the locus. II. To prove that AX is the required locus, prove that a. Every point in AX may be the center of a circle tangent to the sides of the angle, AB and A C, and b. The center of every circle tangent toAB and A C lies in AX. III. Discussion. AX may be considered as the path of a circle that • expands and contracts as it rolls or slides between the rays AB and AC. Find the locus of the center of a circle which Ex. 2. Is tangent to each of two intersecting lines. Ex. 3. Passes through two given points. Ex. 4. Is tangent to a given line at a given point. Ex. 5. Is tangent to a given line and has a given radius. Ex. 6. Is tangent to a given circle at a given point. Ex. 7. Is tangent to a given circle and has a given radius. Ex. 8. Passes through a given point and has a given radius. CONSTRUCTION OF CIRCLES 189. Make the constructions called for in the follov^^ing exercises. The center of the required circle is the inter- section of two loci. The discussion is often interesting. Ex. 1. Construct a circle that shall be tangent to a given line at a given point and pass through a second given point. Ex. 2. Construct a circle tangent to the sides of an angle and to one of these sides at a given point. Ex. 3. Construct a circle that shall be tangent to a given circle at a given point and pass through a second given point. Construct a circle of given radius that shall Ex. 4. Be tangent to each of two given intersecting lines. Ex. 5. Pass through a given point and be tangent to a given line or to a siven circle. LOCI 155 SUPPLEMENTARY EXERCISES EXERCISES INVOLVING CONCURRENT LINES 190. To the tests for concurrent lines given in this chapter should be added Th. 49, given in chapter iv. Each triangle has four sets of concurrent lines. The intersection of each set has a special name as shown below. I. The medians centroid or center of gravity II. Perpendicular bisectors of the sides . . . . circumcenter III. The altitudes orthocenter IV. Bisectors of the angles incenter Note. Be prepared to prove the theorems on which any of the following exercises depend. Review Th. 49. 1. The centroid, incenter, circumcenter, and orthocenter of an equilateral triangle coincide. What facts do you know con- cerning this point? 2. In Fig. 273, ABC is an equilateral triangle. AYf BZy and CX are equal distances laid off on the medians AF^ BG, and CE respectively. Prove that XYZ is an equilateral triangle. ' " Fig'' 273 3. If on the sides of ZA equal distances AB and AC are laid off, perpendiculars erected to the sides of the angle at B and C will intersect on the bisector of ZA. 4. The bisector of two exterior angles of a triangle and of the opposite interior angle are concurrent. 5. The medians and diagonals of a square are concurrent. 6. Construct A ABC, given ^5 = 5.8 cm., the median from A = 7.S cm., and the median from J5 = 6.4 cm. Suggestion. See Th. 49. 7. Construct A ABC, given AC, AB, and the median from A (Fig. 274). 8. Construct A ABC, given the three / ^. medians. fy" Suggestion. Reduce this exercise to the preceding ^ by means of Th. 49. Fig. 274 156 PLANE GEOMETRY EXERCISES INVOLVING CONSTRUCTION OF CIRCLE3 191. Notice (1) that a circle can be circumscribed about a polygon if there is a point equally distant from the ver- tices, that is, if the perpendicular bisectors of the sides are concurrent, and (2) that a circle can be inscribed in a poly- gon if there is a point equally distant from the sides, that is, if the bisectors of the angles are concurrent. 1. Construct a circle of given radius that shall pass through two given points. 2. Given the base of an isosceles triangle and the radius of the circumscribed circle, to construct the triangle. 3. Can a circle be passed through four arbitrary points? Why? 4. When will the perpendicular bisectors of the sides of a triangle meet on one of the sides of the triangle? When will they meet within the triangle? When, will they meet without the triangle? Why? 5. Can a circle be circumscribed about a parallelogram? a. rectangle? In case the circumscribed circle is possible, give analysis, directions, and proof. In case the circumscribed circle is not possible, show why. 6. Answer the questions in Ex. 5 for an isosceles trapezoid and for a trapezoid. 7. How many circles can be constructed tangent to each of three intersecting lines? Discuss all possible cases. 8. Can a circle be constructed tangent to each of three arbi- trary lines? Discuss all possible cases. 9. Inscribe a circle in a given rhombus. 10. Inscribe a circle in a given kite. 11. Circumscribe about a given circle an isosceles right triangle. 12. Construct in full Fig. 275. Fig. 275 13. Prove that the center of the circle circumscribed about an equilateral triangle is also the center of the inscribed circle. LOCI 157 B Fig. 276 /^' \ V.J 14. Construct the inscribed, the circumscribed, and the three escribed circles of an equilateral triangle. Prove that the radius of the inscribed circle is ^4 the radius of the circumscribed circle and }4 the radius of the escribed circles. 15. Make the complete drawing for the molding shown in Fig. 276. The arc ^40 is ^t tangent to line AX at A. The arc BO is tangent to line BY at B. Both arcs pass through O, the mid-point of segment AB. Are the arcs AO and BO tangent to each other? Note. Compound curves may be made from tangent circles as shown in Fig. 277. These may be used for moldings as shown in Fig. 276, or for other architectural details. p 277 16. Show by a complete drawing how to construct Fig. 278- Two diagonal streets meet at A. The corner building has a front that is an arc of a circle of a given radius tangent to each of the streets. If Z /I is 60° and the radius of the circle 50 ft., what will be the length of AB and BC d B and C are the points of tangency? A sector of a circle is a figure bounded by two radii and the subtended arc. 17. Construct a circle inscribed in a sector of a given circle. c B 18. Fig. 279 shows a decorated rafter design. ^ Z A CB is a right angle. A DB is tangent to the sides of ZC. Construct the small circle tangent to the sides of ZC and to ADB. ^ Fig. 279 19. Fig. 280 shows a decorated tile design. ABCD\ is a square with its diagonals. Construct (D O and O' inscribed in the A A DC and ABC respectively. Construct the quadrants with A and C as centers tangent to (D and 0'. p^^ 280 Note. The possibilities of this figure as a design unit may be seen by drawing several figures of the same size, like Fig. 280, and placing them together in various positions. Fig. 278 158 PLANE GEOMETRY 20. Fig. 281 shows a window and rafter is an isosceles right triangle. Construct O O inscribed in A ABC and (D X and Y tangent to the sides of the triangle and to 0. ABC Fig. 281 21. Inscribe a trefoil in a given circle (Fig. 282). Note. The three small circles are tangent to the large circle and to each other. 22. Show how to construct circles which are tangent to each of two concentric circles. 23. Any point in the perpendicular bisec- tor of the segment joining the centers of two equal circles may be used as the . center of a circle tangent to each of the two given circles. Use various positions of the two given circles. Note. The perpendicular bisector mentioned in Ex. 23 is a part of the locus of the center of a circle tangent to each of two equal circles. The remainder of the locus is beyond the province of elementary geometry, but may be readily found by experiment. 24. In Fig. 283, DF and BC are concen- tric with point A as center. DE and A C are concentric with point B as center. DF and DE are tangent at D. AD = DB. AF and BE are drawn with D as center and J/2 AB as radius. Construct O tangent to AC, BC, FD, and DE. _25. In Fig. 284, A ABC is equilateral. AB, BC, and AC are drawn with AB as radius and C, A , and B as centers respectively. Construct O tangent to AB, BC, and AC. Note. Fig. 284 is from a church window design. Fig. 282 Fig. 283 a Fig. 284 LOCI 159 MISCELLANEOUS EXERCISES 192. 1. What is the locus of the mid-points of all equal chords of a circle? 2. What is the locus of the mid-points of a series of parallel chords? 3. What is the locus of the mid-points of segments drawn from a given point to a given line? t4. Find the locus of the vertices of triangles having a given base and a given vertex angle. What is this locus if the given angle is a right angle? Construct a triangle, given 5. The base, the vertex angle, and the median from the ver- tex to the base. 6. The base, the vertex angle, and the altitude. 7. The base, the vertex angle, and one base angle. 8. The base, the vertex angle, and one side. 9. Find the locus of the mid-points of segments drawn to the circle from a fixed point (a) without the circle, (b) on the circle, (c) within the circle. 10. A circle of radius 5 inches contains a moving chord AB, length 8 inches, which is divided into four equal parts by the points P, Q, R. Determine the loci of P, Q, and R. — College En- trance Examination Board, Plane Geometry Examination, 1913. 11. A series of parallelograms stand on the same base and between the same parallels. Find the locus of the intersection of the diagonals. 12. From any point in the base of a triangle straight lines are drawn parallel to the sides. Find the locus of the intersection of the diagonals of all the parallelograms that can be thus formed. 13. Find the locus of the points at which two equal segments of a straight line subtend equal angles. 160 PLANE GEOMETRY 14. Find the locus of the extremities of tangents to a circle that have the same length. 15. Find the locus of points from which tangents to a given circle meet at a given angle. 16. Find the locus of points of contact of tangents drawn from a fixed point to a system of concentric circles. 17. Construct a series of circles tangent to each other at the same point (Fig. 285). Find the locus of points of contact of tangents drawn to these circles from any point in the common tangent. Fig. 285 18. Two circles are tangent to a given straight line at two given points and are also tangent to each other. Find the locus of points of tangency of the two circles. Suggestion. Let A and B he the two given points, C the point of tangency of the two circles. Prove that CO is always equal to 14 A B (Fig. 286). 19. In Fig. 287, AACB is a right triangle with the right angle at C. BCDE is the square con- structed on side BC. Find the locus of the vertex D asC moves about the semicircle BCA . Use Ex . 4 . Fig. 287 20. Upon a line segment AB an arc of a circle containing 240° is constructed and in the arc any chord CD having an arc of 60° is drawn. Find the locus (a) of the point of intersec- tion oi AC and BD, (b) of the point of intersection oi AD and BC. — College Entrance Examination Board, Plane Geometry Exami- nation, 1906. 21. Let A and B be two fixed points on a given circle and P and Q the extremities of a variable diameter of the same circle. Find the locus of the point of intersection of the straight lines AP and BQ. — College Entrance Examination Board, Plane Geometry Examination, 1908. , 22. From a given point on a circle draw the chords that are bisected by a given, chord. Is it always possible to draw such chords? Give reasons for your answer. — College Entrance Exami- nation Board, Plane Geometry Examination, 1907. CHAPTER IX Ratio and Proportion MEASUREMENT OF SEGMENTS 193. To measure a segilient is to find the number of times that it contains another segment which is taken as a unit. In measuring the segment two methods are possible. By the first method the unit is actually laid down successively on the segment to be measured. This method might be used in finding the length of a room if nothing but a yardstick were at hand. By the second method another segment, upon which the unit and its subdivisions are already marked, is laid beside the segment to be measured. The number found is called the measure number, the measure, or the length of the segment. 194. A segment is said to be measured exactly if it will contain the unit without remainder. A segment that is not measured exactly may be measured approximately. In considering the theoretical measurement of segments we have the following cases: First: The unit chosen may be contained in the given segment without remainder. The length of the segment is then an integer. The segment is measured exactly. Illustration 1. If the unit chosen is a segment of one inch, it may be contained in a given segment 5 times with no remainder. The length of the given segment is 5. Second: The unit chosen may not be contained in the given segment without remainder. In such cases it may happen that some fraction of the unit can be found that will measure the segment exactly. If one inch is the unit 161 162 PLANE GEOMETRY chosen, 3^ in., J^ in., 34 in., or .1 in. may be used. If such a unit can be found, the segment is said to be measured exactly, tn such a case the length will be an integer when expressed in terms of the new unit, but a fraction when expressed in terms of the old unit. In either case the seg- ment has been measured exactly. Illustration 2. If the unit chosen is one inch, it may be contained in a given segment 7 times with a remainder less than one inch; but when 3^ inch is chosen as a unit the measure may come out exactly 31. The length is 31 quarter-inches, or 7% in.; or Illustration 3. The unit chosen, one inch, may be contained in a given segment 3 times with a remainder less than one inch; a smaller unit, .1 in,, maybe contained in the segment 32 times with a remainder less than .1 in.; a still smaller unit, .01 in., may be contained in the segment 324 times with a remainder less than .01 in.; but the unit .001 in. may be contained exactly 3247 times. The length of the seg- ment is 3247 thousandths of an inch, or, as it is usually written, 3.247 in. Third: It may happen that no subdivision of the unit can be found that will measure the segment exactly. In such a case we may obtain an approximate measure. By subdividing the unit used the approximation may be made as close as desired. Illustration 4. The unit chosen, one inch, may be contained in the given segment 5 times with a remainder less than one inch. In this case 5 would be an approximate length of the segment. If we choose .1 in. as a unit, it may be contained in the given segment 56 times with a remainder less than .1 in. We now have 5.6 in. as an approximate length. But if we should choose .01 in. as a unit, we could get a still closer approximation. It might happen that .01 in. would be contained in the segment 562 times with a remainder less than .01 in. The approximate length is now 5.62 in. This process might be continued indefinitely. 195. In actual practice an exact measurement can never be obtained. We cannot be sure that a segment is exactly 7 in. or 7J4 in. long. In trying to measure the segment, the end will fall between two marks on the scale. Either of these gives an approximation to the length of the segment, one a little too small and one a little too large. It is the RATIO AND PROPORTION ' 163 usual practice to use the nearest one as the approximate length of the segment. Often extremely close approxima- tions are necessary, but the degree of accuracy depends upon the fineness of the scale used and the definiteness of the end of the segment to be measured. Ex. 1 . Draw a segment 3)4 in. long. Measure it in centimeters and millimeters. Make two approximate measures, one as close as possible but a little too small, the other as close as possible but a little too large. From each result compute the number of centi- meters to an inch and the number of inches to a centimeter. Com- pare your result with the government standard equivalent (p . 300) . Ex. 2. Draw a segment 5.6.cm. long. Make two approximate measures of this segment in inches and sixteenths of an inch. From your results make the computations called for in Ex. 1 . RATIOS DEFINITIONS 196. The ratio of two numbers is the relation expressed by dividing one number by the other. The ratio of a to 6 is written in two forms — a :b or|-. It is read the ratio of a to b. Two numbers are involved, the first term or divi- dend, also called* the antecedent, and the second term or divisor, also called the consequent. When we say that the ratio of 12 to 4 is 3, or ^ = 3, we mean that 12 is 3 times 4; when we say that the ratio of a to b is r, or -J =r, we mean that a is r times 6, or a = br. The quotient r is sometimes called the value of the ratio. It is the common practice to use the term ratio to mean either the indicated relation -f or the quotient r; either ^ or 3. In any case the two numbers or terms a and b are always involved. Two ratios that have the same value are said to be equal. Exercise. Express the following ratios in decimals of three places : 5^ n 45 12 ^ 2 _3 9' 27' 59' 25' 19' 24' 44' 164 PLANE GEOMETRY RATIO OF SEGMENTS 197. By the ratio of two segments is meant the ratio of their measures when expressed in the same unit. Exercise. Draw two segments, one 2 cm. and one 3 cm. long. Measure each in inches and sixteenths of an inch or in inches and tenths of an inch. Find an approximate ratio. 198. Two segments are said to be commensurable if they can be measured exactly by a common unit of measure. Two segments are said to be incommensurable if there is no common unit that will measure each exactly. We shall later prove that the side and the diagonal of a square are incommensurable. If the side of a square is one inch, the diagonal is V2~ inches, an irrational number. An irrational number is a number that cannot be expressed as an integer or as the quotient of two integers. The ratio of two commensurable segments is an integer or a fraction. The ratio of two incommensurable segments is an irra- tional number. Other illustrations of incommensurables will be met later. Two especially may be mentioned: 1. The side of an equilateral triangle and the altitude of the same triangle are incommensurable. 2. The diameter of a circle is incommensurable with the circumference of the circle. Note. The nature of the decimals that correspond to rational and to irrational numbers is interesting and should be noted. Fractions are rational and when reduced to decimals give decimals that either terminate or repeat, for example: He = -0625 >^ = .333+ or .3 K = .1666+ or .16 H = .142857142857 or .1-12857 Irrational numbers, on the other hand, give decimals which neither terminate nor repeat. An inexact root like V2~or V 3" is an irrational number, but not the only kind of an irrational number. Another example is the number called T (pi) (see §§ 298, 301, and 311). RATIO AND PROPORTION 165 ASSUMPTIONS INVOLVING RATIOS 199. The following assumptions will be used; As. 56 expresses the fundamental characteristic of ratios: As. 56. Multiplying or dividing both terms of a ratio by the same number does not change the value of the ratio. As. 57. Ratios equal to the same ratio are equal. As. 58. Equal ratios may be substituted for equal ratios. THEORY OF PROPORTION DEFINITIONS ^ 200. A proportion is an equality of ratios; that is, if two ratios are equal, the four numbers involved are in proportion. A proportion may be written in two forms, a:b = c:d or r- = t' and is read, a is to 6 as <; is to d, or the ratio a to 6 a equals the ratio c to d. The extremes of the proportion are a and d. The means are b and c. Since in dealing with ratios we are dealing with numbers, the laws of algebraic equations apply to proportions. Exercise. Find the value of x in each of the following: ^' 7 ~ X 55~64 ""• 6 2 FUNDAMENTAL THEOREMS OF PROPORTION 201. Theorem 90. If four numbers are in proportion, the product of the means is equal to the product of the extremes. The proof is left to the pupil. Theorem 91. If the product of two numbers equals the product of two other numbers, either pair of factors may be made the extremes and the other pair the means of a proportion. Hypothesis: ay = bx. Conclusion: -r = Suggestion. Divide both sides of ay = bx by by. 12 a X 166 PLANE GEOMETRY Ex.1. Given ay = bx,prove- = - -7 = -»and -^ = -. X y a X a Ex. 2. Derive at least two proportions from each of the following equations: a. ab = xy c. {x-\-y) (x^y) = ab b. aia+b)=:x(x+y) d. {a-1) {x+l) = {a-{-l) (x-l) Theorem 92. If three terms of one proportion are equal respectively to three corresponding terms- of another pro- portion, the fourth terms are equal. Hypothesis: -t = - and -r = - > Conclusion: x = y. The proof is left to the pupil. TRANSFORMATIONS OF PROPORTIONS 202. Theorem 93. If four numbers are in proportion, the first is to the third as the second is to the fourth; that is, they are in proportion by mean alternation. Hypothesis: — = — • Conclusion: — = — . by X y Suggestion. First prove that ay = bx; then use Th. 91. Theorem 94. If four numbers are in proportion, the fourth is to the second as the third is to the first; that is, they are in proportion by extreme alternation. Hypothesis: — = —-• Conclusion: ■^ = — ' by b a Let the pupil give the proof. Theorem 95. If four numbers are in proportion, the second is to the first as the fourth is to the third; that is, they are in proportion by inversion. Hypothesis: — = — - Conclusion: _-=-2_. by ax Let the pupil give the proof. Ex. 1. Verify proportion by mean and extreme alternation .. 8 28 15 9 , a2 ac and mversion ^ri j^ = ^^ ^ = ^, and -^ = y^^ RATIO AND PROPORTION 167 Theorem 96. If four numbers are in proportion, the first plus the second is to the second as the third plus the fourth is to the fourth; that is, they are in proportion by- addition. This is sometimes called proportion by compo- sition. Hypothesis: — =— • Conclusion: a+b x+y y rooj: STATEMENTS b y 2. ay==bx. 3. ay-\-by = bx-\-by. 4. y{a+b)=b{x+y). f, a-j-b x-^y b y Let the pupil give all reasons. Theorem 97. If four numbers are in proportion, the first minus the second is to the second as the third minus the fourth is to the fourth ; that is, they are in proportion by subtraction. This is sometimes called proportion by division. 7 . ax ^ . . a—b x—y Hypothesis: -r- = — Conclusion: —r- = • Ex. 2. Verify proportion by alternation, inversion, addition, and subtraction by means of the following: ]^-?L ?9_1)? ^_^ ^-^ 32~48' 35~28' by~ bx' by~ ab' Q, X Ex.3. If -T = ~' make the following transformations: a. First by mean alternation and that result by addition. b. First by extreme alternation and that result by subtraction. c. First by inversion and that result by addition. d. First by addition and that result by extreme alternation. e. First by subtraction and that result by inversion. 168 PLANE GEOMETRY Ex. 4. li -r = — , prove each of the following: a. b. a-\-b _x+y a X a a+b X x-\-y a-^b x-y d. e. /. a±b_b_ X-\-y~ y a-{-b _ a x+y~ X a — b a x—y X a — b b ^' x-y~J L ^+^ x-\-y ti, , — a—b x~y ' —-^ ^' b^ ~'y^ RATIOS OF SEGMENTS MADE BY PARALLELS FUNDAMENTAL THEOREM 203. Theorem 98. If three parallels cut two transversals, the segments on one transversal have the same ratio as the corresponding segments on the other transversal. Fig. 288 Hypothesis: AX\\BY\\CZ. AB XY Conclusion: The ratio of -g^ = the ratio of y^ • Case A: When AB and BC are commensurable. Analysis and construction: J rr. . ,. ,. AB ^ . XY i. io snow the ratio -^ = the ratio y^» use the meas- ures of the segments AB, BC, XY, and YZ. II. The measures of AB and BC are to be assumed. III. To find the measures of XY and YZ, divide AB and BC into segments equal to their common unit of measure and draw lines through the points of division parallel to CZ, dividing XY and YZ into segments. RATIO AND PROPORTION 169 Verification: I. a. Let the common unit of measure of segments AB and BC be segment p. b. Let the measure of AB = m. (In the figure m = 3.) c. Let the measure of BC = n. (In the figure w = 4.) d. The ratio of f77^= — * (In the figure — = -t'^ BC n ^ *=* w 4 II. a. The Hnes parallel to CZ divide XY into m equal segments and YZ into n equal segments (Th. 45). b. One of these segments may be taken as the unit of measure oi XY and YZ. ^^ . .XY m c. The ratio of -rr^=—' YZ n III. .*. the ratio of -^^ = the ratio of y-y • Note 1. For I d and II c see § 197, the ratio of two segments. Note 2. Since we assume that the number of divisions on X F and YZ are respectively equal to the number of divisions on AB and BC, the formal reasoning above has been called a verification rather than a proof. Case B: When AB and BC are incommensurable. Since it is not possible in this case to express the lengths oi AB and BC in integral or fractional terms of the same unit, the argument given for Case A cannot be used. Case B will be assumed without proof. The proof is possible, but too difficult for this course. In the next six exercises the letters refer to Fig. 288. Ex.1. AB = 7, BC=9, XF = 17K, find FZ. Ex.2. AB = 12, XF=15, FZ= 18, find 5C. Ex.3. AB = S, BC=yl2,XY = 5, find YZ. Ex.4. AB = 3}^, BC=IK YZ = 7, find XF. Ex.5. AB = 2^J^,XY = 5, FZ = 7, find 5C. Ex. G. BC=2.3, XF = 5.7, FZ = 9, find .45. 170 PLANE GEOMETRY APPLICATION OF THEOREM 98 TO TRIANGLES 204. Theorem 99. If a line is parallel to the base of a triangle, the ratio of the segments on one side equals the ratio of the corresponding segments on the other side. Cor. If a line is parallel to the base of a triangle, one side is to either of its segments as the other side is to its corresponding segment. Suggestion. Use Th. 96. Ex. 1. In Fig. 289 prove that ^=^ £A-^ 9A-9R CE be' cb~eb' cb~ce' Ex. 2. Prove Th. 99 if the parallel cuts the Fig 289 sides of the triangle extended. Ex. 3. Verify Th. 98 and Th. 99 and its cor-, by measuring each of the segments in Figs. 288 and 289 in inches and sixteenths (or tenths) of an inch, also in centimeters and millimeters, and finding the ratios from the measurements. In the next eight exercises the letters refer to Fig. 289. Ex.4. CD = S}^, DA = 3H, CE = 2}4, find £5. Ex.5. DA=2y2, CA = 12}4, CB = 9H, find CE and EB. Ex.6. CD = m, CA=25, CB = 15, find CE and EB. Ex. 7. Z>^.= 3.6, EB = 2A, CB = 10.S, find CD. Ex.8. CD = 2^j^; DA =3, CE = 5, find EB. Ex.9. DA=Q, CE = 7, EB = 7, find CA, Ex. 10. CD = 28, DA = 14, CB = 9S, find CE and EB. Ex. 11. DA=2^2', CA=8, EB = 15, find EC. Ex. 12. Name the pairs of segments having equal ratios when a line parallel to the bases of a trapezoid ^^ cuts the non-parallel sides and the diagonals. Ex. 13. Show how Th. 99 may be used to ^^ ^^ ^ find an inaccessible distance (see Fig. 290). Fig. 290 Ex. 14. Draw a figure showing four parallels cutting two transversals. Name the equal ratios formed. RATIO AND PROPORTION 171 EQUAL RATIO TEST FOR PARALLELS 205. Theorem 100. If a line divides the sides of a tri- angle so that one side is to one segment as a second side is to its corresponding segment, the line is parallel to the third side of the triangle. Fig. 291 Hypothesis: AABC is any triangle with DE so drawn CA CB Conclusion: DE \\ AB. Analysis and construction: I. To prove DE \\ AB, prove that DE coincides with a line II AB, II. .*. construct DX from D || AB and prove that DE coincides with DX. III. To prove that DE coincides with DX, prove that E falls on X. IV. To prove that E falls on X, prove that CE = CX. M rr. ,u .rT7 r^ u CA_CB CA _CB V. To prove that CE = CX, show t^ - ^ and Yy) 'rx' . The proof is left to the pupil. For V use Th. 92. Cor. If a Hne divides the sides of a triangle so that the ratio of the segments on one side is equal to the ratio of the segments on the other, the line is parallel to the third side of the triangle. Exercise. If a line divides the non-parallel sides of a trapezoid into segments having the same ratio, will the line be parallel to the base? Give proof, 172 PLANE GEOMETRY CONSTRUCTION OF PROPORTIONAL SEGMENTS 206. Ex. 1. By algebra divide 120 into parts that shall be in the ratio of 7:8. Ex. 2. Using Prob. 7, § 111, find ^f of a given segment. Ex. 3. By algebra divide 144 into three parts that shall be in the ratio of 4, 5, and 9. Problem 14. To divide a segment into two segments that shall be in the same ratio as two given segments. Fig. 292 Suggestion. If XY is the given segment, show how to construct h k the figure so that you can prove that — = t" ' Segments lettered alike are equal. Give proof. Problem 15. To divide a given segment into segments proportional to any number of given segments. Y^ » ;i n Y Z' Fig. 293 Suggestion. It is necessary to construct the figure so that you .. .h k n can prove that — = -7- = — " a b c Ex. 4. Show that Prob. 15 may a be solved as follows (Fig. 294) : From c A draw ray / and from B draw m\\l. Lay off the given segments on / and 7n as shown in the figure and join the pomts of division. Fig. 294 J--- RATIO AND PROPORTION 173 CONSTRUCTION OF FOURTH PROPORTIONALS 207. Ex. 1. Given three segments a, b, and c, construct a fourth segment x so that -7 = - (Fig. 295). « ^ * c Suggestion. The construction is based on o^*:::— 2 — ^ — ^ ^ Th. 99. How are the segments a, b, and c laid off on the sides of Z 0? Ex. 2. Using the three segments given pj^ 295 a b in Fig. 295, construct x so that — = — , so ^ ' c X ^ be , c b ^i. J. b a * c a that - = -, so that- = -, so that - = - so that-r = ~- ax ax ex ox Ex. 3. Do any of the figures called for in Ex. 2 give the same value for a:? Why? The fourth term of a proportion in which the other three terms are the three given numbers taken in order is called the fourth proportional to the three given numbers; for example, if t=-, ^ is the fourth proportional to a, b, and c. Ex. 4. Find the fourth proportional to a. 21, 5, and 4 e. 6H, 8Hj and 5 e. a+l, a, and a+4 b. 5a, 3a, and 2b d. a, a+1, and Qa^ f. Qj^, 26, and 35 Problem 16. To construct a fourth proportional to three given segments. Analysts: Let a, b, and c represent the given segments and x the fourth proportional. To construct a foiuth proportional to a, 6, and c, construct X so that I = - . (See Fig. 295.) Ex. 5. , Using three given segments a, b, ,and c, find a fourth proportional to a, c, and b; to b, c, and a; to c, a, and b. Are the words "in order" essential in the definition of the fourth propor- tional? Why? 174 PLANE GEOMETRY Ex. 6. Find by geometry a fourth proportional to the segments whose lengths are given below. In each case verify by measure- ment and computation. a. 4.2 cm., 2.5 cm., 37 cm. b. 3.5 cm., 4.9 cm., 2.5 cm. c. 3 cm., 3.5 cm., 4.2 cm. Ex. 7. If a, b, and c represent three given segments, find a fourth segment x so that {a) x=^-; (b) x = — -; (c) x = %-- a a b MISCELLANEOUS EXERCISES INVOLVING RATIOS AND PARALLELS CD 1 208. 1. In Fig. 296, ^ = 3* If DE\\AB and DF\\CB, prove that DE = y^ AB. What is the ratio of DF to CJ5? ^ 'f~^ Fig. 296 2. Construct between two sides of a triangle a segment that shall be parallel to the third side and equal to % of the third side. 3. If, in Fig. 297, ED \\ CB and CF \\ EB, AD_AB AD_AB ^ AB _AF prove ^^-^^ AB-Jf' ^^^ DB-Jp- A A nalysis: To prove the two ratios equal, prove them each equal to a third ratio. 4. A line is drawn through the intersection of the medians of a triangle cutting two of the sides of the triangle and parallel to a third. In what ratio are these sides divided? Why? c 5. In Fig. 298, O is any point within A ABC, A'B'\\AB from A', an arbitrary point in AO, and intersects OB at B' . B'C \\ BC from B' and intersects OC at C. A' and C are joined. Prove that ^'C'MC. - ^^^^^^ - 6. Would Ex. 5 be true if point were outside A^^C? Draw the figure and give the proof. RATIO AND PROPORTION 175 7. In Fig. 299, CO is an arbitrary segment from C to AB and X is a point on CO. If ■:r^ is equal to a given ratio ^ (for example, ^^), find the locus of point X as O moves along the line AB. 8. In Fig. 300, O is the mid-point oi AB, AK = liAB, and KH \\ OC. What is the ratio of CH to CA? Answer this question \i AK=% AB. /^^ (f — ^ Fig. 300 9. If, in Fig. 301, AB = BC, AD = CE, and FB^ BG, prove that AC\\DE \\ FG. 10. Prove by Th. 99 that if a line is parallel to the base of a triangle and bisects one side, it bisects /^^V^—f^ the other also. Fig. 301 11. Prove the converse of the theorem quoted in Ex. 10 by Th. 100. Cor. 12. Prove by Th. 98 that if a line is parallel to the bases of a trapezoid and bisects one side, it bisects the other side also. Can the converse of this be proved by proportion? 13. Given any angle and P, any point within it. Draw a line through P meeting the sides of the angle in two points M, N, such that MP = 2PN. — College Entrance Examination Board, Plane Geometry Examination, 1912. 14. Show that a carpenter's steel square may be used to solve problems in proportion. Fig. 302 shows a steel square graduated to half-inches. ACD is a frame made of two pieces of wood hinged at C. AC can slide on the long arm of the square. Find 12 9 a number x so that — = — Place ACD so 16 X that the outer edge of CD is on 6 on the long p^^ 3Q2 arm (12 half -inches) and 4>^ on the short arm. Without changing the angle of adjustment of the frame, move the frame to the left until the outer edge of CD passes through 8 on the long arm. How is the value of x found? Why? 176 PLANE GEOMETRY SIMILAR TRIANGLES TEST I FOR SIMILAR TRIANGLES 209. Theorem 101. If two triangles have the angles of one respectively equal to the angles of the other, the corre- sponding sides have equal ratios. Hypothesis: AABC and ADEF are any two triangles with ZA=ZD, ZB=ZE, and ZC= ZF. Conclusion : — — = — — = — - . DE EF FD Analysis and construction: AT? T^ C I. To prove r—-=——, use a line parallel to a third side of /\DEF. II. .*. place /lABC upon ADEF with point B on point E, AB along DE, and BC along EF. Then prove A'a II DF. Let the pupil complete the analysis and give the proof. T^ • 1 ^ BC CA It IS necessary also to prove — - = — — • EF FD Exercise. If a and a', b and b', c and c' are corresponding sides of mutually equiangular triangles, find 1. ^j'andc', if o =12, 6=18, c =24, anda' = 20. 2. yandc', if a =3M, 6=4, c =5>i and a' = 25. 3. b and c, it a = lOH, a' = 16, b' = 20, and c' = 27. 210. Two polygons are said to be similar if a. The angles of one are equal to the corresponding angles of the other and b. Corresponding sides have equal ratios. RATIO AND PROPORTION 177 Theorem 101 gives us our first test for similar triangles. We will state it formally as Theorem 102. Two mutually equiangular triangles are similar. Why? In popular language, similar figures have the same shape. All enlargements and drawings to scale are practical examples of similar figures. Ex. 1. Prove that all equilateral triangles are similar. Ex. 2. Are all isosceles triangles similar? Why? When are two isosceles triangles similar? Ex. 3. If two isosceles triangles have equal vertex angles, the legs have the same ratio as the bases. Ex. 4. Are all right triangles similar? Why? When are two right triangles similar? Ex. 5. Is a square similar to a rectangle? Why? Ex. 6. Construct two rectangles that are similar. TESTS FOR EQUAL PRODUCTS AND EQUAL RATIOS 211. At the beginning of the course in geometry consid- erable time was spent on the use of congruent triangles in proving segments and angles equal. So important is this that when it is necessary to prove two segments equal we often look first for a pair of congruent triangles. In this chapter we are studying especially equal ratios, which are just as important as equal segments. The tests for equal ratios are as important as the tests for equal segments. When two ratios are to be proved equal, the following possi- bilities must be considered: A. Our fundamental methods for proving ratios equal are: 1. By parallels and transversals. 2. By similar triangles. B. Before either of these mechods can be applied it is often necessary to find a third ratio to which each of thei given ratios can be proved equal. 178 PLANE GEOMETRY The use of similar triangles in proving ratios equal is of considerable importance. The following considerations are often helpful : First: To select the proper triangles: The definition of similar figures says that corresponding sides have equal ratios. This gives — = 77 and by alternation ir = 77 * If. 1 c e then, we are to prove two ratios equal, say -7- = y , we may choose the triangles so that one of them shall have c and e as sides and the other shall have d and / as sides, or so that one of them shall have c and d as sides and the other shall have e and / as sides ; that is so that the numerators shall be sides of one triangle and the denominators sides of the other, or so that the terms of one ratio shall be sides of one triangle; and the terms of the other ratio sides of the other. Second: To select the corresponding sides of a pair of similar triangles: In Fig. 303 the triangles are so placed that corresponding sides can be selected immediately by inspection. When the triangles are not thus conveniently placed it is necessary to remember that corresponding sides are always opposite equal angles. The corresponding sides should be selected carefully from the equal angles as illus- trated in the proof to Ex. 1 on p. 179. Notice that equal angles may be designated by the same numbers, as Z2 and Z2'. When it is required to prove two products equal, a pair of equal ratios may be obtained from the equal products by Th. 91 and the ratios proved equal as explained above. Note. If we say that corresponding sides of similar triangles have a b equal ratios, the ratios should be read ~' '^ ~y' If> however, we say that corresponding sides of similar triangles are proportional, we may use either -^ = -77 or -?- = ■^- Similarly in Fig. 289 if we say the sides are a divided proportionally, we may use the ratios given in Th. 99 and Cor. or either of the forms in Ex. 1, § 204. RATIO AND PROPORTION 179 EXERCISES INVOLVING THE USE OF TEST I FOR SIMILAR TRIANGLES 212. 1. The diagonals of a trapezoid divide each other into segments that have the same ratio (see Fig. 304). Analysis: X y To prove — = — prove the angles of ADOC equal y respectively to the angles of AOBA. ^^^' ^O^ Proof: STATEMENTS 1. Z1=Z1'. 2. Z2=Z2'. 3. ADOC=ZAOB. . X (opposite Z2) _ y (opposite Zl) ^ z (opposite Z2') zy (opposite Zl') 2. In Fig. 305, is any point in segment AB. y^ Any line is drawn through point O not perpen- \^ dicular to AB. From A and B perpendiculars j^ are drawn meeting this line at points F and X. Fig, 305 3. A BCD is a parallelogram with its diagonal AC. BX is a line drawn through B intersecting AC at Y and ^D at X. Prove ;, ^ BY BC '^"' XY = A-X 4. In Fig. 306, ABC is an isosceles triangle. y^^^^^ Z 1 = Z 2. Prove that b^ = cm. ^^'.k..J^.B Analysis: F^c. 306 I. To prove b^ = cm, prove that -r= — tn If the product of two segments equals the square of a third segment, the last segment is called a mean propor- tional between the other two. In Ex. 4, h'^ = cm, 6 is a mean proportional between c and m. If — =-r' —=—iX^ = ab, OC d % or X = Va6, :!c is a mean proportional between a and h. Why ? 5. Investigate the case, Fx. 4, in which A A > Z.C. 180 PLANE GEOMETRY 6. In Fig. 307, AABC is a right triangle with Z5 a rt. Z. DE is drawn ± AC from any point in AB. Prove AB 'AD = AE 'AC. 7. Investigate the case, Ex. 6, in which point D is on AB extended. (. Fig. 307 8. Investigate the case, Ex. 6, in which point D is on BA extended. 9. Prove that the parallel sides of a trapezoid have the same ratio as the segments into which one diagonal is divided by the other. c 10. In Fig. 308, AABC is isosceles and BX = BA. ^> Prove that c is a mean proportional between AC and AX, 11. In Fig. d09,CX±AB and BY±AC. Prove ^, AC CX ^^^^AB = BY' 12, In Fig. 309, prove that BO BY=^BA • BX. 13. In Fig. 310, lines h and k are parallel and are cut by the pencil of rays from point 0. Prove that a _c ~b~l' Suggestion. Prove that each ratio is equal to a third ratio. A X B Fig. 309 Fig. 310 14. What ratios would be equal if h and k (Fig. 310) were on opposite sides of point O ? 15. In Fig. 311, AABC is a right triangle with Z C a right angle. CD±AB from C. Prove y AACD^ACBD and read the ratios of corre- ^. spending sides. Fig. 311 RATIO AND PROPORTION 181 16. In Fig. 312, ^B is a diameter of OO, BD tangent circle a.t B. AD is any line from A cutting the circle at E and the tangent at D. Prove that AB \s2i mean proportional between AE and AD. 17. In Fig. 312 draw BE. Prove that BE is a mean proportional between AE and ED. 18. In Fig. 313, CZ) is a diameter perpendicular to chord AB. Prove that ^C is a mean proportional between CE and CD. 19. In Fig. '313 prove that /IE is a mean propor- tional between CE and ED. 20. In Fig. 313 prove that CE-ED = AE^EB. 21. The cross-section of a street surface is the arc of a circle; the distance from curb to curb is 30 ft.; the rise of the center of the street above the gutter is 7 inches. What is the radius of the circle? 22. In Fig. 314, XF II ^5 and FZ II BC. Prove that b~l' Fig. 313 Fig. 314 23. Draw a square A BCD and the diagonals AC and BD. Let E, F, G, and H be the mid-points of the sides AB, BC, CD, and DA respectively. Join each vertex to the mid-points of the two non-adjacent sides, that is, join A to points F and G, and so on. Find pairs of similar triangles and read the ratios of corresponding sides. 24. Fig. 315 shows three concurrent lines XY\\AB, and YZ\\BC. Prove that j = y • 25. In Fig. 316 the circles are tangent at X, AB and CD are drawn through the point of tan- gency, meeting the circles as shown. Prove that cb = ad. (See §175, Ex.20.) 26. Investigate the case, Ex. 25, in which the circles are tangent internally. Fig. 316 13 182 PLANE GEOMETRY IMPORTANT SPECIAL CASES INTERSECTING CHORDS 213. Theorem 103. If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other. Fig. 317 Hypothesis: Circle O is any circle with the chords AB and CD intersecting at X so that a and h are the segments of AB and c and d the segments of CD. Conclusion : ab = cd. The analysis, construction, and proof are left to the pupil. In the next six exercises the letters refer to Fig. 317. Ex. 1. Find b, if a = 12, ^ = 2%, and c=15. Ex. 2. Find a and b,iiAB = 22,d = 8, and c = 12. Ex. 3. Find a and d, if AB = 19, 6 = 10, and c = 6. Ex. 4. Find d, if a = 5>^, <; = 43^, and b = 5. Ex.5. I^mda,iid = 8li,c = 2%,sindb = 3y5, Ex. 6. Find c and d,iiAB = 2Q,b = 8, and c = d. 214. If point C is between A and B on line AB, AC and C-5 are said to be segments oiAB and ^B is said to be divided internally at C. In Fig. 318, AB is divided ^ c ^ internally at C. ylC+C5 = A5. i -5" — c' If point C is on line AB but not between A and 5, AC and -BC are still said to be segments of AB. AB is said to be divided externally at C. In Fig. 318, AB is divided externally at C. AC'-BC' = AB if C is on AB extended. BC'-AC'=BA if C is on BA extended. RATIO AND PROPORTION 183 INTERSECTING SECANTS 215. Theorem 104. If two secants intersect without a circle, the product of one secant and its external segment is equal to the product of the other secant and its exter- nal segment. Fig. 319 Hypothesis: Circle O is any circle with the two secants h and k intersecting without the circle at E so that a and b are the external segments of h and k respectively. Conclusion: ah = bk. The analysis, construction, and proof are left to the pupil. Ex. 1. The following data refer to Fig. 319; c and d are the ' internal segments of h and k respectively. Find the length of the segments required. a. Find c and d, if A=15, a = 7, and ^ = 35. b. Find b, if a = 9, k= 12, and c = 4. c. Find a, if Prove by proportion. 186 PLANE GEOMETRY SEGMENTS MADE BY THE BISECTOR OF AN EXTERIOR ANGLE OF A TRIANGLE 218. Theorem 107. The bisector of an exterior angle of a triangle divides the opposite side externally into segments that have the same ratio as the other two sides of the triangle. Fig. 323 Hypothesis: AABC is any triangle with CO bisecting the exterior ZACE and dividing BA externally into segments r and 5. Let BC = a, CA = b, and CD = b'. Conclusion : L-^. s h The analysis, construction, and proof are left to the pupil. Discussion. Theorem 107 is not true for the vertex angle of an isosceles triangle. In the next four exercises the letters refer to Fig. 323. Ex. 1. Find a, if r = 24, 5 = 6, and 6 = 5. Ex. 2. Find r,iia= 18, 6 = 5, and ^ = 7. Ex. 3. Find r and s, if AB=S, a=12, and & = 6. Ex. 4. Find r and s, if AB = 4:, a = 9, and b = 6. 219. If two points divide a segment internally and exter- nally in the same ratio, the segment is said to be divided harmonically by the two points. ^^^ Ex. 1. In Fig. 324, ABC is any triangle. ^ CX bisects ZACB and CX' bisects the exterior /.BCD. Prove that AB is divided harmoni- ^ ^ ^ cally at X and X'. ^'^- ^^4 Ex. 2. Show how to divide any given segment harmonically. Suggestion. Construct any triangle on the segment AB as base. RATIO AND PROPORTION 187 PROPORTIONAL SEGMENTS IN RIGHT TRIANGLES 220. Theorem 108. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the perpendicular is a mean proportional between the seg- ments of the hypotenuse. Fig. 325 Hypothesis: A ABC is any triangle with ZC = rt. Z, p± AB from C. AB = c\ m and n are the segments of c. Conclusion: ^ is a mean proportional between m and n. The analysis and the proof are left to the pupil. Theorem 109. If a perpendicular is drawn from the ver- tex of the right angle of right triangle to the h3rpotenuse, either leg is a mean proportional between the whole hypote- nuse and the segment adjacent to that leg. The analysis and the proof are left to the pupil. Exercise. Construct Fig. 325 so that (a) m = Z cm. and « = 7 cm.; (6) 6 = 6 cm. and c = 9 cm.; (c) 6 = 2.4 cm. and w = 1.2 cm. In each case measure the remaining segments and compare them with the results obtained by computation. CONSTRUCTION OF MEAN PROPORTIONALS 221. Problem 17. To construct a mean proportional between two given segments. Solution I. The solution may be based on Fig. 326, by making the hypotenuse of the right triangle equal to the sum of the two given segments. Construct the right triangle by means of a semicircle. Solution II. The solution may be based on ^ Fig. 326, by making c equal to the longer and m equal to the shorter of the given segments. Can you invent a solution based on Fig. 321? Fig. 326 188 PLANE GEOMETRY Ex. 1. If a and_b are two given segments, cons truct a segment X so that (1) x= ylah; (2) x= ^2ab; {S)x= V^^a^; (4) x = }^ ^|ab, Ex. 2. JTaking any given length to represent 1, find segments equal to V2, V8, V12. Measure the results and compare them with the approximate square roots of 2, 8, and 12. — 2 X Suggestion. If x=yl2,x'^ = 2. Then " = ^* RELATION BETWEEN THE SIDES OF A RIGHT TRIANGLE 222. Theorem 110. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Fig. 327 Hypothesis: AABC is any right triangle with ZC a rt. Z , c the hypotenuse, and a and b the legs. Conclusion: a^-\-b^=c'^. Analysis and construction: I. To prove a^-\-¥ = c^, find a value for a^ and a value for b^ and add. II. The terms a^ and b^ above suggest the use of the mean proportional theorem. III. .*. draw a perpendicular from C to AB and find a^ and b^ in terms oi AB and its segments. Proof: STATEMENTS 1. a^ = nc and b^ = mc. 2. a^-]-b^ = nc-]-mc={n-\-'m)c. 3. a2 +62 = ^2. Let the pupil give the reasons. For 2 use: the sum of numbers having a common factor is the common factor multiplied by the sum of the coefficients. RATIO AND PROPORTION 189 Note. Th. 110 is one of the most important theorems of geometry and one of tlie most frequently used. More than one hundred proofs are known. It is called the Pythagorean theorem. Pythagoras, a Greek, is supposed to have given a general proof, although the fact was believed to be true much earlier. We do not know the nature of the proof that Pythagoras gave, but it is probable that it was something like the one given above. Pythagoras settled in Crotona, Southern Italy, where he founded a brotherhood, the members of which were pledged to secrecy. They spent their time in the study of philosophy, ethics, and mathematics and discovered many important theorems in geometry. Pythagoras died about 501 B.C. Exercise. What is the hypotenuse of a right triangle if the perpendicular sides are 3 and 4? Note. The fact that a triangle whose sides are 3, 4, and 5 is a right triangle and the relation 3^+42 = 52 were known to the Egyptians more than three thousand years before the time of Pythagoras. The Egyptians used ropes knotted at equal distances. These were stretched about three poles so as to form a. right triangle. The Egyptians called the men who knew how to use these ropes rope- stretchers. Surveyors use similar methods to-day. How many knots must there be in the rope and how is it used to construct a right triangle? The pyramids of Egypt have an angle nearly equal to an acute angle of a triangle whose sides are 3, 4, and 5. The Chinese and the Hindus probably knew about this right triangle at a very early date. APPLICATIONS OF PYTHAGOREAN THEOREM 223. Ex. 1. What is the diagonal of a rectangle whose sides are 25 ft. and 60 ft.? Ex. 2. The diagonal of a rectangle is 50 ft.; one side is 14 ft. Find the other side. Ex. 3. Find all of the segments in Fig. 328, if ^ a. « = 16, p=12. d. 6=10, a = 24. b. m= 8, 6=17. e. c=50, 6 = 30. c. p= 9, w=12. /. w = 28, » = 63. Ex. 4. The radius of a circle is 11 in. Find the length of a tangent drawn from a point 61 in. from the center. 190 PLANE GEOMETRY Ex. 5. In a given circle let r represent the radius, c a chord, and d the distance of the chord from the center. Find the missing terms as indicated below: a, d = 8, c = 32, r=? c. d=12, c=?, r = 36. b. d=?, c=28, r=18. d. d = 5, c=?, r=45. Ex. 6. One side of a square is 12. Find the sides of an isosceles triangle formed by joining the mid-point of one side to the opposite vertices. Ex. 7. One side of a square is 14. Find the sides of an isosceles triangle formed by joining one vertex to the mid-points of the sides not passing through that yertex. Ex. 8. If one side of a square is 8, find the diagonal. Find the diagonal if one side is 6; 10; 12; 20; 5. Ex. 9. If a diagonal of a square is 15, find the side. Ex. 10. If the hypotenuse of an isosceles right triangle is 12, find the side. Theorem HI. If one side of a square is s, its diagonal is s V2. If the diagonal of a square is d, the side is ^/id yl2. Ex. 11. The base of an isosceles triangle is 12; the equal sides are 18. What is its altitude? Ex. 12. Find one of the legs of an isosceles triangle if the alti- tude is 8 and the base is 12. Ex. 13. Find the altitude of an equilateral triangle one of whose sides is 4. Find the altitude if the side is 5; 10; 12; 5. Ex. 14. If the altitude of an equilateral triangle is 14, find the side. Find the side if the altitude is 6; 8; 7>^; a. Theorem 112. If one side of an equilateral triangle is s, its altitude is V2 s ^3. If the altitude is a, one side of the equilateral triangle is ^/s a VJ. Ex. 15. One side of a rhombus is 24; one of its angles is 60°. Find the diagonals. Ex. 16. One angle of a rhombus is 60°. The longer diagonal is 12. Find the side. RATIO AND PROPORTION 191 APPLICATIONS OF EQUAL RATIOS MISCELLANEOUS EXERCISES 224. 1. Fig. 329 shows proportional compasses used to enlarge or reduce drawings to scale. How must the instrument be adjusted so that b is twice a ? so that b is three times a ? 2. The shadow of a tree is 36 ft. at the same time that the shadow of an 8-ft. pole is 5 ft. How high is the tree ? Fig. 329 3. How may the height of a flagpole be found by noticing just when the length of the shadow of a certain post is equal to the height of the post? Why? Note. It is said that the Greek Thales astonished the Egyptians by measuring the heights of the pyramids from their shadows. Whether he used the method of Ex. 2 or the special case mentioned in Ex. 3 is not known. Thales lived from about 640 to 548 B.C. and introduced the study of geometry into Greece. x 4. An 8-ft. pole is placed at B (Fig. 330). How '' '^ may point A be located and what lines must be meas- y ured in order to find the height of the tower CX? Xb ^ Why? Fig. 330 5. Can you use the method of Ex. 4 in determining the length of a flagpole placed on the comer of a building? Could the length be determined by using the special method of Ex. 3? 6. Show that an inaccessible distance AB (Fig. 331) may be obtained by the following method: Make c AC LAB. Take C, any point on .4 C from which B is visible. Make CD ± CB. Find E, the point ^' at which CD and BA would intersect. What lines must be measured? 7. Show how the measurement suggested in Ex. 6 may be performed practically by the aid of a pole and a carpenter's steel square. Note. In Ex. 6 stakes are set up at the points indicated and the figure laid out on the ground. In Ex. 7 the figure is set up in a vertical plane. 192 PLANE GEOMETRY 8. Show that an inaccessible distance AB (Fig. 332) may be obtained as follows: Make CA ± AB and extend AC until CD is some convenient part oi AC i}4 or H). Make DE±AD. Find E, the point at which BC and DE would intersect. What measure- ment must be taken? Is it necessary that AB and DEhe±AD? Why? Fig. 332 9. Before the invention of the telescope an instrument called the cross-staff was sometimes used to measure inaccessible heights and distances. The cross-bar c was made to slide Tp^--^ up and down the staff a (Fig. 333). Show how this -^ ^ ''^ instrument could be used to find the width of the Fig, 333 stream R. How could it be used to find the height of a steeple? V TRIGONOMETRIC RATIOS 225. From §210 we know that if AABCc^AA'B'C (Fig. 334) ^ = y or T = T}' Why? The second pro- ^ b h' portion may be translated thus: The ratio of two sides of one of two similar triangles equals the ratio of the two corre- sponding sides of the other. In right triangles we have: If two right triangles have an acute angle of one equal to an acute angle of the other, the ratio of any two sides of one is equal to the ratio of the two corresponding sides of the other. Ex. 1. In Fig. 335, AABC and A'B'C are similar right triangles. Apply the proportion given above. Obtain three pairs of equal ratios. Note. In Fig. 335 the triangles are lettered so that side a is opposite A A, side b is opposite Z.B, side c is opposite Z C. Angle C is the right angle. For convenience in the discussion that follows, the right triangles will be lettered in this way. RATIO AND PROPORTION 193 The ratio — is the same for all right triangles having the c same acute angle A. Why? The same may be said of the ratios — and -r • We may say, therefore, that I. If an acute angle of a right triangle is known, we can find the ratios of the sides. XL If the ratio of any pair of sides of a right triangle is known, we can find the angle. These same ratios are of so much importance in relation to the /.A that they have been given special names. The names of the ratios are as follows: a side opposite Z.A . „ , . c , a — or — :; — ^-^-: IS called sine oi Z.A and is written c hypotenuse sin A. h side adjacent to ZA . ,, - . r , a — or T-^ — IS called cosine of /.A and is c hypotenuse written cos A . a side opposite ZA . „ . ^ 4. r y a j • -r or -T-; TT-^ — 7— TH IS called tangent 01 Z A and is side adjacent to ZA written tan A . Ex. 2. Construct two right triangles with angle A = 50° in each but with sides of different lengths. In each triangle measure a, ft, I -I , . a b . a and c and compute the ratios — , — , and -r- Ex. 3. Follow the directions given in Ex. 2 for triangles with angles of 40°, 35°, 62°. Compare the results obtained in Exs. 3 and 4 with the tables given on page 299 showing the values of these ratios for angles of all degrees from 1° to 90°. The ratios in the tables are given approximately with three figures. With your crude methods of measuring you cannot expect to be as accurate. Q Ex. 4. Construct a right triangle so that the ratio -r is %. Measure ZA with a protractor and compare the result with the table. Look for the tangent that is nearest H and note the angle. 194 PLANE GEOMETRY 226. It is important to notice that every acute angle has a particular value for each of the three ratios. If the angle is given, the ratios, sine, cosine, and tangent can be read from the table; for example, sin 23° is .391. Ex. 1. Find from the table sin 36°; cos 42°; tan 27°. sin 49°; cos 15°; tan 76°. So also if the ratio is given, the angle can be found from the table. Ex.2. What is Z^ if sin ^ = .515 ; sin ^ = .966 ; sin ^ = .777. cos ^ = .961 ; cos yl = .839 •,cosA = .292. tan A = .325; tan ^ = 1.00; tan ^ =3.73. If the given ratio is not found in the table, use the one nearest to it; for example, if sin yl = . 239, Z A is about 14°, if tan A = 1 . 56, Z A is about 57°. Ex.3. Find Z^ when sin A = .472 ; cos ^ = .395 ; tan ^ = .726. 227. The trigonometric ratios are used to find the sides and angles of right triangles. Ex. 1. In A ABC (Fig. 336) it is known that a = 23 and (; = 30; find ZA. Solution: Select the ratio involving the opposite side and the / hypotenuse. Fig. 336 sin A = — c _23 "30 = .767 (by division). .-. Z.4 =50" about (from table). RATIO AND PROPORTION 195 Ex.2. Find side b of the AABC when /.A =35° and c=42 (Fig. 337). Solution: Select the ratio involving the adjacent side and hypotenuse. y^ cosA=—- Fig. 337 c cos A =cos 35° = .819 (from table). .•.A=.819. Let the pupil complete the solution. Ex. 3. Find side b when a = 42 and Z^=65°. Solution: Select the ratio involving the opposite side and the adjacent side (Fig. 338). tan A =-r ' tan A =tan 65° = 2.14 (from table). .•.2.14=^- Why? Let the pupil complete the solution. These exercises illustrate the general method which may- be stated in words: To find any particular part of a right triangle, select the ratio formula involving that part and the two known parts; form an equation and solve it for the unknown part. Ex. 4. If Zyl=64° and 6 = 31, find c. Ex. 5. li ZA= 32° and a = 54, find b. Ex.6. If Zyl = 19°andc=16, finda. Ex. 7. If a = 35 and c = 47, find Z^l. Ex. 8. If 6 = 52 and c = 73, find ZA. Ex. 9. If a = 62 and 6 = 26, find ZA. 228. The methods illustrated above are especially used in solving problems involving heights and distances. A tape for measuring distances and some instrument for measuring angles are needed to get the necessary data T 196 PLANE GEOMETRY For measuring angles the surveyor uses a transit which is really two protractors, a leveling tube, and a telescope for easy and accurate seeing. For accurate work he measures angles to the nearest minute or even closer and uses more extended ratio tables. Rough approximations may be made with instruments that any ingenious pupil can make. A good protractor, a plumb line, and a couple of pegs for sights are all that is needed for a rough angle measurer. In Measuring Implements of Long Ago Mr. W. E. Stark describes some ancient forms of such instruments. In finding heights the angle of elevation is ^ used. If B is the top of a tower (Fig. 339), the Z CAB is the angle of elevation. Notice that j^— ^, line AC is horizontal. Fig. 339 It is suggested that after solving the following exercises the pupil apply his knowledge to some practical problems of his own devising. Ex. 1. A tower stands on level ground. The angle of elevation of its top at a point 160 ft. from its base is 43°. Find the height of the tower. Ex. 2. What is the angle of elevation of the sun if a tree 32 ft. high casts a shadow 18 ft. long? Ex. 3. What is the height of a balloon if its angle of elevation is 19° when seen from a place 10 miles from a point directly below it? Ex. 4. The length of a string attached to a kite is 300 ft. Find the height of the kite if its angle of elevation is 56°. Note. Ex. 4, of course, assumes that the string is straight, which is never really true. Will the height found be too i/rea,t or too small? Ex. 5. A perpendicular cliff 650 ft. high subtends an angle of elevation of 48°. How far is it to the base of the cliff ? . Ex. 6. Show how to measure an angle A indirectly without the use of a protractor by measuring the segments marked on Fig. 340. If ZA is given, how is the remainder of the^.- figure constructed? Fig. 340 RATIO AND PROPORTION 197 SUMMARY AND SUPPLEMENTARY EXERCISES 229. SUMMARY OF IMPORTANT POINTS IN CHAPTER IX A. Tests. I. To prove two products equal, use the factors of one as the extremes and the factors of the others as the means of a proportion and prove the ratios equal (Th. 91 and §211). II. To prove two ratios equal, look for a. Two similar triangles (§211 and Ths. 101 and 102). b. Two transversals cut by three parallels (§211 and Th. 98). c. A line parallel to the base of a triangle (§211 and Th. 99). d. Two ratios equal to a third ratio (§211 and As. 57). B. Algebraic equations indicating constructions. I. If T = - , ax = bc, X or X = — , construct x, a a fourth proportional to a, 6, and c (§207). I. If ^ = ^, x' = ab, X b or :*; = -- ylab, construct x, , a mean proportional between a and b (def. Ex. 4. §212; §221). EXERCISES INVOLVING EQUAL RATIOS AND EQUAL PRODUCTS 230. Note. Be prepared to prove the theorems on which any of the following exercises depend. 1. Make a review diagram for Ths. 102, 105, and 106. 2. In Fig. 341, is an arbitrary point in AB. ^ XY and ZW are any two lines through point 0. If ^F and BX are perpendicular to XY, and'*'^^^^^ ^"^ AW and BZ perpendicular to ZW, prove that \/ »' AY AW J ,^, BX = -BZ' . ^"-'^^ 14 198 PLANE GEOMETRY 3. In Fig. 342, A BCD is a parallelogram and Y is an arbitrary point in DC extended. Prove that AO^ = 0X -OY. p^^, 342 4. Draw any triangle and its three altitudes. Find all possible pairs of similar triangles and read the ratios of corresponding sides. 5. In Fig. 343, P is the mid-point of the arc CD. PA and PB are arbitrary chords intersecting chord CD at X and Y respectively. Prove that PY - PB= j PA • PX. 6. Fig. 344 is a square with its diagonals A C and BD. AE bisects Z BAC. The other segments are similarly drawn. Find pairs of similar tri- angles and read the ratios of corresponding sides. Suggestion. First prove that AE, CG, and DB are concurrent and that AE and CF are parallel. What other segments must be proved concurrent? What ones must be proved parallel? 1 7. If two parallel lines are cut by a pencil of rays, correspond- ing segments on the parallels have equal ratios. Investigate two cases. t8. State and prove the converse of Ex. 7. 9. In Fig. 345, ABC is any triangle. AX = BY. XW II BC and YZ \\ AC. Prove that WZ \\ AB. 10. If, in Fig. 345, AX = BY, XW\\CB, and / WZ II ^B, is FZ ll^C? Give proof. ^ ^ Y Fig. 345 11. The common tangent to two circles divides the segment joining the centers into segments that have the same ratio as the radii. Investigate different cases. • 12. In Fig. 346, CO bisects ABC A. Prove a r -P that -7- = — by prolonging i5C so that CD = CA and joining AD. 13. If, in Fig. 346, bisects ZBCA. — J-, prove that CO B >■ O ^ A Fig. 346 RATIO AND PROPORTION 199 14. In Fig. 347, CO bisects the exterior /.ACE. — = — by making CD = CA and joining AD. BO = r and AO = s. 15. If, in Fig. 347, j = —, prove that CO bisects Z.ACE. Prove that Fig. 347 16. In Fig. 348, AB is a diameter of OO, CB and AD are tangents at the ends of the diameter. If AC and BD intersect on the circle at E, prove that AB BC ' AD. 17. In Fig. 349, ABC is an isosceles triangle inscribed in OO. Any line is drawn from C cut- ting AB at E and the circle at D. Prove that AC^ =CD'CE. 18. Investigate the case, Ex. 17, in which CE cuts AB extended. 19. If one of the parallel sides of a trapezoid is double the other, the diagonals trisect each other. 20. In Fig. 350, A BC is any triangle inscribed in the circle. • CQ bisects ZC. Prove that CA • CB = CP • CQ. A\ 21. If two chords intersect within a circle so that one of them is bisected by the other, half of the first chord is a mean proportional between the segments of the second chord. 22. Use Ex. 21 to construct a mean proportional to two given segments. EXERCISES INVOLVING THE PYTHAGOREAN THEOREM 231. 1. Make a review diagram for Th. 110. 2. The radius of a circle is 48 ft. Find the distance between two chords which are 72 ft. and 36 ft. respectively. What two cases are possible? 200 PLANE GEOMETRY 3. In Fig. 351, ED is a perpendicular bisector of the chord AB. In each case given below construct the figure to scale from the data given. Compute the lengths of the segments required and verify your results by measurement. a. CE = i, AB = 20. Find AE,EO, and AD. b. AB = S6, £0 = 30. Find EC, ^£, and ^Z>. c. AE = 26, CE= 10. Find AB, EO, and A D. d. AE = Q1, AB = 120. Find CE, ED, and AD. 4. The radius of a circle is 12 in. Find the length of a tangent drawn from a point 13 inches from the center. 5. In Fig. 352, AX and BX are tangent to OO from point X. AB is the chord joining the points of contact. In each case given below construct the figure to scale from the data given. Compute the lengths of the segments re- quired and verify your results by measurement. a. AX = 6, OX = 10. Find^Oand^^. b. AX = 40, A0 = 9. Find OZ and ^5. c. A0=15, 0X = S9. Find ^Z and ^5. d. A0 = 5, AB = 8. Find ^X and OX. 6. The radii of two concentric circles are 9 and 15 respectively. Find the length of a chord of the outer circle which is a tangent of the inner. 7. In the middle of a pond 10 ft. square grew a reed. The reed projected one foot above the surface of the water. When blown aside by the wind, its top reached to the mid-point of a side of the pond. How deep was the pond? (An old Chinese problem.) 8. The length of the common chord of two intersecting circles is 16 in., the radii are 10 and 17 in. respectively. Find the dis- tance between the centers. 9. The span of a circular arch is 120 ft. If the radius of the circle of which it is a part is 720 ft., find the height of the middle of the arch. 10. Find the altitude of an isosceles trapezoid if the parallel sides are 40 in. and 58 in. respectively and the non-parallel sides are 41 in. RATIO AND PROPORTION 201 11. Two parallel chords in a circle are one inch apart. Find the radius of the circle if the chords are 8 and 6 inches long respectively. . Is there more than one solution of this problem? Suggestion. Find two expressions each equal to the square of the radius of the circle and form an equation. Solve the equation. 1 2. Divide a given segment in the ratio of 1 to V2. 13. Fig. 353 shows an isosceles right triangle. XY is parallel to ^C and so constructed as to equal "* jf CY. Show how to construct XY and find the ratio of CY to YO (see §86, Ex. 12). 14. Fig. 354 shows an isosceles right triangle with its inscribed circle. Find the ratio of CX to XO and hence the length of XO ii AB = ^. 15. In Fig. 355 the arcs AC and BC are drawn with AB a.s radius and B and A as centers respec- tively. The circle O is tangent to AC, CB, and to the semicircle. If AB = s and the radius of i d OO is r, find r in terms of 5 and construct the ^^^- ^^^ figure. A church window design Suggestion. OB=s-r, 0D = }4s-\-r, DB = ^s, OB^ = 0^ -^-Dl^ - Make the substitutions and solve the equation. 16. Find the shortest path that an insect can take (without flying) from one corner of a room to the diagonally opposite corner if the room is 15 ft. long, 12 ft. wide, and 10 ft. high. 17. Choose two points, A and B, upon a given straight line, and two other points, C and Z>, upon a straight line perpendicular to AB. Prove that the hypotenuse of a right triangle whose legs are equal to ^C and BD is equal to the hypotenuse of a right triangle whose legs are equal to ^Z> and BC. — College Entrance Examination Board, Plane Geometry Examination, 1910. 18. A BCD is a rhombus with A and C as opposite vertices. is a point within the rhombus such that OB = OD. Prove that A , O, and C are on the same straight line, and that OA • OC = 2 2 AB —OB . — College Entrance Examination Board, Plane Geom- etry Examination, 1916. 202 PLANE GEOMETRY 19. A sloping embankment rises from a level field. One end of a prop, 20 ft. long, rests on the ground 16 ft. from the foot of the embankment, and the other end rests 9 ft. up the embankment, measured along its sloping side. How high is the upper end of the prop above the level field? Result in feet to one decimal. — College Entrance Examination Board, Plane Geometry Exami- nation, 1914. 20. If, in Fig. 356, AB = AC = \, show that ^ BC= yJ2. If AD = BC, what is the length of BD} If AE = BD, what is the length of BE? Show how the figure may be continued so as to construct segments equal to V5, V6, V7, etc. c Fig. D E 356 21. In Fig. 357, A BCD is a square. D is the center and DB is the radius for the arc BE ; A is the center j> c e g and ^£ is the radius for the arc EF; D is the center and DF is the radius for the arc FG, etc. Find the length of DE, AF, DG, AH, etc., if AB = 1. Fig. F 357 EXERCISES INVOLVING THE TRIGONOMETRIC TABLES 232. 1. Find the legs and the altitude of an isosceles triangle if the base is 24 and each acute angle is 49°. 2. One of the equal sides of an isosceles triangle is 45 and each base angle is 68°. Find the base and the altitude. 3. The distance across a stream may be found as follows (Fig. 358): Lay off AC ± AB, extending AC to some point from which B is visible. Measure AC and angle C. Find ^15 if ^C = 300 ft. and ZC = 56°. 4. Find the distance AC across a pond as shown in Fig. 359. CB is perpendicular to AC, Z5 = 34°, C5 = 165ft. Fig. 359 RATIO AND PROPORTION 203 5. A chord of a circle is 8 in. It subtends at the center of a circle an angle of 36°. Find the radius of the circle and the dis- tance of the chord from the center. 6. An angle at the center of a circle of radius 6 ft. is 40°. Find the length of the subtended chord and the distance of the chord from the center of the circle. 7. Prove that in any triangle -7-= ~. — b (see Fig. 360). sin £> Suggestion. From the figure find sin A and sin B and divide one equation by the other. ^ 8. Find the sides of a triangle if Z^=42°, ZB = 65°, and AB = 8. Use the formula obtained in Ex. 7. 9. Find the sides of a triangle if Z^=68°, ZC = 49°, and a = 25. 10. Two observers IX miles apart observe at the same moment the altitude of the base of a thundercloud that is between them. If the angles are 42° and 61°, how high was the cloud? Note. Ex. 10 illustrates a method actually used by weather bureau men. The two observers are in telephonic communication, select some singular part of the cloud that neither can fail to recognize, and take the observation at a stated time by the watch. MISCELLANEOUS EXERCISES 233. Note. Be prepared to prove the theorems on which any of the following exercises depend. 1. In drawing a certain map all segments are reduced in the ratio . What lengths will represent the sides of a 100 000 coimty which is a rectangle 25 miles long and 18 miles wide? 2. In Fig. 301, ABC is an iscsceles triangle. ^ CE-=EB = CF = FA. FG, CO, and EH are A. AB f^^'^^k from F, C, and £ respectively. If ^ = -^, what '^ g o h b CO '^ ^-^ — '- ^- BO 8' Fig. 361 is the value of II? ' ^'"""^ IS^'''^'' 204 PLANE GEOMETRY 3. In Fig. 362, A BCD is a square. AK = BF== CG = DH, XY and ZW are medians. GN is par- allel to AD from G. li DG=H DC, what is the ratio of HM to MK ? If HM = % HK, what is the ratio of DG to DC? D G Z C 7 I / Fig. 362 From a Pompeian mosaic 4. Restate Ths. 103 and 104 so that two ratios, rather than two products, are to be proved equal. In Th. 103 the segments of the chords, and in Th. 104 the segments of the secants, are inversely or reciprocally proportional . Why? 5. In Fig. 363, 0)0 and X intersect at points A and B. AD and AC are tangent to (DX and respectively at point A. Prove that AB is a mean proportional between DB and BC. 6. Fig. 364 shows the outline of a roof truss. k^^tv. h ABC is an isosceles triangle. The equal sides ^S<1\ /fS^ are each divided into 3 equal parts. CD, EF, a f d k b and HK are ± AB. If AC=35, and CD=^ AB, Fig. 364 From a roof truss design find the length of AB, CD, EF, and ED. 7. In Fig. 365, ABC is any inscribed triangle. Z1=Z2. Frove that AC' RB = CR' AT and that a^ AC'TB = CT-AR, Fig. 365 8. Two tangents each 24 in. long are drawn from the same point to. a circle of radius 7 in. Find the length of the chord joining the points of contact. t9. A perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse. Prove that the ratio of the squares of the legs equals the ratio of the adjacent segments of the hypotenuse. c 10. In Fig. 366, ABC is any inscribed triangle. CN is perpendicular to AB and CD is a diameter. Prove that (1) CB - AN = CN • DB; (2) AC - NB = CN' AD; (3) AC- CB = CD'CN. RATIO AND PROPORTION 205 11. Fig. 367 shows a diagram of the roof of a barn. DE A_ AF. The rafters AD and FD make equal angles ^ with DE. If ^£ = 26 ft., AD = ZO ft., and Z?C = 13 ft., find CB. CB is ± ^F. Pig. 367 12. Find by geometry two segments whose sum is equal to a segment 5 cm. long and whose ratio is 3:7. Find by algebra a 3 the value of a and b if a-{-b = 5 and -r=='^' 7 13. Find by geometry two segments whose difference is 1.5 cm., and whose ratio is 5 to 8. Find by algebra the value of a and 6 if a ^ 6 = 1.5 and -r= -3- 8 ^ 14. In Fig. 368, ABC is an isosceles triangle. CE=y3 CB, CF^VsCA. FG, CO, and EH are ± AB from F, C, and E respectively. Prove that if EFGH is a square CO=OB. What must be the A G O H B Fig. 368 ratio of CO to OB if CE = )4 CB and EFGH is From a roof truss a square? Construct the figure in each case. design 15. Inscribe a square in a given triangle. Suggestion. The construction is suggested in Fig. )9. To prove that HG = HE, prove £^ = ^. HG HE Fig. 369 16. Solve Ex. 15 by the construction shown in Fig. 370. A Fig. 370 Fig. 371 Fig. 372 17. Show how to inscribe a square in a sector (1) as shown in Fig. 371 ; (2) as shown in Fig. 372. 18. Inscribe a square in a given semicircle. Suggestion. Solve by at least two methods; use the method employed for Fig. 372; also that suggested by Fig. 373. A ^ c K n Pig. 373 206 PLANE GEOMETRY 19. Let ABC be a triangle with a right angle at C. Draw CD and CE equally inclined to CB, and meeting AB (or AB pro- longed) in D and E respectively. Let M be the mid-point of AB. Prove that MB is a mean proportional between MD and ME. — College Entrance Examination Board, Plane Geometry Examina- tion, 1910. 20. The distance between two parallel chords on the same side of the center of a given circle is 6 cm. If the chords are 36 cm. and 48 cm. respectively, find the radius of the circle. What would be the distance between the chords if they were on opposite sides of the center? 21. If two circles are tangent externally and a segment is drawn through the point of contact terminated by the circles, the chords intercepted in the two circles have the same ratio as the radii. 22. Fig. 374 represents a gable over an equilateral Gothic arch. The arcs CA and CB are drawn with B and A as centers and A B as radius. The sides of the gable DE and DF are tangent to the sides of the arch from a point in the common chord GC extended. a. Prove that DE = DF. b. Construct the figure so that AB = 6 cm. and DC = 4cm. c. Construct the figure so that AB = Q cm. and DE= 10 cm. d. Construct the figure so that AB = Q cm. and Z£Z)F = 30' 23. Prove the Pythagorean theorem by means of Fig. 375. Suggestion. Let a represent the distance OB, b repre- sent the radius of O O, and c represent the half chord perpendicular to ^ C at B. or &2_a2 = c2. Fig. 374 Prove that {b—a) {b+a) 24. Construct a segment x so that x = ^ ^ab; x 2^2 x = — , where a, b, and c are given segments. RATIO AND PROPORTION 207 25. Show that the following construction will give graphically the solution of the equation x^ — 2x = 24: Construct a circle whose diameter is 2. _At any point on the circle construct a tangent whose length is V24. From the end of the tangent draw a secant which passes through the center of the circle. The entire secant will be one of the roots of the given equation. 26. Ovals are of frequent use in landscape gardening and other branches of engineering. Agreeable ovals may be laid out as follows (Fig. 376): Let AB be the total length of the desired oval. Lay ofi on AB two equal intersecting circles with any radius. Draw the common chord CC and extend CC\ making CD equal to a diameter of the circles. Through D draw lines through the centers of the circles O and O' intersecting the circles at X and Y. With DX as radius draw XY: In a similar manner draw ZW . Show that the circles are tangent at X . If ^0' = HAB and AB==lo ft., find DD' . 27. In Fig. 377, ABC is any triangle inscribed in OO. CX is the altitude to AB, A Y is the altitude to CB. OK is the perpendicular from the center of the circumscribed circle to A B. H is the intersection of the altitudes. Prove OK = 14 CH. Fig. 377 28. Find the locus of points from which the distances to two given intersecting lines are in a given ratio. 29. Investigate the case, Ex. 28, in which the two given lines are parallel. 30. Two equal circles intersect in such a manner that the common chord is equal to the segment joining the centers. If the common chord is 2 in., find the radius of the circles and the width of the oval formed. If the radius is 3 in., find the common chord and the width of the oval formed. 31. Find the hypotenuse of a right triangle if the legs are (1) n and K(""~l); (2) » and(>^«)2— 1. Verify in each case by substituting numbers for n. CHAPTER X Area and Equivalence INTRODUCTORY MEASURING SURFACES 234. To measure the surface inclosed by the sides of a polygon is to find how many times it contains another sur- face chosen as a unit of measure. The area of a polygon is the measure number of the surface of the polygon. It is the common practice to use as a unit of surface a square whose side is a unit of length. Thus if the unit of length is an inch, the unit of surface is a square whose side is an inch and is called a square inch. While any segment may be used as a unit of length with its corresponding unit of surface, it is most convenient prac- tically to use one of the recognized standard units of length, such as the inch, foot, yard, mile, tenth of an inch, centi- meter, etc., with their corresponding units of surface — square inch, square mile, etc. The particular unit chosen depends upon the surface to be measured. EQUIVALENT POLYGONS 235. Ex. 1. In Fig. 378, AABC is isosceles and CD is perpendicular from C to AB. Show that A a and h are congruent. Draw the AABC; cut out Aa and b and place them together so that AD coincides with DB so as to form AXYZ. Are AABC and XYZ congruent? Do they cover the same extent of surface? 208 AREA AND EQUIVALENCE 209 Two polyp^ons that cover the same extent of surface are called equivalent polygons. The symbol ( = ) is used for equivalence. Since area is the measure of surface, 1. Equivalent polygons have equal areas. 2. Polygons with equal areas are equivalent. Since congruent polygons can be made to coincide, they may be made to cover the same surface and are equivalent. Congruent polygons are the simplest examples of equivalent polygons. It does not follow that equivalent polygons are always congruent. The following exercise gives illustrations of polygons that may be made to cover the same surface but are not necessarily congruent. Ex. 2. Construct two congruent right triangles that are not isosceles. Cut them out and place them together in different positions so as to form two isosceles triangles, a rectangle, a kite, two oblique parallelograms, and other polygons, all of which are equivalent. Make careful drawings of these figures. The use of cross-section paper is suggested. 236. It is evident that combinations of congruent polygons will give polygons that are equivalent but not necessarily congruent. The following definitions are necessary: If two polygons are so placed that a side of one falls upon a side of the other, but neither polygon overlaps the other, the polygon inclosed by the entire perimeter is the sum of the two polygons. If one polygon is placed entirely within another, the space between the perimeters is the difference between the polygons. A polygon is bisected by a segment if the segment divides it into two equivalent parts; for example, the diagonal of a parallelogram bisects the parallelogram. Similarly, a polygon may be trisected or may be divided into any number of equivalent parts. 210 PLANE GEOMETRY One figure is transformed into a second if the second is equivalent to the first. 237. The following assumptions will be used in the discus- sion of equivalent figures: As. 59. If equivalent polygons are added to equivalent polygons, the results are equivalent polygons. As. 60. If equivalent polygons are subtracted from equiv- alent polygons, the results are equivalent polygons. As. 61. If equivalent polygons are divided into the same number of equivalent polygons, each part of one is equivalent to any part of the other. As. 62. Polygons equivalent to the same polygon or to equivalent polygons are equivalent. 238. We have then the following preliminary test for equivalent polygons : I. To prove two polygons equivalent, prove that they are made up of parts congruent in pairs. II. To construct two polygons equivalent, construct them of parts congruent in pairs. y Ex. 1. Given AABC with D the mid- ^/..[\e._..^f point of AC. DF\\AB and BF || AC. Prove / \ "y''' AABC=OJABFD (Fig. 379). a^ i/ Fig. 379 A nalysis: I. To prove AABC=OJABFD, prove that they are the sums of congruent parts. II. .-.prove ABED + Al=ABED + All. III. .*. prove AI ^ AIL Ex. 2. Transform a trapezoid into a paral lelogram (Fig. 380). A E B Fig. 380 A nalysis: To transform A BCD into a parallelogram, construct the parts of the parallelogram congruent to the parts of A BCD. AREA AND EQUIVALENCE 211 Ex. 3. Transform a triangle into a rectangle. Ex. 4. Transform a trapezoid into a rectangle. Ex. 5. Any segment through the intersection of the diago- nals of a parallelogram and terminated by the sides divides the parallelogram into two equivalent parts. Ex. 6. Show how to bisect a parallelogram by a line (1) perpen- dicular to the base and (2) parallel to the base. 239. MEASUREMENT OF POLYGONS FUNDAMENTAL ASSUMPTION As. 63. The number of units of area in a rectangle is equal to the product of the number of units of length in the base and altitude. If S represents the area of a rectangle, b the length of its base, and a the length of its altitude, As. 63 may be stated as a formula, S = ab. The assumption will be discussed under two heads : A. When the sides of the rectangle are both commensurable with a given unit of length. In this case the unit of length can be applied an integral number of times to both the base and the altitude of the rectangle. The assumption is evident at once. The unit of length chosen may be contained in the base m times and in the altitude n times, if m and n are whole numbers. By drawing the proper lines the rectangle may be divided into n rows with m unit squares in a row. Illustration 1- Suppose the unit chosen is a square centimeter and the rectangle is 3 cm. long and 2 cm. wide (Fig. 381). The unit of length is contained in the base 3 times and p, , , ,0 in the altitude 2 times. By drawing seg- ments through the points of division par- allel to the sides, the rectangle is divided ~ into 2 rows with 3 sq. cm. in a row, or into 2X3 sq. cm., or 6 sq. cm. The measure number of the surface is 6; the area of the a" surface is 6 sq. cm. 3 cm. Fig. 381 212 PLANE GEOMETRY Sometimes one or both of the sides of the given rectangle are not exactly divisible by the unit chosen, but are divisible by some aliquot part of this unit. In this case this part of the chosen unit may be taken as a new unit of length, and a square whose side is this new linear unit may be con- sidered as the unit of area. The assumption is then evident as above. Illustration 2. Suppose the unit chosen is a square inch. In Fig. 382 one inch is not contained exactly in either AB or AC. One quarter-inch is, however, exactly contained in both AB and A C. One quarter-inch may be used as a convenient linear unit. The measure number of ^5 is 5 and of A C is 3. A square quarter-inch may be considered as the unit of area. By drawing the proper lines we can show that the rectangle con-^ sists of 3 rows with 5 units of surface in a Fig. 382 row. The measure number of the area is 15. In this case we may express the unit, the measure of the sides, and the area in fractional terms of a larger unit. The unit one quarter-inch is M in. The measure of AB, 5 quarter-inches, is ^ in. The measure of ^C, 3 quarter-inches, is ^ in. The area, 15 square quarter-inches, is i^e sq. in. B. When one or both sides of the rectangle are incom- mensurable with the chosen unit. In this case it is not possible to measure one side or perhaps both sides of the rectangle in integral or fractional terms of the chosen unit. Since the ratio of two incommensurable segments is an irrational number (§ 198), these sides may be expressed in irrational terms of the chosen unit, and are measured approximately. From these approximate lengths an approximate area is computed by the rule contained in the assumption. We have seen (§194) that by subdividing the unit of length we can obtain approximate measures for the sides of the rectangle that are as close as we choose to make them. It is evident that the approximation for the area may also be m.ade as close as we choose to make it. AREA AND EQUIVALENCE 213 Illustration 3. Suppose the unit chosen is one square centimeter. In Fig. 3S3,AB is 2 cm. and is commensurable with the unit; AC is equal to the diagonal of a square whose side is 2 cm. and is incommensurable with the unit. The length oi AC cannot be expressed in integral or fractional terms of the unit. We know, how- ever, that we can express i4 C as 2 V2 cm. Accord- ing to our_ assumption,, therefore, the area of R is 2X2 V2"or 4 V2'sq. cm. While the length of AC cannot be expressed exactly, an approximate length can be found for it. From this approximate length an approximate area can be found for R. Fig. 383 These approximate values can be made as close as_we choose. Suppose that 1.4 is taken as the approximate V2, then 2X1 .4 cm., or 2.8 cm., is the approximate length of AC, and 2X2.8 sq. cm., or 5.6 sq. cm., is the approximate area of R. In Fig. 383, the area of the rectangle ABGH represents this approximate area of R. Suppose, again, that 1.41 is taken as the approximate V2, then 2X 1 . 41 cm., or 2 .82 cm., is the approximate length of A C, and 2X2 . 82 sq. cm., or 5.64 sq. cm., is the approximate area of R. Although this approximation cannot be represented on the figure, we know that it is closer than the other but a little less than the area of R. In this case the following points should be noted : 1. The side or sides that are incommensurable with the unit cannot be expressed in integral or fractional terms of that unit. 2. The side or sides that are incommensurable with the unit can be expressed in irrational terms of that unit. In illustration 3, AC is expressed as 2V2 cm. 3. It can be proved that As. 63 is true for those cases in which the sides of the rectangle can be expressed only in irrational tenns of the chosen unit. The proof is, however, too difficult for this course. 4. Sometimes when the sides of the rectangle can be expressed only by irrational numbers, it is possible to express the area of the rectangle by rational numbers. Illustration 4. Connect the mid-points of the sides of a square whose side is 4 cm. Find one side of the square so formed and its area. 15 214 PLANE GEOMETRY 240. Practical measurements. When it is desired to use the rule contained in As. 63 to compute the area of a given rectangle from measurements actually made, an approximate area only is possible. We have seen, § 195, that the exact length of a given segment cannot be obtained in terms of a unit chosen in advance. Since the measures of the sides must of necessity be approximate, the area must of necessity be approximate also. The approximate area may, however, be made as close as we choose if only the divisions of the scale are made sufficiently small. EXERCISES INVOLVING AREA OF RECTANGLES 241. 1. Draw a rectangle whose sides are 3.4 cm. and 2.6 cm. Find the approximate area of this rectangle in inches, measuring (1) to the nearest inch, (2) to the nearest half inch, (3) to the nearest quarter inch, (4) to the nearest sixteenth inch. Make an accurate drawing for each approximation and compare it with the given rectangle. 2. Find the area of a rectangle whose sides are SHe in. and 4^2 in. 3. If the area of a rectangle is 321^6 sq. in. and one side is 3M in., find the other side. 4. Find the area of a walk 3 ft. 6 in. wide which completely surrounds a lot 300 ft. X 500 ft. The dimensions of the lot are taken on the inside of the walk. >■' 7^^" Fig. 384 n 5. Fig. 384 shows forms of columns in cross-section. The outside measures and the width are the same in each case. Find the area of each cross-section shown. Note. The area of the cross-section is an important element in determining the strength of the column. AREA AND EQUIVALENCE 215 6. Using any three given segments a, b, and c, construct a rectangle whose sides are (a+b) and c. Show how this figure illustrates geometrically the algebraic identity c(a-\-b)=ac+bc. Suggestion. c(a-\-b)=ac-\-bc may be translated into geometry: The rectangle whose side3 are c and {a-\-b) may be divided into two rectangles whose sides are a and c, and b and c, respectively. 7. Illustrate geometrically the identity c{a — b) = ac — bc. Suggestion. c{a — b)=ac — bc may be translated into geometry: The rectangle whose sides are c and {a — b) may be obtained by cutting a rectangle whose sides are b and c from a rectangle whose sides are a and c. 8. Using any two given segments a and b, construct a square on the segment {a-\-b) and illustrate geometrically the identity Suggestion. Translate {a-\-bY = a'^-\-b'^-\-2ab into geometry. Draw the segments necessary to divide the square on a-\-b into the required parts. 9. Illustrate geometrically the identity {a — bY = a^-\-b^ — 2ab. Suggestion. Translate {a — bY=a^-\-b'^ — 2ab into geometry. Con- struct a figure formed by adding a square whose side is a to a square whose side is b. Show how two rectangles may be cut from this so as to leave the desired result. What will be the sides of these rectangles? 10. Translate into geometry and illustrate by a figure: a. {a-\-b) {d + c) = ad^bd-\-ac-\-bc. b. {a^-l)^) = {a-b) (a + 6). 11. If the length of a rectangle is 3 ft. more than the width and the area is 70 sq. ft., find the dimensions. 12. If the length of a rectangle is 2K times the width and the area is 360, find the dimensions. 13. The length of a rectangle is 14 ft. more than its width. If its diagonal is 26 ft., find the dimensions and area. 14. Find the dimensions of a rectangle if its perimeter is 38 in. and its area 84 sq. in. 15. Find the dimensions of a rectangle if its diagonal is 13 in. and its area 60 sq. in. 216 PLANE GEOMETRY MEASUREMENT OF THE PARALLELOGRAM 242. Theorem 113. The area of a parallelogram is the product of the base and altitude. V / p / / a / I b B Fig. 385 Hypothesis: P is a /Z7 with base h and altitude a. Conclusion: Area P = ab. Analysis and construction: I. To prove area P = ah, compare P with a rectangle, R, that has h for base and a for altitude. II. .*. construct rectangle R with b for base and between the parallels AB and CD. III. To prove area P= area R, prove that P and R are equivalent. IV. .-. prove /\ADF ^ ABCE. Proof: STATEMENTS REASONS I. AADF^ABEC. I. Let the pupil give de- tails in full. II. 1. Pis equivalent to i?. II. 1. As. 60. 2. .-.area P = area i?. 2. See §235. III. Area R = ab, III. Why? IV. .-.area P = a6. IV. Why? Ex. 1. Prove Th. 113 for the case in which points E and F both fall on CD extended. Ex. 2. Construct a parallelogram whose sides are 5 cm. and 8 cm. and one of whose angles is 60°. Find its altitude and area. Ex. 3. Construct a parallelogram whose sides are 6 cm. and 9.4 cm. and one of whose angles is 30°. Find its altitude and area. Ex. 4. Construct a parallelogram with sides 4.2 cm. and 5.9 cm. and one angle 45°. Find its altitude and area. AREA AND EQUIVALENCE 217 MEASUREMENT OF THE TRIANGLES 243. Theorem 114. The area of a triangle is one-half the product of the base and altitude. Fig. 386 Hypothesis: ABC is a A with base h and altitude a. Conclusion: Area ABC = \ ah. Analysis and construction: I. To prove that area ABC = \ ah, compare A ABC with a parallelogram whose sides are AB and BC. II. .*. construct Let the pupil give the construction. Proof: STATEMENTS 1. A ABC is equivalent to J the parallelogram. 2. Area EJ=ah. 3. .'. area A = i a6. Let the pupil give the reasons. Ex. 1. Find the area of a triangle whose base is 6 ft. 3 in. and whose altitude is 2 ft. 7 in. Ex. 2. Find the altitude of a triangle if its area is S% sq. ft. and its base is 6 ft. 7 in. Ex. 3. Find the area of an isosceles triangle if its base is 32 ft. and one leg is 34 ft. Ex. 4. Find the area of an equilateral triangle if each side is 4 in.; 6 in.; 8 in. Ex. 5. The base of a triangle is 3 ft. more than its altitude. Find the base and the altitude if the area is 90 sq. ft. Ex. 6. The base of a triangle is 3 times its altitude. Find the base and the altitude if the area is 336 sq. ft. 218 PLANE GEOMETRY MEASUREMENT OF THE TRAPEZOID 244. Theorem 115. The area of a trapezoid is equal to one-half the product of the altitude and the sum of the bases. Fig. 387 Hypothesis: ABCD is a ZZ\ with bases b and b' and alti- tude a. Conclusion: Area ABCD = | a{b-\-b') . Analysis and construction: I. To prove area ABCD = l a{b+b'), divide ABCD into two triangles and add their areas. II. .'. construct Let the pupil complete the analysis. Proof: STATEMENTS 1. Area ABC = i ab. 2. Area ADC = ia'b\ 3. Area ABC+avea, ADC = i ab+i a'b'. 4. a = a'. 5. Area ABCD = \ a{b+b'). Let the pupil give the reasons. Ex. 1. One base of a trapezoid is 3 ft. more than the other. If its altitude is 6 ft. and its area 81 sq. ft., find the bases. Ex. 2. The area of a trapezoid is 96 sq. ft., its altitude 8 ft., one base 9 ft. Find the other base. Ex. 3. The area of a trapezoid is the product of the altitude and the median drawn between the non-parallel sides. " Ex. 4. There are other methods of drawing i) Q construction lines in the trapezoid ABCD (Fig. /j "\ 388) so as to divide it into parts whose areas j. f e may be found and added. Prove Th. 115 by Fig. 38S means of Fig. 388 and suggest other possible figures. AREA AND EQUIVALENCE 219 MEASUREMENT OF IRREGULAR POLYGONS 246. The finding of the area of a field shaped like an irregular polygon is one of the important problems that a surveyor must solve. One method frequently used is to divide the field into triangles and apply the method for finding the areas of the triangles. Ex. 1. Compute the area of the field shown in Fig. 389 from the following data: AC = 270 It. ;/ = 104ft. k = 82it. Ex. 2. Compute the area of the field shown in Fig. 390 from the following data: BD=U ft. 6 in. ^£=16 ft. 3 in. PM = 13 ft. CI = 9 ft. 4 in. BL= 11 ft. 2 in. EK=\0 ft. 6 in. Sometimes the field to be surveyed is bounded on one side by a stream, the shore of a lake, or a curved road. In such cases a straight line is run near the curved boundary, and the inclosed figure is cut into trapezoidal shaped figures by offsets run perpendicular to the line, as shown in Fig. 391. In such cases care should be taken that the curved boundaries of the figures are as nearly straight as possible. Ex. 3. Find the area inclosed between the fence AF (Fig. 391) and the river from the following data: yl 5 = 150 ft. DE = 250 ft. CC = 180 ft. 5C=140ft. £F=100ft. DD' = 95it. CD=160ft. BB'=UO(t. ££' = 120 ft. Pig. 391 Ex. 4. The distances along the straight line (AF, Fig. 391) are usually made equal, as the computation is then much easier. Compute the area referred to in Ex. 3 if the offsets are run as indicated below. Draw the figure to scale. Distances on ylF 100 200 300 400 500 600 Length of offsets 55 76 83 80 50 42 65 220 PLANE GEOMETRY Ex. 5. Find the area of the field shown in Fig. 392. ^5 = 300 ft., BC = UO ft., ZB is a right ,^ angle, DE= 125 ft. Offsets are run every yV \\ 50 ft. from D to A. ^D = 310 ft. The y^ j\c lengths of the offsets are ^^-'^^^^^^'^^^^ at D . . at 150 ft. . . . 20 Z^^!^"'"''"^ at 50 ft. . . 22 at 200 ft. . . . ^^^=^^— '^ at 100 ft. . . 26 at 250 ft. ... 18 Fig. 392 Ex. 6. Find the area shown in Fig. I \ -^ 393, using the dimensions given. A BCD is I \ 7 a rectangle. AE = BF. i — \ *^' / - Fig. 393 Ex. 7. Two streets intersect at right angles. A f1||||pi||r" third street cuts the other two at angles of 30° and 60°. LiF The shortest side of the triangular park left is 200 feet. ^B'^ If the streets are 60 feet wide each, find the area of W pavement at their intersection. / 1 Fig. 394 The area of an irregular polygon may be found approxi- mately by weighing. Cut the figure and also a square unit from the same sheet of paper or cardboard and weigh them both. The areas have the same ratio as the weights. Archi- medes used this method to find the areas of certain figures. Surveyors sometimes use it to-day. EQUIVALENT POLYGONS TESTS FOR EQUIVALENCE 246. The following notations will be used: For A ABC or EJABCD, S for area, b for base, and a for altitude. For AA'B'C or EJA'B'CD', S' for area, h' for base, and a' for altitude. A preliminary test for equivalent polygons was given in §238. Others are given in the assumptions of §237. The following tests are now evident and are here stated formally for convenience : AREA AND EQUIVALENCE 221 Test a: for equivalent triangles or parallelograms. Th. 116 is really a corollary of Ths. 113 and 114. Theorem 116. Two parallelograms or two triangles are equivalent if 1. They have equal bases and are between the same parallels. 2. a = a^ and b = b\ 3. ab = a'h'. Test b: for triangles and parallelograms. Th. 117 is a corollary of Ths. 113 and 114. Theorem 117. If a triangle and a parallelogram have equal bases and equal altitudes, the triangle is equivalent to half the parallelogram. By the assumptions of §237 any two polygons are equivalent if they are sums, dijBferences, or equal parts of equivalent polygons. EXERCISES IN TRANSFORMATION 247. 1. Find in Fig. 395 four equivalent parallelograms. 2. How many equivalent parallelo- >/ jkh g f ed c grams can be constructed with their bases \ ,' \ \^ on two given parallels? \\/-^\',C^ 3. Using any OJABCD, construct two Fir 'iQ'i parallelograms equivalent to it, (1) using AB as the common base, and (2) using AD a.s the common base. 4. What equivalent parallelograms can be found in Fig. 396? Segments with the same letters are equal. The lines are par- allel as indicated: k\\h, m \\ n, r \\ s. 5. Transform EJABCD into a paral- lelogram having its base equal to AB and one side equal to a given segment. Is the problem always possible? Fig. 396 6. Transform OJABCD into a parallelogram having its base equal to AB and one angle equal to a given angle. Is this problem always possible? 222 PLANE GEOMETRY Fig. 397 7. How many equivalent triangles can you find in Fig. 397? The segments lettered alike are equal. The lines are parallel as in- dicated: k II h and w || ». Transform a given ^ABC into 8. A triangle having the base equal to AB and one angle equal to a given angle. 9. A triangle having the base equal to BC and one side equal to a given segment. 10. An isosceles or a right triangle having the base equal to (1) AB, (2) BC, (3) AC. 11. A triangle having two sides equal to given segments. Suggestion. Two transformations are necessary. 248. The transformations in the previous section may be performed by means of an algebraic analysis. Problem 18. To transform a given parallelogram into a rectangle which shall have a given segment as its base. Fig. 398 Given the EJ ABCD with h its base and a its altitude, and b' the given segment. Let P represent the CO ABCD. To transform CJ ABCD into a rectangle with h' as its base. Analysis and construction: Let R represent the rectangle, a' its altitude. I. To construct R = P, construct a' so that ah = a'h'. IL .*. construct a' so that -r — ~/' h a in. .'. construct a fourth proportional to 6', 6, and a. Let the pupil make the construction in full and give proof. AREA AND EQUIVALENCE 223 Problem 19. To transform a given parallelogram into a square. Analysis and construction: I. To construct the square = P, construct x so that ab = x^, where x is the unknown side of the square. II. .'.construct x sl mean proportional between a and b. Ex. 1. Transform a given triangle into (1) a rectangle having a given base, (2) a square. Ex. 2. Transform a given parallelogram into an isosceles triangle having a given base. Ex. 3. Construct a square which shall be equivalent to three times a given square. Ex. 4. Construct on a given base an isosceles triangle which shall be equivalent to twice a given square. 249. Problem 20. To transform a given quadrilateral ABCD into a triangle with the base in the line AB and the vertex at point D. Fig. 399 Given the quadrilateral ABCD. To transform it into a triangle with its base in the line AB and its vertex at D. Analysis and construction: I. Since the base is to be in the line AB and the vertex at D, AABD will be a part of the required triangle. .*. transform ADBC into a triangle with DB for one side and another side on AB extended. II. .'. construct a line from .C \\ BD and join DE. Let the pupil name the required triangle and give analysis for the proof and then the proof. 224 PLANE GEOMETRY Ex. 1. Transform a given quadrilateral A BCD into a triangle that shall have a. Its base in the line AB and vertex at C. b. Its base in the line BC and vertex at A . c. Its base in the line BC and vertex at D. d. Its base in the line AD and vertex at C. e. Its base in the line DC and vertex at B. Problem 21. To transform a given polygon into a triangle. A B Fig. 400 The analysis, directions, and proof are left to the pupil. The figure suggests the construction. Let the pupil extend the method so as to transform a given hexagon into a triangle. Ex. 2. Construct a square equivalent to a given 4-side. Ex. 3. Construct on a given base a rectangle equivalent to a given 4-side. Ex. 4. Construct on a given base an isosceles triangle equiva- lent to a given 4-side. EXERCISES INVOLVING EQUIVALENT FIGURES 250. 1. The median of a triangle divides it into two equiva- lent parts. 2. Divide a given triangle into three equivalent parts by seg- ments drawn from the vertex to the base. 3. The diagonals of a parallelogram divide it into four equiva- lent triangles. 4. Given the CJABCD with X any point in the diagonal AC. Prove that AAXD = AAXB (see Fig. 401). 5. Investigate the case, Ex. 4, in which ^'^ point X is on ^C extended. Fig. 401 AREA AND EQUIVALENCE 225 6. Given A BCD, any quadrilateral, with E and F the mid- points oi AB and CD respectively. Prove AECF=AAFD-^AEBC (Fig. 402). Suggestion. Draw AC. 7. Two triangles are equivalent if two sides of one are equal respectively to two sides of the other and the angles included by these sides are supplementary. 8. In Fig. 403, A BCD is a trapezoid with its diagonals AC and DB. Prove AAOD = ABOC. Suggestion. First compare AABD and A ABC or a.^ AADCand. ABDC. Fig. 403 9. In Fig. 404, AE\\DB. Prove AABE = AAED; ABDE = ABDA; AABO=AEDO\ AADC = BCDE. . , Fig. 404 10. Show how to divide a triangle into four equivalent parts. Can this be done in more than one way? 11. If one base of a trapezoid is twice the other, the diagonal divides the trapezoid into two parts one of which is double the other. 12. A BCD is a parallelogram. E and F are the mid-points o( AB and AD respectively. Prove that CF, CA, and CE divide the parallelogram into 4 equivalent parts. 13. Show how to divide a parallelogram into six equivalent parts by lines drawn from the same vertex. 14. Show how to divide a parallelogram into three equivalent parts by lines drawn from the same vertex. 15. An isosceles right triangle is equivalent to K of the square constructed on its hypotenuse. 16. The figure formed by joining the , mid-points of the sides of a square is equivalent to J4 of the square. Suggestion. Construct the medians of the given square. 17. Is Ex. 16 true for any parallelogram? 226 PLANE GEOMETRY THE SQUARE ON THE HYPOTENUSE OF A RIGHT TRIANGLE 261. Theorem 118. The square constructed on the hjrpotenuse of a right triangle is equivalent to the sum of the squares constructed on the other two sides. E M D Fig. 405 Hypothesis: ABC is a rt. A with ZA = 1 rt. Z and the m I, II, and III constructed on the sides AB, AC^ and BC respectively. Conclusion: nUl=\JI+ nil. Analysis and construction: I. Dili can be proved equivalent to DI+DII by dividing Dili into two parts equivalent to DI and nil respectively. II. One way of dividing D III into the two parts required is by drawing a line from A ± BC and proving BEML = D I and CDML = D II. III. To prove CDML = D II, compare them with con- gruent triangles. ly. The triangles desired may be obtained by joining AD and BK. Then prove 1. AACDmBCK. 2. AACD = HCDML. 3. ABCK = }4 OIL AREA AND EQUIVALENCE 227 V. To prove AACD ^ ABCK, prove AC = CK, BC = CD, ZACD=ZKCB, Vr. To prove AACD = H CDML, show that CD is the common base and CL is equal to the altitude of each. VII. To prove ABCK = M □ II, show that CK is the com- mon base and AC is equal to the altitude of each. VIII. To prove AC equal to the altitude of ABCK, prove that HAB is a straight Hne. Let the pupil give the analysis to prove that BEAfL^DI, full details of the proof, and the conclusion. For other methods of proof see §255; §254, Ex. 23; §233, Ex. 23. Problem 22. To construct a square equivalent to the sum of two given squares. Fig. 406 Given D I and D II constructed on the segments a and b respectively. To construct a square equivalent to ni+ nil. The solution is left to the pupil. Problem 23. To construct a square equivalent to the difference between two given squares. The solution of this problem is left to the pupil. Construct a square equivalent to Ex. 1. The sum of three given squares. Ex. 2. The sum of two given parallelograms. Ex. 3. The diderence between two given triangles. Ex. 4. The sum of a given triangle and a given rectangle. Ex. 5. If a and b are two given segments, construct x so that ;c= Va'^-l-62; so that x= yla^-b'^. 228 PLANE GEOMETRY SUMMARY AND SUPPLEMENTARY EXERCISES 252. SUMMARY OF IMPORTANT POINTS IN CHAPTER X A. Formulae obtained. I. Area of rectangle = a6 (§239). II. Area of parallelogram = a& (§ 242) . III. Area of triangle = 3^ a6 (§243). IV. Area of trapezoid = M a (b+b') (§ 244). V. If a and b are the legs and c the hypotenuse of a right triangle, a'^-^b^ = c'' (§251). For area of triangle see §253, Ex. 41 and Ex. 44. Ex.41. S = }i be sin A. Ex. 44. S= ^s{s — a) (s—b) (s — c) where a, b, and c are sides of A. For areas of irregular polygons see §245. 263. EXERCISES INVOLVING NUMERICAL COMPUTATIONS Note. Be prepared to prove the theorems on which any of the following exercises depend. 1. A rectangle whose base is 81 ft. has the same area as a square whose side is 36 ft. Find the difference between the perimeters. 2. The altitudes of two triangles are equal and their bases are 25 ft. and 40 ft. respectively. What is the base of a triangle equivalent to their sum having an altitude 2}^ times as great? 3. The base and altitude of a triangle are 18 ft. and 24 ft. respectively. At a distance of 10 ft. from the base a line is drawn parallel to the base. Find the area of the two parts into which the triangle is divided. 4. Find the altitude of a triangle with base 21 in. which has the same area as a parallelogram whose base is 18 in. and whose altitude is 15 in. 5. The area of a rhombus is ^ the product of the diagonals. 6. Find the area of a rhombus if the sum of the diagonals is 12 ft. and their ratio is 3 : 5. Use Ex. 5. ' 7. A rectangular field is 30X80 ft. It is surrounded by a road ■of uniform width the entire area of which equals the area of the field. Find the width of the road. AREA AND EQUIVALENCE 229 8. The legs of a right triangle are 15 ft. and 20 ft. A perpen- dicular is drawn from the vertex of the right angle to the hypotenuse. Find the areas of the two triangles formed. 9. The area of a rectangle is 120 sq. ft.; one side is 8 ft. Find the diagonal. 10. The perimeter of a rhombus is 100 ft.; the shorter diagonal is 14 ft. Find the area. Use Ex. 5. 11. The area of a rhombus is 2184 sq. ft.; the shorter diagonal is 26 ft. Find the longer diagonal and one side. Use Ex. 5. 12. A house is 45 ft. long, 32 ft. wide, 24 ft. to the roof, and 30 ft. to the ridgepole. Find the number of sq. ft. in the entire exterior surface. 13. Find the area of a right triangle if its perimeter is 84 ft. and its sides are in the ratio 3:4:5. 14. The height of a lean-to roof is 8 ft.; the span is 15 ft. Find the area of the roof if the length is ^ 21 ft. (see Fig. 407). Fig. 407 15. The parallel sides of an isosceles trapezoid are 30 ft. and 40 ft. respectively; the non-parallel sides are each 13 ft. Find the altitude and the area. JO 16. The area of a kite is K the product of the diagonals. 17. In Fig. 408, /LA = l rt. Z. ^5 = ^Z)and ^C is the perpendicular bisector of DB. If ^5 = 4 in. and AC = ^}i in., find the area of A BCD and the length of DC. 18. The design shown in Fig. 409 is symmet- ric with regard to both diagonals of the square. If AY = )4 AB and OX = }iOA, find the area of the. kite AYXZ and hence of the Maltese cross shown, if ^5 = 6 in. 19. Upon the diagonal of a rectangle 11 ft. wide and 60 ft. long a triangle is constructed whose area equals the area of the rectangle. Find its altitude. A Y Fig. 409 16 230 PLANE GEOMETRY D IT T C V m^ s \V ^W^ ^w z A y 3 20. One side of a right triangle is 8 in. and the hypotenuse is 17 in. Find the area of an equilateral triangle constructed on the, third side. 21. The diagonals of two squares are 10 ft. and 15 ft. respec- tively. Find the diagonal of a square equivalent to their sum. 22. Fig. 410 represents a square with each side ^'r-"""-'^-^^ divided into 8 equal parts. What fraction of the given square is each of the irregular figures shown? If yl 5 is 6 in., find the length of the sides of each figure. ^ ^ Fig. 410 Note: Modern standard tiles are made in certain definite sizes and shapes, all of which are* derived directly from the six-inch square. Fig. 410 shows three such tiles. 23. Fig. 411 is a square with the lines drawn symmetric with respect to both medians. If AB=\2 in. and AX = ?> in., find the area of the shaded portion. 24. Find the area of an equilateral triangle if each side is 7 in.; 9 in.; 12 in.; if each side is s. 25. If the area of an equilateral triangle is 60, find one side. 26. Find the area of a regular hexagon if each side is 4 in. 27. The sides of a regular hexagon are extended until they intersect. Find the area of the star formed if each side of the hexagon is 3 cm. 28. Upon the altitude of an equilateral triangle as a side a second equilateral triangle is constructed. Find the area of the second triangle if a side of the first is 6 in. 29. Find the area of an isosceles right triangle if the hypotenuse is 6 in.; 4 in.; 8 in.; 10 in.; if the hypotenuse is a. 30. In Fig. 412, ABC is an isosceles right triangle. ED is parallel to BC. F is the mid-point of ED. FG and FH are parallel to AB and AC respectively. If C5 = 9 in. and ^Z) = 4>^ in., find the areas of the various parts into which the figure is divided. Fig. 411 From a parquet floor design. AREA AND EQUIVALENCE 231 Fig. 414 31. Fig. 413 represents a square with each side divided into 4 equal parts and the points joined as indicated. If AB = 12 in.» find the area of' the star. i>( g- T lO 32. Draw a figure similar to Fig. 413, dividing L-fe^<^-J each side into 3 equal parts. Find the area of the star formed if each side of the square is 6 in. p-^-^^^3^^. 33. A square and an equilateral triangle have each -*^ ^ " — ^ a perimeter of 60 ft. Find the ratio of their areas. ^^^' ^^^ 34. Fig. 414 shows a square with each side divided into three equal parts and the points joined as indi- cated. If each side of the original square is 12 in., find the area of the eight-pointed star and of the irregular octagon in the center. Note. The designs shown in Figs. 413 and 414 are extensively used in industrial ornament. Fig. 415 is from a parquet floor design. See Fig. 181a. -Dciw ^ ^(? 35. Fig. 415 shows a square with AX and CY each one-fourth the diagonal AC. If the points are joined as indicated, prove that XBYD is a rhombus and find its area. AB = Q in. Fig. 415 36. If the parallel sides of a trapezoid are 8 in. and 20 in. and its altitude is 6 in., find the area of the two triangles formed by extending the non-parallel sides until they intersect. 37. In Fig. 410, XF = 50, ZF = 20, ZW = ob, y4X = 63, YD =120, BZ = SO, aF = 00. Find the area of A BCD. AX, DY, BZ, and CW are each perpendicular to XW. x rz — ht Fig. 416 38. A BCD is a quadrilateral with AC and D rt. A and ZA = }4 Tt. Z. If BC = 8 in. and ^Z>=15 in., find the area of the quadrilateral. 39. The area of one rectangle is 3 times the area of a second. If the base of the first is twice the base of the second, find the relation between their altitudes. 40. The area of a certain rectangle is 3K times the area of a triangle. If the base of the triangle is twice the base o^ the rectangle, find the relation between their altitudes. 232 PLANE GEOMETRY 41. lie and b represent two sides of a triangle and A the included angle, prove that the altitude upon side ft is c sin ^ , that the altitude upon side cis b sin A, and that the area is >2 be sin A. Find the area of AABC, using the trigonometric tables, if 42. .4j5 = 8.3cm., ^C = 3.9cm., ZA=37°. 43. ^5 = 9. 6cm., ^C = 5. 4cm., ZA=54°, t44. If a, b, and c are th e sides of a triangle, the area of the triangle is -^sis — a) {s — b) (s — c), ^^ where s = }4{a+b-\-c). Fig. 417 A nalysis: I. To find the area of a triangle in terms of the three sides, find the area in terms of the base and altitude and express the altitude in terms of the three sides by means of the Pythago- rean theorem. II. Ave3. = y2ch (Fig. 417). III. h^ = b^-x\ IV. h^ = a^-{c-xy. Solve the two equations for h and x and substitute the value of h in the area formula. Outline of algebraic proof: i2-x2 = a2_(c-x)2 b^-x^ = a'^-c^+2cx-x'^ \ 2c (2bc-b^-c^+a^) (2bc+b^+c^-a^) 4c2 ,,, (a-b+c) (a+b-c) (b+c+a) (b+c-a) " ~ 4c^ This may be put in a more concise form by noting that a+b—c = a-\-b+c—2c a-b+c = a+b+c-2b b-\-c—a = a-\-b-{-c—2a We may let a-\-b-^c=2s for convenience. Then a-\-b-c=2s-2c = 2{s-c) a-b+c = 2s-2b = 2{s-b) b-{-c-a = 2s-2a = 2{s-a) AREA AND EQUIVALENCE 233 Substituting in the value for /t', ^, 2(s-b) . 2{s-c) . 25 . 2{s-a) ^ 4c^ , 2 V5(5-a) (s-b) (s-c) n = /. area of A = 5 ch= "^ s{s—a) (s — b) {s—c), where s = ^{a-^b-{-c). Note. What changes must be made in the details above if h falls outside the triangle? Note. This formula was given by a noted Greek engineer and surveyor, Hero of Alexandria (about 100 B.C.). 45. Find the areas of triangles whose sides are : a. 13, 14, 15 c. 15, 18, 21 b. 8, 10, 12 d. 24, 33, 41 46. Compute from the following data the area of the field shown in Fig. 418, using the formula given in Ex. 44. ^ A B = 210 ft. DE = 305 it. BC = 210 it. EA =225 it. CD = 210 it. BD = SMit. AD = ^25 it. * EXERCISES INVOLVING EQUIVALENT POLYGONS 254. Note. Be prepared to prove the theorems on which any of the following exercises depend. 1. Make a review diagram for Th. 117. 2. The area of a triangle is one-half the product of the perimeter and the radius of the inscribed circle. 3. The area of any polygon circumscribed about a circle is one- half the product of the perimeter and the radius of the inscribed circle. 4. Given the isosceles right A ABC with OF and GE drawn from the mid-point oi AC parallel to AB and BC respectively, prove area EBFG=}/2 area ABC (see Fig. 419). " F J. 419 5. The line joining the mid-points of the parallel sides of a trapezoid divides it into two parts that have equal areas. 234 PLANE GEOMETRY 6. Given A^ BC with A" an arbitrary point on the median CO. Prove AACX equivalent to ABCX. 7. The three medians of a triangle divide it into 6 equivalent triangles. 8. Given AADC and DEE with AD = DB, DE = EC, ZADC+ZBDE = 2 rt.A, prove area DBE = y> area ADC (Fig. 420). 9. Given A ABC with CD, an arbitrary segment from C to AB, divided into three equal parts at points X and Y, compare area AYB, area AXB, and area ACB (Fig. 421). 10. Construct a square whose area shall be H of the area of a given square. 11. In. Fig. 422, ABC is an isosceles right tri- angle. AE = EB; BF = FC; £G and Ffi^ are per- pendicular to AC. Compare areas AEG, EBF, EFHG, and ABC. Fig. 421 * 12. In Fig. 423, ABC is an isosceles A. AC and CB are each divided into 3 equal parts. CD, FM, and GN are A_ AB. Compare the areas of ^ fJg.^423 ^ the various figures formed. King-rod and Queen-rod truss design 13. Given the quadrilateral A BCD with the ^ diagonal BD bisected at X, prove that AX CD is equivalent to AXCB (see Fig. 424). 14. Transform quadrilateral A BCD into quad- rilateral ABCE so that (1) Z ECB shall be equal to a given angle, (2) side AE shall be equal to a given segment. Is the problem always possible? Fig. 424 15. A quadrilateral is bisected by one of its diago- nals if the second diagonal is bisected by the first. 16. Given AABC with EF\\AB. equivalent to ABCE (Fig. 425). Prove AACF Fig. 425 AREA AND EQUIVALENCE 235 17. In Fig. 426, A BCD is a square with its diagonals and medians. The semi-medians, OE, OF, OG, and OH, are bisected at X, F, Z, and W, respectively, the square. Prove that the star formed is 18. Construct in a given square a star that shall be ^i of the given square. 19. If through any point on the diagonal of a parallelogram segments are drawn par- allel to the sides, the parallelograms on oppo- site sides of the diagonal are equivalent (see Fig. 427). 20. In Fig. 428, ABC is any triangle, [sj AC ED and BFEC have the common side CE and have AC and BC as bases. Prove that if d Z?F is joined, ABFD is a parallelogram equivalent to the sum of [sJ AC ED and BFEC. p^^ ^28 Suggestion. Extend EC and show that CO ABFD can be broken up into parts equivalent to EJACED and CO BFEC. 21. In Fig. 429, ABC is any triangle. ACFG ^ and BCED are any parallelograms on the sides AC and BC respectively. GF and DE are extended to meet at M. On AB a parallelogram is constructed with one side AH equal and paral- lel to MC. Prove that CJABKH is equivalent to CJ ACFG-]- CJ BCED. Note. Ex. 21 is known as Pappus' theorem (about 300 a.d.). 22. Show hoUr to construct a parallelogram equivalent to the sum of two given parallelograms. 23. Deduce the Pythagorean theorem from Pappus' theorem. D_ c 24. In Fig. 430, A BCD is a square with its diago- nals and medians. HE, EF, FG, and GH are each divided into 3 equal parts and the points are joined as indicated. Compare the area of the shaded portion with the area of the square A BCD. a e^ Fig. 430 : JSL. : 236 PLANE GEOMETRY 25. In Fig. 431, ABCD is a square with its diagonals and medians. OX, OY, OZ, and OW are each H of the corresponding semi-diagonal, ^'f^^f^^ Compare the area of the star with the area of the square ABCD. a ji b Fig. 431 Note. Figs. 426, 430, and 431 are from parquet floor designs. 26. If the mid-points of two adjacent sides of a parallelogram are joined, the triangle formed is equivalent to M the parallelo- gram. 27. If the mid-points of two sides of a triangle are joined to any point of the base, the quadrilateral formed is equivalent to 3^ the triangle. 28. The triangle formed by joining the mid-point of one of the non-parallel sides of a trapezoid with the extremities of the opposite side is equivalent to >2 the trapezoid. 29. Given the EJ ABCD and 0, any point within the parallelo- gram, prove that /\AOB -\- ADOC is equivalent to K EJABCD. 30. If in Ex. 29 point were without the OJ ABCD, what would be the relation between AAOB and DOC and EJ ABCD} 31. In Fig. 432, AABC is isosceles. CD is perpendicular to AB. AB, AC, and CB are each divided into 3 equal parts and the points are joined as indicated. Compare the areas of the triangles formed. t32. Construct a rectangle so that its area shall be equal to the area of a given square ^" and the sum of its base and altitude equal to ; I a^ I a given segment (see Fig. 433). I ' A nalysis: Let X represent the base and y the altitude of the rectangle. To construct a rectangle so that its area is equal to the area of the X a square, construct x and y so that xy = a^, or so that - = - • /, x-\-y must be divided into two segments so that a is a mean pro- portional between x and y. Directions and proof are left to the pupil. F Fig. 432 From a roof truss design Fig. 433 AREA AND EQUIVALENCE 237 t33. Construct a rectangle so that its area shall be equal to the area of a given square and the difference between its base and altitude equal to a given segment (see Fig. 434). Suggestion. Let a be one side of the square and x and y the base and altitude of the rectangle. An analysis similar to that for Ex. 32 will show that a will be the mean proportional between two segments x and y whose difference is -45 (see Fig. 434). 134. If two triangles have an angle of one equal to an angle of the other, the ratio of the areas equals the ratio of the products of the sides that include the equal angles (see Fig. 435). Fig. 434 Fig. 435 area ABC be . „,^, = Tj-y ' compare the area of each triangle wi area A B'C be ^ th A nalysis: I. To prove the area of a third triangle having the same altitude. TT . J /-D/ J c J 4.U ^- .area ABC , . area AB'C II. . . draw LB and find the ratio of . „,>; and of t-^ttt^, area A B'C area A B'C and multiply. Let the pupil give the proof. 35. If s and s' represent the areas of the A ABC and A B'C shown in Fig. 435, find 1. c',iib =24, b' = 15,c=lS, and s = s'. 2. c', if b =56, 6' = 35, c = 70, and s = 2s'. 3. b, if 6'= 18, c' = 28, c = 45, and s=r2s'. 36. Given a triangular field A BC, show how to divide ^/jy it into two equivalent parts by running a fence from point X on one side (see Fig. 436). Fig. 436 Suggestion. Let Y be the point at which the fence meets BC. Represent ^ C by b, XC by b\ CF by a', and 5C by a. Point Y and .'. a' are to be found, struction in full. Find a' so that ab_ a'b' 1 Why? Give con- 238 PLANE GEOMETRY EXERCISES INVOLVING THE PYTHAGOREAN THEOREM 256. 1. In the details of the proof of Th. lis as given above, prove that ACBF and ABE have equal bases and equal altitudes instead of proving them congruent (Fig. 437). Fig. 437 A nalysis: I. To prove that ACBF and AABE have equal bases and equal altitudes, use BF and AB as bases and prove the altitudes equal. drop perpendiculars from E and C to ^B and BF extended respectively and prove ABEP^ ABRC. II. 2. Give the details for the proof for Th. 118 (1) if square CE is constructed on the opposite side of CB from that shown in Fig. 437; (2) if square AF is constructed on the opposite side oi AB from that shown in the figure; (3) if both CE and AF are on opposite sides of CB and AB respectively from thovse shown in the figure. 3. Show how Fig. 438 may be used to prove the Pythagorean theorem. A EH is the given right triangle. Extend AE and AH, making AB = AD = a-\-b, Complete the square on a+b. Make BF = CG = DH = a. Join E, F, G, and H. From E and F draw lines parallel to AD and AB respec- tively. Prove that EFGH is a square equivalent to a^-\-b^ 438 4. Show how Fig. 439 may be used to prove the Pythagorean theorem. ABE is the given right x*,^ triangle. Construct the square on AB. Make AK = BE. JoinDK. Construct A DCF and A C5G congruent to AABE. Prove KHFD and BGHE squares whose sum is equivalent to c^. Fig. 439 5. Show how Fig. 439 may be constructed by constructing squares BGHE and KHFD first. 6. If a and b are two given segments, show by geometry that '4a^ + b'^ is not a+b. AREA AND EQUIVALENCE 239 7. If a, b, and c are three given segments, construct a fourth segment x so that (I) x= ^|a^^¥Tc^, (2) x= yja^ + f^-c^. 8. Construct a square whose area shall be H of the difference between the areas of two given squares. 9. Construct on a given base a right triangle whose area shall be equal to the sum of the areas of two given triangles. 10. AB is the diameter of a semicircle. D is a point on ^5 so located that AD=/i AB. DE is ± AB a.t D and cuts the semi- circle at E. Prove that the square on ^£ is H the square on AB. 1 1 1 . If ^ BC is any triangle with Z A an S. acute angle, and sides a, 6, and c are the' \.y^ 'V sides opposite AA, B^ and C respectively, y^ l'' \ a'^ = ¥-\-c'^ — 2cp where p is the projection of ^^^ — ^.L—. — I A^ 6 upon c (Fig. 440). Fig. 440 Outline of algebraic proof: Let h be the altitude on c. a^ = h^-{-(c-p)^ ^ Note. The distance between the feet of the perpendiculars drawn to a given line from the ends of a segment is called the projection of the segment upon the line. 12. Give the proof for Ex. 11 for a figure in which h falls on AB extended. ^^, tl3. If ABC is any triangle with Z A any >''^^/i^ ''^ obtuse angle, and sides a, b, and c are opposite ^^ / I A A, B, and C respectively, a^ = b'^-\-c^-{-2cp b c :/—-p—J where p is the projection of b upon c (Fig. 441). ^^^^- ^^^ Outline of algebraic proof: Let h be the altitude upon c. a2 = A2 + (c+p)2 h2 = b^-p2 ... a^ = i,i^p2^(^c-{-pr- 14. State and prove the converse of the proposition that the square on the hypotenuse of a right triangle equals the sum of the squares of the other two sides. 15. If the sides of a triangle are as given below, is the largest angle right, acute, or obtuse: (1) 20,39,36? (2) 15,30,39? (3) 15, 36, 39 ? CHAPTER XI Similarity INTRODUCTORY DEFINITION 256. Similar polygons have been defined as polygons that have 1. The angles of one equal to the corresponding angles of the other, and 2. Corresponding sides proportional. By the ratio of similitude of two similar polygons is meant the ratio of any two corresponding sides. For convenience corresponding sides will be lettered alike, as AB and A'B\ BC and B'C. TESTS FOR SIMILAR POLYGONS TEST I FOR SIMILAR TRIANGLES 257. The first test for similar triangles is given in Th. 102. Two mutually equiangular triangles are similar. Let the pupil review the proof for Th. 102. Ex. 1. Two triangles are similar if two angles of one are equal respectively to two angles of the other. Ex. 2. Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. Ex. 3. Two isosceles triangles are similar if an angle of one is equal to a corresponding angle of the other. Ex. 4. Construct two similar triangles with the ratio of similitude %. Ex. 5. Two triangles similar to a third are similar to each other. 240 SIMILARITY 241 TEST n FOR SIMILAR TRIANGLES 258. Ex. 1. Given ABC any triangle, and A'B' of AA'B'C, construct AA'B'C so that Z^ = Z^' and AB AC A'B' A'C Theorem 119. Two triangles are similar if an angle of one is equal to an angle of the other and the ratios of the including sides are equal. Hypothesis: A'B' A'C' Conclusion: Fig. 442 In AABC and AA'B'C, ZA= ZA' and AABCooAA'B'C. Analysis and construction: L To prove AABC co AA'B'C , prove AB = ZB' and /:c=zc'. II. To prove ZB= ZB', place AABC upon AA'B'C so that the sides of Z A are collinear with the sides of ZA' and prove BC\\B'C'. III. To prove BC\\ B'C, prove ^, = ^,. The proof is left to the pupil. Ex. 2. Construct AA'B'C similar to AABC if the sum of sides A'B' and B'C is given. Ex. 3. If a segment is drawn parallel to the base of a triangle and terminated by the sides, the triangle formed is similar to the given triangle. Give two proofs. Ex. 4. Solve Ex. 3 if the parallel to the base cuts the sides of the triangle extended. 242 PLANE GEOMETRY TEST III FOR SIMILAR TRIANGLES 259. Ex. 1. Given A ABC any triangle. Construct AA'B'C A 7? J^r* A (^ SO that jr^'^'^X'^AX^' ^^^^^ '^'^' °^ A^'j5'C'. Theorem 120. Two triangles are similar if the corre- sponding sides have equal ratios. Fig. 443 Hypothesis: In AABC and A A 'B'C\ - = -=--> cab Conclusion: AABC ~ A A 'B'C Analysis and construction: I. To prove AABC<^ AA'B'C , compare each with a third triangle. II. On CA construct CX = b\ on CB construct CY = a\ and prove 7i ~ Tg and Tz ^ Tg where T^ = AABC, T2 = AA'B'C\ and T^ ^AXYC. III. To prove T2 ^Tg, prove c' = z. (Use Th. 4.) IV. To prove c' = z, compare two proportions containing z and c'\ t- = -» from A Ti and 7„, and 7- = - given. be ^ be b' z V. To prove ^ =-> prove T^ 00 Tg. (Use Th. 119.) Let the pupil give the proof. For IV use Th. 92. Ex. 2. Construct a triangle similar to a given triangle with its perimeter equal to a given segment. Ex. 3. Prove that two triangles are similar if two sides and the median to one of them are proportional to the corresponding parts of the other. SIMILARITY 243 CONSTRUCTION OF SIMILAR TRIANGLES 260. Problem 24. Upon a given segment corresponding to a side of a given triangle, to construct a triangle similar to the given triangle. GENERAL TEST FOR SIMILAR POLYGONS 261. Theorem 121. Two polygons are similar if diago- nals drawn from two corresponding vertices divide the polygons into the same number of triangles similar each to each and similarly placed. E^A—P Fig. 444 Hypothesis: Polygons P and P' are divided into triangles by the diagonals drawn from the corresponding vertices A and A' so that AI~ AI', AII~ AlF, etc. Conclusion: Polygon P co polygon P'. Analysis: I I. ZB= IB\ /.C= AC\ etc. To prove Pc^^P/, prove < ^^ <^ _^_^ ( 'a'~~h'~7' To prove r> = -/. prove each ratio equal to 777^,- be yi C Let the pupil complete the analysis and give the proof. CONSTRUCTION OF SIMILAR POLYGONS 262. Problem 25. Upon a given segment corresponding to a given side of a given polygon, to construct a polygon similar to the given polygon. Can this problem be solved in more than one way.? 244 PLANE GEOMETRY PROPERTIES OF SIMILAR POLYGONS COMPOSITION OF SIMILAR POLYGONS 263. Theorem 122. If two polygons are similar, diago- nals drawn from two corresponding vertices divide the polygon into the same number of triangles similar each to each and similarly placed. The analysis and the proof are left to the pupil. RATIOS OF CORRESPONDING SEGMENTS 264. Theorem 123. The ratio of corresponding altitudes of two similar triangles equals the ratio of the bases. Fig. 445 Hypothesis: In AABC^ AA'B'C, h and h' are corre- sponding altitudes and c and c' are the bases. Conclusion : ri^-r h c Analysis: To prove ti = ~ third ratio — ; • h c compare -t-, and — with the Let the pupil complete the analysis and give the proof. Ex. 1. Prove that in two similar triangles the ratio of the following segments equals the ratio of similitude of the triangles: a. Bisectors of corresponding angles. h. Corresponding medians. c. The radii of the circumscribed circles. d. The radii of the inscribed circles. Ex. 2. Are there any other segments in two similar triangles whose ratio is equal to the ratio of similitude of the triangles? Give proofs. SIMILARITY 245 Ex. 3. In two similar polygons the ratio of corresponding diagonals is equal to the ratio of similitude of the polygons. Ex. 4. If b and b' are the bases of two triangles and a and a' the corresponding altitudes, show that -, = rr if the triangles are similar and — - = -— if they are equivalent. a 265. Theorem 124. In a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. TT , . a c e Hypothesis: T" = "7 = "7 = etc. Conclusion: = — . • • 6-f-fi+/+ctc. b Proof: STATEMENTS a c a __ e 2. ab = ab be =ad be = af 3. ab+bc-{-be-\-etc. =ab-{-ad-{-af -^etc. 4. b{a-{-c+e+etc.) =a{b+d-{-f +etc.). a + c + e 4-etc. _ _a ^ ^' b +^H-/+etc.~ b ' Let the pupil give all reasons. h AD ^ h DB Ex. 1. Prove Th. 123 by provmg that Jji^'^r^t and ^/ = ^/^/' Theorem 125. The ratio of the perimeters of two similar triangles is equal to the ratio of similitude. Let the pupil give the analysis and the proof. Use Th. 124. The series of equal ratios is obtained from the ratios of corresponding sides. Ex. 2. Corresponding diagonals of two similar polygons have the same ratio as the perimeters. 17 240 PLANE GEOMETRY RATIOS OF AREAS 266. Theorem 126. The areas of two similar triangles have the same ratio as the squares of the bases or the squares of the altitudes. Fig. 446 Hypothesis: In AABC^ l\A'B'C , h and h' are corre- sponding bases and a and a' the altitudes. Area ABC ^^_a^ 52 ^2 Analysis: To prove _^^ \'i'Dtnt ^TJ'i.^~''i" prove each Conclusion. ArQ2i A' B'C area ABC " ^^"^" area A '^'C h'^ a'' ratio equal to a third ratio. Proof: STATEMENTS I. 1. Kve2iABC=y2ah. 2. AreaA'^'C' = 3^a'6'. area ABC ah X 0. ••areaA'^'C' a'h'' II. 1. a b a' b'' 2. a a b a a b b b '''' a'' b'~b' ' b' thatis,^, = -,^, = p-.. III. .-. area ABC 6^ a^ areaA'Br'"6'2 a"^ Let the pupil give all reasons. SIMILARITY 247 Ex. 1 . In two similar triangles let s and 5' represent the areas, b and b' the bases, and a and a' the altitudes. Find a. s', if 5 =45 and a= ^ a'. b. a\ if a =16 and s = Hs'' c. b', if ^>= 10 and 5' = 35. Ex. 2. The sum of the areas of two similar triangles is 78 sq. ft. Two corresponding sides are 10 ft. and 15 ft. Find the areas and the altitudes. 267. Theorem 127. The areas of two similar polygons have the same ratio as the squares of two corresponding sides. Fig. 447 Hypothesis: In polygons P and P', a and o' are corre- sponding sides. Area P _a^ Conclusion : Area P' Analysis and construction: area P a I. To prove — '■ — tt, = -?.>, break the polygon up into tri- ^ area P a^ angles that are similar in pairs and find the ratio of the sums of the areas of the triangles; that is, draw diagonals from corresponding vertices and prove A-fB+C+ete. a2 A'+B'+C'+etc. a'2 11. To prove ^,_^^,^^,^^^^ = a2 A B = ^„ prove ^,= ^, C -etc., that is, prove ., a^ B a2 a'2' B' a'2' '^^" and then use Th. 124. Let the pupil give the pi oof. 248 PLANE GEOMETRY Ex. 1. If the area of one polygon is twice that of a similar polygon, what is the ratio of the bases? Ex. 2. The area of one polygon is three times the area of a similar polygon. If the base of one is 15, find the base of the other. Ex. 3. Two corresponding sides of two similar polygons are 5 and 9 respectively. If the area of the first is 30, find the area of the second. Ex. 4. The sum of the areas of two similar polygons is 910. The ratio of similitude is %. Find the area. SUMMARY AND SUPPLEMENTARY EXERCISES 268. SUMMARY OF IMPORTANT POINTS IN CHAPTER XI A. Tests for similarity. I. To prove two triangles similar, prove that a. They are mutually equiangular (§257). h. Their corresponding sides have equal ratios (§259). c. An angle of one equals an angle of the other, and the ratios of, etc. (§258).' II. To prove two polygons similar, prove that diagonals drawn from corresponding vertices, etc. (§261). B. Properties of Similar polygons. I. Angles are equal and corresponding sides have equal ratios (§256). IL Two similar polygons may be divided into triangles, similar in pairs by, etc. (§263). III. Ratios of corresponding segments in similar poly- gons equal the ratio of similitude of the polygons (§§264,265). IV. Ratio of the areas of similar polygons equals the square of the ratio of similitude of the polygons (§§266, 267). SIMILARITY 249 EXERCISES INVOLVING SIMILAR POLYGONS 269. 1. Make review diagrams for Ths. 122 and 127. If the segments which join the vertices of a polygon with a given point are divided in the same ratio from the given point and the points of division joined in the same order as the vertices of the polygon, the polygon so formed and the given polygon are radially placed. The point may be without the polygon as in Fig. 448a, or within the polygon as in Fig. 4486. In each case 7yTi = 7JW}^^^^- "^^^ radial point is called the center of similitude of the polygons. D o<- FiG. 4486 2. Two polygons that are radially placed are similar. Give the proof for the two cases. 3. Show how a polygon may be constructed similar to a given polygon A BCD with a given segment A'B' corresponding to side AB and with any arbitrary point O as a center of similitude. Note. The center of similitude of two similar figures may be of use in enlarging or reducing drawings. 4. Make a drawing and give a proof for Ex. 3 with point O falling (1) on point A; (2) on AD. 5. If the perimeter of an equilateral triangle is 66, find the sides of a similar triangle with half the altitude. 6. On one side and on the altitude of an equilateral triangle as bases construct similar triangles. What is the ratio of their areas? 7. On one side and on the diagonal of a square construct equilateral triangles. What is the ratio of their areas? 8. Construct a triangle similar to a given triangle whose area, shall be ^4 the area of a given triangle; ^2 the area; 14 the area; }i the area. 250 PLANE GEOMETRY t9. If the hypotenuse of a right triangle is twice the shorter side, one of the acute angles of the triangle is 60°. 10. The triangle A BC has a right angle at C, and D trisects AC so that AD = 2DC. It is then found that AD = BD. Find the size of angle A. — College Entrance Examination Board, Plane Geometry Examination, 1907. fll. If two triangles have their corresponding sides parallel each to each, the triangles are similar. tl2. If two triangles have their corresponding sides perpen- dicular each to each, the triangles are similar. 13. Fig. 449 shows two quadrilaterals with their sides respectively parallel. If AC \\ A'C, prove the quadrilaterals similar. 14. Construct a quadrilateral similar to a given ^^ quadrilateral having one diagonal equal to a given segment. Fig. 449 15. The perimeters of two similar triangles are 21 and c The altitude of the first is 8. Find the altitude of the second. 16. In Fig. 450, ABC is an isosceles triangle. CD J_ from C to the base AB. PQ is the per- pendicular bisector of AC. Prove that A A DC is similar to AAPQ and find the length of ^Q in terms of ^Z> and CD. "* 17. In Fig. 451, ABCD is a square. AE = BF= CG = DH and the points are joined as indicated. Find all pairs of congruent triangles and of similar ''^ triangles formed. Read the ratios between corre- sponding sides of the similar triangles. If AB = S6 cm. and AE=y3AB, find AX, XY, and YF. 18. The equal sides of an isosceles triangle are 13, the base is 10. Find the sides of a similar triangle whose altitude is 4. 19. What is the area in acres of a portion of a map that covers 4^2 sq. in.? The scale of the map is 1 inch = l mile. 20. The sides of a polygon are 10, 15, 9, and 22. Find the sides of a similar polygon if its perimeter is 140. SIMILARITY 251 t21. If similar polygons are constructed on the sides of a right triangle as corresponding sides, the area of the polygon constructed on the hypotenuse is equal to the sum of the areas of the polygons on the other two sides (see Fig. 452). Analysis: Fig. 452 To prove that C=i4-fB, find^ and -^ in terms of the sides of the triangle, and add. t22. To construct a polygon equivalent to one of two given polygons and similar to the other. Fig. 453 Analysis: I. Suppose it is required to construct a polygon C equivalent to B and similar to i4. A a* II. Since Cis to be ^^1 — =— . A A III. Since C is to be equivalent to 5, 7; = ^ . IV. As A and B are not similar, we cannot compare their areas by comparing squares of corresponding sides. .'. reduce A and B to equivalent squares. Let A be equivalent to a square whose side is m. Let B be equivalent to a square whose side is n, __ _,, A m» V. Then^=-. VL .•.^=^. a n VII. .*. to find a', find a fourth proportional to m, n, and a. Let the pupil give directions in full, make the drawing, and give the proof. CHAPTER XII Regular Polygons DEFINITION 270. A regular polygon has been defined as a polygon with all of its sides and all of its angles equal. Ex. 1. A regular quadrilateral is a square. Ex. 2. An equilateral triangle is a regular polygon. CONSTRUCTION OF REGULAR POLYGONS GENERAL THEOREMS 271. Theorem 128. If a circle is divided into n equal arcs, the chords joining the points of division form a regular polygon. The analysis and proof are left to the pupil. 272. Theorem 129. If a circle is divided into n equal arcs, the tangents drawn to the points of division form a regular polygon. Fig. 434 Hypothesis: OO is divided into the equal arcs WX, XY, YZ, etc., and tangents AB, BC, CD, etc., are drawn to the points of division, X, Y, Z, etc. Conclusion: A BCD etc. is a regular polygon. 252 REGULAR POLYGONS 253 Analysis and construction: I. To prove that ABCD etc. is a regular polygon, prove AA= Z.B= ZC; = ctc., and >lB = 5C: = CD = etc. II. To prove that Z A = ZB= ZC = etc., and that AB = BC = CD = etc., join WX, XY, YZ, Z\\ etc.. and prove AWXA m AXYB ^ AYZC m, etc. Suggestion. Notice that AX = BY= CZ = etc By congruent triangles. BX = CY = ZD =etc By congruent triangles. AB = BC= CD = etc Equal segments added to equal segments. The proof is left to the pupil. CONSTRUCTION OF THE INSCRIBED SQUARE AND RELATED POLYGONS 273.. Problem 2G. To inscribe a square in a circle. No. 1 No. 2 No. 3 Fig. 455 Analysis and construction (Fig. 455, No. 1) : I. To inscribe a square in a circle, divide the circle into 4 eqiial arcs. II. .'.divide the perigon about the center of the circle into 4 equal angles. III. .'.construct Let the pupil complete the directions and give the proof. 274. Problem 27. To inscribe a regular octagon in a circle (Fig. 455, No. 2). Ex. 1. What other regular inscribed polygons can be obtained from the inscribed square? Ex. 2. Show how to construct regular circumscribed polygons of 4, 8, and 16 sides (Fig. 455, No. 3). 254 PLANE GEOMETRY CONSTRUCTION OF THE REGULAR INSCRIBED HEXAGON AND RELATED POLYGONS 276. Problem 28. To inscribe a regular hexagon in a circle. No.S Fig. 456 Analysis and construction (Fig. 456, No. 1) : I. To inscribe a regular hexagon in a circle, divide the circle into 6 equal arcs. II. .'.divide the perigon about the center of the circle into 6 equal angles. III. Since H of 360 is 60, construct 6 angles of 60° each at the center of the circle. IV. .*. construct ..... Let the pupil complete the directions and give the proof. Ex. 1. Prove that the side of the regular inscribed hexagon is equal to the radius of the circle. Ex. 2. Inscribe an equilateral triangle in a circle. Ex. 3. Inscribe a regular 12-side in a circle. Ex. 4. What other regular inscribed polygons may be obtained from the regular inscribed hexagon? Ex. 5. Show how to construct regular circumscribed polygons of 3, 6, and 12 sides. Note. Regular polygons are in very common use for towers, spires, bay and dormer windows, hoppers, nuts, and the like. They are extensively used in ornament. Even the less common forms, such as polygons with 7, 9, 11, or even 13 sides, occasionally occur. REGULAR POLYGONS 255 CONSTRUCTION OF THE REGULAR INSCRIBED DECAGON AND RELATED POLYGONS 276. The construction of the regular inscribed decagon depends directly upon the following problem. Problem 29. To divide a given segment so that the larger portion is a mean proportional between the whole segment and the smaller portion. Fig. 457 Given the segment a. Ci X To find a segment x so that — = X a — x Analysis and construction: I. To find X so that — = a —X solve for x. 1. a'^—ax = x'^ 2. x'^-^ax = a'^. 3. x^+ax+{\aY = a^-^(Aa)\ 4. {x+\ay = a^+{\a)\ Equation 4 suggests the Pythagorean theorem. II. .*. construct a right triangle with a and \a for perpen- dicular sides. The hypotenuse will be x-\-\a. III. .'. X will be the difference between the hypotenuse and \a. Outline of proof: 1. (^ + ia)2 = a2 + Ga)^ 2. x'^ + ax + {\ay = a'' -Vilay. 3. x^-^-ax 4. x'' i2_ ax = a{a — x) 5. ^ = ^^- x a —X Let the pupil give the reasons. 256 PLANE GEOMETRY If a segment is so divided that the larger portion is a mean proportional between the whole segment and the smaller portion, the segment is said to be divided into extreme and mean ratio. Problem 30 may be stated : To divide a segment into extreme and mean ratio. Note. The division of a segment into extreme and mean ratio is often called the Golden Section. It was known to the Pythagoreans. It is an interesting fact, often used in the theory of design, that that division of a rectangle which is the most pleasing to the majority of people is the one that most closely approximates the golden section. It might be used in designs for doors or windows (Fig. 458) . ^^°- ^^^ Exercise. Show that if — = —3- the ratio — is approximately 8:13. 277. Problem 30. To inscribe a regular decagon in a circle. Fig. 459 Given circle 0. To construct a regular inscribed decagon. Analysis: I. yo construct AB, a side of the decagon, construct Z0= Ho of 4 rt. ^ or Vs of 2 rt. A. :. construct A ABO isosceles so that ZA=2Z0. II. The problem is reduced to the construction of AAOB, The fact that ZA = 2Z0 suggests drawing the bisector of Z A. A ABC and ABO would then be similar and .*. -r^ = -^^ or AB =0B ' BC. REGULAR POLYGONS 257 Construction: 1. Divide the radius of the circle, OB, into extreme and mean ratio at C. IL Use the larger segment, OC, as a side of the decagon. in. .-. make AB = OC. Join OA and prove ZO = M of a Proof: STATEMENTS makeAB = OC. straight angle. 1. OB s OC OC CB 2. OB AB AB BC 3. AABC^AABO. 4. AABC is isosceles. 5. .*. A A BO is isosceles. 6. :.AC = AB = OC. 7. .*. AOC is isosceles. 8. Z2=Z0. 9. .-. Z3 = Z0. 10. /. Zi4=twice ^O. 11. Z.4 + Z5+ZO = 5XZO = 180°. 12. .-. ZG = H 180° = Mo 360°. Let the pupil give the reasons. Ex. L Show that the radius of a circle may be divided into extreme and mean ratio and therefore that a side of the regular inscribed decagon may be obtained as follows: Draw ^5 and CD, two diameters perpendic- ular to each other. Let E be the mid-point of OA . With E as center and EC as radius cut OB at F. OF is a side of the decagon (Fig. 460). Ex. 2. Construct with ruler and compasses Fig. 460 angles of 36°, 18°, 54°, 72°. Ex. 3. What angles can you construct with ruler and compasses without the aid of a protractor? 258 PLANE GEOMETRY 278. Problem 31. To inscribe a regular pentagon in a circle. The analysis and the proof are left to the pupil (Fig. 461, No. 1). Ex. 1. What other regular inscribed polygons can be obtained* from the regular inscribed decagon? Ex. 2. Show how to construct regular circumscribed polygons of 5 and 10 sides. 279. Problem 32. To inscribe a regular pentadecagon in a circle. Suggestion. The central angle must be 24° : 60" -36° = 24° (Fig. 461, No. 2). Exercise. What inscribed and circumscribed regular polygons may be obtained from the regular inscribed pentadecagon? How? CONSTRUCTION OF OTHER REGULAR POLYGONS 280. Note. From the time of Euclid until 1796 it was supposed that the regular polygons mentioned in §§273-279 were the only ones that could be constructed with ruler and compasses. This includes polygons with 3 • 2", 4 • 2", 5 * 2", and 15 • 2'* sides. The smaller poly- gons not included in this set are those of 7, 9, 11, 13, 14, 17, 18, 19 sides. In 179G, however, Karl Friedrich Gauss, then a young man of nineteen in one of the German universities, proved that regular polygons with a prime number of sides could be inscribed in a circle by means of a ruler and compasses if and only if the prime number was of the form 2" + l. That no polygons of 7, 9, 11, 13, 19, etc., sides can be constructed in the given manner follows from Gauss's proof. A polygon of 7 sides can be constructed by the use of a parabola and a circle; one of 9 sides by the use of a hyperbola and a circle. REGULAR POLYGONS 259 281. While regular polygons of 7, 9, 11, 13, etc., sides cannot be constructed exactly with ruler and compasses, methods for constructing them approximately are frequently given in courses in mechanical drawing. The accuracy of these methods can be tested by trigonometry. PROPERTIES OF REGULAR POLYGONS THE CIRCUMSCRIBED CIRCLE 282. Theorem 130. A circle can be circumscribed about any regular polygon. Fig. 462 A BCD etc. is any regular polygon. A circle can be circumscribed about polygon Hypothesis: Conclusion: A BCD etc. Analysis: I. To prove that a circle can be circumscribed about ABCD etc., prove that a point can be found which is equally distant from the vertices. XL .'.construct perpendicular bisectors of any two con- secutive sides, BC and CD. Let the bisectors meet at O and prove that OB = OC = OD = OE = etc. III. To prove that OS = 0C, prove IV. To prove that OC = OZ), prove ..... V. To prove that OE = OB, prove ABOC mADOE. Let the pupil complete the analysis and give the proof. Cor. The radius of the circumscribed circle of a regular polygon bisects the angle through whose vertex it passes. 260 PLANE GEOMETRY THE INSCRIBED CIRCLE 283. Theorem 131. A circle can be inscribed in any regular polygon. Fig. 463 Hypothesis: ABODE etc. is any regular polygon. Conclusion: A circle can be inscribed in polygon ABCDE etc. Analysis and construction: I. An inscribed circle will be tangent to the sides of the polygon, .'.the perpendiculars from the center to the sides of the polygon must be radii and .*. equal, .'.construct the circumscribed circle, draw the per- pendiculars OX, OY, OZ, etc., from the center to the sides, and prove OX = OY = OZ — etc. The proof is left to the pupil. Suggestion. Show that Th. 131 may be proved by Th. 67 or by Th. 130 Cor. and Th. 85. 11. III. PROPERTIES DEPENDENT UPON THE CIRCUMSCRIBED AND INSCRIBED CIRCLES 284. The center of the circumscribed and of the inscribed circle of a regular polygon is called the center of the polygon. The radius of the circumscribed circle of a regular polygon is called the radius of the polygon. The radius of the inscribed circle of a regular polygon is called the apothem of the polygon. By the central angle of a regular polygon is ineant the angle between two consecutive radii. REGULAR POLYGONS 261 Cor. I. The central angle of a regular polygon of n sides is Vn of 360°. Cor. n. The radius of a regular polygon bisects the angle between two consecutive apothems, and the apothem bisects the angle between two consecutive radii. Cor. Ill The radius of a regular polygon bisects the arc between the points of contact of the inscribed circle. Ex. 1. Any two diagonals of a regular pentagon are equal. Ex. 2. Two diagonals from the same vertex of a regular pentagon trisect the angle at that vertex. Ex. 3. Show how to divide a regular hexagon into two con- gruent isosceles trapezoids, three congruent rhombuses, or six congruent equilateral triangles. Ex. 4. On a given base construct a regular hexagon without constructing the circumscribed circle. Ex. 5. A principal diagonal of a regular hexagon passes through the center of the circumscribed circle and is parallel to a pair of opposite sides (Fig. 464). Ex. 6. If a regular polygon has an even number of sides, the diameter of the circumscribed circle drawn from any vertex passes through the opposite vertex. Ex. 7. If a regular polygon has an even number of sides, the opposite sides are parallel. Ex. 8. If any regular polygon has an odd number of sides, the diameter of the circumscribed circle drawn from any vertex is a perpendicular bisector of the opposite side. Ex. 9. A side of an inscribed equilateral triangle bisects the radius drawn to the mid-point of the subtended arc. Ex. 10. The central angle of a regular polygon is the supple- ment of the interior angle of the polygon. Ex. 11. The area oC a square circumscribed about a circle is twice the area of the square inscribed in the same circle. Ex. 12. If squares are described outwardly on the sides of a regular hexagon, the outer vertices of the squares are the vertices of a regular duodecagon. 18 262 PLANE GEOMETRY THE ANGLE OF A REGULAR POLYGON 285. Theorem 132. Each angle of a regular polygon of ., . 271-4 . . n sides is rt. A. n The proof is left to the pupil. THE AREA OF A REGULAR POLYGON 286. Theorem 133. The area of a regular polygon is one-half the product of the perimeter and the apothem. Analysis: To find the area of a regular polygon draw the radii and add the areas of the triangles formed. SIMILAR REGULAR POLYGONS TEST FOR SIMILAR REGULAR POLYGONS 287. Theorem 134. Two regular polygons of the same number of sides are similar. Hypothesis: The two regular polygons O and G' h::vc the same number of sides. Conclusion: Polygon 0~ polygon 0\ Analysis: I. To prove polygon 0^^ polygon 0', prove ZA= LA', , I. . r>/ . .ABBCCD ZB= ZB' = etc., and J7^f = ^T^f = -^^f = ^^^- ^^ ^ AB BC AB A'B' ^ IL To prove -^7^, = ^,^' pro ve -^^ = -^^ = 1 . The proof is left to the pupil. REGULAR POLYGONS 263 RATIO OF CORRESPONDING SEGMENTS 288. Theorem 135. If two regular polygons have the same number of sides, the ratio of the perimeters is equal to the ratio of the radii or of the apothems. Fig. 466 Hypothesis: In the two regular polygons O and O' with the same number of sides, r and r' are the radii, a and a' the apothems, p and p' the perimeters, and 5 and s' are sides. Conclusion: —, = -,'=—,• p' r' a' Outline of proof: . s _r J'~? r__a ?~a' Let the pupil make an analysis and give all the reasons in the proof. Cor. The ratio of the perimeter to the diameter of the inscribed or of the circumscribed circle is the same for all regular polygons of the same number of sides. RATIOS OF AREAS 289. Theorem 136. If two regular polygons have the same number of sides, the ratio of the areas is equal to the ratio of the squares of the radii or of the apothems. The analysis and the proof are left to the pupil. 264 PLANE GEOMETRY SUMMARY AND SUPPLEMENTARY EXERCISES 290. SUMMARY OF IMPORTANT POINTS IN CHAPTER XH A. Construction of regular polygons. I. To construct a regular inscribed or circumscribed polygon of n sides, divide the circle into n equal arcs by constructing an angle ^/n of 360° at the center of the circle (§§271 and 272). II. To construct regular 4-, 8-, or 18-sided polygons, construct two perpendicular diameters (§§273, 274). III. To construct regular 3-, 6-, or 12-sided polygons, construct a central angle of 60° by means of an equilateral triangle (§ 275). IV. To construct regular £-, 10-, or 15-sided polygons, divide the radius of the circle into extreme and mean ratio (§§277, 278, 279). B. Properties of regular polygons. I. A circle can be circumscribed about, etc. (§282). II. A circle can be inscribed in, etc. (§ 283). III. The radius of the circumscribed circle of a regular polygon bisects, etc. (§282). 2w— 4 IV. Each angle of a regular polygon is rt. A ft (§285). V. The central angle of a regular polygon is V« of 360° (§284). C. The area of a regular polygon is ^ per. X apothem (§286). D. Similar regular polygons. I. Regular polygons are similar if they have, etc. (§287). II. The ratio of the perimeters equals, etc. (§288). III. The ratio of the areas equals, etc. (§289). For similar polygons in general see §268. REGULAR POLYGONS 265 Fig. 467 EXERCISES INVOLVING PROPERTIES OF AND SPECIAL CONSTRUCTIONS FOR REGULAR POLYGONS 291. 1. In a regular hexagon the secondary diagonals FD and AC are parallel to each other (Fig. 467). 2. In Fig. 467, prove that UVWXYZ is a regu- lar hexagon. Note. The star shown in Fig. 467 is extremely com- mon. It seems to be an ancient symbol of Deity. It is used in such modern instances as the policeman's star and many trademarks. 3. The area of the inscribed equilateral triangle is half the area of the regular hexagon inscribed in the same circle. 4. Show that the area of a regular hexagon inscribed in a circle is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles. 5. The central angle of a regular octagon is 45°. 6. Given a side of a regular octagon, to construct the circumscribed circle. Suggestion. See Fig. 468. In order to construct Zl=22>^^ construct Z2=45°. Fig. 468 7. Given the secondary diagonal AB oi a regular octagon (Fig. 469), to construct the octagon. Suggestion. Since the number of degrees in Zl is known, AAEB may be constructed. The problem is then to construct DCllAB so that DC=AD. Note. Ex. 7 is a problem which might occur in building octagonal bay windows. ADCB would be the outline of the window and AB the line on the house. Fig. 469 8. Show that a regular octagon may be inscribed in a square by the following method (Fig. 470): With O as center and OF, the half median, as radius cut the diagonals at F, Z, W, and X. At F, Z, W, and X construct perpendiculars to the diagonals. J/ -*j>-- Fig. 470 266 PLANE GEOMETRY 9. Show that a regular octagon 'may be inscribed in a square by the following method (Fig. 471): With the j^ ^ s_o vertices of the square as centers and the semi- '^'^^^ diagonal as radius cut the sides of the square at Xy Y, Z, etc., and join the points as indicated. Pig. 473 Note. Exs. 8 and 9 are extremely useful. They can be used as shown in Fig. 472 to construct a tiled-floor pattern composed of regular octagons and squares, or as in Fig. 473 to cut a square timber down into an octagonal one. The cuttings at the sawmill would be along the lines AB and BX, DC and CY, and so around the timber. Ex. 9 gives the common method for cutting out an octagonal table top. 10. Fig. 474 shows an inscribed regular pentagon with its diagonals. Prove {\)AX = XB = BY^Qtc. {2)AB = AY. {Z) AC ^ is divided into extreme and mean ratio at F. (4) ^ F is divided into extreme and mean ratio at X. CB Suggestion for CB=AY. (3). Prove that ^ CY and 11. From the foregoing exercises show how to construct the diagonal of a regular pentagon, given one side, and therefore how to construct the regular pentagon, given one side. 12. Show that if AD \^ taken as the radius of a circle, AB will be a side of the regular inscribed decagon (Fig. 474). 13. Prove that VWXYZ is a regular pentagon (Fig. 474). 14. The segments joining the mid-points of the sides of a regular pentagon in order form a regular pentagon. Note. The star shown in Fig. 474 is called the pentagram star and was used as a symbol of recognition among the Pythagoreans, an ancient Greek brotherhood that studied geometry. They called it Health. It is also the star used in the flag of the United States. REGULAR POLYGONS 267 292. EXERCISES INVOLVING THE MEASUREMENT OF REGULAR POLYGONS 1. Show that if a square is inscribed in a circle of radius 1 the side of the square is V2, and that if the radius is R the side of the square is i? V2. What is the area of the inscribed square? 2. Find one side of a regular octagon inscribed in a circle of radius 1, without the tables. Suggestion. In Fig. 475, ^45 is a side of the inscribed square and ^4 C of the_inscribed regular octagon. If A0 = l,AD = D0 = yzyl2. Find CD and .'. ^ C. Fig. 475 3. Find one side of the regular inscribed octagon if the radius of the circle is R. 4. Find one side and the apothem of an inscribed equilateral triangle if (1) the radius of the circle is 1 ; (2) the radius is R. 5. Find one side of a regular duodecagon in- scribed in a circle of radius 1, without the tables. Suggestion. In Fig. 476, ^5 is a side of the inscribed regular hexagon and BC a. side of the inscribed regular duodecagon. If BO = l, AB = 1, and BD = }4, then 0D = }4^^' Find DC and .: BC. Fig. 476 6. Find one side of the regular inscribed duodecagon if the radius of the circle is R. 7. Find a side of the regular circumscribed hexagon if the radius of the circle is 1 ; if it is R. 8. Find one side of the regular inscribed decagon if the radius of the circle is 1. Suggestion. Let s represent one side of the decagon. Show that s may be obtained from the equation — = - — - . 9. Find one side of the regular inscribed decagon if the radius of the circle is R. 10. Using the side of the regular inscribed decagon found in Ex. 9 and the radius of the circle as 1, find a side of the regular inscribed pentagon. Suggestion. In Fig. 477, BC is a mean proportional between DC and CE. BE can be found from BC and CE. Fig. 477 268 PLANE GEOMETRY 11. Find the ratio of the side of a square inscribed in a circle to the side of an equilateral triangle inscribed in the same circle. 12. ABCDEF is a regular hexagon. X, Y, and Z are the mid- points of AF, BC, and DE respectively. Prove that XYZ is an equilateral triangle and find its area if AB = 20 in. STAR POLYGONS 293. Fig. 474 shows a star polygon of five points and Fig. 467 one of six points. There are one or more regular star polygons related to each of the regular convex polygons. They can be constructed by dividing the circle into n equal parts and joining each point of division to the ^th one from it, where k is an integer greater than one and less than - • Fig. 478 shows another illustration. (See also Fig. 258.) These figures have been studied wherever geometry has been studied. They abound in cut- glass designs. The five-pointed star mentioned in the note, §291, Ex. 14, and the six-pointed star mentioned in the aote, §291, Ex. 2, are of special importance. Exercise. In Fig. 478 the circle is divided into 16 equal parts and each point joined to the 7th from it. The following questions apply to Fig. 478. 1. How many sets of equal angles are there? 2. Find the number of degrees in one angle of each set. 3. Prove that AH = HB = BK = etc. 4. How many other sets of equal segments are there? 5. How many regular polygons of 16 sides can be formed by joining corresponding intersections? Give proof for each. 6. Can regular octagons be formed by joining corresponding intersections? How? 7. How many squares are in the figure? CHAPTER XIII Measurement of the Circle THE CIRCUMFERENCE OF THE CIRCLE DEFINITION 294. The length of a straight-line segment was defined as the number of times a certain straight-line segment taken as a unit can be applied to the segment to be meas- ured. It is at once evident that we cannot measure a circle in this way. We shall, however, assume that the circle can be measured in terms of a straight-Hne unit. The measure of the circle is called its length or its circumference. GENERAL METHOD 295. The perimeter of an inscribed or of a circumscribed polygon of many sides may be used as an approximation to the length of the circle. If a regular hexagon is inscribed in a circle and the arcs between its vertices are bisected, the chords joining the points of division form a regular inscribed polygon of 12 sides. By the same process regular polygons of 24, 48, 96, 192 sides may be obtained, and so on indefinitely. If such a succession of polygons is constructed, one is soon found which can with difficulty be distinguished from the circle. Instead of the regular inscribed hexagon, the inscribed square might have been used as the starting point. Ex. 1. The perimeter of a regular inscribed poly- gon is less than the perimeter of the regular polygon of double the number of sides inscribed in the same circle (Fig. 479). Fig. 479 269 270 PLANE GEOMETRY If a regular hexagon is circumscribed about a circle and the arcs between the points of tangency are bisected, the tangents drawn at the points of division form a regular circumscribed polygon of 12 sides. By the same process, regular circumscribed polygons of 24, 48, 96, 192 sides may be obtained, and so on indefinitely. Just as in the series of regular inscribed polygons, so in this series of regular cir- cumscribed polygons, a polygon is soon found which can with difficulty be distinguished from the circle. The start- ing point of this series might have been the circumscribed square instead of the regular circumscribed hexagon. The perimeters of the polygons in these series have been computed. The computations can be made as indicated in the next two problems. Ex. 2. The perimeter of a regular circumscribed polygon is greater than the perimeter of the regular polygon of double the number of sides circum- scribed about the same circle (Fig. 480). Fig. 480 296. Problem 33. To find a side of a regular inscribed polygon of 2n sides, given a side of a regular inscribed polygon of n sides and the radius of the circle. Fig. 481 Given a side AC (or a) of a regular polygon of n sides inscribed in a circle of radius r. Required to find a side (x) of a regular inscribed polygon of 2w sides. MEASUREMENT OF THE CIRCLE 271 Analysis (segments are lettered as shown in Fig. 481) : I. To find X in terms of the known lengths use x^ = d^-\-{i a)-, a is known; fi must be found. TT a> ^ J ^ J I ri^ known. II. To find a, use a = r — V i . . r i [ y must be found. III. To find y, use J" = r^ — {\ ay. a and r are known. Let the pupil make the ntlmerical computation when ^ C is a side (1) of the inscribed sciuare, (2) of the regular inscribed hexagon. Use for the radius of the circle (1) \, (2) r. The following f ormula gives the value of x in terms of r and a : jc= V 2r'^ — r \'4r'-— a^. 297. Problkm 34. To find a side of a regular circum- scribed polygon of 2n sides, given a side of a regular circum- scribed polygon of n sides and the radius of the circle. Fig. 482 Given a side AB (or a) of a regular polygon of n sides circiunscribed about a circle of radius r. Required to find a side {%) of a regular circumscribed polygon of 2w sides. Analysis (the segments are lettered as shown in Fig. 482) ; I. To find X, use Th. 106. '' — 2 ^ a and r are known. y 2^'~2^ [>'tobe found. II. To find y, use y'^=-{\ ay-\-r-. Let the pupil make the numerical computation when AB \s, o. side of the circumscribed square. Use for the radius of the circle (I) i ; (2) r. The general formula is x= - — ; — , -7 o .. • 2r+ V4r2-f a-^ 272 PLANE GEOMETRY PERIMETERS OF REGULAR POLYGONS INSCRIBED IN OR CIRCUMSCRIBED ABOUT A CIRCLE WITH DIAMETER ONE 298. The methods indicated in Probs. 33 and 34 have been used in obtaining the results given in the tables below. TABLE I Number OF Sides Perimeter OF Inscribed Polygon Perimeter OF Circumscribed Polygon 6 3. 3.464101 12 3.105828 3.215390 24 3.132628 3.159659 48 3.139350 3.146086 96 3.141031 3.142714 192 3.141452 3.141873 • 384 3.141557 3.141662 768 3.141583 3.141610 1536 3.141590 3.141597 3072 3.141592 3.141594 6144 3.141592 3.141593 TABLE II Number OF Sides Perimeter OF Inscribed Polygon PiiRIMETER OF Circumscribed Polygon 4 2.828427 4.000000 8 3.061667 3.313708 16 3.121445 3.182597 32 3.136548 3.151724 64 3.140331 3.144118 128 3.141277 3.142223 256 3.141513 3.141750 512 3.141572 3.141632 1024 3.141587 3.141602 2048 3.141591 3.141595 4096 3.141592 3.141593 8192 3.141592 3.141592 MEASUREMENT OF THE CIRCLE 273 Table I shows the perimeters of a series of inscribed poly- gons and of a series of circumscribed polygons that start with the regular inscribed and circumscribed hexagon respectively. Table II shows the perimeters of two series that start with the inscribed and circumscribed square respectively. The diameter of the circle is 1 in each case. Can you come to any conclusion by comparing results ? In higher mathematics it is proved that the perimeters of the polygons in Tables I and II approach a definite num- ber. This number cannot be found exactly. It has been named w (pi). COMPUTATION OF PERIMETERS OF REGULAR POLYGONS INSCRIBED IN OR CIRCUMSCRIBED ABOUT ANY CIRCLE 299. Problem 35. To find the perimeters of regular polygons inscribed in or circumscribed about any circle. Solution: I. Let d and d' represent the diameters of two circles. p and p' represent the perimeters of two regular polygons of the same number of sides inscribed in or circumscribed about the two circles. II. Then ^ = ^ (see Th. 135 Cor.). p a ^, = d,whend' = l. P .'. p = dp' if the diameter of circle about p' is 1. The perimeter of a regular polygon of any number of sides inscribed in or circumscribed about a circle can be found from the equation p = dp\ if the perimeter of a similar polygon inscribed in or circumscribed about a circle of diameter 1 is known. The perimeters of polygons inscribed in or circumscribed about any circle and similar to those in Tables I and II may be found by multiplying the diameter of the given circle by the proper number in the tables. 274 PLANE GEOMETRY LIMITING VALUES OF PERIMETER OF INSCRIBED AND CIRCUMSCRIBED POLYGONS 300. As. 64. The limit of the perimeters of a series of regular polygons inscribed in or circumscribed about the same circle as the number of sides is increased indefinitely is the same. This limit doe? not depend upon the number of sides of the initial polygon nor upon the method of increasing the number of sides. This limit is ird. This assumption is proved in higher mathematics. LENGTH OF THE CIRCLE 301. The assumptions in §300 lead us to the following definition : The length or the circumference of a circle is defined as the limit of the perimeters of a series of regular polygons inscribed in or circumscribed about a circle as the number of sides is increased indefinitely. The perimeter of any one of the polygons may be regarded as an approximation to the length of the circle. From As. 64 and the definition above, we have Theorem 137. The circumference of a circle of diameter d is Tvd, If c is the circumference, d the diameter, r the radius, We have .*. c= ird or c = 2Trr. Note. The number tt is a very important number in mathematics. It is an irrational number (see §198), but very unHke such irrational numbers as V2 or V3. We found it possible by means of a straight- edge and compasses to construct a straight-line segment to represent V2 (Ex. 2, §221). The ancient Greeks could do this. They failed, however, in the attempt to construct a straight-line segment to repre- sent TT. In modern times it has been proved that it is impossible to construct such a segment for tt by the use of straightedge and com- passes. It can, however, be done by means of an instrument called an integraph, invented by a Pole about 1878. MEASUREMENT OF THE CIRCLE 275 AREAS OF CIRCLES, SECTORS, AND SEGMENTS THE AREA OF THE CIRCLE 302. As. 65. The limit of the areas of a series of regular polygons inscribed in or circumscribed about the same circle as the number of sides is increased indefinitely is the sam:'. This limit is one-half the product of the radius of the circle and its circumference. This assumption is proved in higher mathematics. 303. We shall define the area of a circle as the limit of the areas of a scries of inscribed or circumscribed regular polygons as the number of sides is increased indefinitely. 304. From §§302 and 303 we have Theorem 138. The area of a circle is one-half the product of its radius and circumference. Using A for the area of the circle, r for the radius, c for the circumference, A = y2cr = H r ' 2Trr=Trr^ since c = 27rr. Note. We know that tt represents a definite number, although we cannot find that number exactly. We have learned that a straight- line segment can be constructed which is equal to the circumference of a given circle, although it cannot be constructed with compasses and straightedge. We will, therefore, assume that there is a triangle, square, or rectangle equivalent to any given circle, although these figures cannot be constructed with straightedge and compasses. AREAS OF SECTORS 305. As. 66. The area of a sector has the same ratio to the area of a circle of which it is a part as the angle of the sector has to four right angles. Using 5 for the area of the sector, a for the angle of the sector, 7rr- for the area of the circle, Solvingfor 5, 5 = -^- 270 PLANE GEOMETRY AREAS OF SEGMENTS 306. A circular segment is a figure boimded by an arc and its chord. The relation between circular segments and sectors may be shown by constructing the sector having the same arc as a given circular segment. It is at once evident (Fig. 483) that the area of a circular segment can be obtained if the Fig. 483 areas of the associated sector and triangle can be obtained. RATIOS AND CIRCLES 307. The following theorems follow at once from the formulae :. Theorem 139. The ratio of the circumference to the diameter is the same for all circles. If c = wd, then -r = tt. a Theorem 140. The ratio of the circumferences of two circles equals the ratio of their diameters or of their radii. Using c and ci for the circumferences of two circles, d and di for their diameters, r and ri for their radii, we have: c _Ci ^ c _d d di " Ci di ^. d 2r r c r Smce -r = ?r~=~"> •'•~r = — di 2ri n c^ ri Theorem 141. The ratio of the areas of two circles equals the ratio of the squares of their radii or of their diameters. Using A and Ai for the areas of two circles, r and n for their radii, d and di for their diameters, we have: A 7rr2 Ai Trn^ A r' " A^ n' r d r'~ Smce - = j-» —7, n di Ti dr . A d' "Ai di" B. For two circles : c d ; — C\ rfi c r : = — Ci n A r2 d" — - — Ai' u' d,' MEASUREMENT OF THE CIRCLE 277 SUMMARY AND SUPPLEMENTARY EXERCISES 308. FORMULAE OBTAINED IN CHAPTER XIII A. For one circle: c = 2Trr = ird A = 7rr2 c C. For sectors: _ wr-a EXERCISES INVOLVING THE MEASUREMENT OF THE CIRCLE 309. 1. Find the circumference and the area of a circle if the diameter is 4 ft.; 6 ft.; 25 ft. 2. Find the radius of a circle if the circumference is 10 ft. ; 25 ft.; 63 ft. 3. Find the radius of a circle if the area is 20 sq. ft.; 56 sq. ft.; 96 sq. ft. 4. If a wheel is 3 ft. in diameter, how many revolutions will it make in going a mile? 5. What is the diameter of a wheel that makes 250 revolutions in traveling one mile? 6. A locomotive is running at the rate of 40 mi. per hour. If the driving wheels are 80 in. in diameter and the truck wheels are 36 in., how many revolutions per minute does each make? 7. The 72-in. drive wheel of a locomotive makes 250 turns per minute. What is the rate of the locomotive in miles per hour? 8. Two pulleys, one 36 in. and one 15 in. in diameter, are connected by a belt. The 36-in . pulley makes 200 turns per minute. How many turns does the 15-in. pulley make per minute? 9. It is the general practice not to permit the rim of a fly wheel to have a greater velocity than a mile a minute. How many revolutions per minute may be allowed an 8-ft. fly wheel? 19 278 PLANE GEOMETRY 10. Fig. 48-4 shows the construction of certain types of trefoils. Find the entire area if one side of the equilateral triangle is 5 ft. 11. Find the ratio of the area of a circle to the area of the inscribed square and to the area of the Fig. 484 Fig. 485 circumscribed square. 12. One side of the equilateral triangle in Fig. 485 is 7 ft. Show how the shaded portion is formed and find its area. 13. Show how the shaded figure in the square in Fig. 486 is formed and find its area if one side of the square is 6 in. 14. Show how the four leaf-shaped figures in Fig. 487 are formed and find the sum of their areas if one side of the square is 6 in. 15. If the equatorial diameter of the earth is 7926 mi., what is the length of a degree of longitude at the equator? Fig. 487 16. (a) Construct the locus of the center of a circle, radius one- half inch, which rolls around an equilateral triangle, altitude two inches, (b) Compute to two decimals the area inclosed by the locus and the perimeter of the locus. — College Entrance Examina- tion Board, Plane Geometry Examination, 1914. 17. A circle of radius 2 in. rolls around a square whose side is 4 in. Construct the locus of the center of the circle, and find to two decimal places both the length of the locus and the area inclosed by it. — College Entrance Examination Board, Plane Geometry Examination, 1915. 18. If a circle is constructed on each of the sides of a right triangle as diameter, the area of the circle on the hypotenuse is equal to the sum of the areas of the other two circles. 19. Find by geometrical construction the diameter of a pipe the area of whose cross-section is equal to the sum of the areas of the cross-sections of two given pipes. MEASUREMENT OF THE CIRCLE 279 20. A 12-in. water pipe branches into three equal pipes whose combined capacity is the same as that of the 12-in. pipe. If the quantity of water carried depends upon the area of the cross-section of the pipes, what must be the diameter of each of the three pipes? 21. Semicircles are constructed on the sides of a right triangle as shown in Fig. 488. Show that the sum of the areas of the two crescents is equal to the area of the triangle. ^ ^ Fig. 488 Note. This problem is due to Hippocrates (about 470 B.C.). His solution is the first case of the area of a curvilinear figure proved equal to the area of a rectilinear figure. 22. A circular grass plot 12 ft. in diameter is cut by a straight gravel path 3 ft. wide, one edge of which passes through the center of the plot. What is the area of the remaining grass plot? — College Entrance Examination Board, Plane Geometry Examina- tion, 1911. 23. The diameters of two circular pulleys are respectively 12 ft. and 2 ft., and the distance between their centers is 10 ft. Find the length of the shortest string which will go around the pulleys, correct to three significant figures. — College Entrance Examina- tion Board, Plane Geometry Examination, 1910. 24. There are many reasons why an egg- shaped sewer is more satisfactory than a circu- lar sewer. The cross-section of the sewer in Fig. 480 is made up of four circular arcs A By BC, CD, and DA. The center of arc AB is F; of arc yl D is P; of arc DC is X. The fig- ure i^ symmetric with respect to line XY. Fig. 489 If ;?= 1 .2 ft., r= .7 ft., ^ = 2.0 ft., Z P = 30°, find the entire circum- ference of the sewer. AB is a, semicircle. 25. Find the length of a circular railway curve of radius K mile and central angle 30°. 26, The rail of a street-car track is 12 ft. from the curb. In rounding a square corner it cannot be closer than 3 ft. from the curb. What is the radius of curvature of the required curve? How 6 far back must the curve begin? (Fig. 490.) Fig. 490 280 PLANE GEOMETRY 27. To relieve the jolt that always comes in passing from a tangent to a sharp curve, many railway curves are made up of arcs of different curva- ture. Find the length of the compound curve in Fig. 491. ^O(±5^)=300 ft. C is center of AM. CO = C'O = OD=150 ft. C is center of NB. ZP = 60°. Z) is center of ikfiV. The figure is symmetric with respect to OP. 28. Find the area of a circular segment if its arc is 60°; 45^ 29. In Fig. 492, the arcs AC and BC are constructed with B and A as centers respec- tively and AB as radius. Show that the entire area of the circular figure may be found by subtracting the area of the triangle from twice ^ the area of the sector. What is the number Fig. 492 of degrees in the angle of the sector? Find the area if AB = 8 ft. 30. The figure shown in Fig. 493 is a form ^ frequently used in church window designs. Show how it is constructed and find its area if ^J5 = 6 ft. 31. Find the area of the ring between two con- centric circles if the radii of the circles are 5 in. and ^' 3 in. respectively. Fig. 493 32. Find a formula for the area of the ring between con- centric circles if the radii are R and r respec- tively. • 33. Fig. 494 shows two concentric circles. Show how to find the segment AB so that the circle constructed on AB as diameter shall have the same area as the ring. Fig. 494 34. The circumferences of two concentric circles differ by 6 in. Compute the width of the ring between the two circles correct to three significant figures. — College Entrance Examination Board, Plane Geometry Examination, 1909. 35. Compare the area of a circle with the area of a square if the perimeter of the square equals the circumference of the circle. MEASUREMENT OF THE CIRCLE 281 36. Show how Fig. 405 is formed and find the area of HKLMFEH and of HKLMFGH if one side of the square A BCD is 6 in. 37. If the sum of the radii of two given circles is equal to the radius of a third circle, prove that Fig. 495 the circumference of the third circle is equal to the From an old r 1 • c r 1 • . , Roman sum of the circumferences of the two given circles. pavement 38. Construct a circle whose circumference is equal to the difTerence between the circumferences of two given circles. 39. Show how Fig. 490 is formed and find its area if a side of the hexagon is 5 ft. 40. In Fig. 497 the diameter of the circle was divided into three equal parts. Show how the arcs a shown are drawn and prove that they divide the circle into three equal parts. Fig. 496 Fig. 497 41. Show how Fig. 498 is formed, area of OABCD if 05 = 5 ft. Find the 42. Show how to divide a circle into four equal parts by the arcs of circles only. Can this prob- lem be solved in more than one way? Note. Ex. 42 is said to be a problem which Napoleon once pro- posed to his staff. The figure formed by dividing the circle into two equal parts (Fig. 498) is the trademark of one of the western railroads. 43. In Fig. 499, C is the mid-point of radius ^0 of OO. CX is perpendicular to AO at C. Prove that the circle with O as center and OX as radius has half the area of the circle with radius AO. 44. Show how to divide a circle into three equal parts by concentric circles. Fig. 439 282 PLANE GEOMETRY 45. It is desired to construct a half-mile track. The start and finish are to be straightways intersecting at right angles at the goal. The rest of the track is to be an arc of a circle tangent to the two straightways. Find the radius of the arc and the length of the arc in feet; also the area inclosed by the track in acres. (Results to be correct to two decimals.) — College Entrance Exami- nation Board, Plane Geometry Examination, 1914. APPROXIMATION CONSTRUCTIONS 310. We have seen that numbers can be found which more or less closely approximate the value of tt, although a number cannot be found which is exactly equal to ir. Similarly, segments can be constructed whose lengths more or less closely approximate the value of tf, although no segment can be constructed with ruler and compasses whose length exactly represents the value of tf. The following exercises give some of these constructions. They are used by draftsmen in obtaining the development of cylinders, by carpenters in obtaining the dimensions of veneers for semicircular heads of doors and windows, and by mechani- cal tradesmen generally. The computations for the true lengths of the segments often involve considerable geome- try, and are necessary in order to ascertain the degree of approximation obtained. Ex. 1. In Fig. 500, ACB is a semicircle with /. e £ AB its diameter. AABD is equilateral, with AB \ as one side. EF is a tangent to the semicircle parallel to AB. DA and DB are extended to meet the tangent EF at F and E respectively. Using AB = Q, find the difference between the length of EF and of the semicircle ACB. Find the ratio of this difference to the length of A CB. Ex. 2. In Fig. 501, ACB is a semicircle with AB its diameter. CO ± bisector of AB. ZOCZ) = 30°. Find the difference between the length oi AD and of the arc AC, using AB = 5. Find the ratio of this ^ o n b difference to the length of the arc AC. ^^^- ^^^ MEASUREMENT OF THE CIRCLE 283 Ex. 3. In Fig. 502, Z ^0£ is a central angle of 45° in circle O. OELAB. Using 0£ = 8, find the length of DE and the difference between WE -{-DE and the circumference. Find the ratio of this difference to the circumference. Ex. 4. Inscribe a square in a given circle. Find the value of three times the diameter of the ' circle plus one-fifth of the side of the square, using the diameter of the circle as d. Find the difference between this sum and the circumference of the circle. Find the ratio of this difference to the circumference of the circle. (From Ball's Mathematical Recreations and Essays.) Ex. 5. Determine the value of w by the following experiment: Wrap a paper about a cylinder. Stick a pin through the over- lapped paper. Measure the distance between the two pinholes and the diameter of the cylinder and compute w. 311. We do not know who made the first attempt to compare the area of a circle with the area of a square. But the first record that has so far been found is on an Egyptian papyrus by Ahmes (about 1700 B.C.). He says, "Cut off }i of a diameter and construct a square on the 256 remainder." This implies that the area of the circle is {% d)^ or-^r'; 81 256 that is, the approximation to w is -^^j- =3. 16049 -|-. This is a much ox better approximation than was used by some other ancient peoples. In the countries of Asia 3 was commonly used. Both 3 and 33^^ are found in the Bible (I Kings 7: 23; Daniel 7: 25). Archimedes (287-212 B.C.) used inscribed and circumscribed polygons of 96 sides and showed that TT was less than Sl'j and greater than 3 ^^f i . His method was practically the only method used for the next 2,000 years until the invention of the calculus in modern times. S} area ABD, prove the altitude of AABOthe altitude of AABD. II. .*. draw CX and DY±AB from C and D respectively; draw DE\\AB from D, intersecting CX at E, and prove that E lies between C and X\ that is, prove CX>EX. III. To prove CX> EX, join EA and EB and prove AOAE. IV. To prove AOAE, prove AC +CB> A E+EB. V. To prove AC+CB>AE-\-EB, prove AD+DB>AE-\-EB. VI. To prove AD-{-DB>AE+EB, prove area AEB = area ylD5. The proof is left to the pupil. Suggestion. Show that AEB is an isosceles triangle. Since area AEB=avea,ADB, AD+DB>AE+EB. MAXIMA AND MINIMA 287 POLYGONS IN GENERAL PRELIMINARY THEOREM 316. Theorem 145. Of all polygons having all sides but one equal to given segments taken in order, that with the greatest area can be inscribed in a semicircle with the undetermined side as diameter. Fig. 508 Hypothesis: The polygon ABCDEF is the maximum polygon that can be formed with all sides but one equal to the segments AB, EC, CD, DE, and EF taken in the order given. Conclusion: ABCDEF can be inscribed in a semicircle with ^ F as diameter. Analysis and construction: I. To prove that ABCDEF can be inscribed in a semi- circle with A Fas diameter, prove that any vertex, as D, lies on the semicircle. 11. To prove that D Hcs on the semicircle, join DA and DF and prove /LADF=^l rt.*Z. III. To prove AADF^l rt. Z, prove area ADF a maxi- mum. The proof is left to the pupil. Suggestion for step III. Use indirect proof. If area ADF is not a maximum, we may increase or decrease A ADF by sliding points A and /^ along the line XFuntilarca/lD/''is a maximum. If, as points A and Fmove along line .Y F, figures A BCD and DEFrviwixxn unchanged, the area of ABCDE would be increased, which is contrary to tlie hypothesis. 288 PLANE GEOMETRY TEST FOR MAXIMUM AREAS 317. Theorem 146. Of all polygons that have their sides equal respectively to given segments taken in order, that which can be inscribed in a circle has the greatest area. Hypothesis: The polygon ABODE is inscribed in a circle and polygon A'B'C'D'E' cannot be inscribed in a circle. AB = A'B\ BC = B'C\ CD = CD\ etc. Conclusion: Area ABODE > area A'B'C'D'E'. Analysis and construction: To prove area ABODE > area A'B'O'D'E', Draw diameter AX. Join OX and DX. Construct AD'X'O' ^ADXO. Compare area ABOX with area A'B'O'X' and area AEDX with area A'E'D'X'. Subtract area DXO from area ABOXDE and area D'X'C from area A'B'0'X'D'E\ The proof is left to the pupil. Discussion. Might one part of ABODE have the same area as the corresponding part of A'B'O'D'E'l Could both parts of ABODE have the same area as the corresponding parts of A'B'O'D'E'l Why? MAXIMA AND MINIMA 289 THE MAXIMUM AREA WITH GIVEN PERIMETER 318. Theorem 147. Of all polygons with a given perime- ter and a given number of sides, that with the maximum area is regular. Fig. 508 Suggestion. To prove P regular, prove that it can be inscribed in a circle and is equilateral. To prove AC=CB, prove AABC a, maxi- mum by indirect proof. THE MINIMUM PERIMETER WITH GIVEN AREA 319. Theorem 148. Of all polygons with the same area and the same number of sides, the regular polygon has the least perimeter. Fig. 509 Analysis and construction: I. To prove per. Parea P\ compare P and P' with a third polygon. II. .*. join E, any point in AB, with C and construct ACDE^AACE and prove (1) area P'=area BCDE; (2) area P>area BCDE. III. To prove area P>area BCDE, prove per. P = per. BCDE. Discussion and conclusion: In the same manner it can be proved that a regular polygon of five sides has a greater area than a square of same perimeter and that a regular hexagon has a greater area than a regular pentagon of same perimeter, and so on. 321. Since the area of a circle has been defined as the limit of the areas of a series of regular inscribed polygons as the number of sides is increased indefinitely, we have: As. 67. The area of a circle is greater than the area of any polygon of equal perimeter. In higher mathematics we prove : As. 68. Of all figures having the same perimeter, the circle has the maximum area. MAXIMA AND MINIMA 291 THE MINIMUM PERIMETER WITH GIVEN AREA 322. Theorem 150. Of all regular polygons with the same area, that having the greatest number of sides has the least perimeter. Fig. 511 Analysis and construction: I. To prove per. P per. Q. III. To prove per. P'>per. Q, prove area P'>area Q. The proof is left to the pupil. 323. Since the circumference of a circle has been defined as the limit of the perimeters of a series of regular inscribed polygons as the number of sides is increased indefinitely, we have: As. 69. The perimeter of a circle is less than that of any polygon of the same area. In higher mathematics we prove : As. 70. Of all figures having the same area, the circle has the minimum perimeter. Note. As. 70 has an important application in engineering. The flow of water in an aqueduct or sewer is checked by the friction of the water on the walls. The friction is proportional to the perimeter of the cross-section. It is desirable to keep this perimeter as small as possible. The circle, then, is the best form to meet this condition. In case it is not desirable because of expense and other considerations to use a circle, regular polygons are often used. NOTES ON ARITHMETIC AND ALGEBRA FRACTIONS 324. The following fundamental law of fractions under- lies all operations that involve fractions. Multiplying or dividing numerator and denominator of a fraction by the same number does not alter the value of the fraction. A. The sum or difference of two or more fractions that have a common denominator is the sum or difference of the numerators divided by the common denominator. Two or more fractions that have not a common denomi- nator must be reduced to a common denominator before adding or subtracting. To reduce fractions to a common denominator, apply the fundamental law given above. Add and subtract the following: 1 ^ + ^ ^' 12^18 ^' 24^36 6 B. The product of two fractions is the product of the numerators divided by the product of the denominators. Where possible, divide numerator and denominator by com- mon factors. Multiply the following: 16 15 be a — b x — 1 292 b c 5.4 ^+f: y 4v- 4. ±+l_± be ac ab ^- :.-hl X-: NOTES ON ARITHMETIC AND ALGEBRA 293 C. The quotient of one fraction divided by a second is the product of the first multiplied by the reciprocal of the second. Divide the following: ,. 14,33 2. ?^^6a6 3. "'-"* • "-* 20 25 8*3 — "• Uab^ 2a ROOTS 325. The rule for square root is based on the algebraic formula {a-^by = a^-\-2ab-}-b\ Notice that a^-{-2ab-\-b^ may be written a^-\-b{2a-{-b). The method is illustrated below: Illustration 1. Find V 694. 563 694.56,3 | 26.3 4_ 2(20) = 40 294 40+6 = 46 276 2(260) = 520 1856 520+3 = 523 1569 287 In the illustration above, notice: (1) The number was divided into periods of two figures each, counting to the left and to the right from the decimal point. (2) The largest square under 6 is 4. The 4 was sub- tracted from 6 and the next period annexed. This gave a remainder of 294. The square root of 4, or 2, was written as the first figure in the root. (3) A zero was placed after the 2, making 20. The 20 was doubled, making 40. The 40 is used as a trial divisor for the remainder 294. The next figure of the root is either 6 or 7. The 6 is added to the 40, making 46. The 46 is multiplied by 6, giving 276. The 276 is subtracted from 294, leaving 18. The next period is annexed, giving 1856. (4) The process above is repeated at each step of the work; thus, a zero is placed after 26 and the result doubled, giving the 520. The work is then continued as above. 20 294 PLANE GEOMETRY In general we may say: Annex a zero to the part of the root already found and double the result. To this result add the next figure of the root. Multiply the result by the last figure of the root found. Show that this statement may be regarded as a trans- lation of b{2a+b) in the formula {a-\'b)^ = a^+b{2a-hh). Find the square root of the following: 1. 1369 4. 106276 7. 6 2. 3744 5. 3 8. 15 3. 2304 6. 5 9. 7 Because of the frequent occurrence of the square roots of 2 and 3 in geometry work, the application of the following law should be noted : 326. The square root of a product is the product of the square roots of the factors. Illustration 2. 36 = 4 X 9 /. V36 = Vi X V9 This law is used most conveniently for inexact -square roots when one factor is a perfect square. Illustration 3^ 18 = 9X2 .*. V 18 = V9 X V2 = 3 V2 Notice that V2 occurs when the side of a square and n diagonal of the square are used in the same exercise. Illustration 4. 12 = 4X3 .-. Vr2 = Vi X V3 = 2 V3 Notice that V3 occurs when the side of an equilateral tri- angle and its altitude occur in the same exercise. Illustration 5. 20 = 4 X 5 /. V20 = Vi X V5 = 2 VS The V5 occurs in connection with the regular decagon and pentagon. Find the value of the following correct to three decimal places. Apply the law given above. 1. V8 4. Vl08 7. Vl28 10. Vl50 13. V54 2. Vl8 5. V32 8. V75 11. Vl25 14. V45 3. V27 6. V80 9. V320 12. V98 15. Vl80 NOTES ON ARITHMETIC AND ALGEBRA 295 16. V20 18. V48 20. V50 22. V288 24. V243 17. V96 19. V72 21. V300 23. Vl62 25. V242 327. The square root of a fraction. A. If the denominator is a perfect square: Find the square root of the numerator and of the denominator sepa- rately and divide the first result by the second. 144 12 Illustration 6. _ 625 25 Illustration 7. -4/- = 3/2 V3 B. If the denominator is not a perfect square, two methods are suggested: (1) The fraction may be reduced to a decimal and the square root of the result found. (2) Numerator and denominator may be multiplied by some number that will make the denominator a perfect square' and method A above used. Illustration 8. To find \'^ either 1. Find square root of .333333-}-; or 2. Write H = % and use Vz Vs. Find the value of the following correct to three decimal places: ■•n's "-vt Wf -Vi- WE '■<} «i • H)^ is added to the left side to make the left side a perfect square. It is added to the right side to preserve the balance of the equation. (H • H^y or ^He, is obtained by squaring half the coefficient of x. Notice that in step (2) the equation is divided by 3 to make the first term ic^, which is a perfect square. Solve the following equations: 1. a;2+3:c=18 3. 3:^2-ll:x: = 2 2. 2x''-x = \b 4. 2x''-\-bx =17 330. To solve a system of equations consisting of two equations containing two unknowns, eliminate one of the unknowns and solve the resulting equation for the other. A. When both equations are of the first degree, eliminate by addition or subtraction. 5:r-4y = 6.5 Illustration Ai. Solve for aj and y: ,- , - oo or 5:c-4y = 6.5 (1) 7jc+53; = 38.25 (2) 35;c-28y = 45.5 (1) X 7 35:c+25>'= 191.25 (2) X 5 -53y= -145.75 Subtract the third ^ = 2.75 equation from the second Notice that x may be found by multiplying equation (1) by 5 and equation (2) by 4 and adding the results or by substituting 2.75 for y in either equation (1) or (2) and solving the result for x. 298 PLANE GEOMETRY B. When one of the equations is of the first degree and one of the second, solve the first-degree equation for one of the unknowns in terms of the other unknown and substitute in the other equation. \x -hy =8 . . . |x2+3;2 = 34 . . . x=8—y . 64-163;+2/ = 34 2^2-16^+30 = Illustration 5. Solve for x and y: Solve (1) for x, Substitute 8 — 3; for :v in (2) (1) (2) (3) y^-8y+15 = (y-5) (3'-3)=0 y = D and y = 3 To find X, substitute the values of y in (3) : y= 5 y- = 3 x=8-y x=8-5 =8-5 =8-3 = 3 =5 \ x=3 ^ x=5 The solutions are i ^ in [y==5 [y=^ Solve the following systems for x and y : 1. 2x+33;=16 2. bx-2y = U 3. 2x^-j-y^ = 57 5.r-2>; = 21 xy = 20 x-y=l TABLES 331. TABLE OF SQUARE ROOTS 1 2 3 4 5 6 7 3 9 1.000 1.414 1.732 2.00c 2.236 2.449 2.645 2.828 3.000 1 3.162 3.316 3.464 3.605 3.741 3.872 4.000 4.123 4.242 4.358 2 4.472 4 . 582 4.690 4.795 4.898 5.000 5.099 5.196 5.291 5.385 3 0.477 5.567 5.656 5.744 5.830 5.916 6.000 6.082 6.164 6.244 4 6.324 6.403 6.480 6.557 6.633 6.708 6.782 6.855 6.928 7.000 5 7.071 7.141 7.211 7.280 7.34S 7.416 7.483 7 . 549 7.615 7.681 G 7 . 745 7.810 7.874 7.937 S.OOO 8.062 8.124 8.185 8.246 8.306 7 8.366 8.426 8.485 8.544 8.602 8.660 8.717 8.774 8.831 8.888 8 8.944 9.000 9.055 9.110 9.165 9.219 9.273 9.327 9.380 9.433 9 9.486 9.539 9.591 9.643 9.695 9.746 9.797 9.848 9.899 9.949 NOTES ON ARITHMETIC AND ALGEBRA 299 332. TABLE OF SINES, COSINES, AND TANGENTS Deg. Sine Cosine Tangent Deg. Sine Cosine Tangent 1 .017 .999 .017 46 .719 .695 1.036 2 .035 .999 .035 47 .731 .682 1.072 3 .052 .999 .052 48 .743 .669 1.111 4 .070 .998 .070 49 .755 .656 1.150 5 .087 .996 .087 50 .766 .643 1.192 6 .105 .995 .105 51 .777 .629 1.235 7 .122 .993 .123 52 .788 .616 1.280 8 .139 .990 .141 53 .799 .602 1.327 9 .156 .988 .158 54 .809 .588 1.376 10 .174 .985 .176 55 .819 .574 1.428 11 .191 .982 . 194 56 .829 .559 1.483 12 .208 .978 .213 57 .839 .545 1 540 13 .225 .974 .231 58 .848 .530 1.600 14 .242 .970 .249 59 .857 .515 1.664 15 .259 .966 .268 60 .866 .500 1.732 16 .276 .961 .287 61 .875 .485 1.804 17 .292 .956 .306 62 .883 .469 1 881 18 .309 .951 .325 63 .891 .454 1 963 19 .326 .946 .344 64 .899 .438 2.050 20 .342 .940 .364 65 .906 .423 2.144 21 .358 .934 .384 66 .914 .407 2.246 22 .375 .927 .404 67 .921 .391 2.356 23 .391 .921 .424 68 .927 .375 2.475 24 .407 .914 .445 69 .934 .358 2 605 25 .423 .906 .466 70 .940 .342 2.747 2d .438 .899 .488 71 .946 .326 2.904 27 .454 .891 .510 72 .951 .309 3.078 28 .469 .883 .532 73 .956 .292 3.271 29 .485 .875 .554 74 .961 .276 3.487 30 .500 .866 .577 75 .966 .259 3.732 31 .515 .857 .601 76 .970 .242 4.011 32 .530 .848 .625 77 .974 .225 4.331 33 .545 .839 .649 78 .978 .208 4.705 34 .559 .829 .675 79 .982 .191 5.145 35 .574 .819 .700 80 .985 .174 5.671 36 .588 .809 .727 81 .988 .156 6.314 37 .602 .799 .754 82 .990 .139 7.115 38 .616 .788 .781 83 .993 .122 8.144 39 .629 .777 .810 84 .995 .105 9.514 40 .643 .766 .839 85 .996 .087 11.430 41 .656 .755 .869 86 .998 .070 14.301 42 .669 .743 .900 87 .999 .052 19.081 43 .682 .731 .933 88 .999 .035 28.636 44 .695 .719 .966 89 .999 .017 57.290 45 .707 .707 1.000 - 300 PLANE GEOMETRY 333. Units of Length English 12 inches (in.) = 1 foot (ft.) 3 feet = 1 yard (yd.) 5J yards = 1 rod (rd.) 320 rods or 5280 ft. = 1 mile (mi.) Metric 10 centimeters (cm.) = 1 decimeter (dm.) 10 decimeters = 1 meter (m.) 1000 meters = 1 kilometer (km.) 1 meter = 39. 37 in. 1 kilometer = . 62 of a mile 1 foot = 30.48 centimeters 1 mile =1.6093 kilometers 334. Units of Surface English 144 square inches (sq. in.) = 1 square foot (sq. ft.) 9 square feet = 1 square yard (sq. yd.) 30M square yards = 1 square rod (sq. rd.) 160 square rods = l acre (A.) 4840 square yards = 1 acre (A.) 640 acres = 1 square mile (sq. mi.) Metric 100 square centimeters = 1 square decimeter 100 square decimeters = 1 square meter OUTLINE SUMMARY PARALLELS AND PERPENDICULARS Tests for parallels : Two straight lines in the same plane are parallel if; 1. The alt. int. angles are equal §61, Th. 2. The corresponding angles are equal §63, Th. 10 3. The int. angles on the same side of the transversal are sup §63, Th. 11 4. They are perpendicular to the same line §63, Th. 12 5. They are parallel to the same line §66, Th. 13 A line is parallel to one side of a triangle if: 6. It passes through the mid-points of the other two sides : §114, Th. 48 7. It divides the other two sides proportionally §20o, Th. 100 and Cor. A line is parallel to the bases of a trapezoid if: 8. It passes through the mid-points of the legs §118, Th. 51 Construction of parallels §64, Prob. 6 Tests for perpendiculars: 1. If a ray starts from a point in a straight hne §18 2. A line perpendicular to one of two parallels §70, Th. 17 3. A line tangent to a circle is perpendicular to . §145, Th. 69 4. If any two circles intersect the line of centers. . , . §151, Th. 73 5. If two equal circles intersect the line of centers and the common chord §152, Th. 74 Cor. 6. An angle inscribed in a semicircle §162, Cor. II Construction of a perpendicular: 1. To a line from a point in the line. §43, Prob. 3; §167, Prob. 9 2. To a line from a point not in the line §44, Prob. 4; §167. Prob. 10 3. To a segment bisecting the segment §45, Prob. 5 CONGRUENCE Tests for congruent triangles : Any two triangles are congruent if: 1. Two sides and the included anglj §3o, Th. 1 2. Two angles and the included side §36, Th. 2 3. Three sides §39, Th. 4 301 302 PLANE GEOMETRY Two right triangles are congruent if: 4. The hypotenuse and an acute angle §81, Th. 22 5. The hypotenuse and a side §82, Th. 23 Test for congruent parallelograms: Two parallelograms are congruent if two sides and the included angle §98, Th. 35 General test for congruent figures : Any two figures are congruent if they can be made to coincide .... §33 Tests for equal angles : Two angles are equal if they are: 1. Sums, differences, equal multiples, or equal parts of equal angles § §30, 47 2. Right angles or straight angles §§22, 29 3. Supplements or complements of equal angles §§24, 25, 29 4. Vertical angles §§28, 29 5. Corresponding angles of congruent triangles §38 6. Base angles of an isosceles triangle §37, Th. 3 7. Alt. int. angles of parallel Hnes §68, Th. 14 8. Corresponding angles of parallel lines. §69, Th. 15 9. The third angles of two triangles that have two angles of one equal respectively §75, Cor. II 10. Angles with their sides parallel right side to right side . §73, Ex. 3 11. Angles with their sides perpendicular §86, Exs. 38, 39 12. Opposite angles of a parallelogram §95, Th. 32 13. Ceatral angles subtended by equal arcs §135, As. 50 14. Central angles subtended by equal chords §136, Th. 61 15. Angles measured by equal arcs §173, II 16. Corresponding angles of similar figures §§210, 256 17. Angles of a regular polygon §§78, 270 Construction of equal angles §40, Prob. 1 ; §41, Prob. 2 Tests for equal segments: Two segments are equal if they are: 1. Sums, differences, equal multiples, or equal parts of equal segments §§30, 47 2. Radii of the same or equal circles §§12, 29 3. Sides of an isosceles triangle §34 4. Corresponding sides of congruent triangles §38 5. Opposite sides of a parallelogram §95, Th. 31 6. Parallel or perpendicular segments between par- allels §104, Ths. 39, 40 OUTLINE SUMMARY 303 Equal segments are formed when: 7. The diagonals of a parallelogram intersect §95, Th. 33 8. A series of parallels cutting equal segments on one transversal intersect a second transversal §111, Th. 45 9. Perpendiculars are drawn from a point in the bisector of an angle to the sides of the angle. . . . §179, Th. 85 10. A point in the perpendicular bisector of a segment is joined to the extremities of the segment §180, Th. 86 11. A radius is perpendicular to a chord §137, Th. 63 12. Perpendiculars are drawn from the center of a. circle to two equal chords §141, Th. 67 13. Two tangents are drawn to a circle from a point without §146, Th. 70 14. Three terms of one proportion are equal respec- tively to three terms of another proportion. .. §201, Th. 92 Division of a segment into equal parts §111, Prob. 7 Tests for equal arcs: Two arcs are equal if: 1. They have equal central angles §135, As. 49 2. They have equal chords §136, Th. 62 3. A radius is perpendicular to' the chord of an arc . . §137, Th. 63 4. They are intercepted by parallel chords §166, Th. 80 5. They are intercepted by a chord and a tangent parallel to it §171, Th. 82 6. They measure equal angles §173, II Tests for equal chords : Two chords are equal if: 1. They have equal central angles §136, Th. 61 2. They have equal arcs §136, Th. 62 3. They are equally distant from the center §140, Th. 66 SIMILARITY Tests for similar triangles : Two triangles are similar if: 1. They are mutually equiangular §210, Th. 102; §257 2. An angle of one equals an angle of the other and the sides including the angle are proportional .... §258, Th. 119 3. Corresponding sides are proportional §259, Th. 120 304 PLANE GEOMETRY Tests for similar polygons : Two polygons are similar if: 1. The angles of one are equal respectively to the angles of the other and the corresponding sides are proportional §256 2. Diagonals from corresponding vertices divide the polygons into triangles that are similar and simi- larly placed §261, Th. 121 3. They are regular polygons of the same number of sides §287, Th. 134 Properties of similar figures: 1. The corresponding angles are equal. §§210, 256 2. Corresponding sides have equal ratios §§210, 256 3. Diagonals from corresponding vertices divide the polygons into triangles that are similar and simi- larly placed §263, Th. 122 4. The ratio of corresponding segments equals the ratio of simiHtude §264, Ths. 123, 125 5. The ratio of the areas equals the square of the ratio of similitude §266, Th. 126; §267, Th. 127 Equal ratios and circles: 1. The ratio of the circumferences of any two circles equals the ratio of the diameters or of the radii . . §307, Th. 140 2. The ratio of the areas of any two circles equals the ratio of the squares of the diameters or of the radii §307, Th. 141 Tests for equal ratios or equal products §211 Two ratios or two products are equal when : 1. Parallels cut two transversals §203, Th. 98; §204, Th. 99 and Cor. 2. Polygons are similar §210 3. Two ratios are equal to a third ratio §199, As. 57 Construction of proportional segments: 1. The division of a segment into parts proportional to any number of given segments §206, Prob. 15 2. The fourth proportional to three given segments §207, Prob. 16 3. The mean proportional to two given segments .§221, Prob. 17 Important cases of equal ratios occur when: 1. Two chords intersect within a circle §213, Th. 103 2. Two secants intersect without a circle. §215, Th. 104 OUTLINE SUMMARY 305 3. A secant and a tangent intersect without a circle. §216, Th. 105 4. A line bisects an angle of a triangle §217, Th. 106 5. A line bisects an exterior angle of a triangle §218, Th. 107 6. A perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse §220, Ths. 108, 109 EQUIVALENCE Tests for equivalence §246, Ths. IIG, 117 Two parallelograms or two triangles are equivalent or a triangle is half of a parallelogram if: 1. They have the same base and the same altitude. 2. The product of the base and altitude of one, etc. Any two polygons are equivalent if they are: Sums, differences, or equal parts of equivalent figures. Construction of equivalent figures: 1. To transform a parallelogram into a rectangle on a given base §248, Prob. 18 2. To transform a parallelogram into a square §248, Prob. 19 3. To transform a polygon into a triangle. . . . §249, Probs. 20, 21 4. To construct a square equal to the sum of two squares §251, Prob. 22 5. To construct a square equal to the difference between two squares §251, Prob. 23 MEASUREMENT Meastirement of angles: 1. Of central angles §157, As. 53 2. Of inscribed angles §161, Th. 77 3. Of an angle formed by a chord and a tangent. . §170, Th. 81 4. Of an angle formed by two chords that intersect. §164, Th. 78 5. Of an angle formed by two secants, two tangents, or a secant and a tangent ..§165, Th. 79; §172, Ths. 83, 84 6. By trigonometric ratios §225 Angle-sums: The sum of 1. Adj. angles on one side of a st. line having a common vertex is 2 rt. -4 §§26, 29 2. Adj. angles about a point is 4 rt. A §§26, 29 3. Int. angles on one side of a transversal when two parallels are cut b}' a third st. line is 2 rt. zi §69, Th. 16 4. The interior angles of a triangle is 2 rt. ^ §74, Th. 18 306 PLANE GEOMETRY The sum of 5. Two angles of a triangle is equal to the opposite exterior angle §76, Th. 19 6. The acute angles of a right triangle is 1 rt. Z . . . . §75, Cor. Ill 7. The interior angles of a polygon of n sides is 2{n—2) rt. A §79, Th. 20 8. The exterior angles of any polygon is 4 rt. A §80, Th. 21 Measurement of polygons: 1. The area of a rectangle is ab §239, As. 63 2. The area of a parallelogram is ah §242, Th. 113 3. The area of a triangle is ^ a& §243, Th. 114 " y2 he sin ^ . . . .^. . . ^. . . §253, Ex. 41 " ^|s{s-a){s- h) {s'-c) . . §253, Ex. 44 4. The area of a trapezoid is 3^o(6+6') §244, Th. 115 5. The area of a regular polygon is % per. X apothem §286, Th. 133 6. For the area of irregular polygons see §245 Measurement of circles and sectors : 1. The circumference of a circle is 2itr §301, Th. 137 2. The area of a circle is tt^-^ §304, Th. 138 3. The area of a sector is -rr-TTr §305, As. 66 3b0 ELEMENTARY FIGURES Properties of triangles: 1. The sum of the angles of a triangle is 2 rt. A §74, Th. 18 2. Thd angle opposite the greater side of a triangle. §129, Th. 55 3. The side opposite the greater angle of a triangle. . §128, Th. 54 4. The medians are concurrent §115, Th. 49 5. The perpendicular bisectors of the sides are con- current §184, Th.87 0. The altitudes are concurrent §184, Th. 88 7. The bisectors of the angles are concurrent §185, Th. 89 Construction of triangles: Two sides and an angle opposite one §55, Ex. 4 Properties of isosceles triangles: 1. Two sides are equal §34 2. The base angles are equal §37, Th. 3 3. Bisector of vertex angle, the altitude, and the median to the base coincide §50, Th. 6; §51, Th. 7; §84, Th. 25 OUTLINE SUMMARY 307 Tests for isosceles triangles : A triangle is isosceles if : 1. Two sides are equal §§34, 83 2. Two angles are equal §83, Th. 24 Properties of right triangles: 1. The acute angles are cornplcnienLary §75, Cor. Ill 2. The median from the vertex of the right angle is one-half the hypotenuse §116, Th. 50 3. If a and b are the legs and c is the hypotenuse, a2+62=c2 §222, Th. 110; §251, Th. 118 Properties of parallelograms: 1. The opposite sides are parallel §95 2. The opposite sides are equal §95, Th. 31 3. The opposite angles are equal §95, Th. 32 4. The diagonal bisects the parallelogram §95, Th. 30 5. The diagonals bisect each other §96, Th. 33 Properties of special parallelograms §§105-110 Tests for parallelograms: A quadrilateral is a parallelogram if: 1. Each side is parallel to its opposite §99 2. One side is equal and parallel to its opposite §100, Th. 36 3. Each side is equal to its opposite '. . §101, Th. 37 4. The diagonals bisect each other §102, Th. 38 Properties of regular polygons: 1. The sides and angles are equal §§78, 270 2. A circle can be circumscribed about the polygon . . §282, Th. 130 3. A circle can be inscribed in the polygon §283, Th. 131 4. The radius bisects the angle §282, Th. 130 Cor. 2»7 —4 5. Each angle is rt. A §285, Th. 132 n Tests for regular polygons : A polygon is regular if : 1. The sides and angles are equal §§78, 270 2. A circle is divided into equal arcs and a. The points of division are joined §271, Th. 128 b. Tangents are drawn to the points of division §271, Th. 129 Construction of regular polygons §§273-280, 290 308 PLANE GEOMETRY INEQUALITIES Tests for unequal segments : 1. The sum of two sides of a triangle, etc §126, Ass. 37, 38 2. If one angle of a triangle is greater than another, etc §128, Th. 54 3. The perpendicular is the shortest distance, etc. . . . §130, Th. 56 4. If from a point in a perpendicular to a line oblique segments are drawn cutting off unequal distances, etc §130, Th. 57 5. If from a point in a perpendicular to a line two unequal oblique segments are drawn, etc §130, Th. 58 6. If two triangles have two sides of one equal to two sides of the other but the included angles, etc. . . §131, Th. 59 7. If from a point within a triangle segments are drawn to the extremities of one side §127, Th. .53 Tests for unequal angles: 1. The exterior angle of a triangle, etc §125; §58, Th. 8 2. If one side of a triangle is greater than another, etc §129, Th. 55 3. If two triangles have two sides of one equal to two sides of the other but the third side of one, etc. . . §132, Th. 60 INDEX [References are to page numbers) Abbreviations 19 Acute angle 9 Addition or composition. . . . 167 Addition: of angles 7 of polygons 209 Adjacent angles 7 Algebraic analysis 222 Algebraic equations indicating constructions 197; 227, Ex. 5 Algebraic notation in proof s . 51 Ahmcs 283 Alternate exterior angles .... 49 Alternate interior angles. ... 49 Alternation: extreme 166 mean 166 Altitude: length of, in equi- lateral triangle 190 of parallelogram 85 of trapezoid 84 of triangle 70 Altitudes of triangle, concur- rent 150 ratio of 244 Analysis, to make an 25 Angle 6 acute 9 arms of 6 bisector of 8, 31 central, in circle 108, 124 central, of regular poly- gon 260, 261 complement of 12 degree of 11 designation of 6 division of 31 exterior, of triangle 48, 61 21 309 formed by rotation.. 6 included 22 inscribed in an arc 128 inscribed in a circle 126 measurement of 11, 137 obtuse 9 of elevation 196 of sixty degrees 60 of regular polygon 262 re-entrant 63 right 9 right and left sides of 58 sides of 6 size of 6 straight 9, 16 supplement of 12 trisection of 32 vertex of 6 vertex, of isosceles triangle. 20 Angles: addition of 7 adjacent 7 alternate exterior 49 alternate interior 49 complementary 12 congruent 7 consecutive of parallelo- gram 79 construction of equal 7, 30 corresponding, of congru- ent figures 20 corresponding, of lines cut by a transversal 49 equal, definition of (See Equal angles) 11 exterior, of lines cut by a transversal 49 310 INDEX Angles {continued) : interior non-adjacent 48 interior, of lines cut by a transversal 49 made by parallels and transversals 54 opposite 15 opposite, of parallelo- gram 79» 80 subtraction of 8 sum of, in polygon ....... 64 sum of, in triangle 59 supplementary 12 symmetric 75 vertical 15 Angles and parallels 54, 57 Angle-sums 16, 305 Antecedent 163 Apothem 260 Apothems, ratio of 263 Approximate constructions. . 282 Approximate measure : of length 161, 162 of surface of rectangle .212, 214 of heights and distances. . . 196 Approximate value of 7r..272, 283 Arab 98 Arc 5 degree of 123 intercepted, of central angle 108 intercepted, of inscribed angle . 126 major 108 measurement of. ..108, 123, 137 minor 108 Archimedes 220, 283 Architecture, exercises from gable {See Carpentry ; Church windows; Gothic arch ; Mouldings ; Rafters ; Roof trusses; Steel beams) Arcs: congruent 108 equal {See Equal arcs) .... 108 Area 208 assumptions concerning 211, 275, 290 maximum 284, 290 Area: of circle 275, 277 of irregular polygons 219 of kite 229, Ex.16 of parallelogram 216 of rectangle 211 of regular polygon 262 of rhombus 228, Ex. 5 of sector of circle 275 of segment of circle 276 of trapezoid 218 of triangle 217, 228 Areas, ratio of 237, Ex. 34; 246, 247, 263, 276 Assumption 35 Assumptions concerning; angles 16, 17 angle-sums i6 area 211, 275, 290 circles 16, 109, 118, 124, 125, 274, 275, 290, 291 congruence 20 equal angles 16 equivalence 210 inequality 48, 99, 100 location of lines, rays, and segments 15 location of points 15 maximum area 290 minimum perimeter 291 parallels 53 perimeters 274, 291 ratios 165 sectors 275 segments 16, 17 straight angles 16 INDEX 311 Axial symmetry 76 relation to central sym- metry 78 Axis of symmetry 76 of circles 1 18, 1 19 Base of an isosceles triangle . 20 Bases : of parallelogram 79 of trapezoid 84 Bisection: of an angle 8, 31 of a polygon 209 of a segment 4, 35 Bisector of an angle: as locus 146 construction of 8, 31 segments made by . . . . 185, 186 Bisectors: of angles of a triangle, concurrent ... . 150 perpendicular (See Perpen- dicular bisector) Carpentry, exercises from 31, 129, 157, I75» 265, 266 Center: of a circle 5 of gravity 155 of regular polygon 260 of similitude 249 of symmetry 77 of symmetry of parallelo- gram 81 Centers, line of {See Line of centers) 117 Centers of circles, loci of . . . 158 Central angle : of a circle {See equal angles; Unequal angles) 108 measure of 124 Central angle of a regular polygon 260 measure of 261 Central symmetry 77 relation to axial symmetry 78 test for 77 Chinese 189, 283 Chord 5 common {See Common chord) 117 fundamental theorem of . . 1 1 1 Chord and tangent, measure of angle made by 135 Chords: equal, tests for {See Equal chords) 303 intersecting {See Intersect- ing chords) parallel 132 Church windows 140, 158, 201, 280 Circle 5 arc of 5 area of 275, 277 central angle of 108 chord of 5 circumference of 269, 274 circumscribed 120, 151 circumscribed, of regular polygon 259 construction of. . 151, 154, 156 definite location of 113 diameter of. 5 escribed 151 inscribed. 120, 151 inscribed angle of 126 inscribed, of regular poly- gon 260 measurement of 277, 306 radius of 5 sector of 157, 275 segment of 276 tangent to 114, 134 Circles: assumptions concern- ing 16, 109, 118, 124, 125, 274» 275, 290, 291 concentric 117 congruent 108 construction of. . .J 51, 154, 156 inequalities in 125 312 inde: Circles {continued) : intersecting Ii8 loci of centers of 153 tangent 117, 119 tangents to 137 Circles and equal ratios 276 Circles and symmetry 109, 118, 119 Circular segments 276 Circumference 269, 274 formula for 274 ratio of, to diameter 276 Circumferences, ratio of 276 Circumscribed circle 120 construction of 151 of regular polygons 259 Circumscribed polygon 120 Coincident rays 5 Collinear rays 5 Commensurable segments. . . 164 Common chord 117 as bisector '. . . 119 bisected 118 Compasses: proportional 191 use of 6, 32 Complement 12 Complementary angles 12 Composition or addition. ... 167 Compound curves 157 Computations: of areas 214, 228, 277 by trigonometric ratios 194, 196, 202 measurement of regular polygons 267 of perimeters of regular inscribed polygons. .270, 272 Concave polygon 63 Concentric circles 117 Conclusion 36 Concrete representation: of points I of straight lines I Coiicurrcnt lines 91 exercises involving 155 special cases of . . . 115, 148, 150 Congruence 20, 301 Congruent angles 7 arcs 108 circles 6, 108 figures 20 parallelograms 81 segments 4 triangles 301 Consecutive angles 79 Consecutive sides 79 Consequent 163 Construction lines 57 Construction: of equal angles 7, 30 of bisector of an angle. . . .8, 31 of circles 151, 154, 156 of circumscribed circle ... . 151 of decagon, regular 256 of equal segments 88 of equivalent figures 210, 222-224, 227 of escribed circle 151 of extreme and mean ratio. 256 of fourth proportional 1 73 of hexagon, regular 254 of inscribed circle 151 of mean proportional 187 of octagon, regular 253 of parallels 53 of pentadecagon, regular. . . 258 of pentagon, regular 258 of perpendiculars 9-10, 33-35, 133 of proportional segments . . 1 72 of regular polygons. . . .252-258 of similar polygons 243 of square 253 of tangents 115, 134, 137 of triangles 2 1 , Ex. 4 ; 24, Ex. ; 47, Ex. 4 INDEX 313 Constructions: by algebraic analysis 222 indicated by equations 197; 227, Ex. 5 Contact, point of 114 Converse theorems 56 Convex polygon 63 Corresponding angles: of con- gruent figures 20 of lines cut by a transversal 49 Corresponding sides: of con- gruent figures 20 of similar figures 176, 240 of similar triangles 178 Corollary 36 Cosine of an acute angle. ... 193 Cross-sections of columns. . . 214 Cut-glass designs 141, 268 Decagon {See Regular deca- gon) 63 Definite location : of circles . . 112 of lines 2, 15 of points 2, 15 of rays 5, 15 of segments 4, 15 Degree : of angle 11 of arc 123 Design, theory and practice of 71, 79, 157, 231, 256 Determination of points 152 Diagonals: of parallelograms 79, 80 of polygons 63 of quadrilaterals 80, 81 of square, formula for 190 Diagrams for review 43 Diameter of circle 5 tests for 1 12, 1 16 Diameters, ratio of 276 Difference : between two angles 8 between two polygons .... 209 Direct proof 51 Distance : between two points 4 from a point to a line 104 Division or subtraction 167 Division of angles 31 Division of segments: exter- nal and internal 182 harmonic 186 in extreme and mean ratio 255, 256 in equal parts 88 in parts proportional to given segments 172 Draftsman's methods, me- chanical drawing 9, 10, 32, 53, 259, 282 Duodecagon {See Regular duodecagon) 63 Egypt, Egyptians I, 69, 189, 191, 283 Elements, Euclid's 45 Elevation, angle of 196 Engineering, problems from {See Architecture ; Sur- veying ; Railroading) 279, 291 Equal angles 1 1 assumptions concerning. . . 16 tests for 16, 37, 302 Equal arcs 108 have equal central angles. . 109 have equal chords no measure equal angles 137 measured by equal angles.. 137 tests for 303 Equal central angles of a circle: have equal arcs. . 109 have equal chords no Equal chords: equally distant from center 113 have equal arcs no have equal central angles. . no 314 INDEX Equal products: exercises involving 179, 197 important cases of 182 tests for 177 Equal ratios: applications of. 191 exercises involving 174, 179, 197 important special cases of 182 series of 245 tests for 177, 304 Equal segments 4 • tests for , 302 Equations 296 Equilateral triangle 20 altitude of, formula for... . 190 angles equal 25 side of, formula for 190 value of each angle 60 Equivalent polygons 209 assumptions concerning. . . 210 construction of 210, 220, 236, Ex. 32; 237, Ex. 33; 251, Ex. 22; 305 exercises involving .... 224, 233 tests for 210, 220, 305 Escribed circle 151 Euclid 45, 137,258 Exact measure 161 Exterior angle of triangle 48 Exterior angles of lines cut by transversal 49 External division 182 Extreme alternation 166 Extreme and mean ratio. . . . 256 Extremes 165 Fixed line, ray, or segment. . 58 Floor designs 69, 79, 97, 157, 204, 230, 235, 266, 281 Formula: for altitude of equilateral triangle 190 for angle-sums 305 for diagonal of square 190 for measurement of circles and sectors 277, 306 for measurement of poly- gons 228, 306 for side of equilateral tri- angle 190 for side of square 190 Fourth proportional 173 construction of 173 Fractions 292 Fundamental assumption: of measurement of poly- gons 211 regarding parallels 53 Fundamental characteristic: of parallelograms 80 of ratios 165 Fundamental relation be- tween arcs and angles. . . 124 Fundamental test: for in- equality 100 for parallelograms. ...'.,. . 82 for parallels 50 f ov special quadrilaterals . . 84 Fundamental theorems of proportion 165 Gable 206 Gauss 258 General assumptions 16 Generation or formation of angles 6 Geometric forms, occurrence of 69,71,79, 106, 158, 231, 236, 254, 265, 266, 281, 291 Geometrical problem {See Construction) 36 Golden section 256 Gothic arch 206, 280 Gravitv, center of I55 INDEX 315 Greek geometry 21, 2>2, 45, 59, 189, 191, 233, 235. 256, 258, 266, 274, 283 Harmonic division 186 Heptagon 63 Hero of Alexandria 233 Hexagon {See Regular hexa- gon) 63 Hindus 189,283 Historical notes: Ahmes 283 Arab 98 Archimedes 220, 283 Chinese 189, 283 Egyptians. . i, 69, 189, 191, 283 Euclid 45. 137,258 Gauss 258 Greeks 32, 59» 274 Hero of Alexandria 233 Hindus 189,283 Hippocrates 279 Moderndiscoveries258,274,283 Pappus 235 Pi(7r) 164, 274» 275. 283 Pythagoras 59, 189 Pythagoreans 189, 256, 266 Thales 27, 191 Trisection of angles 32 Hippocrates 279 Hypotenuse 66 Hypothesis 36 Incenter 155 Included angle 22 Included side 23 Incommensurable segments. . 164 Indirect proof 51 Inequalities {See Unequal angles; Unequal sides); assumptions concerning 48,99 fundamental test for 100 in circles 125 Inscribed angle in an arc. .. . 128 Inscribed angle in a circle. . . 1 26 measure of 126 Inscribed angle in a semicircle 128 Inscribed circle 1 20 construction of 1 51 in regular polygons 260 Inscribed polygon 120 Inscribed regular polygons 253-258 Integraph 274 Intercepted arc 108, 126 Interior angles: of a polygon 64 of a regular polygon 262 of lines cut by a transversal 49 Internal division 182 Intersecting circles 118, 119 Intersecting chords: measure of angle 130 of product of segments of. . 1 82 Intersecting loci, use of: in determination of points . 152 in construction of circles. . 1 54 Intersecting secant and tan- gent : measure of angles of 136 product of segments of . . . . 184 Intersecting secants: measure of angle of 131 product of segments of . . . . 183 Intersecting tangents: are equal 1 1 5 measure of angle of 13^ Inverse proportion 204 Inversion 166 Irrational numbers 164, 274 Irregular polygons, area of.. . 219 Isosceles trapezoid 84 legs of 84 properties of 86 Isosceles triangle 20 base angles of 24 properties of 3^6 tests for ..68,307 31G INDEX Kites. 84 area of 229, Ex. 16 properties of 86 Legs : of an isosceles trapezoid 84 of a right triangle 66 Length : of circle 269 of segment 4, 161 units of 300 Leveling device 41 Limiting values of perimeter of inscribed and circum- scribed regular polygons 274 Limits, exercises involving 40, 57, 67, 103, 134, 136 Line: parallel to base of triangle §9, 90, 170 straight {See Straight line) Line of centers 117 as axis of symmetry 118 as bisector 118 Lines : concrete representa- tion of I concurrent {See Concur- rent lines) 91 construction 57 definite location of 15 parallel {See parallel lines) . 50 perpendicular {See perpen- dicular lines) 9 symmetric 75 Location, definite {See Defi- nite location) Loci 143 complete proofs for 145 finding of 143 intersecting 152, 154 miscellaneous exercises on . 1 59 of centers of circles 1 53 of points 146 of vertices of triangles 159, Ex. 4 Major and minor arc 108 Maximum and minimum. .. . 284 Mean alternation 166 Mean proportional 179 construction of 187 Mean ratio, extreme and. ... 256 Means 165 Measure (measurement): ap- proximate 161 exact 161 of angles ...11, 137, 305 of arcs .■ 123, 137 of circles {See Circles) 269, 277, 306 of polygons {See Polygons) 211-220, 306 of segments 161 of surfaces 208 practical 162, 214 Measure number : of angles . . 11 of a segment 161 of a surface 208 Mechanical drawing 9, 10, 32, 53, 259, 282 Median: of right triangle. ... 92 of a triangle 42 of a quadrilateral 85 Medians of a triangle con- current 91 Mid-point of .segment, deter- mination of 4, 35 Minimum 284 Minor arc 108 Minutes n, 123 Modern discoveries in geom- etry 258, 274,283 Moldings I57 Nature of theorems and proofs 35 Navigation, exercise from... . 71 Numbers: irrational 164, 274 ratio of 163 INDEX 317 Obtuse angle 9 Octagon {See Regular octa- gon) 63 Opposite angles 15 Opposite angles of a parallelo- gram 79. 80 Opposite sides of a parallelo- gram 79. 80 Origin of ray 5 Orthocenter i55 Pappus 235 Parallel chord and tangent ... 136 Parallel chords 132 Parallel lines 50 construction of 53 fundamental assumption regarding 53 tests for .* 53. 301 Parallel rulers 95 Parallel to base of a triangle 89, 90, 170 Parallels and transversals : angles formed by 54. 55 equal segments formed by 88, 98 proportional segments formed by 168 Parallelogram 79. 85 altitude of 85 area of 216 bases of 79. 85 consecutive angles and sides of 79 diagonals of 63, 79, 80, 81 opposite angles and sides of 79 Parallelograms : congruence of 81 properties of 307 test for 82, 83, 307 Pencil of rays 5 Pentadecagon (See Regular pentadecagon) 63 Pentagon (See Regular penta- gon) 63 Pentagram star 266 Perigon 9 Perimeter 63 minimum 285, 289, 291 ratio of, to diameter 263 Perimeters: computation for 270, 271 lengths of 272 ratio of 245, 263 Perpendicular bisector : as locus 147 construction of 10. 35 Perpendicular bisectors con- current 149 Perpendicular lines 9 construction of .9, 10,33-35, ^33 tests for 301 Physics, exercises from. . ig6, 138 Pi (tt) 273 approximate constructions for 282 approximate value of. .273, 283 construction of 274 history of 283 is irrational 164, 274 Point: determination of mid- point 4, 35 of contact of line and circle 114 of contact of tangent circles 1 20 of tangency 114 variable 143 Points: concrete representa- tion of I definite location of 2, 15 determination of 1 52 locus of 146 symmetric 75, 118 Polygon 63 area of 208 318 INDEX Polygon (continued) : bisection of 209 circumscribed 1 20 concave 63 convex 63 diagonal of 63 inscribed 1 20 perimeter of 63 regular (See Regular poly- gon) 64 sides of 63 star (See Star polyrons) 141, 268 surn of angles of 64 trisection of 209 vertices of 63 Polygons: addition of 209 congruence of (See Con- gruent figures; Con- gruent parallelograms; Congruent triangles) .... 20 difference between 209 equivalence of {See Equiva- lence) 220, 305 measurement of 211-219, 228, 306 names of 63 regular (See Regular poly- gon) , 252 similar (See Similar figures) 240 star (See Star polygons) 141, 268 subtraction of 209 sum of 209 transformation of 210, 221 Pons asinorum 46 Practical measurements. .162, 214 Problem, geometrical (See Construction) 36 Products, test for equal (See Equal products) 178 Projection 239 Proof 36 by superposition 36 direct synthetic 36, 51 indirect 51 Properties: of isosceles trape- zoids 86 of isosceles triangles 306 of kites 86 of parallelograms 307 of rectangles 86 of regular polygons 307 of rhombuses 87 of right triangles 307 of squares 87 of similar polygons 304 of trapezoids 93 of triangles • 306 Proportion 165 by addition 167 by alternation 166 by composition 167 by division 167 by inversion 166 by subtraction 167 fundamental theorems of . . 165 inverse 204 reciprocal 204 Proportional compasses (See Compasses) Proportional: fourth 173 mean 179, 187 Proportional segments (See Equal ratios) 178 Construction of 172, 304 special cases of 182, 304 Proportionally, divided.'. ... 178 Proportions, transformations of 166 Protractor 124 Pythagoras 59, 189 Pythagorean theorem 1 89 exercises involving 189, 199, 238 INDEX 3m Pythagorean theorem (cont'd) : proofs for i88; 206, Ex. 22; 226; 235, Ex. 22; 238 related theorems 239, Exs. II, 12; 251, Ex. 21 Pythagoreans 189, 256, 266 Quadrilateral 63 Quadrilaterals, special 84 Radially placed 249 Radii, ratio of 263, 276 Radius: of a circle 5 of a regular polygon. . . 260, 261 Rafter designs, decorated (See Truss).: 157. 158 Railroading, exercises from 142, 270, 280 Ratio: extreme and mean 256 of altitudes 244 of apothems 263 of areas 237, Ex. 34; 246, 247, 263, 276 of circumference to dia- meter 276 of circumferences 276 of corresponding sides. 176, 240 of diameters 276 of perimeter to diameter . . 263 of perimeters 245, 263 of radii 263 of segments made by parallels 168, 170 of similitude 240 of two numbers 163 Ratios: assumptions concern- ing 165 equal (See Equal ratios) fundamental characteristic of 165 trigonometric 192 Ratios and circles 276, 304 Ray 5 fixed 58 origin of 5 Rays: coincident 5 collinear 5 definite location of 15 pencil of 5 Reciprocally proportional 204 Rectangle 85 area of 211 exercises involving area of 214 properties of 86 Rediictio ad ahsurdum 51 Re-entrant angle 63. Regular decagon : construc- tion of 256 exercises involving 267 Regular duodecagon : con- struction of 254 exercise involving 267 Regular hexagon, construc- tion of 254 exercises involving. 69, 265, 267 occurrence of 69, 265, Regular octagon : construc- tion of 253 exercises involving 265, 266, 267 occurrence of 265, 266 Regular pentadecagon, con- struction of 258 Regular pentagon: construc- tion of 258' exercises concerning . . . 266, 267 occurrence of 266 Regular polygon 64 angle of 262 apothem of 260 area of 262 center of 260 central angle of 260 radius of 260 Regular polygons: construc- tion of 253-258, 265 occurrence of 254 320 INDEX Regular polygons {continued) : perimeters of 270-272 properties of. 259, 260, 265, 307 measurement of 262, 267 similar 262 tests for 307 Representation of points and straight lines I Review diagrams 43 Rhombus 85 area of 228, Ex. 5 properties of 87 Rigid figures 41 Right angles {See Perpen- diculars) 9 Right triangle 66 hypotenuse of 66 legs of 66 properties of 307 sixty-degree 250, Ex. 9 Roof trusses, exercises based on 42, 127, 203, 204, 205, 234, 236 parts of 42, Ex. 15 rigidity of 41 Roots, rules for 293-295 table of square 298 Rosettes 142 Rulers, parallel 95 Secant 131 Secant and tangent 136, 184 Secants, intersecting {See Intersecting secants) Section, Golden 256 Sector of circles. 157 area of 275, 306 Segment joining center of circles as bisector 118 bisected {See Line of centers) 119 Segment of circles 276 area of 276 Segment, straight-line 3 bisection of 4. 35 definite location of . ; 4, 15 division of {See Division of segments) fixed 58 length of 4, 161 measure of 161 mid-point of 4» 35 Segments: commensurable. . . 164 congruent 4 equal {See Equal segments) 4 incommensurable 164 proportional 178, 304 ratio of 164 Semicircle 108 measure of angle in 128 Series of equal ratios 245 Sewers 279, 291 Side included 23 Side: of equilateral triangle. . 190 of square 190 Sides: consecutive 79 corresponding {See Corre- sponding sides) of an angle 6 of a parallelogram 79 of a polygon 63 opposite ; 79 right and left, of an angle . 58 Similar figures, or polygons 176, 240 construction of 243 corresponding angles of 176,240 corresponding sides of 176, 140 properties of 244-247, 248, 304 tests for . 240-243, 248, 303, 304 Similar regular polygons . 262, 264 Similar triangles 176 corresponding sides of 178 tests for 240-244, 248, 303 Similitude, center of 249 ratio of 240 INDEX 321 Sine of an angle 193 Sixty-degree right triangle 250, Ex. 9 Sixty-degrees 60 Size of an angle 6 Special quadrilaterals 84 exercises concerning 96 Square 85 construction of inscribed. . 253 diagonal of, formula for. . . 190 side of, formula for 190 properties of 87 Square roots {See Roots) Star pentagram 266 Star polygons: formation of.. 268 occurrence of 141, 265, 266 Steel square, carpenter's 31, 129, 175 Straight angle 9 Straightedge 32 Straight line 1,2 Straight-line segment {See Segment, straight line) Subtraction : of angles 8 of polygons 209 Subtraction or division 167 Sum of angles 7 of a polygon 64 of a triangle 59 exterior, of a polygon 65 interior of parallel lines 55 Sum of polygons 209 Sums, angle 16, 305 Superposition 36 Supplement 12 Supplementary adjacetit angles 13 Supplementary angles 12 Surface: measure of 208 units of 300 Surveying, exercises from 27,46,72, 157, 170, 189,191, 196, 202 Symbols and abbreviations. . 19 Symmetric figures 75 Symmetric points 75, 118 occurrence of 79 Symmetry : axial 76 central 77 relation between axial and and central 78 Symmetry and circles 109, 118, 119 Symmetry and parallelograms 8 1 Synthetic form 36 Tables: of square roots 298 trigonometric 299 Tangency, point of 114 Tangent and chord parallel 136 measure of angle of.» 135 Tangent and secant {See Intersecting secant and tangent) Tangent circles 117 tests for 119 Tangent of an angle 193 Tangent to a circle 114 construction of ... 1 15, 134, 137 tests for 114 Tangents to two circles 138 Tests: for congruence 301, 302 for diameters 112, 116 for equal angles 302 for equal arcs 303 for equal chords 303 for equal products. . . .178, 304 for equal ratios 177, 304 for equivalence 305 for inequality 100 for isosceles triangles 307 for parallels 301 for parallelograms 307 for pehpendiculars 301 for regular polygons ...... 307 for similar polygons 304 322 INDEX Tests (continued) : for similar triangles 303 for special quadrilaterals . . 85 for tangent circles 119 for tangents 114 for unequal angles 308 for unequal segments 308 Thales 27, 191 Theorems 35 converse of 56 proof of 36 Pappus' 235 Pythagorean (See Pythago- rean theorem) 1 89 Tiles (See Floor designs) 230 Transformation of polygons 210, 221 Transformation of propor- tions 166 Transversals and parallels (See Parallels) Trapezoid 84 altitude of 85 area of 218 bases of 84 properties of 93 Trapezoid, isosceles (See Isosceles trapezoid) Triangle 20 altitude of 70 area of 217, 228 center of gravity of 155 centroid of 155 circumcenter of 1 55 construction of 21, Ex. 4; 24; 47, Ex. 4 exterior angle of 48 incenter of 155 median of 42 orthocenter of 155 properties of 306 sum of angles of 59 Triangle: equilateral (See Equilateral triangle) isosceles (See Isosceles tri- angle) right (See Right triangle) Triangles: congruent, tests for 301 equivalent, tests for 305 similar, tests for 303 Trigonometric ratios 192, 202 Trisection of angles 32 Unequal angles, tests for. . . . 308 Unequal segments, tests for. . 308 Units : of length 300 of measure of angles 1 1 of measure of arcs 108, 123 of surface. . 300 Variable point 143 Vertex angle of isosceles tri- angle 20 Vertex of an angle 6 of isosceles triangle 20 Vertical angles 15 of polygons 63 Vertices, loci of 159, Ex. 4 Width of board , 89 Windo# designs 140, 158, 201, 280 YB 35952