UC-NRLF ^B SE7 TS5 PLANE GEOMETRY WITH PROBLEMS AND APPLICATIONS BY H. E. SLAUGHT, Ph.D. ASSOCIATE PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF CHICAGO AND N. J. LENNES, Ph.D. INSTRUCTOR IN MATHEMATICS IN THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY OF THE UNIVERSITY OF Boston ALLYN AND BACON 1910 COPYRIGHT, 1910. BY H. E. SLAUGHT AND N. J. LENNES. 7 )Ci t/V- -7- Noriooofi 53reg8 J. 8, Cashing Co. — Berwick & Smith Co. • Norwood, Mass., U.S.A. PREFACE. In writing this book the authors have been guided by two main purposes : (a) That pupils may gain by gradual and natural processes the power and the habit of deductive reasoning. (b) That pupils may learn to know the essential facts of elementary geometry as properties of the space in which they live, and not merely as statements in a book. The important features by which the Plane Geometry seeks to accomplish these purposes are : 1. The simpUJlcation of the first Jive chapters by the exclusion of many theorems found in current books. These five chapters correspond to the usual five books, and the most important omissions are the formal treatment of the theory of limits, the incommensurable cases, maxima and minima, and numer- ous other theorems, together with the deduction of complicated algebraic formulae, such as the area of a triangle and the radii of the inscribed, escribed, and circumscribed circles, in terms of the three sides. Chapter VI contains a graphic representation of certain im- portant theorems and an informal presentation of incommen- surable cases and limits. The treatment of limits is based upon the graph, since the visual or graphic method appeals more directly to the intuition than the usual abstract processes. Chapter VII is devoted to advanced work and to a review of the preceding chapters. iii IV PREFACE. 2. Tfie subject has been enriched by including many applica- tions of special interest to pupils. Here an effort has been made to include only such concrete problems as come fairly within the observation and comprehension of the average pupil. This led to the omission, for example, of problems relating to machinery and technical industries, which might appeal to an exceptional boy, but which are entirely inappropriate for the average student. On the other hand, free use is made of certain sources of problems which may be easily compre- hended without extended explanation and which involve varied and simple combinations of geometric forms. Such problems pertain to decoration, ornamental designs, and architectural forms. They are found in tile patterns, parquet floors, lino- leums, wall papers, steel ceilings, grill work, ornamental windows, etc., and they furnish a large variety of simple exercises both for geometric construction and proofs and for algebraic computation. They are not of the puzzle type, but require a thorough acquaintance with geometric facts and develop the power to use mathematics. These problems form an entirely new type of exercises, and while they require more space in the text-book than the more difficult "originals" stated in the usual abstract terms, they excel the latter in interest for the pupil and in helping to train his mathematical common sense. Many of these exer- cises are simple enough to be solved at sight, and such solution should be encouraged whenever possible. All the designs are taken from photographs or from actual commercial patterns now in use. By thus showing that the abstract theorems of geometry find concrete expression in a multitude of familiar objects, it is sought to make the subject a permanent part of the pupil's mental equipment. 3. Persistent effort is made to vitalize the content of the defini- tions and theorems. It is well known that pupils often study and recite definitions and theorems without really comprehend- PREFACE. V ing their meaning. It is sought to check this tendency by- giving definitions only when they are to be used, and by imme- diately verifying both definitions and theorems in concrete cases. The figure on page 4 is the basis for a large number of questions of this type. For example, see § 25, Ex. 3 ; § 30, Ex. 1 ; § 34, Exs. 1, 2 ; § 36, Ex. 3 ; §§ 322, 324. In this connection special attention is called to the emphasis placed upon those theorems which are of fundamental impor- tance both in the logical chain and in their immediate use in effecting Constructions and indirect measurements otherwise difficult or impossible. For example, see the theorems on con- gruence of triangles, §§ 31-43, the constructions of §§ 44-58, and the theorems on proportional segments, §§ 243-254. Com- pare especially § 34, Ex. 5, § 244, Ex. 2, and § 254, Exs. 4, 5. The summaries at the close of the chapters, which are to be made by the pupil himself, will vitalize the theorems as no made-to-order summaries can possibly do. 4. TJie student is made to ajyproach the formal logic of geometry by natural and gradual processes. He is expected to gi^ow into this new, and to him unusual, way of thinking. The treat- ment is at the start informal, leading through the congruence theorems directly to concrete applications and geometric con- structions. The formal development then follows gradually and is characterized by a judicious guidance of the student, by questions, outlines, and other devices, into an attitude of mental independence and an appreciation of clear reasoning. This informality of treatment, most frequent in the earlier parts, is used throughout wherever occasion seems to justify it. See §§ 318, 319. An effort has been made to vary the methods of attack and to avoid monotony. Some theo- rems are proved in full; some are outlined; in some, hints and suggestions are given. Any uniform method would make it impossible to leave that to the pupil which he can do for himself, and at the same time to give full assistance where VI PREFACE. that is needed. In no case are questions asked whose answers are implied by the form of the questions. The arrangement of the text is adapted to three grades of courses : (a) A minimum course, consisting of Chapters I to VI, without the problems and applications at the end of each chapter. This would provide about as much material, theo- rems, constructions, and originals as is found in the briefest books now in use. (6) A medium course, consisting of Chapters I to VI, includ- ing a reasonable number, say one half or two thirds, of the applications at the end of each chapter. This would fully cover the college entrance requirements. (c) An extended course, including Chapter VII, which con- tains a complete review, together with many additional theorems and a large number of further applications. This would provide ample work for the strongest high schools, and for normal schools in which more mature students are found or piore time can be given to the subject. Chapter VII gives a complete treatment of the incommen- surable cases, though not based on the formal theory of limits. It is believed that for high school pupils the notion of a limit is best studied as a process of approximation, and that the best preparation for the later understanding of the theory is by a preliminary study of what is meant by " approaches," such as is given in Chapters III and IV. Acknowledgment is due to Miss Mabel Sykes, of Chicago, for the use of a large number of drawings and designs from her extensive collection ; also to numerous commercial and manufacturing houses, both in this country and in Europe, through whose courtesy many of the patterns were obtained. H. E. SLAUGHT. N. J. LENNES. Chicago and Boston, January, 1910. CONTENTS. Chapter I. Rectilinear Figures. PAGE Introduction 1 Notation for Points and Lines 5 Angles and their Notation 7 Triangles and their Notation . 10 Congruence of Geometric Figures 12 Tests for Congruence of Triangles .14 Construction of Geometric Figures . . . . . . .20 Theorems and Demonstrations 26 Inequalities of Parts of Triangles 32 Theorems on Parallel Lines . . ... . . .35 Applications of Theorems on Parallels . . . . . . 40 Other Theorems on Triangles . . 43 Determinations of Loci 50 Theorems on Quadrilaterals 54 Polygons . 66 Symmetry 68 Methods of Attack 72 Summary of Chapter I . . .75 Problems and Applications 76 Chapter II. Straight Lines and Circles. Theorems on the Circle 86 Measurement of Angles 88 Problems and Applications 92 Angles formed by Tangents, Chords, and Secants .... 95 Summary of Chapter II 106 Problems and Applications . . 107 Chapter III. Measurement of Straight Line-segments. Approximate Measurement 112 Ratios of Line-segments 114 Theorems on Proportional Segments 116 vii Vlll CONTENTS, PAGE Similar Polygons 124 Computations by Means of Sines of Angles 136 Problems of Construction 140 Summary of Chapter III . . . 145 Problems and Applications 146 Chapter IV. Areas of Polygons. Areas of Rectangles . . . ' 154 Areas of Polygons . * 156 Problems of Construction 166 Summary of Chapter IV 169 Problems and Applications . . . ... . . .170 Chapter V. Regular Polygons and Circles. Regular Polygons 176 Problems and Applications 183 Measurement of the Circle .188 Summary of Chapter V 194 Problems and Applications 195 Chapter VI. Variable Geometric Magnitudes. Graphic Representation 207 Dependence of Variables . . . 215 Limit of a Variable . ^ * . 217 Chapter VII. Review and Further Applications. On Definitions and Proofs 220 Loci Considerations .......... 226 Problems and Applications 226 Further Data concerning Triangles . 230 Problems and Applications 233 The Incommensurable Cases 238 Problems and Applications 245 Further Applications of Proportion 250 Further Properties of Triangles 259 Problems and Applications . 264 Maxima and Minima ........... 268 Index . . . . . . 277 pla:^e geometry. CHAPTER I. RECTILINEAR FIGURES. INTRODUCTION. 1. Elementary geometry is a science which deals with the space in which we live. It begins with the consideration of certain elements of this space which are called points, lines, planes, solids, angles, triangles, etc. Some of these terms, such as point, line, plane, are here used without being defined in a strictly logical sense. Their meaning is made clear by description and by con- crete illustrations like the following. 2. Certain portions of space are occupied by objects which we call physical solids, as, for instance, an ordinary brick. That which separates a solid from the surrounding space is called its surface. A This may be rough or smooth. If a surface is smooth and flat, we call it a plane surface. A pressed .brick has six plane surfaces called faces. Two adjoining faces meet in an edge. Three edges meet in a corner. The brick is bounded by its six faces. Each face is bounded by four edges, and each edge is bounded by two corners. 1 2 PLANE GEOMETRY. 3. If instead of the brick we think merely of its form and magnitude, we get a notion of a geometrical solid, which has the three dimensions, length, breadth, and thickness. The faces of this ideal solid are called planes. These are flat and have length and breadth, but no thickness. The edges of this solid are called lines. They are straight and have length, but neither breadth nor thick- ness. The corners of this solid are called points. They have position, but neither length, breadth, nor thickness ; that is, they have no magnitude. 4. It is possible to think of these concepts quite inde- pendently of any physical solid. Thus we speak of the line of sight from one point to another ; and we say that light travels in a straight line. The term straight line is doubtless connected with the idea of a stretched string. Of all the lines which may be con- ceived as passing through two fixed points that one .^'^^ ^^ is said to be straight between these points which corresponds most nearly to a stretched string. Likewise a plane may be thought of as straight or stretched in every direction, so that a straight line passing through any two of its points lies wholly in the plane. 5. If one of two intersecting straight lines turns about their common point as a pivot, the lines will continue to have only one point in common until all at once they will coincide throughout their whole length. Hence, Two straight lines cannot have more than one point iji common unless they coincide and are the same line; that is, two points determine a straight line. This would not be so if the lines had width, as may be seen by examining the figures. RECTILINEAR FIGURES, 3 6. EXERCISES. ^ 1. How does a carpenter use a straight-edge to determine whether a surface is a plane ? Do you know of any surface to which this test will apply in one direction but not in all directions ? 2. What tool does a carpenter use in reducing an uneven surface to a plane surface? Why is the tool so named? 3. If two points of a straight line lie in a plane, what can be said of the whole line ? 4. How many points of a straight line can lie in a plane if it con- tains at least one point not in the plane ? - 5. If two straight lines coincide in more than one point, what can you say of them throughout their whole length ? Do you know of any lines other than straight lines of which this must be true ? — 6. How do the material points and lines made by crayon or pencil differ in magnitude from the ideal points and lines of geometry ? 7. A machine has been made which rules 20,000 distinct lines side by side within the space of one inch. Do such lines have width ? Are they geometrical lines ? 8. Of all the lines, straight or curved, through two points, on which one is the shortest distance measured between the two points ? See the figure of § 4. Historical Note. The Egyptians appear to have been the first people to accumulate any considerable body of exact geometrical facts. The building of the great pyramids (before 3000 B.C.) re- quired not a little knowledge of geometric relations. They also used geometry in surveying land. Thus it is known that Rameses II (about 1400 B.C.) appointed surveyors to measure the amount of land washed away by the Nile, so that the taxes might be equalized. The Greeks, however, were the first to study geometry from a logi- cal point of view. Between 600 b.c, M^hen Thales, a Greek from Asia Minor, learned geometry from the Egyptians, and 300 B.C., when Euclid, a Greek residing in Alexandria, Egypt, wrote his Elements of Geometry, the crude, practical geometric information of the Egyp- tians was transformed into a well-nigh perfect logical system. Euclid's " Elements " contains the essential facts of every text- book on elementary geometry that has been written since his time. PLANE GEOMETRY M / \ B A 8 \;\ / c/ R y \ \ / z \ Vr, V D G U W RECTILINEAR FIGURES. 6 NOTATION FOR POINTS AND LINES. 7. A point is denoted by a capital letter. A straight line is denoted by two capital ' letters marking two of its points or by — '^^^ ^, - ? — one small letter. The word line alone i usually means straight line. Thus, the point A, the line AB, or the line /. 8. A straight line is usually understood to be un- limited in length in both directions, while that part of it which lies between twO of its points is ^ called a line- segment, or simply a segment. These points are called the end-points of the segment. Thus, the segment AB or the segment a. Two segments with the same end-points are coincident. —. 9. A part of a straight line, called a ray or half -line, may be thought of as generated by a point starting from a fixed position and mov- ' ' ^ ing indefinitely in one direction. The starting point is called the end-point or origin of the ray. If A is the origin of a ray and B any other point on it, then it is read the ray AB, not tJie ray BA. 10. Two line-segments are said to be added if they are placed end to end so as 4 £ ^ to form a single segment. Thus, segment ^C = segment ^5 + segment BC, or AC — AB+BC. If AC — AB -\- BC, then AC is greater than either AB or BC, and this is written AC > AB and AC > BC. A segment may also be subtracted from a greater or from an equal segment. Thus, \iAC = AB-\- BC, then AB=AC- 5Cand BC = AC - AB. PLANE GEOMETRY. Nl^ A segment is multiplied by an integer n by taking the sum of n sucb segments. Thus, if A C is the sum of n segments each AB, then A C = n • AB. If AG is n times AB^ then AC may be divided by w. Thus, AB = AC -^ n or AB = -. AC. n 11. A broken line is composed of connected line-segments not all lying in the same straight line. A curved line, or simply a curve, is a line no part of which is straight. A curved line or a broken line may inclose a portion of a plane, while a straight line cannot. 12. A circle is a plane curve containing all points equally distant from a fixed point in tlie plane, /^ and no other points. The fixed point is called the center of the circle. Any line-segment joining the center to a point on the circle is a radius of the circle. Any portion of a circle lying between two of its points is called an arc. Evidently all radii of the same circle are equal. Any combination of points, segments, lines, or curves in a plane is called a plane geometric figure. Plane Geometry deals with plane geometric figures. 13. EXERCISES. 1. How many end-points has a straight line? How many has a line-segment? How many has a ray? A circle? 2. Can you inclose a portion of a plane with two line-segments? With three? With four? BECTILINEAR FIGUBES. ANGLES AND THEIR NOTATION. 14. An angle is a figure formed by two rays proceeding from the same point. The point is the vertex of the angle and the rays are its sides. The angle formed by two rays is ^^^^"^^ sHT said to be the angle between them or simply their angle. Two line-segments having a common end-point also form an angle, namely, the angle of the rays on which the segments lie. An angle is determined entirely by the relative directions of its rays and not by the lengths of the segments laid off on them. 15. An angle is denoted by three letters, one at its vertex and one marking a point on each of its sides. The one at the vertex is. read between the other ^ two, as the angle CAB^ or the angle BAC^ not ABC. The one letter at the vertex is also used alone to denote an angle in case ^ no other angle has the same vertex, as, for instance, the angle A. In case several angles have the same vertex, a small letter or figure placed within each angle, together with an arc connecting its sides, is a convenient notation. The sign Z is used for the word angle. Thus in the figure we have Zl, Z2, Z 3, read, angle one, angle two, angle three. a 16. If two rays lie in the same straight line and ex- tend from the same end-point in opposite s^ directions they are said to form a straight | angle. If they extend in the same direc- side > side tion, they coincide and form a zero angle. 8 PLANE GEOMETRY. 17. An angle may be thought of as generated by a ray turning about its end-point as a pivot. Thus Z BA C is generated by a ray rotating from the position AB to the position A C. The rotating ray is usually conceived as moving in the direction opposite to the hands of a clock and the sides of the angle should usu- ally be read in this order. Thus ZBAC, not ZCAB. If the ray continues to rotate until ^-^ it lies in a direction exactly opposite — — '^ * ' to its original position, it generates a straight angle, as the straight angle BAC. If the ray rotates until it reaches its original position, the angle generated is called a peri- gon, that is, an angle of complete rotation. ' The position from which the rotating ray starts is called the origin of the angle. From this point of view two rays from the same point form two angles according as one or the other of the rays is regarded as the origin. In elementary geometry only angles less than or equal to a straight angle are usually considered. The units of measure for angles are one three-hun- dred-sixtieth of a perigon which is called a degree, one sixtieth of a degree called a minute, and one sixtieth of a minute called a second. These are denoted respectively b}^ the symbols °, ', '^ Thus, an angle of 20° 45' 30". A straight angle is therefore an angle of 180° and a perigon is an angle of 360°. 18. Angles are measured by means of an instrument called a protractor, which consists of a semicircular scale with degrees from 0° to 180° marked upon it. An inexpensive protractor made of cardboard or brass may be had at any stationery store. See the figure of § 33. RECTILINEAR FIGURES. 9 19. Two angles are said to be equal if they can be made to coincide without changing the form of either. If a ray is drawn from a point in a §a straight line so that the two angles thus formed are equal, each angle is called /^ a right angle, and the ray is said to be ^ perpendicular to the line. Thus, ii Zl = Z2, each angle is a right angle and AC is then said to be perpendicular to BD. Since the straight angle BAB is composed of Z 1 and Z 2, each of which is a right angle, it appears that a straight angle equals two right angles. See § 39 for the addition of angles in general. An acute angle is less than a right angle. An obtuse angle is greater than a right ./ angle and less than a straight angle. ^Xc^^^^ N.v^^'' Acute and obtuse angles are called oblique angles. A reflex angle is greater than a straight angle and less than a perigon. One line is oblique to another if the angles between them are oblique. A ray which divides an angle into two ^ 2" equal angles is called its bisector. Thus a perpendicular is the bisector of a straight angle. 20. EXERCISES. 1. Since we can always place two straight angles so as to make them coincide, what can we say as to whether or not they are equal? What of two right angles? ^^^ 2. What part of a straight angle is a right angle? 3, How many degrees in a right angle? In two right angles? What kind of an angle is one of 45°? One of 130°? 10 PLANE GEOMETRY. 4. Suppose that in the figure the ray AC rotates about the point A from the position AB to the position AD. What change takes place in Zl? What in Z2? Can there be more than one position oi AC for which Z 1 = Z 2 ? In this way it may be made clear that any angle has one and only one bisector. jT ^ b 5. How many rays perpendicular to BD at the point A can be drawn on the same side of BDl Does the answer to this question depend upon the answers to the questions in Ex. 4 ? How ? 6. Pick out three acute angles, three right angles, and three obtuse angles in the figure on page 4. TRIANGLES AND THEIR NOTATION. 21, If the points A^ 1?, C do not lie in the same straight line, the figure formed by the three segments, AB^ BC, and CA^ is called a triangle. The segments are the sides of the triangle, and the points are its vertices. The symbol A is b used for the word triangle. Each angle of a triangle has one side opposite and two sides adjacent to it. & . Similarly each side of a triangle has one angle opposite and two angles adjacent to it. . The side opposite an angle is often denoted by the corresponding small letter. Thus, in the figure the side a and the angle A are opposite parts, as are angle B and side b and angle C and side c. The three sides and three angles of a triangle are called the parts of the triangle. These six parts are considered as lying in order around the figure, an Z A, side b, Z C, side a, etc. An angle of a triangle is said to be included between its two adjacent sides, and a side is said to be included be- tween the two angles adjacent to it. Thus, in the figure the side a is included between Z B and Z C, and ZA is included between the sides b and c. BECTILINEAR FIGURES. 11 22. A triangle is called equilateral if it has its three sides equal, isosceles if it has at least two sides equal, scalene if it has no two sides equal, equiangular if it has its three angles equal. Select each kind from the figures on this page. 23. A triangle is called a right triangle if it has one right angle, an obtuse triangle if it has one obtuse angle, an acute triangle if all its angles are acute. Select each kind from the figures on this page. The side of a right triangle opposite the right angle is called the hypotenuse in dis- tinction from the other two sides, which are sometimes called its legs. 24. The side of a triangle on which it is supposed to stand is called its base. The angle opposite the base is called the vertex angle, and its vertex is the vertex of the triangle. The altitude of a tri- angle is the perpendic- ular from the vertex to the base or the base produced. Evidently any side may be taken as the base, and hence a triangle has three different altitudes. Vertex Vertex Base 12 PLANE GEOMETRY. 25. EXERCISES. 1. Is every equilateral triangle also isosceles? Is every isosceles triangle also equilateral ? i 2. Is a right triangle ever isosceles? Is an obtuse triangle ever isosceles? Draw figures to illustrate your answers. 3. In the figure on page 4 determine by measuring sides which of the triangles HNP, LKW, IHN, MIJ, KVU, OKJ, LVW, are isos- celes, which are equilateral, and which are scalene. 4. Determine whether /, K, V of the same figure may be the vertices of a triangle ; also whether J, 0, G may be. 5. Pick out ten obtuse triangles in this figure; also ten acute triangles. CONGRUENCE OF GEOMETRIC FIGURES. 26. In comparing geometric figures it is assumed that they may he moved about at will^ either in the same plane or out of it, without changing their shape or size. 27. Two figures are said to be simila r if they have the same shape. This is denoted by the symbol ~, read is similar to. For a more precise definition see §§ 255, 256. Two figures are said to be equivalent or simply equal if they have the same size or magnitude. This is denoted by the symbol =, read is i — i equivalent to or is equal to. L I Two figures are said to be congruent if, | — . without changing the shape or size of either, I I ~ they may be so placed as to coincide . ,, . throughout. This is denoted by the symbol ~ ^, read is congruent to. In the case of line-segments and angles, congruence is determined by size alone. Hence hi these cases we use the symbol — to denote congruence, and read it equals or is equal to. RECTILINEAR FIGURES. 13 28^ It is clear that if each of two figures is congruent to the -^ame figure they are congruent to each other. Hence if we make a pattern of a figure, say on tracing paper, and then make a second figure from this pattern, the two figures are congruent to each other. 29. li AABC ^AA^b'c' , the notation of the triangles c may be so arranged that AB = A'b\ bc = b'c', ca = c'a', z_^— ^^ , and ZC=Zc'. In this case AB is said to correspond to a'b', BC to b'c', CA to c'a\ ZA to Za', etc. Hence, we say that corresponding parts of congruent triangles are equal. 30. EXERCISES. 1. Using tracing paper, draw triangles congruent to the triangles MIN, NHP, OAB, OFE, OKL, UKV, OGL on page 4, and by ap- plying the pattern of each triangle to each of the others determine whether any two are congruent. 2. Find as in § 28 whether any two of three accompanying triangles are congruent, and if so arrange the notation so as to show the corre- sponding parts. 3. Give examples of figures which are similar, equal, or con- gruent, different from those in § 27. _. 4. If two figures are congruent, does it follow that they are equal? Similar ? 5. If two figures are similar, does it follow that they are equal? Congruent? .^ 6. If two figures are equal, are they similar? Congruent? 14 PLANE GEOMETBY, TESTS FOR CONGRUENCE OF TRIANGLES. 31. The method of determining whether two triangles are congruent by making a pattern of one and applying it to the other is often inconvenient or impossible. There are other methods in which it is necessary only to determine whether certain sides and angles are equal. These methods are based upon three important tests for congruence of triangles. 32. First Test for Congruence of Triangles. If two triangles have two sides and the included angle of one equal respectively to tivo sides and the included ' a7igle of the other, the triangles are congruent. ■ This may be shown by the following argument: Let ABC and A'BC be two triangles in which AB = A^B!, AC = A'C, and Z.A = ZA'. We are to show that A ^5C^ A ^'s'c'. Place A ABC upon AA'b'c' so that Z.A coincides with Z^', which can be done since it is given that Z.a=Za', Then point B will coiYicide with b' and G with c', since it is given that AB = A'b' and AC = A'c', Hence, side BC will coincide with b'c' (§ 8). Thus, the two triangles coincide throughout and hence are congruent (§ 27). ' ^' The process just used is called super position . It ^ may sometimes be necessary to move a figure out of its plane in order to superpose it upon another as in the case of the accompanying triangles. BECTILINEAB FIGURES. 15 33. The equality of short line-segments is conveniently tested by means of the dividers or compasses. Place the divider points on the end-points of one segment ^5 and then see whether they will also coincide with the end-points of the other segment A'B'. If so, the two segments are equal. The equality of two angles may be tested by means of the protractor. Place the protractor on one angle i^OC as shown in the figure and read the scale where OC crosses it. Then place the protractor on the other angle y^^^^*'""'"'"'"^*^^^ B'O'C and see whether O'C crosses the scale at the same point. If so, the two angles are equal. 34. EXERCISES. 1. Using the protractor determine which pairs of the following angles on page 4 are equal : HPG, LGW, GWL, AOB, VLW, LVW. 2. By the test of § 32 determine whether, on page 4, AJKU^AG WL, also whetiier A MIH ^AKVW. First find whether two sides of one are equal respectively to two sides of the other, and if so compare the included angles. 3. Could two sides of one triangle be equal respectively to two sides of another and still the triangles not be congruent ? Illustrate by constructing two such triangles. 4. Show by the test of § 32 that two right triangles are congruent if the legs of one are equal respectively to the legs of the other. Can this be shown directly by superposition ? - 5. Find the distance AB when, on account of some obstruction, it cannot be measured directly. Solution. To some convenient point C measure the distances A C and BC. Continuing in the direc- tion A C lay off CA' = AC, and in the direction BC lay off CB' = BC. Then Z 1 = Z 2 (see § 74) . Test this with the protractor. Show that the length AB is found by measuring A'B'. 16 PLANE GEOMETRY. 35. Second Test for Congruence of Triangles. If tiDO triangles have two angles and the included side of one equal respectively to two angles and the in- cluded side of the other ^ the triangles are congruent This is shown by the following argument ; G \cj ' A B A' B' Let ABC and A^B'C be tw^ triangles in which Z.A=:i/-A\ Z.B=^ZEf, and AB=A'Ef. <« We are to show that AABC^Aa'b'c' . Place A ABC upon AA'b'c' so that AB coincides with its equal A'b', making C fall on the same side oi a' B^ as c'. Then ^C will take the direction of A'&, since ZA = Z j.', and the point C must fall somewhere on the ray a'&. Also BC will take the direction of b'c' (Why?), and hence C must lie on the ray b'c'. Since the point C lies on both of the rays A'& and b'c'^ it must lie at their point of intersection & (§5). Hence, the triangles coincide and are, therefore, congruent (§ 27). 36. EXERCISES. 1. In the figure of § 35 is it necessary to move A ABC out of the plane in which the triangles lie ? Is it necessary in the figure here given ? 2. Show how" to measure the height of a tree by using the second test for congruence. Suggestion. Lay out a triangle on the ground which is congruent to A ABC, using § 35. 3. By the second test determine whether A OHG ^ A OJK on page 4. 4. Draw any triangle. Construct another tri- RECTILINEAR FIGURES. 17 angle congruent to it. Use § 35 and also § 32. Use the protractor to construct the angles. 5. Find the distance A C, when C is inaccessible. Let i^ be a convenient point from which A and C are visible. Lay ont a triangle ABC mak- ing Z 3 = Z 1 and Z 4 = Z 2. Show that the dis- tance AC may be found by measuring AC. 6. Show how to find the distance between two inaccessible points A and B. Solution. Suppose that both A and B are visible from C and D. (1) Using the triangle CDA, find the length of AD as in Ex. 5 above. (2) Using the triangle CBD, find DB in the same manner. (3) Using the triangle DBA, find ^i? as in Ex. 5, § 34. 37. The proof of the third test for congruence of triangles involves the following : The angles opposite the equal sides of an isosceles tri angle are equal. Let ABC be an isosceles triangle having AC = EC We are to show that Z A = Z b. Suppose CD divides z. acb so that Z1=Z2. By means of § 32 show that A ACD ^AbCD. Then Z ^ = Z 7; by § 29. The theorems § 35 and § 37 are due to Thales. It is said he used § 35 in calculating the distance from the shore /^o a ship at sea. ' 38. EXERCISE. On page 4 pick out as many pairs of angles as possible which may be shown to be equal by § 37. Test these by using the protractor. 18 PLANE GEOMETRY. 39. Definitions. Two angles which have a common vertex and a common side are said to be adjacent if neither angle lies within the other. /^ i\ Thus, Z 1 and Z. 2 are adjacent, while Z 1 and Z 3 are not adjacent. The sum of two angles is the angle formed by the sides not common when the two angles are placed adjacent. Thus, Z3 = Z1 + Z2. IfZ3 = Zl + Zl2, then we say that Z 3 is greater than either Z 1 or Z 2. This is written Z 3 > Z 1 and Z 3 > Z 2. An angle may also be subtracted from a greater or equal angle. ThusifZ3 = Zl + Z 2, then Z3 -Zl = Z2and Z3 — Z 2 = Z 1. It is clear that : If equal angles are added to equal angles^ the sums are equal angles. Angles may be multiplied or divided by a positive in- teger as in the case of line-segments. See § 10. 40. We may now prove the third test for congruence of triangles, namely : // two triangles have three sides of one equal respec- tively to three sides of the other, the triangles are con- gruent* ' Let ABC and A'B'C' he two triangles in which AB = A'B'j BC=B'C',CA = CA'. We are to show that A ABC ^ AA'b'c', RECTILINEAR FIGURES. 19 Place Aa'b'c' so that A'b' coincides with AB and so that c' falls on the side of AB which is opposite C. (Why is it possible to make A'B' coincide with AB"}) Draw the segment CC'. From the data given, how can § 37 be used to show that in A ACC' Z.1 = Z 2 ? Use the same argument to show that Z 3 = Z 4. But if Z 1 = Z 2 and Z3 = Z4, then Z 1 + Z 3 = Z 2 + Z 4. (§39) That is, ^ACB = Z BC'A. How does it now follow that A ABC ^ A ABC' ? (§ 32) But AABC' ^AA'b'c'. (§26) Hence, AABC^Aa'b'c'. (§28) Make an outline of the steps in the above argument, and see that each step is needed in deriving the next. 41. Definition. If one triangle is congruent to another because certain parts of one are equal to the corresponding parts of the other, then these parts are said to determine the triangle. That is, ant/ other triangle constructed with these given parts will be congruent to the given triangle. 42. EXERCISES. 1. In § 37 show that CD is perpendicular to AB and that AD = DB. State this fully in words. 2. Using § 40, determine which of the following triangles on page 4 are congruent : OJK, HNP, OTH, PRG, JKU. 3. Do two sides ^e^ermme a triangle ? Three sides? Two angles? Three angles? Illustrate by figures. 4. A segment drawn from the vertex of an isosceles triangle to the middle point of the base bisects the vertex angle and is perpendicular to the base. 5. What parts of a triangle have been found sufficient to determine it? In each case how many parts are needed? 20 PLANE GEOMETRY. 43. The three tests for congruence of triangles, §§ 32, 35, 40, lie at the foundation of the mathematics used in land surveying. The fact that certain parts of a triangle determine it shows that it mai/ be possible to compute the other parts when these parts are known. Rules for doing this are found in Chapter III. CONSTRUCTION OF GEOMETRIC FIGURES. 44. The straight-edge ruler and the compasses are the instruments most commonly used in the construction of geometric figures. By means of th^ ruler straight lines are drawn, and the compasses are used in laying off equal line-segments and also in constructing arcs of circles (§ 12). Other common instruments are the protractor (§ 33) and the triangular ruler with one square corner or right angle. The three tests for congruence of two triangles are of constant use in geometrical constructions. 45. Peoblem. To find a point ivhose distances from the extremities of a given seg- \g/^ ment are specified. Solution. Let AB be the given seg- ment and let it be required to find a A b point C which shall be one inch from each extremity of AB. Set the points of the compasses one inch apart. With ^ as a center draw an arc m^ and with S as a center draw an arc n meeting the arc m in the point C. Then every point in the arc m is one inch from A and every point in the arc n is one inch from B (§ 12). Hence C, which lies on both m and w, is one inch from A and also from B. RECTILINEAR FIGURES. 21 46. EXERCISES. 1. In the preceding problem is there any other point in the plane besides C which is one inch distant from both A and B'i If so, show how to find it. 2. Could AB be given of such length as to make the construction in § 45 impossible ? 3. Is there any condition under which one point only could be found in the above construction ? If so, what would be the length of AB'} 4. Find a point one inch from A and two inches from B and dis- cuss all possibilities as above. 5. Given three segments a, 6, c, construct a triangle having its sides equal to these segments. Discuss all possibilities depending upon the relative lengths of the given segments. 47. Problem. To construct an angle equal to a given angle, ivithout using the py^otractor. Solution. Given the angle A. Lay off any distance AB on one of its sides and any dis- tance AC on the other. Draw the segment BC forming the triangle ABC. As in Ex. 5 above, construct a triangle a'b'c' so that A'b' = AB, b'c' = BC, A'c' = AC. Show that A ABC ^ A a'b'c' by one of the tests, and hence that Z A = Z. A', being corresponding angles of con- gruent triangles, § 29. In the above construction, would it be wrong to make AB = AC? Is it necessary/ to do so ? 22 PLANE GEOMETRY. 48. Peoblem. To construct the ray dividing a given angle into two equal angles, that is, to bisect the angle. Solution. Given the angle A. To construct the ray bisecting it. On the sides of the angle lay off n. segments AB and AG so that y^ ^^^ / / \m AB = AC. ' . /^n A^^ — ^ With B and C as centers and '-S with equal radii construct arcs m and n meeting at I). Draw the segments CD, BD, and AD. Now show that one of the tests for congruence is applicable to make AACD ^ A ABD. Does it follow that Z 1 = Z 2 ? Why ? 49. EXERCISES. 1. Is it necessary in § 48 to make AB = ACi In this respect com- pare with the construction in § 47. * 2. Is any restriction necessary in choosing the radii for the arcs m and n? Is it possible to so construct the arcs m and n, still using equal radii for both, that the point D shall not lie within the angle BA C ? In that case does the ray A D bisect ABACi 3. By means of § 48 bisect a straight angle. What is the ray called which bisects a straight angle? In this case what restriction is necessary on the radii used for the arcs m and n ? 4. By Ex. 3 construct a perpendicular to a line at a given point in it. 5. Construct a perpendicular to a segment at one end of it without prolonging the segment and without using the square ruler. Suggestion. Let AB he the' given segment. Construct a right angle A'B'C as in Ex.'4. Then as in § 47 construct Z^ 5 C= ZA'B'C. 50. Definition. A line which is perpendicular to a line- segment at its middle point is called the perpendicular bi'sector of the segment. RECTILINEAR FIGURES. 23 51. Peoblem. To construct the perpendicular bisec- tor of a given line-segment. \1^ Solution. Let AB be the given segment. ^45 As in § 45, locate two points, C and D, each of which is equally distant from A and B. /f Draw the segment CD meeting AB in O. Then CD is the required perpendicular bisector of AB. To prove this, show that AACD ^ABCD. Hence Z3 = Z4. (Why?) By what test can it now be shown that AAOC^ABOC? Hence Z 1 = Z 2. (Why ?) Therefore CO (or CD) is perpendicular to AB (Why?) and also A0= OB (Why?). It has thus been shown that CD is perpendicular to AB and bisects it, as was required. 52. The steps proved in the above argument are : (a') AACD ^ A BCD. (5)Z3 = Z4. (c^ AAOC^ABOC. (d) Z 1 = Z 2, and AO = BO. Study this. outHne with care. What is wanted is the last result (d). Notice that (d) is obtained from (c), (c) from (b), and (b) from (a). Thus each step depends on the one preceding, and would be impos- sible without it. To understand clearly the order of the steps in a proof as shown by such an outline is of great importance in master- ing it. 53. EXERCISES. 1. In the construction of § 51, is it necessary to use the same radius in locating the points C and D ? 2. Name the isosceles triangles in the figure §51: (a) if the same radius is used for locating C and D, (b) if different radii are used. 24 PLANE GEOMETRY. 54. Problem. To construct a perpendicular to a give7i straight line from a given point outside the line. Solution. Let / be the given line, and P the given point outside it. With P as a center draw an arc cutting the line I in two points, A and B. With A and B as centers, and with equal radii, draw the arcs m and n intersecting in C. Draw the line PC cutting I in the point D. Then the line PC is the perpendicular sought. To prove this, draw the segments PA^ PB^ CA^ CB. Complete the proof by showing that PC is the perpen- dicular bisector of AB^ and hence is perpendicular to I from the point P. 55. Problem. To construct a triangle ivhen two sides and the included angle are given. Solution. Let h and c be the / ^)/ given sides, and A the given angle. As in § 47, construct an angle a' equal to /-A. On the sides of Z A' lay off A'B A'c = h. Connect B and C. Then A'bc is the required triangle. (Why ?) RECTILINEAR FIGURES. 25 56. Problem. Coiisti^uct a triangle ivhen hvo angles and the included side are given. Solution. Let Z A and ZB he the given angles, and c the given side. Construct Z A' = Z.A. On one side of Z a' lay off A^b' = e. At b' construct Z A'b'k equal to Ab. Let b'k meet the other side of Za' at C. Then ^'^'c is t'he required triangle. (Why ?) 57, EXERCISES. 1. If in the preceding problem two different triangles are con- structed, each having the required properties, how will these triangles be related? Why? 2. If in the problem of § 5.5, two different triangles are constructed, each having the required properties, how will these triangles be related? Why? 3. If two triangles are constructed so that the angles of one are equal respectively to the angles of the other, will the triangles neces- sarily be congruent ? 4. If two different triangles are constructed with the same sides, how will they be related ? Why ? 5. Construct an equilateral triangle. Use § 37 to show that it is also equiangular. 58. We have now seen that the three tests for the con- gruence of triangles are useful in making indirect measurer ments of heights and distances when direct measurement is inconvenient or impossible, and also in making numerous geometric constructions. It will be found, as we proceed, that these tests are of increasing usefulness and importance. 26 PLANE GEOMETRY. THEOREMS AND DEMONSTRATIONS. 59. A geometric proposition is a statement affirming cer- tain properEtes of geometric figures. Thus : " Two points determine a straight line " and " The base angles of an isosceles triangle are equal " are geometric propositions. A proposition is proved or demonstrated when it is shown to follow from other known propositions. A the,oxem is a proposition which is to be proved. The argument used in establishing a theorem is called a proof. 60. In every mathematical science some propositions must he left unproved^ since every proof depends upon other propositions which in turn require proof. Propositions which for this reason are left unproved are called axioms. While axioms for geometry may be chosen in many dif- ferent ways, it is customary to select such simple propo- sitions as are evident on mere statement. 61. Among the axioms thus far used are the following : Axioms. I. A figure may he moved about in space without changing its shape or size. See § 26. II. Through two points one and only one straight line can be drawn. See §§8, 32. III. The shortest distance between two points is meas- ured along the straight line-segment connecting them. Thus one side of a triangle is less than the sum of the other two. IV. // each of two figures is congruent to the same figure, they are congruent to each other. See §§ 28, 40. V. // a, h, c, d are line-segments (or angles) such that a = b and c = d, then a-\-c = b + d and a — c = b — d* In the latter case we suppose a'^c, hy-d. See §§10, 39. RECTILINEAR FIGURES. 27 VI. // a and b are line-segments (or angles) such that a=h, then a xn=b xn and a-^n=h-^n; and if a>h, then a xn>b xn and a^n>h-hn, n being a "positive integer. See §§ 10, 39. Note. An equality or an inequality may be read from left to right or from right to left. Thus, a > 6 may also be read 6 < a. Other axioms are given in §§ 82, 96, 119, and in Chapter VII. Certain other simple propositions may be assumed at present without detailed proof. These are called prelimi- nary theorems. PRELIMINARY THEOREMS. 62. Two distinct lines can meet in only one point. For if they have two points in common, then by Ax. II they are the same line. 63. All straight angles are equal. § 20, Ex. 1. 64. All right angles are equal. See Ax. VI. 65. Every line-segment has one and only one middle point. See § 51, where the middle point is found by construction. 66. Every angle has one and only one bisector. See § 48, where the bisector is constructed. 67. One and only one perpendicular can be drawn to a line through a point whether that point is on the line or not. See § 20, Exs. 4, 5; § 49, Ex. 4; § 54. 68. The sum of all the angles about a point in a straight line and on one side of it is two right angles. 69. The sum of all the angles about a point in a plane is four right angles. In §§ 68, 69 no side of one angle is to lie inside another. 28 PLANE GEOMETRY. 70. Definitions. Two angles are said to be complementary if their sum is one right angle. Each is then called the complement of the other. Thus, Z a and Z h are complementary angles. Two angles are said to be supple- mentary if their sum is two right angles. Each is then said to be the supplement of the other. Thus, Z 1 and Z. 2 are supplementary angles. Two angles are called vertical angles if the sides of one are pro- longations of the sides of the other. Thus, Z 1 and Z 3 are vertical angles, and also Z 2 and Z4. 71. ' EXERCISES. ,x ( 1. What is the complement of 45°? the supplement? 2. If the supplement of an angle is 140°, find its complement. 3. If the complement of an angle is 21", find its supplement." 4. Find the supplement of the complement of 30°. ) y 5. Find the angle whose supplement is five times its complement. 6. Find the angle whose supplement is n times its complement. 7. Find an angle whose complement plus its supplement is 110°. d\G 8. If in the first figure Ah = 2 Aa, and Zc=:Za + Z ft, find each angle. 9. If in the second figure Z.h = lZ.a, Zc = Za + Zb, and Z d = Q Z a, find each angle. 10. If in the third figure Zb= Ze, Zc^ Za + Zh, Zd = 2Zb, and Ze =\Zd^ find each angle. RECTILINEAR FIGURES. 29 PRELIMINARY THEOREMS. 72. Angles ivhich are comjolemeiits of the same angle or of equal angles are equal. For they are the remainders when the given equal angles are sub- tracted from equal right angles. Ax. V. 73. Ajigles ivhich are suppleme7its of the same angle or of equal angles are equal. For they are the remainders when the given equal angles are sub- tracted from equal straight angles. 74. Vertical angles are equal. \ They are supplements of the same angle. 75. If tioo adjacent angles are supplementary, their exterior sides are in the same straight line. For the two angles together form a straight angle. 76. If tioo adjacent angles have their exterior sides in the same straight line, they are supplementary. For a straight angle is equal to two right angles. 77. EXERCISES. 1. Prove that if one of the four angles formed by two intersecting straight lines is a right angle, then all are right angles. 2. Show that the rays bisecting two complementary adjacent angles form an angle of 45°. 3. Find the angle formed by the rays bisecting two supplementary adjacent angles. Prove. 4. Find the angle formed by the rays bisecting two vertical angles. Prove. 5. The sum of two adjacent angles is 74°. Find the angle formed by their bisectors. 6. The angle formed by the bisectors of two adjacent angles is 37° 18'. Find the sum of the adjacent angles. 30 PLANE GEOMETRY. ON THE NATURE OF A DEMONSTRATION. 78. A theorem consists of two distinct parts, hypothesis and conclusion. In a geometrical theorem, the hypothesis specifies cer- tain properties which the figures in question are assumed to possess. The conclusion asserts that certain other prop- erties*belong to the figures whenever the assumed proper- ties are present. The hypothesis and conclusion are often intermingled in a single statement, in which case they should be explic- itly separated before making the proof. For example, in the theorem of § 37, The angles opposite the equal sides of an isosceles triangle are equal, the hypothesis is, " Two sides of a triangle are equal," and the conclusion is, " The angles opposite them are equal." 79. If the hypothesis consists of several parts, these should be tabulated and then checked off as the demon- stration proceeds. If the theorem is properly stated, each part of the hypothesis will be used in the proof. For instance, in the theorem of § 32, the hypothesis is : AB —A'B', AC= A' C% and Z A = Z A'; and the conclusiou is: A ABC ^ A A'B'Cy. It will be found on examining the proof that each part of the hypothesis is needed and used in the course of the demonstration. If the conclusion could be proved without using every part of the hypothesis, then the parts not used should be omitted from the hypothe- sis in the statement of the theorem. 80. In the proof of a theorem no conclusion should be taken for granted simply from the appearance of the figure. Each step in a proof should be based upon a definition, an axiom, or a theorem previously proved. It will then follow that the theorem is as certainly true as are the simple, unproved propositions with which we start, and upon which our argument is based. RECTILINEAR FIGURES, 31 Historical Note. The Egyptians showed no knowledge of a logical demonstration, nor did the Arabians, who studied geometry quite extensively. The Greeks developed the process of demonstra- tion to a high state of perfection. They were fully aware, moreover, that certain propositions must be admitted without proof (see § 60). Thus Aristotle (384-322 B.C.) says : *' Every demonstration must start from undemonstrable principles. Otherwise the steps of a demon- stration would be endless." Euclid divided unproved propositions into two classes: axioms, or "common notions," which are true of all things, such as, " If things are equal to the same thing they are equal to each other"; and postulates, which apply only to geometry, such as, " Two points determine a line." The best usage in modern mathe- matics is to adopt the one word axiom for both of these, as in § 60. Much practice is needed in writing demonstrations in full detail. This should be done in the shortest possi- ble sentences, usually giving a separate line to each state- ment, followed by the definition, axiom, or theorem on which it depends. For this purpose the following symbols and abbrevia- tions are convenient: Z, A, angle, angles. A, i^, triangle, triangles. ^ ^ r parallelogram, [^parallelograms. □, [s], rectangle, rectangles, j right angle, [right angles, [straight angle. rt.Z, rt.^. St. Z, St. A, [straight angles. rt.A,rt.A,|"^^^*^'^^^g^"' [right triangles. O, ©, circle, circles. ''^, ^, arc, arcs. = , is equal, or equivalent, to. ~, is similar to. ^, is congruent to. >, is greater than. <, is less than. ^, is less than or equal to. ^, is greater than or equal to. II, parallel, or is parallel to. . r perpendicular, or [is perpendicular to. lis, parallels. Js, perpendiculars. .'. , therefore or hence, ax., axiom, th., theorem, def., definition, cor., corollary, alt., alternate, ext., exterior, int., interior, hyp., by hypothesis. 32 PLANE GEOMETIiY, INEQUALITIES OF PARTS OF TRIANGLES. 81. Definition. If one side of a triangle is produced, the angle thus formed is called an exterior angle of the triangle. Thus, Z 1 is an exterior angle of the triangle ABC. A 82. Axiom VII. If a, h, c are line-segments {or angles) such that a>h and h^c,or such that a^h and h>c, then a > c. The proof of the following theorem is shown in full detail as it should be written by the pupil or given orally, except that the numbers of paragraphs should not be required. 83. Theorem. An exterior angle of a triangle is greater than either of the opposite interior angles. Given the A ABC with the exterior angle DBC formed by pro- ducing the side AB. To prove that Z. DBG > Z c and also Z dbc > Z a. Proof: Let E be the middle point of BC. Find E by the construction for bisecting a line-segment (§ 51). Draw AE and prolong it, making EF= AE^ and draw BF. In the two A ACE and FBE, we have by construction CE = EB and AE = EF. RECTILINEAR FIGURES. 33 Also /. CEA == /. BEF. (Vertical angles are equal, § 74.) .-. A ACE ^ A FBE, (Two triangles which have two sides and the included angle of the one equal respectively to two sides and the included angle of the other are congruent, § 32.) .-. /.C= Z FBE. (Being angles opposite equal sides in congruent triangles, § 29.) But ZDBO Z FBE. (Tf an angle is the sum of two angles it is greater than either of them, §39.) .-. Zdbc> Z C. (Since Z DEC > Z FBE and Z FBE =:ZC, Ax. VII, § 82.) (%^-[) In order to prove Z DBC > Z A, prolong CB to some point G. Then Zabg = Zdbc. (Vertical angles are equal, § 74.) Now bisect AB, and in the same manner as before we may prove ZABG>ZA. .'. Zdbc> Za. (Since Z DBC = Z ABG and Z ABG >ZA, Ax. VII, § 82.) For the second part of the proof let H be the middle point of AB. Draw CH and prolong it to X, making cri = HK. Let the student draw the figure for the second part of the proof and give it in full. Hereafter more and more of the details of the proofs will be left for the student to fill in. When reference is made to a paragraph in the text or when the reason for a step is called for, the complete state- ment of the definition, axiom, or theorem should be given by the student. 34 PLANE GEOMETRY. 84. Theokem. If two sides of a 'triangle are un- equal, the angles opposite these sides, are unequal, the greater angle being opposite the greater side. Given ^ABC in which AC>BC. To prove tliat Z ABC > Za, Proof : Lay off CD = CB and d raw BD. Now give the reasons for the follow- ing steps: ^ (1) Z ABC > Z DBC. (2) ZdBC = ZCDB. (3) ZCDB>ZA. (4) .-. ZABO ZA. (§39) (§37) (§ 83) (§82) 85. Theorem. // two angles of a triangle are un- equal, the sides opposite them are unequal, the greater side being opposite the greater angle. Given A ABC in which ZB>ZA. To prove that h > a. Proof: One of the following three statements must be true: (1) h = a, (2) ha. But it cannot be true that b = a, for in that case Zb =Za, (§ 37) contrary to the hypothesis that Z b > Z A. And it cannot be true that h < a, for in that case Zb a. BECTILINEAB FIGURES. 35 86. The above argument is called proof by exclusion. Its success depends upon being able to enumerate all the possible cases^ and then to exclude all but one of them by showing t^iat each in turn leads to some contradiction. 87. EXERCISES. 1. The hypotenuse of a right triangle is greater than either leg. 2. Show that not more than two equal line-segments can be drawn from a point to a straight line. Suggestion. Suppose a third drawn. Then apply §§ 37, 83, 84. 3. Show by joining the vertex A of the triangle ABC to any point of the side BC that ZB + ZC <2 rt. A. Use § 83. 4. If tioo angles of a triangle are equal, the sides opposite them are equal. Use §§ 84, 86. 5. Either leg of an isosceles triangle is greater than half of the base. Suggestion. Use § 42, Ex. 4, and Ex. 1. THEOREMS ON PARALLEL LINES. 88. A straight line which cuts two straight lines is called a transversal. The various angles formed ar^ named as follows: Z4 and Z5 are alternate-interior angles; ^ also Z 3 and Z 6. Z 2 and Z 7 are alternate-exterior angles ; also Z 1 and Z8. Z 1 and Z 5 are corresponding angles ; also Z 3 and Z 7, Z 2 and Z 6, Z 4 and Z 8. Z 3 and Z 5 are interior angles on the same side of the trans vei*sal ; also Z4 and Z 6. 89. Definition. Two complete lines which lie in the ^same plane and which do not meet are said to be parallel. Two line-segments are parallel if they lie on parallel lines. 36 PLANE GEOMETRY. 90. Theorem. // two lines cut by a transversal have equal alternate interior angles, the lines are parallel. Given the lines /i and I2 cut by t so that Z1 = Z2. To prove that Zj il l<^. p^Z 1^ Proof : Suppose the lines l^ and l^ were to meet on the right of the trans- — versal. Then a triangle would be / formed of which Z 1 is an exterior angle and Z 2 an opposite interior angle. This gives an exterior angle of a triangle equal to an opposite interior angle, which is impossible. (Why?) Repeat this argument, supposing l^ and l^ to meet on the left side of the transversal. Hence Zj and Zg cannot meet and are parallel (§ 89). 91. The type of proof used here is called an indirect proof. It consists in showing that something impossible or contradictory results if the theorem is supposed not true. 92. Theorem. // two lines cut by a transversal have equal corresponding angles, the lines are parallel. Given the lines /i and 4 cut by t so that Z 2 = Z 3. _ To prove that l^ || l^. Proof : Quote the authority for each of the following steps : Z3 = Z2. Z3 = Z1. (§74) (Ax. IV) . Z 1 = Z 2. .-. Z1I1Z2. (§90) RECTILINEAR FIGURES. 37 93. Theorem. // two lines cut by a transversal have the sum of the interior angles on one side of the trans- versal equal to two right angles, the lines are parallel, A Given k and l^ cut by t so that Z4 + Z2=2rt. zi. To prove that \ II l^. Proof: Z 4 is supplementary to Zl and also to Z 2. (Why ?) .-. Z1 = Z2. (Why?) .-. /ill^2- (Why?) 94. EXERCISES. 1. Show that if each of two lines is perpendicular to the same line, they are parallel to each other. 2. Let ABC be any triangle. Bisect BC at Z). (^ ^ Draw AD and prolong it to make DE — AD. /NT \\ Draw CE. Prove CE II AB. /^/^LIFOBg\^^i2J^Cr7L7iV^E^2? FIGURES, 39 99. Theorem. If two 'parallel lines are cut by a transversal, the sum of the interior angles on one side of the transversal is two right angles. Suggestion. Make use of the preced- ing theorem and give the proof in full. Of what theorem is this the converse ? 100. EXERCISES. 1. state and prove the converse of the theorem in § 92. 2. Prove that if two parallel lines are cut by a transversal the alternate exterior angles are equal. Draw the figure. 3. State and prove the converse of the theorem in Ex. 2. 4. If a straight line is perpendicular to one of two parallel lines, it is perpendicular to the other also. 5. Two straight lines in the same plane parallel to a third line are parallel to each other. Suppose they meet and then use § 96. 6/5 1^11 D 6. If l^ II ^2 II h and if Zl = 30°, find the other angles in the first figure. 7. If l^Wl^, how are the bisectors of Zl and Z3 related? Of Z3 andZ4? OfZlandZ2? Use § 102 for the last case. 3^ 8. If Zj 11/3, and ^0=05, show that i)0=OC. t^ ^' State this theorem fully and prove it. - aJ^'^ 1^ 9. If /JU2andZ2 = 5Zl, findZ4andZ3. ^ 10. If two parallel lines are cut by a transversal, the sum of the exterior angles on one side of the transversal is two right angles. 11. State and prove the converse of the preceding theorem. 40 PLANE GEOMETRY. APPLICATIONS OF THEOREMS ON PARALLELS. 101. Problem. Through a given "point to construct a line parallel to a given line. Given the line / and the point P outside of it. To construct a line \ through P \\ to I. Construction. Through P draw any line making a con- venient angle, as Z 1 with I. / Through P draw the line Zj, making , pA ^ Z2 = Z1(§47). Then ?! II?. Proof : Use the theorem, § 92. Hereafter all constructions should be / described fully as above, followed by a proof that the con- struction gives the required figure. 102. Theorem. The sum of the angles of a triangle is equal to two right angles. ^ Given A ABC with Z 1, Z 2, Z 3. c \ To prove that \\^ \ Zl + Z2 + Z3 = 2rtZs. \^ ^\4\, Proof: Prolong ^j5 to some point Z). ^ B Through B draw BE \\ AC. Then Z5 + Z4 + Z3=2rtZs. (Why?) But Z4 = Z2and Z5 = Z1. (Why?) Hence, replacing Z.6 and Z4 by their equals, Zl and Z2, wehaveZl + Z2 + Z3 = 2rtZs. Historical Note. This is one of the famous theorems of geome- try. It was known by Pythagoras (500 B.C.), but special cases were known much earlier. The figure used here is the one given by Aristotle and Euclid. As is apparent, the proof depends upon the theorem, § 97, and thus indirectly upon Axiom VIII. The interdepend- ence of these two propositions has been studied extensively during the last two centuries. RECTILINEAR FIGURES. 41 103. Theorem. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. The proof is left to the student. Compare this theorem with that of § 83. 104. Definition. A theorem which follows very easily from another theo- rem is called a corollary of that theorem. U.g. the theorem in § 103 is a corollary of that in § 102. 105. EXERCISES. 1. Find each angle of an equiangular triangle. 2. If one angle of an equiangular triangle is bisected, find all the angles in the two triangles thus formed. 3. If in SiAABC,AB =^Cand ZA = Z 5 + Z C, find each angle. 4. If in the figure AB = AC and Zi = 120°, find Z1,Z2, Z3. 5. If one acute angle of a right triangle is 30°, what is the other acute angle? If one is 41° 23'? 6. If in two right triangles an acute angle of one is equal to an acute angle of the other, what can be said of the remaining acute angles. What axiom is involved? 7. If in two right triangles the hypotenuse and an acute angle of one are equal respectively to the hypotenuse and an acute angle of the other, the triangles are congruent. Prove in full. 8. Can a triangle have two right angles? Two obtuse angles? Can the sum of two angles of a triangle be two right angles? What is the sum of the acute angles of a right triangle? 9. If two angles of a triangle are given how can the third be found? If the sum of two angles of one triangle is equal to the sum of two angles of another, how do the third angles compare? 10. Prove the theorem of § 102, using each of the figures in the margin. The first of these figures was used by Pythagoras. 42 PLANE GEOMETRY. 106. Definition. An angle viewed from the vertex has a right side and a left side. 107- Theorem. If two angles have their sides i^espectively parallel, right side to right side, and left side to left side, the angles are equal. Given Z 1 and Z 2 such that a II a' and 6 li &' . To prove that Z 1 = Z 2. Proof : Produce a and b' till they meet, forming Z. 3. Complete the proof. Why do a and h' meet when produced ? Make a proof also by producing h and a' till they meet. 108. Theorem. If two angles have their sides re- spectively perpendicular, right side to right side, and left side to left side, the angles are equal. Given Z 1 and Z 2 such that al.a' and h ± b'- To prove that Z1 = Z2. Proof : Produce a and a^ till they meet in O. Through o draw a line parallel to h. Now show that Z 2 = Z 3 since each is the complement of Z4. Complete the proof. 109. EXERCISES. If two angles have their sides respectively parallel, or perpendicular, right side to left side, and left side to right side, the angles are supple- mentary. In each figure Zl = Z3. (Why?) Historical Note. The theorems of §§ 107, 108, 109 are not found in Euclid's Elements. RECTILINEAR FIGURES. 43 OTHER THEOREMS ON TRIANGLES. 110. Theoeem. // in two right triangles the hypote- nuse and one side of one are equal respectively to the hypotenuse and one side of the other, the triangles are congruent, ^ b' b AiA \ -'A' Given the right A ABC and A'EfC, having -45= A '5** and BC=BfC. To prove that A ABC ^ A A'b'c'. Proof : Place the triangles so that j^ and B^c' coincide, and so that A and A' are on opposite sides of BC. Then AC and CA' lie in a straight line (Why ?), and A aba' isosceles (Why?). Hence, show that Z A = /. A' and complete the proof. 111. Corollary. If from a point in a perpendicular to a straight line equal oblique segments are drawn to the line, these cut off equal distances from the foot of the perpendicular, and make equal angles with the perpendic- ular and with the given line. ' Give the proof in full, as of an independent theorem. 112. Theorem. // from a point in a perpendicular to a line oblique segments are drawn, cutting off equal distances from the foot of the perpendicular, these seg- ments are equal, and make equal angles with the perpen- dicular and with the given line. Give the proof in full, using A ABA^ of § 110. 44 PLANE GEOMETRY. 113. Theorem. The 'perpendicular is shorter than any oblique segment from a point to a line. Suggestion. Show by § 102 that the angle opposite the oblique segment in the triangle formed is greater than either of the other angles, and then make use of § 85. 114. The distance from a point to a line means the shortest distance, and hence is measured on the perpen- dicular. 115. Theorem. If from a point in a perpendicular to a line segments are drawn cutting off unequal dis- tances from the foot of the perpendicular, the segments are unequal, that segment heing the greater ivhich cuts off the greater distance. Given AO ± BC and 0C> OB. To prove that AC> AB. Proof : Let ob' = OB and draw AB'. Then, Zob'A< rt. Z, and .-. Z AB'C > rt. Z. Also Z OCA < rt. Z. .-.aoab'. .'. AC > AB since AB' = AB. Give reasons for each step. llal6. Theorem. If from a point in a perpendicular to a line unequal segments are drawn, these cut off unequal distances from the foot of the perpendicidar, the greater segment cutting off the greater distance. Suggestion. Using the figure of § 115 and the hypoth- esis that AC > AB, show that two of the following state- ments are impossible: (1) OC = 0B ', (2) OC < OB ; (3) OC > OB. See § 86. BECTILINEAR FIGURES. 45 117. Theorem. // in two triangles two sides of the one are equal respectively to two sides of the other, hut the included angle of the first is greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given A ABC and A'B'C in which AB = A'B\ BC = B'C' and ZB>ZB'. To prove that AOA'c'. Proof: Place Aa'b'c' on A ABC so that a'b' coincides with its equal AB and C' is on the same side of AB as C. Let BD bisect Z c'bc, meeting AC in D. Draw DC' . Then ABDC' ^ABDC. (Wby?) .\DC' — DC and AD -K DC' = AC. (Why ?) But , AD + DC'>AC'. (Ax. Ill, § 61) AOAC'. (Ax. VII, §82) 118. Theorem. // in two triangles two sides of the one are equal to two sides of the other hut the third side of the first is greater than the third side of the second, then the included angle of the first is greater than the included angle of the second. Suggestion. Using the figure of § 117 and the hypothesis that AC>a'c', show that two of the three following statements are impossible: (1)ZB = Zb'; (2)Z5Zb'. See § 86. 46 PLANE GEOMETRY. 119. Axiom IX. // a, h, c, d are line-segments (or angles) such that a>h and c — d or such that a>h and c>d, then a-{- c>h + d. Also if a>h and cSd, then a — c> h — d, provided a>c and h>d.' (See §§ 10, 39.) 120. Theorem. The sum of the segments drawn from a point within a triangle to the extremities of one side is less than the sum of the other two sides. c Given the point within the A ABC. To prove that A0+ OB <:AC -\- CB. Proof : AO + OE 5, the arc will cut AK only once on the right of J., and hence only one triangle will be found. Repeat this construction, making a separate figure for each case. Make the construction when Z ^ is a right angle. Are all three cases possible then? Make the construction when Z ^ is obtuse. What cases are possible then ? 48 PLANE GEOMETRY. 123. EXERCISES. •-— 1. A carpenter bisects an angle A as follows: Lay off AB = AC. Place a steel square so that BD = CD as shown in the figure. Draw the line A D. Is this method correct ? Give proof. Would this method be correct if the square were not right-angled at Z) ? 2. In the triangle ABC, AC = BC. The points D, E, F are so placed that AD = BD and ^F = BE. Compare DE and DF. Prove your conclusion. 3. If in the figure Z 2 - Z 1 = 15°, and Z 4 = 120, find each angle of the triangle. -=^ 4. If in A^BC, AC = BC, and if ^C is extended to D so that AC =^ CD, prove that DB is perpendicular to AB. 5. In A ABC, CA = CB, AD = BE. Prove A ADB ^ A ABE. 6. In the triangle KLN, NM is perpendicular to KL, and KM - MN - ML. Prove that KLN is an isosceles right triangle. -^7. If in the isosceles A ABC a point D lies in the base, and Z 1 = Z 2, determine whether there is any position for D such that DE = DF. 8. If the bisectors AD and BE of the base angles of an isosceles triangle ABC meet in 0, what pairs of equal angles are formed? What pairs of equal segments ? Of congruent triangles? 9. If the middle points of the sides of an equi- lateral triangle are connected as shown in the fig- ure, compare the resulting four triangles. BECTILINEAR FIGURES. 49 I 10. Triangle ABC is equilateral. AD - BE = CF. Compare the triangles DBF, ECF, FAD. AD B 11. Two railway tracks cross as indicated in the figure. What angles are equal and what pairs of angles are supplementary? State a theorem involved in each case. O 12. In a A ABC does the bisector oi ZA also bisect the side BC (]) ii AC = BC but AC. yi 2\ To prove that (a) every point in I is / equidistant from A and B; (5) every / point which is equidistant from A and B / lies in I. L Proof : (a) Let P be any point in I. Draw TA and BB. Then BA^BB. (§112) (5) Let P be any point in the plane such that BA — BB. Bisect Z ABB with the line BD. Now show that A ABB ^ A BBD. And hence that AD = BD and Z ADB = Z BDB. Hence BD is the perpendicular bisector of AB^ and since there is only one such, this is the line I (§ 67). Thus the perpendicular bisector of the segment fulfills the two requirements for the locus in question. Steps (a) and (5) together show that a point not on the line I is unequally distant from A and B. D 52 PLANE GEOMETRY. 129. Problem. To find the locus of all points at a given distance from a given straight line. Solution. Given the line / and the segment a. Construct a perpendicular to I at some point A and lay off AB = a. ^ Through B draw Zj II L Then l^ is a part of the locus required. b p ^ Proof: (1) To show that any point P in Zj is at the distance a from I. Draw PA and also let fall PK±l. a k Now A ABP ^ A AKP and PK = AB = a. (Why ?) (2) To show that any point P in the plane above I whose distance from Z is a lies in Zj. Draw a ± from P to Z, meeting I and Zj in K and p' re- spectively. Then by (1) EP' = a. But KP = a. Hence P and P' coincide and P lies on Z^. .-. Zj is a part of the locus sought. Let the student find another line which is also a part of the locus. Note. In (1) and (2) above, great care is needed in keeping the hypothesis clearly in mind. 130. EXERCISES. 1. Find the locus of all points in the plane equally distant from two parallel lines. 2. Find the complete locus of all points in the plane equally dis- tant from two intersecting lines. 3. Find the locus of all points in the plane equally distant from a fixed point. 4. Find the locus of all points in the plane equally distant from two fixed points. RECTILINEAR FIGURES. 53 ^ c 131. Theorem. The bisectors of the three angles of a triangle meet in a 'point Given AD, BE, and CF bisecting Z^, Z5, Z C respectively of the triangle ABC. To prove that AD^ BE^ and CF meet in a common point. Proof : Let AD and BE meet in some point, as O. Then O is equidistant from AB and AC. Why? Also O is equidistant from AB and BC. (Why ?) .-. O is equidistant from AC and BC. (Wliy ?) Then O lies on the bisector of Z C. (Why ?) That is, CF passes through the point o, and thus the three bisectors meet in a common point. 132. Theorem. The three perpendicular bisectors of the sides of a triangle meet in a point. Given FH, DG, and EK perpendicular bisectors of the sides AB^ BC, and CA of i^ABC. To prove that FH^ DG, and EK meet in a point. Proof: The proof is exactly similar to that of § 131. 0V5 " 133. EXERCISES. 1. In § 131 how do we know that ^Z) and 5jE; meet? Suggestion. Show that AD and BE cannot be parallel. 2. In § 132 how do we know that FH and GD meet? 3. Do the bisectors of the angles always meet inside the triangle ? 4. Do the perpendicular bisectors of the sides always meet inside the triangle? Draw figures to illustrate the various cases. 54 PLANE GEOMETRY, THEOREMS ON QUADRILATERALS. 134. Definitions. If no three of the points -4, iJ, C, D lie in the same straight line, the figure formed by the four segments AB^ BC^ CD, DA, is called a quadrilateral. The segments are the sides of the quadrilateral a!iad4he points are its vertices. Two sides are adjacent if they meet in a vertex,' as ^J5 and BC. Otherwise they are opposite, as AB and CD. Two vertices are adjacent if they lie on the same side, as A and B. Otherwise they are opposite, as A and C. A diagonal of a quadrilateral is a segment joining two opposite vertices, as ^C and BD. Quadrilaterals which have a reentrant angle, such as Zbcd, and those in which two sides intersect, such as AB and CD, are not considered here. Rhomboid Rhombus Rectangle Square 135. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. A rhomboid is a parallelogram whose angles are oblique. A rhombus is an equilateral rhomboid. A rectangle is a parallelogram whose angles are right angles. A square is an equilateral rectangle. The side on which a parallelogram is supposed to stand is called its lower base, and the side opposite is its upper base. RECTILINEAR FIGURES. * 55 136. A trapezoid is a quadrilateral having only one pair of opposite sides parallel. An isosceles trapezoid is one in which the two non-parallel sides are equal. In jjj trapezoid the two parallel sides are the upper and lower bases. 137. The altitude of a parallelogram or a trapezoid is the perpendicular distance between its bases, and its diameter is the segment joining the middle points of the other sides. 138. EXERCISES. 1. Name each of the following quadrilaterals on page 4 : IHPN^ IHGO, AEDB, COED, JMPG, RSED, KLWV, JIBC. To deter- mine whether opposite sides of these figures are parallel, use the pro- tractor for measuring the necessary angles. 2. Is every rectangle a parallelogram ? Is the converse true ? 3. Is every rectangle a square ? Is the converse true ? 139. Theorem. Opposite sides of a parallelogram are equal. 2) c A B Given ABCD, a parallelogram; that is, AB II CD and AD II BC- To prove that AB = CD and BC= AD. Proof : Draw the diagonal AC. Prove A ABC ^ A ACD and compare corresponding sides. What determines which are corresponding sides ? 140. EXERCISES. 1. Show that a diagonal of a parallelogram divides it into two congruent triangles. 2. Give the proof in § 139, using the diagonal BD. 56 ' PLANE GEOMETRY. 141. Theorem. The diagonals of a 'parallelogram bi- sect each other, n c A B Given O ABCD with its diagonals meeting at the point O. To prove that OC = OA and OB = OD. Proof : In the triangles AOB and COB^ determine whether sufficient parts are equal to make them congruent, and if so compare corresponding parts. Give all the details of the proof. Can the proof be given by using the AAOD and BOd If so, give it. 142. Theorem. // a quadrilateral has hath pairs of opposite sides equal, the figure is a parallelogram. D ' G A B Given a quadrilateral ABCD in which AB = CD and AD = BC. To prove that AB II CD and AD II BC. Proof : Draw the diagonal AC. In the A ABC and ADC determine whether any test for congruence applies, and if so compare corresponding angles. 143. EXERCISES. 1. By use of the last theorem the question of Ex. 1, § 138, can be answered by measuring sides instead of angles. Verify the results by this process. 2. Which two of the theorems §§ 139, 141, 142 are converse? State in detail the hypothesis and conclusion of each. RECTILINEAR FIGURES. 57 144. Theorem. If a quadrilateral has one pair of opposite sides equal and parallel^ the figure is a parallel- ogram. ^ jy 2"^.. A B Given. (State all the items given in the hypothesis.) To prove. (State what needs to be proved in order to show that ABCD is a parallelogram.) Proof : From the data given prove that A ABC ^ A BBC. Use corresponding angles of these congruent triangles to show that the other two opposite sides are parallel. Hence show that the figure is a parallelogram. Write out this demonstration in full. Could the theorem be proved equally well by drawing the other diagonal ? If so, draw it and give the proof. 145. EXERCISES. 1. Prove that if the diagonals of a quadrilateral bisect each other it is a parallelogram. What is the converse of this proposition ? 2. Show that if two intersecting line-segments bisect each othet, the lines joining their extremities are parallel. 3. The parallel lines Z, and l^ are cut by a trans- versal AB. AC and AD bisect Z2 and Zl respec- tively. Prove that A CBA and DBA are isosceles. Compare the segments CB and BD. 4. If in an isosceles right triangle ABC the bisectors of the acute angles meet at 0, find how many degrees in the Z 1 thus formed. 58 PLANE GEOMETRY. 146. Theokem. // the two diagonals of a 'parallelo- gram are equal, the figure is a rectangle. Given O ABCD with AC = BD. To prove that Zl = Z 2 = Z3 = Z4 = rt.Z. Proof : Show that A ABD ^ A ABC. .'. Zl = Z3. (Why?) But Zl + ^3 = 2rt.Zs. (Why?) .-. Zl = Z3 = rt.Z. (Why?) In like manner prove that Z 2 = Z 4 = rt. Z. Hence the figure is a rectangle (§ 135). 147. EXERCISES. 1. State and prove the converse of the theorem, § 146. 2. Show that a diameter of a parallelogram passes through the in- tersection of its diagonals. See § 137. Suggestion. Show that it bisects each diagonal. 3. Prove that the diagonals of a square are perpendicular to each other. 4. Prove that the diagonals of a rhombus are perpendicular to each other. 5. Does the same proof apply to Exs. 3 and 4 ? 6. Are the diagonals of a square equal? Is this true of a rhom- bus? Prove each answer correct. 7. Do the diagonals of a square bisect each other? Is this true of a rhombus? Of a trapezoid ? 8. Show that if two adjacent angles of a parallelogram are equal, the figure is a rectangle. BECTILINEAB FIGURES. 59 148. Theorem. Two parallelograms having an angle and the two adjacent sides of one equal respectively to an angle and the two adjacent sides of the other are con- gruent ^ ^ ^, ^, Given [U ABCD and A'tlCD' such that AB = A'B\ AD = A'Df andZ^ = A'. To prove that O ABCD ^ O a'b^c'b^. Proof: Apply O ABCD to CJ a'b'c'd' so as to make Z.A coincide with Z.A', AB falling on A'b', and AD on A'D'. Then BC takes the direction b'c\ since Z.B=/.b\ being supplements of the equal angles A and A'. (Why ?) C falls on c\ since BG = jR'o', being segments equal to the equal segments AD and A'd'. (Why ?) Hence CD coincides with c'd'. (Why ?) Therefore O ABCD ^ O a'b'g'd\ since they coincide throughout. 149. EXERCISES. 1. Are two parallelograms congruent if they have a side and two adjacent angles of the one equal respectively to a side and two adja- cent angles of the other ? Draw figures to illustrate your answer. 2. Are two parallelograms congruent if they have four sides of one equal to four sides of the other? Show why, and draw figures to illustrate. 3. Compare the theorem of § 148 and Exs. 1 and 2 preceding with the tests for congruence of triangles. 4. Prove that the opposite angles of a parallelogram are equal. 5. State and prove the converse of Ex. 4. 60 PLANE GEOMETRY. OTHER THEOREMS APPLYING PARALLELS 150. Theorem. // a line bisects one side of a triangle and is parallel to the base, then it bisects the other side and the included segment is equal c to one half the base. Given a line / il AB in A ABC such y^ — ^ ^» that AD =^ DC. To prove that BE = EC and de=Iab. Proof : Draw EF through E jjarallel to CA. Now show that AFED is a parallelogram, and that A dec ^ A FBE, from which * CE = EB, DE = FB, DE = AF, and AB = AF-{- FB = 2de, or DE = I AB. State the reasons for each step. 151. Theorem. A segment connecting the middle points of two sides of a triangle is parallel to the third side and equal to one half of it. Given A ABC and the segment DE such that AD - DC and CE = EB. See figure in § 1.50. To prove that DE II AB and DE — ^ AB. Proof : Since each of the sides AC and BC has but one middle point (§ Q>k>)., it follows that there is but one seg- ment DE bisecting both these sides. But by § 150 a certain segment parallel to the base fulfills this condition. Hence, DE is parallel to the base AB. Then, DE = 1 AB as in § 150. RECTILINEAR FIGURES. 61 152. Theorem. If a segment is parallel to the bases of a trapezoid and bisects one of the non-parallel sides, then it bisects the other also and is equal to one half the sum of the bases. d ^c eZ. A B Given the trapezoid ABCD in which AE = ED and EFllAB. To prove that BF= FC and EF= ^(AB + DC}. Proof : Draw the diagonal AC meeting EF in O. In A ACD,AO = OC and E0 = ^ DC. (Why?) In A ABC,BF = FC and OF^^AB. (Why ?) Adding, EO+OF = EF= |(^Z? + DC). (§ 61, V) Prove this theorem also by drawing the diagonal BD. State a similar theorem for a parallelogram and make a proof for that case which is simpler than the above. 153. Theorem. // a segment connects the middle points of the two non-parallel sides of a trapezoid, it is parallel to the bases and equal to one half their sum. Proof : The argument is similar to that in § 151. Give it in full. 154. EXERCISES. 1. Does a theorem similar to tliat of § 153 hold for a parallelogram ? If so, state it and give a simpler proof in this case. 2. If in the figure DE, FG, HI, etc., are parallel to AC and if FEy HG, JI, etc., are parallel to BC, find the sum of BD, DE, EF, FG, GH, etc. How, if at all, does the length of AB enter into the solution? Suggestion. Produce GF, IH, KJ, to meet BC, forming ZI7. Then show that BD + EF + GH, etc., ^ BC, and similarly ED+ GF+ IH, etc., = AC. 62 PLANE GEOMETRY. 155. Theorem. // a series of ^parallel lines intercept equal segments on one transversal, they intercept equal segments on every transversal. Given the parallel lines k, Uj /g, 4, cutting the transversal ti soth&.tAB = BC=CD. To prove that on the transversal t^ EF= FG = GH. Proof : The figure ACGE is a parallelogram or a trapezoid according as t^ II t^ or not. in either case BF., which bisects AC, also bisects EG (§152). .■.EF=FG. Similarly, in the figure DHFB, FG = GH. .'. EF= FG= GH. 156. Theorem. The three altitudes of a triangle meet in a point, n tA' Y . Outline of Proof: Through each vertex of the given triangle ABC draw a line parallel to the opposite side, forming a triangle A'b'c'. RECTILINEAR FIGURES. 63 Show that ACA'b^ and AB'cb are parallelograms, and hence that b'c = AB = CA'. That is, C is the middle point of A'b'. In the same manner show that A and B are the middle points of b'c' and a'c' respectively. Also show that the altitudes of A ABC are the perpendicular bisectors of the sides of A a' b'c', and therefore meet in a point (§132). 157. Definition. A segment connecting a vertex of a triangle with the middle point of the opposite side is called a median of the triangle. 158. Theorem. The three medians of a triangle meet in a point which is two thirds the distance from each ver- tex to the middle point of its opposite side. A H B Given A ABC with medians BD and AE meeting in O. To prove that the median from C also passes through o, and that AO = ^AE, BO = ^ BD, and CO = | CH. Outline of Proof : Taking F and G?, the middle points of OB and OA respectively, use §§ 151 and 144 to show that the figure GFED is a parallelogram, and hence that 1)0= OF = FB and EO = 0G= GA. That is, O trisects AE and BD. In the same way we find that AE and CH meet in a point o' which trisects each of them. Hence o and o' are the same point. Therefore the three medians meet in a point which trisects each of them. 64 PLANE GEOMETRY. 159. EXERCISES. 1. If one angle of a parallelogram is 120°, how many degrees in each of the other angles ? 2. If the angles adjacent to one base of a trapezoid are equal, then those adjacent to the other base are equal. 3. If the angles adjacent to either base of a trapezoid are equal, then the non-parallel sides are equal and the trapezoid is isosceles. 4. In an isosceles trapezoid the angles adjacent to either base are equal. 5. Divide a segment AB into three equal parts. ^ ^^^ Suggestion. From A draw a segment AC^ ^ and on it lay off three equal segments, AD^ DE, j^- EF (§ 33). JDraw FB and construct EG and DH each parallel to FB. Prove AH = HG = GB. 6. To cut braces for a roof, as shown in the figure, a carpenter needs to know the angle DBC when the angle DAB is given, it being given that AB = AD. Show how to find this angle. (See § 123, Ex. 14.) 7. If each of the perpendicular bisectors of the ^ sides of a triangle passes through the opposite vertex, what kind of a triangle is it ? If it is given that two of the perpendicular bisectors of sides pass through the opposite vertices, what kind of a triangle is it? If only one? 8. Find the locus of the middle points of the segments joining a vertex of a triangle to all points on the opposite side. 9. Given a line I and a point P not in the line. Find the locus of the middle points of all segments drawn from P to l. 10. The length of the sides of a triangle are 12, 14, 16. Four new triangles are formed by connecting middle points of > the sides of this triangle. What is the sum of the sides of these four triangles ? 11. Draw any segment PA meeting a line Z in ^. Lay off AB on I. With P and B as centers and with AB and ylP as radii respectively, strike arcs meet- ing in C. Draw PC, and prove it parallel to l. RECTILINEAR FIGURES. 65 12. If the side of a triangle is bisected by the perpendicular upon it from the opposite vertex, the triangle is isosceles. 13. State and prove the converse of this theorem. 14. If in a right triangle the hypotenuse is twice as long as one side, then one acute angle is 60° and the other 30°. Suggestion. Let D be the middle point of AB. Use Ex. 12 and the hypothesis to show that A A CD is equilateral. Prove the converse by drawing CD so as to make ZBCD = ZB. 15. A stairway leading from a floor to one 12 feet above it is con- structed with steps 8 inches high and 10 inches ^^ .^ wide. What is the length of the carpet re- quired to cover the stairway, allowing 10 inches for the last step, which is on a level with the upper floor. 16. A stairway inclined 45° to the horizon- tal leads to a floor 15 feet above the first. What is the length of the carpet required to cover it if each step is 10 inches high ? If each is 12 inches? If each is 9 inches? Can this problem be solved without know- ing the height of the steps? Is it necessary to ^i know that the steps are of the same height? 17. If the vertex angle of an isosceles triangle is 60°, show that it is equilateral. 18. By successively constructing angles of 60° divide the perigon about a point into six equal angles. (This is possible because 360 ^ 6 = 60.) With as a center construct a circle cutting the sides of these angles in points A, B, C, D, F, F. Draw the segments AB, BC, etc. Show by Ex. 17 that each of the six triangles thus formed is equilateral. Show also that ABCDEF is equiangular and equi- lateral, that is, AB = BC, etc., and Z ABC = Z BCD, etc. 66 PLANE GEOMETRY. POLYGONS. 160. Definitions. A polygon is a figure formed by a series of segments, AB, BC^ CD, etc., leading back to the starting point A. The segments are the sides of the polygon and the points A, B, C, D, etc., are its vertices. The angles A, B, C, Z), etc., are the angles of the polygon. A E Convex. F Concave. Equiangular. Equilateral. Regular. A polygon is convex if no side when produced enters it. Otherwise it is concave. Only convex polygons are here considered. A polygon is equiangular if all its angles are equal and equilateral if all its sides are equal. A polygon is regular if it is both equiangular and equi- lateral. A segment connecting two non-adjacent vertices is a diagonal of the polygon. The perimeter of a polygon is the sum of its sides. 161. Theorem. The sum of the angles of a 'polygon having n sides is (2 n — 4) right angles. Proof : Connect one vertex with each of the other non-adjacent vertices, thus forming a set of triangles. Evidently the sum of the angles of these triangles equals the sum of the angles of the polygon. RECTILINEAR FIGURES. 67 Now show that if the polygon has n sides there are (w — 2) triangles. The sum of the angles of one triangle is 2 rt. A. Hence, the sum of the angles of all the tri- angles, that is, the sum of the angles of the polygon, is 2(7i-2) = (2n-4) rt. A 162. Theorem. The sum of the exterior angles of a polygon, formed by producing the sides in succession, is four right angles. Outline of Proof : The sum of both exterior and interior angles is 2 n rt. A. (Why?) The sum of the interior angles is (2 w — 4) rt. A. (Why ?) Hence, the sum of the exterior angles is 4 rt. A. (Why ?) Write out the proof in detail, using the figure. 163. EXERCISES. 1. What is the sum of the angles of a polygon of 3 sides? of 4 sides? of 5 sides? of 6 sides? of 10 sides? of 18 sides? 2. Find each angle of a regular polygon of 3 sides, 4 sides, 5 sides, 6 sides, 8 sides, 14 sides, n sides. 3. Construct a regular triangle, thus obtaining an angle of 60". 4. Construct a regular quadrilateral. What is its common name? 5. Prove that a regular hexagon ABODE F may be constructed as follows : Let A be any point on a circle with center 0. With A as center and OA as radius describe arcs meeting the circle in B and in F. With B as center and the same radius describe an arc meeting the circle in C, and so for points to D and E. See § 159, Ex. 18. 68 PLANE GEOMETRY. SYMMETRY. 164. Two points A and A' are said to be symmetric points with respect to a line Z if Z is the per- pendicular bisector of the segment AA'. A figure is symmetric with respect to an axis I if for every point P in the figure there is also a point P' in the figure such that P and P^ are symmetric points with respect to I. This is called axial symmetry. Two separate figures may have an axis of symmetry between them. 165. Theorem. Two figures which are symmetric with respect to a line are congruent. Given two figures F and F' symmetric with respect to a line /. To Prove that f^f'. Proof: This is evident, since, by folding figure F over on the line I as an axis, every point in F will fall upon a corresponding point in f' (Why ?). 166. Corollary. // points A and A^ and also B and B' are symmetric with re- spect to a line I, then the segments AB and A^B^ are symmetric with respect to I. 167. EXERCISES. 1. How many axes of symmetry has a square ? A rectangle? A rhombus? An isosceles trapezoid? 2. If a diagonal of a rectangle is an axis of symmetry, what kind of a rectangle is it? RECTILINEAR FIGURES. 69 3. If a triangle has an axis of symmetry, what kind of a triangle is it? Assume that the axis passes through one vertex. 4. If a triangle has two axes of symmetry, what kind of a triangle is it ? 5. How many axes of symmetry has an equilateral triangle? 6. How many axes of symmetry has a regular pentagon (five-sided figure) ? 7. How many axes of symmetry has a regular hexagon ? 8. Show that one figure of § 40 has an axis of symmetry. State this as a theorem. 168. Problem. Given a polygon P and a line I not meeting it, to construct a 'polygon P' such that P and P' shall he symmetric with respect to I. Solution. Let A, B, C, D, E be the vertices of the given polygon P, and / the given line. Construct A' ^ b', c\ 1>', e' symmetric respectively to J., J?, C, 7), E with respect to the line I. Then the polygon P' formed by joining the points A' ., b\ c', d', e\ a' in succession is symmetric to P. Proof : Give the proof in full. Figures having an axis of symmetry are very common in all kinds of decoration and architectural construction. 70 PLANE GEOMETRY. 169. Two points A and A' are symmetric with respect to a point O if the segment A A' is bisected by o. A figure is synimetric with respect to a point O if for every point P of the figure there is a point p' also in the figure such that P and p' are symmetric with respect to O. Such a figure is said to have central sym- ^^ metry with respect to the point. The point /\;;--.(0// is called the center of symmetry. A circle / /^''' ^-^ has central symmetry. '^^ Two separate figures may have a center of symmetry between them. 170. EXERCISES. 1. Prove that if A and A' and also B and B' are symmetric with respect to a point 0, then the segments AB and A'B' are symmetric with respect to 0. 2. Prove that if the triangles ABC and A'B'C are symmetric with respect to a point 0, then they are congruent. 171. Problem. Given a polygon P and a point outside of it, to construct a polygon P' symmetric to P with respect to 0. Solution". Construct points symmetric to the vertices of P. Connect these points, forming the polygon P', and prove this is the polygon sought. D' \P ^V D" BECTILINEAR FIGURES. . 71 172. Theorem. // a figure has two axes of symmetry at right angles to each other, their point of intersection is a center of symmetry of the figure. Outline of Proof : It is to be shown that for every point P in the figure, a point p" also in the figure can be found such that PO = P"o and POP" is a straight line. Draw PP' ± Z, p'p" ± V, p"p"' ± I and connect the points P'" and P. Now use the hypothesis that I LV, and each an axis of symmetry, to show that pp'pf'p'" is a rectangle ^" of which DD" and b'd"' are the diameters. Hence O, the inter- section of DD" and d'd'", bisects the diagonal PP", making p" symmetric to P with respect to O, § 147, Ex. 2. Give the proof in full. 173. EXERCISES. 1. Has a square a center of symmetry? has a rectangle? 2. If a parallelogram has a center of symmetry, does it follow that it is a rectangle ? . 3. Has a trapezoid a center of symmetry? has an isosceles trapezoid ? 4. If two non-parallel straight lines are symmetric with respect to a line I, show that they meet this line in the same point and make equal angles with it. (Any point on the axis of symmetry is regarded as being symmetric to itself with respect to the axis.) 5. If segments AB and A'B' are symmetric with respect to a point 0, they are equal and parallel. 6. Has a regular pentagon a center of symmetry ? See figure of Ex. 6, § 167. 7. Has a regular hexagon a center of symmetry ? Ex. 7, § 167. 8. Has all equilateral triangle a center of symmetry ? 72 . PLANE GEOMETRY. METHODS OF ATTACK. 174. No general rule can be given for proving theorems or for solving -problems. In the case of theorems the following suggestions may be helpful. (1) Distinguish carefully the items of the hypothesis and of the conclusion. It is best to tabulate these as suggested in § 79. (2) Construct with care the figure described in the hy- pothesis. The figure should be as general as the terms of the hypothesis per- mit. Thus if a triangle is called for but no special triangle is men- tioned, then a scalene triangle should be drawn. Otherwise some particular form or appearance of the figure may lead to unwarranted conclusions. (3) Study the hypothesis ivith car,e and determine whether any auxiliary lines may assist in deducing the properties required by the conclusion. Study the theorems previously proved in this respect. A careful review of these proofs will lead to some insight as to how they were evolved. 175. Direct Proof. The majority of theorems are proved by passing directly from the hypothesis to the conclusion by a series of logical steps. This is called direct proof. It is often helpful in discovering a direct proof to trace it backward from the conclusion. Thus, we may observe that the conclusion C follows if statement B is true, and that B follows if A is true. If then we can show that A is true, it follows that B and C are true and the theorem is proved. Having thus discovered a proof, we may then start from the beginning and follow it directly through. As an example consider the following theorem: RECTILINEAR FIGURES. 73 The segments connecting the middle points of the opposite sides of any quadrilateral bisect P each other. s< Given segments AB and CD connect- ing the middle points of the quadrilat- eral PQRS. To prove that AB and CD bisect each other. Proof : Draw the diagonals PR and SQ and the segments AD, DB, BC, CA. Xow AB and CD bisect each other if ADBC is aO, and ADBC is a Oil AD W CB and A C II DB. But AD II CB since each is II SQ. See § 151. And A C II DB since each is II PR. Hence ADBC is aOand AB and CD bisect each other. Notice that the auxiliary lines PR and SQ divide the figure into triangles and this suggests the use of § 151. 176. Indirect Proof. In case a direct proof is not easily found, it is often possible to make a proof by assuming that the theorem is not true and showing that this leads to a conclusion known to he false. As an example consider the follow- ing theorem : A convex polygon cannot have more than three acute angles. Proof : Assume that such a polygon may have four acute angles, as Z 1, Z2, Z3, Z4. Extend the sides forming the exterior angles 5, 6, 7, 8. Since Z 1 + Z 5 = 2 rt. ^, Z 2 + Z 6 = 2 rt. A, etc., and since Z 1, Z2, Z 3, Z 4 are all acute by hypothesis, it follows that Z 5, Z 6, Z 7, Z8 are all obtuse, and hence Z 5 4- Z 6 + Z 7+ Z 8 > 4 rt. zi. But this cannot be true since the sum of all the exterior angles is exactly 4 rt. ^ by § 162. Hence the assumption that Zl, Z2, Z3, Z4 are all acute is false. That is, a convex polygon cannot have more than three acute angles. 74 PLANE GEOMETRY. The proof by the method of exclusion (§ 86) involves the indirect process in showing that all but one of the pos- sible suppositions is false. 177. The solution of a problem often involves the same kind of analysis as that suggested for the discovery of a direct proof (§ 175). For instance, consider the following problem : To draw a line parallel to the base of a triangle such that the segment included hetwee^i the sides shall equal the sum of the segments of the sides between the parallel and the base. Given the A ABC. c To find a point P through which to draw DE WAB y\ so that DP+ PE = AD-^r EB. pX^P \ ?<7 Construction. Draw the bisectors oiAA and B ..g^^^V"^ r^ meeting in point P. A. B Then P is the point required. Proof : Z 1 = Z 3 since DE WAB, and Zl = Z2 since BP bisects Z B. .-. Z 3 = Z 2 and PE = BE. (Why ?) Likewise in A A DP, AD =DP. Hence DP+ PE = AD+ BE. (Ax. Y.) Or DE = AD + BE. This construction is discovered by observing that a point P must be found such that PE = BE and PD = AD. This will be true if Z3=:Z2 and this follows if Zl = Z2, while at the same time Z6 =Z5 and Z 4 = Z 5. Hence the bisectors of the base angles will determine the point P. Having thus discovered the process, the construction and proof are made directly. 178. In general the most effective help is a ready knowl- edge of the facts of geometry already/ discovered, and skill in applying these will come with practice. It is important for this purpose that summaries like the following be made by the student and memorized: BECTILINEAR FIGUBES. 75 179. (fl^) Two triangles are congruent if they have: (1) Two sides and the included aiigle of the one equal to the corresponding parts'of the other. (2) Two angles and the included side of the one equal to the corresponding parts of the other. (3) Three sides of the one equal respectively to three sides of the other. In the case of right triangles : (4) The hypotenuse and one side of one equal to the corre- sponding parts of the other. (5) Two segments are proved equal if: (1) They are Homologous sides of congrueyit triangles. (2) They are legs of an isosceles triangle. (3) They are opposite sides of a parallelogram. (4) They are radii of the same circle. SUMMARY OF CHAPTER I. 1. Make a summary of ways in which two angles may be shown to be equal. 2. Make a summary of ways in which lines are proved parallel. 3. What conditions are sufficient to prove that a quadrilateral is a parallelogram ? 4. Make a list of problems of construction thus far given. 5. Make a list of definitions thus far given. Which of the figures defined are found on page 4? 6. Tabulate all theorems on (a) bisectors of angles and segments, (6) perpendicular lines, (c) polygons in general, (d) symmetry. ' 7. What are some of the more important applications thus far given of the theorems in Chapter I ? 76 PLANE GEOMETRY. Border, Parquet Flooring. PROBLEMS AND APPLICATIONS. 1. Divide each side of an equilateral triangle into three equal parts (§ 159, 5), and draw lines through the division points as shown in the figure. Prove (a) The six small triangles are equilateral and congruent to each other. (First prove them equiangular.) (b) The two large triangles are congruent. (c) The inner figure is a regular hexagon. 2. Let Zj and h be parallel lines, with A E per- pendicular to both. Lay off segments AB, BC, CD, . . . and EF, FG, GH, . . . each equal to AE. Connect these points as shown in the figure. H Tile Floor Border. Parquet Floor Border. Prove (a) EKA, EFK, A KB, BLC, etc., are congruent right isosceles triangles. (h) FLBK, etc., are squares. Bisect AB, BC, . . . EF, FG . . . and join points by lines parallel to J&5 and ^irrespectively. Notice that the resulting figure forms the basis for the floor border to the right. 3. Let Zj h be parallel lines, with MN perpendicular to both. Through F, the middle point of MN, draw lines KE and A G, each making an angle of 30° with MN. Q H G M E D Parquet Floor Border. Prove FE = AFhj showing /\ANF^/\ FEM. Also prove KF = FG and that ^ FKA and FEG are equilateral. BECTILINEAR FIGURES. 77 Lay off segments AB, BO, ED, DP, KJ, etc., each equal to AK. Connect points as shown in the figure. Prove ABCO^A CPD ^ A AFK, etc. Prove that ABCDEF and JKFGHl are regular hexagons. 4. A network of congruent equilateral tri- angles is constructed in a rectangle ABCD as shown in the figure. • (a) How many of these triangles meet in a common vertex? (b ) Does this n umber of equilateral triangles exactly cover the whole plane about the vertices? Why? (c) Do the triangles that meet in one point form a regular hexagon ? (d) At what angle to the horizontal lines are the oblique lines that form sides of the triangles? e.g. what is the angle DCK'i TTTTTl rWTTTT (e) Compare the lengths of CK and AK (see § 159, Ex. M). Tile Flooring. To construct this network when a side a of the triangles is given, proceed as follows : From the vertex J. of a right angle lay off segments equal to a along one side AK. With one of these division points K as center and with a radius equal to twice AK strike an arc meeting the side AC at C. Draw CD parallel to AB and lay off segments on it equal to a. Connect these points as shown in the figure, and through the intersection points draw the horizontal lines, thus fixing the division points on ^ C and BD. Prove that the resulting triangles are equilateral and congruent to each other. Suggestion. Use in order the converse of § 159, Ex. 14 ; § 144, and the fact that a diagonal divides a parallelogram into two congruent triangles. Notice how this construction is studied. The figure is first sup- posed constructed and its properties tabulated. Some of these (c) and (d) are then used in making the construction. This method is of very general application in problems of construction. Note. This is the method by which a designer would construct a network of congruent regular triangles. 78 PLANE GEOMETRY, 5. Construct a network of triangles as in Ex. 4, using pencil and then ink in parts of the lines, making a set of regular hexagons as shown in the figure. How many such hexagons meet in a point? Will this number ,9 exactly cover the plane about the point? Why? ^ / \ f \ / \ y-H-H-< T"^^T" E¥3 Tile Flooring. 6. Point out how the figure constructed under Ex. 5 may be made the basis for the two tile floor patterns here given. 7. Given a rhombus whose acute angles are 60 °. Show how to cut off two triangles so the resulting figure shall be a regular hexagon. What fraction of the whole rhombus is thus cut off? What fraction of the pattern is black ? 8. In a parallelogram, prove : («) The diameters divide it into four congruent parallelograms. {h) The segments joining the midpoints of the sides in order form a parallelogram, and the four triangles thus cut off may be pieced together to form another parallelogram congruent to the one first formed. (c) The diameters and diagonals all meet in a point. Suggestion. Prove that a diameter bisects a diagonal. Do these propositions hold for a square? What part of each of the three large squares in the adjoining figure is white? 9. The octagons (eight-sided polygons) in the figure are equiangular and the small quadrilaterals are squares. %v^ • Find the angles of the irregular hexagons. If the octagons were regular, would the hexagons be equiangular ? equilateral ? reg- ular? Tile Pattern. ^^"•^'J!li|iiii 7'ili:i!! l'yTil XT :^ .U. Tile Border. BECriLINEA R FIG URES. 79 Tile Pattern. 10. In the figure the octagons are equi- angular. (a) Will two such octagons and a square completely cover the plane about a point ? Why ? (J)) At what angles do the slanting sides of the octagons meet the horizontal line ? Find all the angles of the two small white trapezoids to the right. (c) Could a pattern of this type be constructed by using regular octagons ? 11. Equilateral triangles and regular hexagons have thus far been found to make a complete pavement. What other regular polygons can be used to make a complete pavement? '^ 12. If three regular pentagons (five-sided polygons) meet in a common vertex, will they completely cover the plane about that point? If not, by how great an angle will they fail to do so? V 13. If four regular pentagons be placed about a point, by how great an angle will they overlap? 14. If three regular seven-sided polygons are placed about a point, by how great an angle will they overlap? If two, by how many ^degrees will they fail to cover the plane? In Exs. 12, 13,' 14 it is understood that, so far as possible, the poly- gons are so placed that no part of one of them lies within another; that is, they are not to overlap. 15. Answer questions like those in Ex. 14 for regular polygons of 8, 9, and 10 sides. Can any regular polygon of more than six sides be used to form a complete pavement? Note that the larger the number of sides of a regular polygon the larger is each angle. 16. A carpenter divides a board into strips of equal width as fol- lows : Suppose five strips are desired. Place a steel square in two positions, as indicated in the figure, at such angles that the distance in inches diagonally across the board shall be some multiple of five (in the figure this distance is 15 inches). Mark v ^ the points and connect them as shown in the figure, lines divide the board into equal strips. Prove that these 80 PLANE GEOMETRY. (The black and white figures on this page are parquet floor patterns.) 17. Given an isosceles right triangle ABC with the altitude CO upon the hypotenuse. (a) Show how to draw xyWAC such that xy = Cy. Suggestion, Bisect Z OCA and let the bisector meet AB in x. Draw xy II A C. Prove A xyC isosceles. (b) Prove xyCA s^u isosceles trapezoid. (c) Draw yz II OB and prove yz = yx. Suggestion. Prove A Cyz isosceles. (d) Prove that xyCA and xyzB are congruent trapezoids. 18. ABCD is a square. Lines are drawn as shown in the figure so that Z1 = Z2 = Z3 = ,Z4. (a) Prove A ABy ^ A BCz ^ A CDw ^ADAx. (b) Prove each of these triangles a right triangle. (c) Prove that xyzw is a square. (d) If Z 1 = 45°, what can be said of the figure xyzw ? This ^nd the following design are of Arabic origin. 19. On the sides of a square ABCD the points E, F, G, H are laid ofe so that AE = BF= CG = DH. Ax and zC, By and wD are in the • diagonals of the square and Ey, Fz, Gw, and Hx are parallel to these.^ D G C A E Prove that : (a) AxH, EyB, etc., are congruent right isosceles triangles. (b) AEyx, BFzy, etc., are congruent parallelograms. (c) xyzta is a square. (d) If AB = a, find how long A E should be taken in order that xy shall equal EB ; 'also equal to one half EB. RECTILINEAR FIGURES. 81 G C 20. A BCD is a square, AE = BF= CG = DH and A G, CE, BH, and DF are drawn as shown in the figure. Prove that : (a) A G W EC and BHWFD. (b) Triangles HwA, ExB, etc., are congruent. (c) The four trapezoids A Exw, etc., are congruent. (d) xyzw is a square. D N M G ^m Parquet Flooring. 21. In the square ABCD, AF = AG = HB = BK, etc., and the figure is* completed as shown. (a) Pick out all isosceles triangles. Prove. (/>)" Pick out all congruent triangles. Prove. (c) Has this figure one or more axes of syiiimetry? D M L C A F G B Linoleum Pattern. 22. On the sides of the square ABCD points E, F, G, H, . . . are taken, so that AE = AF = GB = BH = etc. The middle points of EF, GH, KL and MN are connected, form- ing the quadrilateral PR ST. Prove that : (a) AC \^ the perpendicular bisector of EF and KL. (V) FGRP is an isosceles trapezoid. (c) PRT and RST are congruent right triangles. (d) PR ST is a square. 82 PLANE GEOMETRY. 23. Ou the sides of a square ABCD points G, H^ E, F, P, etc., are taken, so that GA = AH = EB = BF=PC, etc. On the diagonals AC and BD points K, L, M, N, are laid off so that OK = OL = 0M= ON. (a) Prove that GAHN = EBFK, etc. (b) Fvovethsit HEKON=FPLOK. _ Suggestion. other. Superpose one figure on the 24. The bisectors of the angles of a rhomboid form a rectangle ; those of a rectangle form a square. ^ ^ Parquet Flooring. 25. In the figure ABCDEF is a regular hex- j agon. ABHG, BCLK, etc., are squares. (a) What kind of triangles are BHK, CLM, ' etc. ? Prove. (b) Is the dodecagon (twelve-sided polygon) HKLMN. . . regular? Prove. ♦ (c) How many axes of symmetry has this dodecagon ? ,^ (d) Has it a center of symmetry ? (e) Are the points S, F, B, K collinear? (That is, do they lie in the same straight line?) Q. xx^' ■IJsi Tile Pattern. 26. If from any two points P and Q in the base of an isosceles triangle parallels to the other sides are drawn, two parallelograms are formed whose perimeters are equal. 27. The middle point of the hypotenuse of a right tri- angle is equidistant from the three vertices. (This is a very important theorem.) 28. State and prove the converse of the theorem in Ex. 27. RECTILINEAR FIGURES. 83 m 29. The bisectors of the four interior angles formed by a transversal cutting two parallel lines form a rectangle. • 30. The sum of the perpendiculars from a point in the base of an isosceles triangle to the sides is equal to the altitude from the vertex of either base angle on the side opposite. 31. Find the locus of the middle points of all seg- ments joining the center of a parallelogram to points on the sides. (See § 159, Ex. 8.) 32. In the parallelogram A BCD points E and F are the middle points oi AB and CD respectively. Show that AF and CE divide BD into three equal segments. y 33. The sum of the perpendiculars to the sides of an equilateral triangle from a point P within is the same for all such points P (i.e., the sum is a constant). Suggestion. Prove the sum equal to an altitude of the triangle. 34. In any triangle the sum of two sides is greater than twice the median on the third side. ' 35. A BCD is a square, and EFGH a rec- tangle. Does it follow th'dt AAEH ^AFCG and AEBF^AHDG'^ Prove. H A E B ■ 36. If a median of a triangle is equal to half the base, the vertex angle is a right angle. 37. State and prove the converse of the theorem in Ex. 36. CHAPTER II. STRAIGHT LINES AND CIRCLES. Outside Point 180. A circle (§ 12) divides the plane into two parts such that any point which does not lie on the circle lies within it or outside it. 181. A line-segment joining any two points on a circle is called a chord. A chord which passes through the center is a diameter. 182. If a chord is extended in one or both directions, it cuts the circle and is called a secant. 183. A tangent is a straight line which touches a circle in one point but does not cut it. An indefinite straight line through a point outside a circle is a secant, a tangent, or does not meet the circle. 184. The portion of a circle included between any two of its points is called an arc (§12). An arc AB is denoted by the symbol AB. A circle is divided into two arcs by any two of its points. If these arcs are equal, each is a semicircle. Otherwise one is called the major arc and the other the minor arc. Unless otherwise indicated AB means the minor arc. In case of ambiguity a third letter may be used, as arc AmB. . 84 STRAIGHT LINES AND CIRCLES. 85 An arc is said to be subtended by the ^Aed Ato chord which joins its end-points. Evi- dently every chord of a circle subtends two arcs. Unless otherwise indicated the arc subtended by a chord means the minor arc. 185. An angle formed by two radii is called a central angle. An angle formed by two chords drawn from the same point on the circle is called an inscribed angle. If the sides of an angle meet a circle the arc or arcs which lie within the angle are called intercepted arcs. If the vertex of the angle is within or on the circle there is only one intercepted arc ; if it is outside the circle there are two inter- cepted arcs, as AB and CD in the figure. 186. If a circle is partly inside and partly outside another circle, then they cut each other. If two circles meet in one and only one point, they are said to be tangent. Arcs of two circles are tangent to each other if the complete circles of which they form a part are tangent to each other. 187. Two circles which can be made to coincide are said to be equal. The word congruent is unnecessary here, since all circles are similar, 188. EXERCISES. 1. Does the word circle as used in this book (§ 12) mean a curved line or the part of the plane inclosed by that line? 2. In how many points can a straight line cut a circle ? 3. In how many points can two circles cut each other ? OS 86 PLANE GEOMETRY. PRELIMINARY THEOREMS ON THE CIRCLE. 189. Radii or diameters of the same circle or of equal circles are equal. 190. If the radii or diameters of two circles are equal, the circles are equal. 191. A diameter of a circle is double the radius. 192. A point lies within, outside, or on a circle, accord- ing as its distance from the center is less than, greater than, or equal to the radius. \ 193. If an unlimited straight line contains a point within a circle, then it cuts the circle in two points. 194. // two circles intersect once, they intersect again. See figure, § 186. 195. // a straight line is tangent to each of two circles at the same point, then the circles do not intersect, hut are tangent to each other at this point. See § 186. 196. // two arcs of the same circle or equal circles can he so placed that their end-points coincide and also their centers, then the arcs coincide throughout or else form a complete circle. -V 197. If in two circles an arc of one can he made to coincide with an arc of the other, the circles are equal. 198. A circle is conveniently referred to by indicating its center and radius. Thus, O OA means the circle whose center is and radius OA. When no ambiguity arises, the letter at the center alone may be used to denote the circle. Thus, O C means the circle whose center is C. STRAIGHT LINES AND CIRCLES. 87 199. Theorem. In the same circle or in equal circles equal central angles intercept equal arcs. Given the equal circles C and C and Z C=Z C. To prove that AB = Jab'. Proof : Place O C on Q & so that C falls on &, and Z c coincides with Z C'. Then A falls on A' and B on b'. (§ 189) Hence AB = A^'. (§ 196) 200. Theorem. In the same circle or in equal circles equal arcs are intercepted by equal central angles. Given O C = O C, AB = A^. (See figure, § 199.) To prove that Zc=Zc' . Proof : Since AB = A'B\ the equal circles can be made to coincide in such manner that the arcs will also coincide. That is, A will fall on A', B on b\ and C on C' . Hence Z c coincides with Z c'. 201. EXERCISES. 1. If two equal circles are so placed that their centers coincide, what can be said of the circles? 2. Can two intersecting circles have the same center? " 3. From a point on a circle construct two equal chords. , 4. Show that the bisector of one of the angles formed by the chords in Ex. 3 passes through the center of the circle. 88 PLANE GEOMETRY. /, 202. Measurement of Angles. If the ^erigon at the cen- ter of a circle be divided by radii into 360 equal angles, these radii will divide the circle into 360 equal arcs accord- ing to the theorem, § 199. Hence, we speak of an arc of 1°, 2°, 3°, etc., and similarly for minutes and seconds. For this reason a central angle is said to be measured hy the arc which it intercepts^ meaning that a given central angle contains a number of unit angles equal to the number of unit arcs in the intercepted arc. 203. Definitions. A quadrant is an arc of 90°. A semicircle is an arc of 180°. A right angle is, there- fore, measured by a quadrant and a straight angle is meas- ured by a semicircle. A sector is a figure formed by two X X^ radii and their intercepted arc. Thus, the sector BCA is formed by the radii CB,CA,2indiBA. 204. EXERCISES. 1. Show how to bisect an arc, using §§ 48, 199. Divide an arc into four equal parts. Into eight equal parts. 2. Show that in the same circle or in equal circles two sectors hav- ing equal angles are congruent. 3. Show that if two sectors in the same circle or in equal circles have equal arcs, the sectors are congruent. 4. How many degrees in the arc which measures 'a right angle? a straight angle? half a right angle? three fourths of a straight an- gle? two thirds of a right angle? two thirds of a straight angle? 5. Show that the diameter is the longest chord of a circle. 6. Show that the two arcs into which the extremities of a diameter divide a circle are equal ; that is, each is a semicircle. Suggestion. Fold the figure over on the diameter. ' 7. Show that by bisecting the angles between two perpendicular diameters, a circle is divided into eight equal parts. STRAIGHT LINES AND CIRCLES. 89 205. Theorem. A diameter perpendicular to a chord bisects the chord and also its subtended arc. Given the diameter EF ± AB at D. To prove that AD = DB and AF — BF. Proof : Draw the radii CA and CB. If it can be shown that A ACD ^ A BCD, then (1) AD=BD (Why?), and (2) Zacd=Z.bcd (Why?); (3) i>=i> (Why?). 206. Theorem. A line perpendicular to a radius at its extremity is tangent to the circle. h ^ Given AB _L CD at D. To prove that AB is tangent to the circle ; that is, does not meet it in any other point than D. Proof : Let E be any point of AB other than D. f^ Draw segment CE. Then CE > CD (Why?). Hence E is outside the circle (§ 192). That is, every point of AB except D is outside the circle, and hence AB is tangent to the circle (§ 183). 90 / PLANE GEOMETRY. 207. Theorem. // a line is tangent to a circle, it is perpendicular to the radius drawn to the point of tan- gency. ^ D E B Given O CD with a line AB tangent to the circle at D. To prove that AB ± CD at B. Proof: If CB is iiot-L^J5, then some other line, as CE^ must be±^B (§ 67), thus making CE< CB (Why?). The point E would then lie within the circle (§ 192), and the line AB would meet the circle in two points (§ 193). But this contradicts the hypothesis tliat AB is tangent to the circle. Hence no other line than CB can be perpendicular to AB from c, and as one such line exists, it must be CB. 208. EXERCISES. 1. What type of proof ^s used in the preceding paragraph ? 2. How are the two theorems immediately preceding related to each other. 3. Show that there is only one tangent to a circle at a given point on it, and that the perpendicular from the center upon the tangent meets it at the point of contact. 4. A perpendicular to a tangent at the point of tangency passes through the center of the circle. 5. The perpendicular bisector of any chord passes through the center of the circle. 6. Two tangents at the extremities of a diameter are parallel. 7. A diameter bisects all chords parallel to the tangei^ts at its extremities, and also bisects the central angle subtended by each chord. STRAIGHT LINES AND CIRCLES. ' 91 8. A diameter bisecting a chord (or its subtended arc) is perpendicular to the chord. 9. The mid-points of parallel chords all lie on a diameter. 10. A tangent to a circle at the mid-point of any arc is parallel to the chord of the arc. 209. Theorem. If two circles meet on the line joining their centers, they are tangent to each other at this point. Given (D CB and CB meeting in a point B on the line C'C. To prove that (D CB and c' B are tangent to each other at B. Proof: (1) When each circle is outside the other. Let D be ani/ point on O c'b other than B. Draw CD and c'd. Then CD + c'd > CB + c'b. (Why ?) But c'b = c'd. .-. CD > CB. (Ax. IX, § 119) .'. D is outside of O CB. (2) WTien O CB is inside of O c'b. Let D be ani/ point on O CB other than B. Draw CD and c'd. Then c'c + CD > c'd. (Why ?) But C'C + CD = C'C + CB = c'b. (Why ?) .-. C'b > c'd. (Ax.VII, § 82) .-. D is within O c'b. Therefore O c'b and O CB have only one point in com- mon and hence are tangent to each other (§ 186). 92 PLANE GEOMETRY. PROBLEMS AND APPLICATIONS. 1. If the distance between the centers of two circles is equal to the sum of their radii, how are the circles related? Construct and prove. 2. If the distance between the centers of two circles is equal to the difference of their radii, how are the circles related ? Construct and prove ? 3. If the distance from the center of a circle to a straight line is equal to the radius, how is the line related to the circle ? Construct and prove. 4. Given two circles having the same center, construct a circle tangent to each of them. Can more than one such circle be con- structed ? What is the locus of the centers of all such circles ? 5. Prove the converse of the theorem in § 209. 6. The straight line joining the centers of two intersecting circles bisects their common chord at right angles. 7. A line tangent to each of two equal circles is either parallel to the segment join- ing their centers or else it bisects this seg- ment. • 8. In the figure AD = DB. Semicircles are constructed on AD, DB, and yl5 as di- ameters. Which semicircles are tangent to each other? 9. In the figure A, B, C, D are the vertices of a square. Show how to construct the entire figure. What semicircles are tangent to each other? This construction occurs frequently in designs for tile flooring. See accompanying figure. This is from a Roman mosaic. STBAIGHT LINES AND CIRCLES. 93 % 10. Given two parallel lines BE and AD, to con- / struct arcs which shall be tangent to each other and one of which shall be tangent to BE at B and the other tangent to AD at A. Solution. Draw AB and bisect this segment at C; construct J- bisectors oi AC and BC. From A and B draw Js to AD and BE respectively, thus locating the points and 0'. Prove that and 0' are the centers of the required arcs. Suggestion. Show that 0, C, and 0' lie in a straight line and use the theorem of § 209. This construction occurs in archi- tectural designs and in many other applications. In the accompanying designs pick out all the arcs that ^^ are tangent to each other and also the points of tangency. ^ S^r^„ ^^^^ 11. On the sides of the equilateral triangle ABC as diameters, semicircles are drawn, as AEFB. Also with A, B, C as centers and AB B,s radius arcs are drawn, ^ as^^, ^. (a) Prove that the arcs AEFB, BDEC, and CFDA meet in pairs at the middle points D, E, F of the sides of the triangle. Suggestion. If the mid- dle points of the sides of an equilateral triangle are joined, what kind of triangles are formed? (h) What arcs in this figure are tangent to each other ? • (c) Has the figure one or more axes of symmetry? This figure and the two follo^ipg occur frequently in church windows and other decorative designs. Fourth Presbyterian Church, Chicago. 94 PLANE GEOMETRY. 11. Construct the design shown in the figure. Suggestion. Divide the diameter AB into six equal parts and construct the three semicircles. On DC and DC as bases construct equilateral triangles with vertices and 0'. With radius equal to CB and centers O and 0' construct circles. AC' D C B (a) Prove that O is tangent to each of the three semicircles. Likewise O 0'. (h) Erect a, ± to AB at D and prove © and 0' tangent to it. (c) Prove circles with centers at and 0' tangent to each other. (c?) Has this figure one or more axes of symmetry? 12. In the figure AB, CD and OD are bisected, and O'O" II AB through E. DO' = DO" = | DB. Circles are constructed as shown in the figure. (a) If ^5 is 4 feet, what is the radius of each circle? LQ"\ E\ (b) Prove that O is tangent to GO' and also to O 0". Suggestion. Show that 00' is the sum of the radii of the two circles. (c) Is Q 0' tangent to the arc ACB and also to the line AB ? (d) Has this figure one or more axes of symmetry ? 13. A BCD is a square. Arcs are constructed with A, B, C, D as centers and with radii each equal to one half the side of the square- The lines AC, BD, MN, and RS are drawn, and the points E, F, G, H are connected as shown in the figure. The arc SN is extended to P, forming a semicircle. The line LP meets SAf in K, and BK meets MN in 0'. (a) Prove that EFGH is a square. {h) Prove that A KLO' and KPB are mutually equiangular and each isosceles. (c) Prove that O'K is tangent to FG and to SN. ^ (c?) How many axes of symmetry has the solid figure? (e) Show that O'K is tangent to RN by drawing O'C and fold- ing the figure over on the axis of symmetry MN. \ \ STRAIGHT LINES AND CIRCLES. 95 210. Theorem. An angle inscribed in a circle is measured by one half the intercepted arc. Given Z DBA inscribed in O CB. To prove that Z DBA is measured by ^ AD. Proof : (1) If one side^ as BD^ is a diameter. Draw the radius CA. Show that Z 2 = | Z 1. But Z 1 is measured by AD (§ 202). Hence Z 2 is measured by ^ AD. (2) If the center C lies within the angle. Draw the diameter BE. Now Z DBA = Z 1 + Z 2. Complete the proof. (3) If the center C lies outside the angle. Draw BE and use the equation Z DBA = Z\ - Z2. 211. It follows from § 210 that if in equal circles two inscribed angles intercept equal arcs, they are equal ; and conversely, that if equal angles are inscribed in equal circles, they intercept equal arcs. 212. EXERCISES. 1. If the sides of two angles BAD and BA'D pass through the points B and Z) on a circle, and if the vertex A is on the minor arc BD and A' is on the major arc BD, find the sum of the two angles. 2. In Ex. 1 if the points B and D remain fixed while the vertex A of the angle is made to move along the minor arc of the circle, what can be said of the angle A ? What if it moves along the major arc ? 96 PLANE GEOMETRY. 213. Theorem. The locus of the vertices of all right triangles on a given hypotenuse is a circle whose diame- ter is the given hypotenuse. ^ ^ ^' Outline of Proof : Let AB be the given hypotenuse. (1) If P is any point on the circle whose diameter is AB^ Z APB = rt. Z. (Why ?) '(2) .If AF^ B is any right triangle with AB as hypote- nuse, then AC=CB = CP'. (See Ex. 27, p. 82.) State the proof in full. 214. PROBLEMS ON LOCI. Find the following loci : 1. The centers of all circles of fixed radius tangent to a fixed line. 2. The centers of all circles tangent to two parallel lines. 3. The centers of all circles tangent to both sides of an angle. 4. The centers of all circles tangent to a given line at a given point. Is the given point a part of this locus ? 5. The vertices of all triangles which have a common base and equal altitudes. I 6. The middle points of all chords through a fixed point on a circle. Use Ex. 8, §208, and then §213. 7. The points of intersection of the diagonals of trapezoids formed by the sides of an isosceles triangle and lines parallel to its base. 8. Two vertices of a triangle slide along two parallel lines. What is the locus of the third vertex if the triangle is fixed in size and shape? ,. *, 9. ^5CZ) is a parallelogram all of whose sides are of fixed length. *The side AB is fixed in position. Find the locus of the middle points of the remaining three sides. STRAIGHT LINES AND CIRCLES. 97 10. Prove that in the same circle or in equal circles equal chords are equally distant from the center. Suggestion. MB = ND. Why? Then prove A BMC = A CND. 11. State and prove the converse of the theorem in the preceding exercise. (What parts of A BMC and CND are now known?) 12. Find the locus of the middle points of all chords of equal length in the same circle. 13. Find the locus of the middle point of a segment AB of fixed length which moves so that its end-points slide along the sides of a right angle. (Use Ex. 27, p. 82.) 14. Find the locus of the points of contact of two varying circles tangent to each other, and each tangent to a given line at a given point. Suggestion. A and B are tlie fixed points, I \\z:'-i^.^S' and P one point of contact of the circles. Draw the common tangent PD. Prove AD = DP and DB = DP. Hence, D is the middle point of AB and DP is constant. That is, the locus is a circle of which AB is a diameter. 15. Find the locus of the centers of all circles tangent to a fixed circle at a fixed point P. Is the fixed point P a part of this locus ? Is the center of the fixed circle a part of it? ^ 16. Find the locus of the centers of all circles of the same radius which are tangent to a fixed circle. % Under what conditions will this locus include the fixed circle itself? The center of this fixed circle ? Will the locus ever contain a circle within the fixed circle? '^ Under what conditions will the locus consist , of two circles, each outside the fixed circle ? i Under what condition does the locus consist V,pf only one circle? 17. In making core-boxes, pattern makers use a square as indicated in the figure to test whether or not the core is a true semicircle. Is this method correct? Prove. 98 PLANE GEOMETRY. 215. Theorem. The arcs intercepted by two parallel chords or by a tangent and a chord parallel to it are equal. Given AB \\ DE and LK II MN. To prove that AD — BE and LB = KB. Proof : (1) Draw chord DB. Compare Z 1 and Z 2, and hence show- that iB = B^. (Why?) (2) Draw the radius CB to the point of tangency. Then CB ± MN and CB J. LK. (Why?) Prove A LCS ^ A KCS, and hence that LB = BK. (§ 199) 216. EXERCISES. 1. Prove that a tangent at the vertex of an inscribed angle forms equal angles with the two sides, if these are equal chords. 2. If the vertices of a quadrilateral lie on a circle, any two of its opposite angles are supplementary. 3. If two chords of a circle are perpendicular to each other, find the sum of each pair of opposite arcs into which they divide the circle. 4. If the vertices of a trapezoid lie on a circle, its diagonals are equal. 5. Two circles intersect at C and D. Diame- ters CA and CB are drawn. Prove that A, D, B lie on a straight line. Suggestion. Prove that ZADC = Z CDB = rt. Z. STRAIGHT LINES AND CIRCLES. 99 217. Theokem. An angle formed by two intersecting ^ chords is measured by one half the sum of the arcs inter- cepted by the angle itself and its vertical angle. Given Z 1 formed by the chords AB and DE. To prove that Zl is measured by | (^AE+ BD). Proof : Through A draw the chord AF || ED. Compare Z 1 and Z 3. Compare AE and DF, also AE + BD and BD + DF. How is Z 3 measured ? Hence, how is Z 1 measured ? 218. EXERCISES. 1. A chord AB ia divided into three equal parts, AC, CD, and DB. OA, OC, OD, and OB are drawn. Compare the angles AOC, COD, and DOB. 2. The accompanying table refers to the figure in § 217. Fill out blank spaces. 3. In a circle C with a diameter AB 21, chord AD \q drawn, and a radius CE \\ A D. Prove that arcs DE and EB are equal. 4. The vertices of a square A BCD all lie on a circle. E is any point on the arc AB. Prove that EC and ED divide the angle A EB into three equal parts. Zl AE BD EB AnD 35° 40° 80° 48° 50 216 40° 50° 60° 60° 54° 190° 45° 90° 180° 34° 108° 164° 100 PLANE GEOMETRY, 219. Theorem. An angle formed by a tangent and a chord drawn from the 'point of tangency is measured by one half the intercepted arc. D/ Given Z 1 formed by tangent BD and chord BA. To prove that Z 1 is measured hy ^ BA. Proof : Draw a chord EF 11 BD intersecting BA in Q. Compare Z 1 and Z 2, also EB and BF. How is Z 2 measured ? Hence, how is Z 1 measured ? Give the proof in full. 220. Definitions. A segment of a circle, or a circle-segf- ment, is a figure formed by a chord and the arc which it subtends. For each chord there are two circle-segments corresponding to the two arcs which it subtends. If a chord is a diameter the two circle- segments are equal. An angle is said to be inscribed in an arc if its vertex lies on the arc and its sides meet the arc in its end-points. Such an angle is also said to be inscribed a^ in the circle-segment formed by the arc and its chord. E.g. Z 1 is inscribed in the' arc APB or in the segment APB. STRAIGHT LINES AND CIRCLES. 101 221. EXERCISES. 1. Show that an angle inscribed in a semicircle is-j; right angle. 2. If the sides of a right angle pass through the extremities of a diameter, show that its vertex lies on the circle. 3. If a triangular ruler J/iVO, right-angled at 0, is moved about in the plane so that two fixed points, A and B, lie always on the sides MO and NO respectively, what path o does the point trace? 4. Draw two concentric circles, having different radii, and show that all chords of the outer circle which are tangent to the inner circle are equal. 5. In an equilateral triangle construct three equal circles, each tangent to the two other circles and to two sides of the triangle. Suggestion. Construct the altitudes of the triangle and bisect angles as shown in the figure. Complete the construction and prove that the figure has the required properties. (a) Has the figure consisting of the triangle and the three circles one or more axes of symmetry? u (b) Has it a center of symmetry? 6. Within a given circle construct three equal circles, each tangent to the other two and to the given circle. Suggestion. Trisect the circle at D, E, c and F by making angles at the center each equal to 120°. Draw tangents at D, E, and F, and prove that A ABC is equilateral. Construct the altitudes and prove that they meet the sides of the triangle at the points of tangency of the given circle with the sides of ABC, and also that they pass" through the center of the given circle. Bisect angles as shown in the figure and prove that the centers of the required circles are thus obtained. • (a) Has the figure consisting of the four circles one or more axes of symmetry? ». (6) Has it a center of symmetry ? 102 PLANE GEOMETBT. \ 22ii. Theorem. The angle formed hy two secants, two ia;ng0nts; x)r:a tangent and a secant, meeting outside a circle, is measured hy one half the difference of the intercepted arcs. Outline of Proof : In each case the given angle is equal to Z 1, and the arc which measures Z 1 is the difference between two arcs, one of which is the larger of the two intercepted arcs and the other is equal to the smaller. For instance, in the first figure, FH = DF ~ DH =z DF — BE. , Give the proof in detail for each figure. 223. EXERCISES. 1. If (in left figure, §222) ZA =17° and EB = 25 ^ find DF. 2. If ZA =37° (in middle figure), find the arcs into which the points B and E divide the circle. 3. With a given radius construct a circle passing through a given point. How many such circles can be drawn ? What is the locus of the centers of all such circles ? 4. Draw a circle passing through two given fixed points. How many such circles are there ? What is the locus of the centers of all such circles ? 5. Construct a circle having a given radius and passing through two given points. How many such circles can be drawn? Is this construction ever impossible? Under what conditions is only one such circle possible ? STRAIGHT LINES AND CIRCLES. 103 224. Problem. To construct a circle through three fixed points not all in the same straight line. Given three points A, B, C not in the same straight line. To construct a circle passing through them. Construction. Let the student give the construction and proof in full. (See § 132.) 225. Definition. The circle OA in § 224 is said to be circumscribed about the triangle ABC and the triangle is said to be inscribed in the circle. 226. EXERCISES. 1. In the construction of § 224 why do DM and EN meet ? 2. Why cannot a circle be drawn through three points all lying in the same straight line ? Make a figure to illustrate this. 3. Show that an angle inscribed in an arc is greater than or less than a right angle according as the arc in which it is inscribed is less than or greater than a semicircle. 4. Prove that the bisectors of the angles of an equilateral triangle pass through the center of the circumscribed circle. 5. Draw a circle tangent to two fixed lines. How many such circles are there ? What is the locus of their centers ? Is the point of intersection part of this locus? Discuss fully. 6. Show that not more than one circle can be drawn through three given points, and hence that two circles which coincide in three points coincide throughout. 104 PLANE GEOMETRY. 227. Problem. To construct a circle tangent to each of three lines, no two of which are "parallel and not all of which pass through the same point. Given the lines /i, h, ly To construct a circle tangent to each of these lines. Construction. Since no two of the lines are II, let \ and Zg meet in ^, l^ and Zg in ^, and Zg and Zj in D, where A^ />♦, and Z) are distinct points. Draw the bisectors of Z^ and AB and let them meet in point C. Then C is the center of the required circle. (See § 131.) Give the proof in full. 228. Definitions. The circle in the construction of § 227 is said to be inscribed in the triangle ABB. Three or more lines which all pass through the same point are called concurrent. Hence the lines Z^, Zg, Zg are not con- current. 229. EXERCISES. 1. Why is the construction of § 227 impossible if /^ Z2, and h are concurrent ? 2. If two of the lines are parallel to each other, show that the con- struction is possible. How many tangent circles can be constructed in this case ? Draw a figure and give the construction and proof in full. 3. Is the construction possible when all three lines are parallel? Why? STRAIGHT LINES AND CIRCLES. 105 4. If two sides of the triangle are produced, as AB and AD in the figure of § 227, construct a circle tangent to the side BD and to the prolongations of the sides AB and AD. This is called an escribed circle of the triangle. 5. How many circles can be constructed tangent to each of three straight lines if they are not concurrent and no two of them are parallel? 6. Draw a triangle and construct its inscribed and circumscribed circles and its three escribed circles. 230. Pkoblem. From a given point outside a circle to draw a tangent to the circle. n Given O CA and an outside point P. To construct a tangent from P to the circle. Construction. Draw CP. On CP as a diameter con- struct a circle, cutting the given circle in the points A and B. Draw tlie lines PA and PB. Then PA and PB are both tangents. Give the proof. 231. EXERCISES. 1. If in the figure of § 2.30 the point P is made to move towards the circle along the line PC until it finally reaches thgjcircle, while PA and PB remain tangent to the circle, describe the motion of the points A and B and also of the lines PA and PB. How does this agree with the fact that through a point on the circle there is only one tangent to the circle ? 106 PLANE GEOMETRY. 2. Can a tangent be drawn to a circle from a point inside the circle ? Why ? 3. Show that the line connecting a point outside a circle with the center bisects the angle formed by the tangents from that point. 4. Why are not more than two tangents possible from a given point to a circle ? 5. The two tangents which can be drawn to a circle from an ex- terior point are equal. ^ 6. In a right triangle the hypotenuse plus the diameter of the inscribed circle is equal to the sum of the two legs of the triangle. 7. If an isosceles triangle inscribed in a circle has each of its base angles double the vertex angle, and if tangents to the circle are drawn through the vertices, find the angles of the resulting triangle. 8. If the angles of a triangle ABC inscribed in a circle are 64°, 72°, and 44°, find the angles of the triangle formed by the tangents to the circle at the points A , B, and C SUMMARY OF CHAPTER II. 1. Make a list of all the definitions involving the circle. 2. State the theorems on the measurement of angles by inter- cepted arcs. 3. State the theorems involving equality of chords, central angles, and intercepted arcs. 4. State the theorems on the tangency of straight lines and circles. 5. State the theorems involving the tangency of two circles. 6. Make a list, to supplement that in the summary of Chapter T, of ways in which two angles or two line-segments may be proved equal. 7. State the ways in which two arcs of the same or equal circles may be proved equal. 8. State the problems of construction given in Chapter II. v^/ 9. Explain what is meant by saying that a central angle is meas- ured by its intercepted arc. 10. State some of the important applications of Chapter 11. (Return to this question after studying those which follow.) STRAIGHT LINES AND CIRCLES. 107 PROBLEMS AND APPLICATIONS. 1. Given two roads of different width at right angles to each other, to connect them by a road whose sides are arcs of circles tangent to the sides of the roads. (a) Make the construction shown in the figure and prove that it has the required properties. ih) Is this construction possible when the given roads are not at right angles to each other ? Illustrate. (c) Can the curve be made long or short at will ? id) Make the construction if the given roads have the same width. 2. Two circles C and C are tangent at the point D. ^^ is a segment through D tern)i- nating in the circles. Prove that the radii CA and CB are parallel. ^ 3. Through a point on the bisector of an angle to construct a circle tangent to both sides of the angle. Construction. Through the given point P draw EP _L to ^P. Lay off EB = EP and at D construct DC ±AD meeting the line A P in C. Then C is the center of the required circle and CD is its radius. Proof : Draw PD and prove that A DEP is isosceles and hence also A PDC. Is it possible to construct another circle having the properties required? If so, construct it. This construction is used in the accompany- ing design in which the shape is determined by fixing the pgint P in ad- vance. Such designs are of frequent occurrence in decorative work such as the steel ceiling panel given here. 108 PLANE GEOMETRY. 4. In an isosceles triangle construct three circles as shown in the figure. Suggestion. First construct the inscribed circle with center O. Let the bisector oi Z.A meet a this circle in a point P. Then use Ex. 3. 5. The angles formed by a chord and a tangent are equal respec- tively to the angles inscribed in the arcs into which the end-points of the chord divide the circle. 6. If a triangle whose angles are 48°, 56°, and 76° is circumscribed about a circle, find the number of degrees in the arcs into which the points of tangency divide the circle. 7. Divide each side of an equilateral triangle into three equal parts (Ex. 5, § 159) and connect points as shown in the figure. Prove that DEFGHK is a regular hexagon. 8. If a circle is inscribed in the triangle of Ex. 7, prove that all sides of the hexagon are tangent to the circle. Suggestion. Show that the perpendicular bisec- tors of the segments HK, KD, DE meet in a point equidistant from these segments. ^ 9. Within a given square construct four equal circles so that each circle is tangent to one side of the square and to two of the circles. Suggestion. First construct the diagonals of the square. In the F 10. figure, A BCD is a rectangle D^ with AD = I AB. E and F are the middle points oi AB and CD respectively. Semicircles are con- structed with E and F as centers and \ AE di& a radius, etc. Ji E B Fan vaulting from Gloucester Cathedral, England. 8TBAIGHT LINES AND CIRCLES. 109 (a) Prove that these quadrant arcs are tangent to each other in pairs and also to the semicircles. (b) Lines are drawn tangent to the arcs at the points where these are met by the diagonals of the squares AEFD and BCFE. Prove that these lines form squares KLMN and X YZ W. (c) Construct the small circles within each of these squares. The above design occurs in fan-vaulted ceilings. The gothic or pointed arch plays a conspicuous part in modern architecture, and examples of it may be found in al- most any city. Its most common use is in church windows. The figure represents a so-called equilat- eral gothic arch. The arcs AC and BC are drawn from B and A as centers respectively, and with AB as a radius. The segment ^B is called the spaw of the arch,., and the point C its a'pex. 11. In the figure AB = DB. ABC, ADE, and DBF are equi- lateral gothic arches. (a) Construct the circle with center tangent to the four arcs as shown. Suggestion. Take X so that DX = XB. With centers A and B and radius AX draw arcs meeting at 0. Complete the con- struction and prove that the figure has the required properties. (b) Prove that DE and DF are tangent to each other. Also BF and BC, and AE and AC. (c) What axis of symmetry has this figure ? 12. A triangle ABC whose angles are 45°, 80°, and 55° is inscribed in a circle. Find the angles of the triangle formed by the tangents at A, B, and C. Door, Union Park Church, Chicago. 110 PLANE GEOMETRY. 13. Inscribe a circle in an equilateral gothic arch ABC. Suggestions. Construct CD ± to AB and ex- tend it to P, making DP = AB. From P con- struct a tangent to A C at L. (a) Prove that A BDP ^ A BLP and hence PL = BD. (b) A OLP ^ABDO and hence OD = OL. Then O OD is the required circle. See § 209. Notice that this figure is symmetrical with re- spect to the line PD, and hence if the circle is proved tangent to ACj we know at once that it is tangent to BC. 14. In the figure ABC is an equilateral gothic arch with a circle inscribed, as in Ex. 1.3. (a) Construct the two equilateral arches GHE and HKF, as shown in the figure. Construction. Draw BK and AG 1. AB. With a radius equal to OD -\- DB, and with as center draw arcs meeting BK and AG in K and G respectively. Draw GK, construct the arches and show that each is tangent to the circle. (b) Do the points E and F lie on the circle ? Suggestion. Suppose KF to be drawn, and compare Z HKF with Z HKO by comparing the sides HK and KF and also GK and KO. 15. Construct an arc passing through a given point B, and tangent to a given line AD at given point D. 16. In the figure ABC is an equilateral arch. BK is f of BD. KBF and A HE are equal equi- lateral arches. Arcs KQ and HQ are tangent to arcs KF and HE respectively. From Lincoln Cathedral, England. (a) Find by construction the center of the circle tangent to A C, BC, KF, and HE, and give proof. STRAIGHT LINES AND CIRCLES. Ill (b) Find by construction the centers of the arcs KQ and HQ. How is this problem related to Ex. 15? 17. Two circles are tangent to each other in- ternally. Find the locus of the centers of all circles tangent to both externally. 18. Two circles are tangent to each other ex- ternally. Find the locus of the centers of all circles f({ tangent to both, but external to one and internal to the other. 19. Two equal circles are tangent to each other externally. Find the locus of the centers of all circles tangent to both. 20. AD'B is an angle whose vertex is outside the circle and whose sides meet the circle in the points A and B, while Z.ADB is an inscribed angle intercepting the arc AB. Prove that Z. ADB > Z. AD'B J provided each of the segments D'A and D'B cuts the circle at a second point. 21. Through two given points A and 5 con- struct a circle tangent to a given line which is perpendicular to the line AB. Is this construction possible if the given line passes through either of the points A or 5? If it meets AB between these points? 22. In kicking a goal after a touchdown in the game of football, the ball is brought back into the field at right angles to the line marking the end of the field. The distance between the goal posts being given, and also the point at which the touchdown is made, find by a geo- metrical construction how far back into the field the ball must be brought in order that the goal posts may subtend the greatest pos- sible angle. CHAPTER III. THE MEASUREMENT OP STRAIGHT LINE- SEGMENTS. 232. A straight line-segment is said to be exactly meas- ured when we find how many times it contains a certain other segment which is taken as a unit. The number thus found is called the numerical measure, or the length of the segment. E.g. a line-segment is 9 in. long if a segment 1 in. long can be laid off on it 9 times in succession. Thus, 9 is the numerical measure, or the length of the segment, when 1 in. is taken as a unit. 233. In selecting a unit of measure it may happen that it is not contained an integral number of times in the seg- ment to be measured. Thus, in measuring a line-segment the meter is often a convenient unit. Suppose it has been applied five times to the segment AB and that the last time the end falls on A^, A^B being less than one meter. Then, taking a decimeter (one tenth of a meter) as a new unit, suppose this is contained ^ 5 S 2 S three times in A^B with a remainder AJi less ^i ^2-B than a decimeter. * Finally, using as a unit a centimeter (one tenth of a decimeter), sup- pose this is contained exactly six times in A^B. Then, the length of ^^ is 5 meters, 3 decimeters, and 6 centi- meters, or 5.36 meters. The process of measuring considered here is ideal. In practice we cannot say that a given segment is contained exactly an integral number of times in another segment. See § 235. 112 MEASUEEMENT OF LINE-SEGMENTS. 113 234. It may also happen that, in continuing this ideal process of measuring as just described, no subdivided unit can be found which exactly measures the last interval, that is, such that the final division point falls exactly on B. E.g. it is known that in a square whose sides are each one unit the diagonal is V'2, and that this cannot be exactly expressed as an integer or a fraction whose numerator and denomi- nator are both integers. By the ordinary process of extracting square root we find V2 = 1.4142 •••, each added decimal mak- ing a nearer approximation. But this process never terminates. Hence, in attempting to measure the diagonal of a square whose side is one meter, we find 1 meter, 4 decimeters, 1 centimeter, 4 milli- meters, etc., or 1.414 meters approximately. It should be noticed, however, that 1.415 is greater than the di- agonal and hence the approximation given is correct within one millimeter. 235. Evidently any line-segment can be measured either exactly or to a degree of approximation, depending upon the fineness of the instruments and the skill of the opera- tor. The word measure is commonly used to include both exact and approximate measurement. For practical purposes, a line-segment is measured as soon as the last remainder is smaller than the smallest unit available. It should be noticed that all practical measurements are in reality only approximations, since it is quite impossible to say that a given distance is, for in- stance, exactly 25 ft. It may be a fraction of an inch more or less. E.g. in the above example 1.414 meters gives the length of the diagonal for practical purposes if the millimeter is the smallest unit available. The error in this case is less than one millimeter. 114 PLANE GEOMETRY. 236. Definition. Two straight line-segments are com- mensurable if they have a common unit of measure. Other- wise they are incommensurable. E.g. two hne-segments whose lengths are exactly 5.27 and 3.42 meters respectively have one centimeter as a common unit of meas- ure, it being contained 527 times in the first segment and 342 times in the second. But the side and the diagonal of a square have no common unit of measure. In the example of § 234, the millimeter is contained 1000 times in the side and 1414 times in the diagonal, plus a remainder less than one millimeter. A similar statement holds for any unit of measure, however small. 237. For the purposes of practical measurement any two line-segments mai/ be considered as commensurable, but for theoretical purposes it is necessary to take account of in- commensurable segments also. The theorems in this chapter are here proved for com- mensurable segments only. They are proved for incom-- mensurable segments also in Chapter VII. RATIOS OF LINE-SEGMENTS. 238. The ratio of two commensurable line-segments is the quotient of their numerical measures taken with respect to the same unit. E.g. if two segments are respectively 3 ft. and 4 ft. in length, the ratio of the first segment to the second is | and the ratio of the second to the first is f. 239. The ratio of two commensurable segments is the same, no matter what common unit of measure is used. E.g. two segments whose numerical measures are 3 and 4 if one foot is the common unit, have 36 and 48 as their numerical measures if one inch is the common unit. But the ratio is the same in both cases, namely : || = |. MEASUREMENT OF LINE-SEGMENTS. 115 240. The approximate ratio of two incommensurable line- segments is the quotient of their approximate numerical measures. It will be seen that this approximate ratio depends upon the length of the smallest measuring unit available, and that the approximation can be made as close as we please by taking the measuring unit small enough. E.g. an approximate ratio of the side of a square to its diagonal is = Another and closer approximation is = ■. In 1.41 141 1.414 1414 this case the numerical measure of one of the segments is exact. An approximate ratio of V2 to VS, in which neither has an exact 1 41 141 measure, is — — = — . Another is 1.73 173 1.732 1732 241. It should be clearly understood that the numerical measure of a line-segment is a number, as is also the ratio of two such segments. Hence they are subject to the same laws of operation as other arithmetic numbers. For example, the following are axioms pertaining to such numbers : (1) Numbers which are equal to the same number are equal to each other. (2) If equal numbers are added to or subtracted from equal numbers^ the results are equal numbers. (3) If equal numbers are multiplied by or divided by equal numbers., the results are equal numbers. It is understood, however, that all the numbers here considered are positive. For a more complete con- sideration of axioms pertaining to numbers, see Chapter I of the Advanced Course of the authors' High School Algebra. 116 PLANE GEOMETRY. 242. A proportion is an equality, each member of which is a ratio. Four numbers, a, 5, c^ and c?, are said to be in proportion, in the order given, if the ratios — and - are d equal. In this case a and c are called the antecedents and h and d the consequents. Also a and d are called the extremes and h and c the means. The proportion - = — is sometimes written a : b = c : d, d and in either case may be read a is to b as e is to d. If D and E are points on the sides of the triangle ABC^ and if w, n, jt?, and q^ the numerical measures respectively of AD, DB, AE, and EC, are such that —=E^ n q then the points D and E are said to divide the sides AB and AC proportionally, that is, in the same ratio. For conveftience it is common to let AB, DB, AE, and EC stand for the numerical measures of these segments, and thus to AD AE write the above proportion, — = — or AD : DB = AE:EC. DB EC THEOREMS ON PROPORTIONAL SEGMENTS. 243. Theorem. If a line is parallel to one side of a triangle and cuts the other two sides, then it divides these sides in the same ratio. Given A ABC in which DE II BC. To prove that— =— . DB EC Proof : Choose some common measure of AD and DB, as AK. Suppose it is con- tained 3 times in AD and 5 times in DB. MEASUREMENT OF LINE-SEGMENTS. 117 Then AD (1) Through the points of division on AD and DB draw lines parallel to 5C, cutting AE and EC. By § 155 these paral- lels divide AE into three equal parts and EC into five equal parts. Hence, ^^ ^ (2) EC =h §241 .-. from (1) and (2) ^ = ^. ^ DB EC For a proof in case AD and DB are incommensurable, see § 410. 244. EXERCISES. 1. If DEW EC in A ABC, compute the segments left blank from those given in the following table : AD DB AE EC AB AC 20 24 15 4 56 42 102 12 408 25 18 342 Required to com- 2. 5 is a point visible from A but inaccessible, pute the distance from A to B. Suggestion. Select some accessible point C from which A and B are both visible. Through E, a point near A and on the line of sight from A to C, draw ED II CB and meeting the line of sight from ^ to 5 at D. Now AE, EC, and AD can be measured. Then _ i)i? and hence ^5 can be computed by § 243. Note. The advantage of this method over that in § 34, Ex. 5,, is that here a small triangle AED is made to do the service, which was there performed by another triangle the same size as ABC. i± i:Li. 118 PLANE GEOMETRY. 245. Theorem. If four numbers m, n, p, q are such that — = -, then it follows that: n q <"S= 1. P (2) ^ = ^. (3) '» + n — =1 p + q (^)m-n p-q^ q n q (5) m-hn_p + q m — n p — q Given n q To prove (1) divide the members of 1 = 1 by those of To prove (2) multiply each member of (a) by -• P To prove (3) add 1 to each member of (a) and reduce each side to a common denominator. To prove (4) subtract 1 from each member of (a) and reduce each side to a common denominator. To prove (5) divide the members of (3) by the members of (4). Write out these proofs in full, giving the reason for each step, and read off the results as applied to the figure. For example, show that (3) gives, when applied ^^ to the figure, _ >^ V AB^AC n^'^^q DB EC /- 1 246. The results in the above theorem are sometimes named as follows : The proportion (a) is said to be taken by inversion in (1), by alternation in (2), by composition in (3), by division in (4), by composition and division in (5). MEASUREMENT OF LINE-SEGMENTS. 119 247. EXERCISES. 1. If — =^, prove that = —^ — , aud hence show in the n q m -\- n p + q above figure that AD AE AB^^A^' 2. If in the figure on page 118 DEWBC, compute the segments indicated by blanks in the accompanying table. 3. If ^ n 9 show that^ + '^ = P n + q 4. If ^ n show that '"^-P = n + q m n ' ^ 5. If ^ = X ^ and 7 m_p y q show that a: = y- \ 6. If !^ = * X = -, show that y x = y 7. State Exs. 4, 5, and 6 in words. AD AB DB AE AC EC 8 12 6 6 10 16 6 9 7 12 10 .10 10 8 18 10 7 14 240 200 380 160 140 20 120 100 50 35 21 14 40 15 30 500 200 400 90 ^4% 70 „ 20 .30 ^ - 800 360 300 27 30 48 30 20 50 27 560 48 8. A triangle is formed by a chord and the tangents to the circle at its extremities. Prove that the triangle is isosceles. |k 9. A triangle with angles A, B, C is circumscribed about a circle. Find the angles of the triangle formed by the chords joining the points of tangency. 120 PLANE GEOMETRY. 248. Theorem. // a line divides two sides of- a tri- angle in the same ratio, it is parallel to the third side. Given the points D and E on the sides of the A ABC such ,^ , AD AE that — = — . DB EC To prove that BE W BC. Proof : Suppose de' is drawn paral- lel to BC. It is proposed to prove that the point E^ coincides with E. AB AC Since BE' II BC, we have AB E'C But by (3), § 245, ^ = ^. - ^ ^ ^ DB EC Now use the proof of Ex. 5, § 247, to show that e'c = EC^ and hence that e' and E coincide, so that BE II BC. Give the proof in full detail. 249. EXERCISES. 1. Show' by § 248 that the Hne joining the middle points of two sides of a triangle is parallel to the third side. Compare § 151. 2. Show by § 243 that the line which bisects one side of a triangle and is parallel to a second side bisects the third side. 3. If in the A ABC a segment Cy connects the vertex to any point y of the base, find the locus of the point x on this segment such that Cx : Cy is the same for all points y. 4. ABCD is a O whose diagonals meet in 0. If ?/ is a point on any side of the O, find the locus of a point x on the segment Oy such that Oy : xy is the same for every such point y. 5. Find the locus of the points of intersection of the medians of all triangles having the same base and equal altitudes. (Use §§ 158, 248.) MEASUBEMENT OF LINE-SEGMENTS. 121 250. Theorem. The bisector of an angle of a triangle divides the opposite side into segments whose ratio is the same as that of the adjacent sides. Given CD bisecting Z C in A ABC, r^ ... AD AC To prove that — = — . DB BC Proof : Through B draw BE II DC. Prolong ^C to meet BE at E. In A ABE ^^4£, (Why?) DB CE ^ '^ ^ Now show that A BCE is isosceles, and hence that BG may be substituted for CE in the above proportion. Com- plete the proof. 251. EXERCISES. 1. Fill in the blank spaces in the table, if in the figure of § 250 CD is the bisector of Z ^ CB. AC CB AD DB AB 8 10 6 20 16 12 35 17 40 3 2 1 7 9 12 12i 8 16 364 200 480 54 65 105 24.5 18.3 32.6 122 PLANE GEOMETUY. 252. Definition. A segment is said to be divided exter- nally by any point which lies on the line of the segment but not on the segment itself. E.g. point C divides the segment AB externally, the parts being ^ C and CB, while point B divides the segment AC in- ternally, the parts being ^5 and BC. • ' ' 253.' Theokem. a line which bisects an exterior angle of a triangle divides the opposite side externally into two segments whose ratio is the same as that of the adjacent sides of the triangle. Given CD, the bisector of the exterior angle at C of the triangle ABC. _, ., , AD AC To prove that — = — . BD BC Proof : Through B draw BE II DC. In A ACD AD^AC (Why?) BD EC \ J y Now show that A EBC is isosceles, and hence that EC = BC. Complete the proof. 254. EXERCISES. 1. Draw a triangle with an acute exterior angle bisected. Using different lettering from that in § 253, prove the theorem again. 2. Compare the proofs in §§ 250 and 253. Give the proof in § 253 for a figure in which AC < BC. MEASUREMENT OF LINE-SEGMENTS. 3. Fill in the blank spaces in the table below if CD is the bisector of the exterior angle BCK. 123 AC CB AD DB AB 7 4 9 14.3 9.6 18 164 48 144 13.7 84 60 32 60 25 56 80 40 4.8 12.5 6 550 600 400 350 200 300 c- / 2 ^K 4. To measure indirectly the distance from an accessible point A to an inaccessible point B by means of § 253. Suggestion. Through C, a point where A and B are both visible, draw CK making Z 1 = Z 2. Produce KC to a point D on the line BA extended. •^' A B What lines must now be measured in order to compute AB1 5. What methods have been used so far for the indirect measure- ment of the distance from an accessible to an inaccessible point? Compare these (a) As to the simplicity of the theory involved. (5) As to the simplicity and ease of the direct measurements re- quired. 6. Divide a given line-segment in a given ratio without construct- ing a line parallel to another. 7. Similarly divide a given line-segment externally in a given ratio. 8. Solve Ex. 3, § 249, if x is on Cy extended. 9. Solve Ex. 4, § 249, if x is on Oy extended. 124 PLANE GEOMETRY. SIMILAR POLYGONS. 255. Two polygons, in which the angles of the one are equal respectively to the angles of the other, taken in order, are said to be mutually equiangular. The angles of the two polygons are thus arranged in pairs of equal angles, which are called corresponding angles. Two sides, one of each polygon, included between cor- responding angles, are called corresponding sides. 256. Two polygons are similar if (1) they are mutually equiangular and if (2) their pairs of corresponding sides are proportional. Two polygons may have property (1) but not (2). For example, a rectangle and a square. Or they may have property (2) and not (1). For example, a square and a rhombus. Hence any proof that two polygons are similar must show that both (1) and (2) hold concerning them. In the case of triangles it will be proved that either property speci- fied in the definition of similar polygons is sufficient to make them similar. 257. Theorem. If two triangles are mutually equi- angular, they are similar. B c Given A ABC and A'B'C, in which Z A = Z.A', ZB= Z B', and ZC = ZC'. MEASUREMENT OF LINE-SEGMENTS. 125 To prove that the other property of similarity holds, 1 .. . AB BC CA namely that = = — — ,. *" A'B' B'C' c'a' Proof: Place A a'b'c' on A ABC with Z A' upon its equal Z A, and b'c' taking the position BE. Now show that BE II BC and hence — =— , , n , . » AB AC that IS, ^'B' A'& In like manner, placing Z ^' upon Z B, show that 4^ = 4^. Give the full details of this proof. 258. EXERCISES. 1. To measure indirectly the distance from an accessible point A to an inaccessible point B. from A to B, and ED ± AD. Let C be the point -27----iL on AD which lies in line with E and B. Now show that S^EDC and BAG are mutually equiangular and hence similar. What segments need to be measured in order to com- pute AB'i Give full details of proof. 2. Prove that two right triangles are similar if they have an acute angle of one equal to an acute angle of the other. 3. Two isosceles triangles are similar if they have the vertical angle of one equal to the vertical angle of the other. 4. Two triangles which have the sides of one respectively parallel or perpendicular to the sides of the other are similar. 5. Show by similar triangles that the segment joining the mid- points of two sides of a triangle is equal to one half the third side. 126 PLANE GEOMETRY. 259. Theorem. // two triangles have an angle of one equal to an angle of the other and the 'pairs of adjacent sides in the same ratio, the triangles are similar. Z A' and AB A'B' AC A'C' Given A ABC and A'B'C in which ZA To prove that A ABC'^AA'b'c'. Proof: Place AA'b'c' upon A ABC with ZA' on Z A, B'c' taking the position BE. Then, AB^AC (Why?) AD AE V J y and hence, BE \\ BC. (Why ?) Now show that A ABE and ABC are mutually equian- gular and hence similar. 260. Theorem. // two tri- angles have their pairs of corresponding sides in the same ratio, they are similar. Given A ABC and A'B'C in AB ^ BC ^ CA A'B' B'C CA' ' B c To prove that Aabc^^ A A' b'c', that is, to prove Z^ = Z ^', Zb = Z b\ Z c=Zg'. C which MEASUREMENT OF LINE-SEGMENTS. 127 Proof : Lay off on AB and AC respectively AD = A'b' and AE = A'c', and draw DE. Now prove, as. in § 259, that AADE^AABC^ and hence, that — = (1^ DE BC ^ ^ B"t' i^ = ~- (Why?) (2) Hence, since AD = A'b', it follows from (1) and (2) that de=b'c', as in Ex. 5, § 247. Now show that A A'b'c'^ A ADE, and hence, that AA^b'^-^AABC. Make an outline of the steps in this proof and show how each is needed for the one that follows. 261. EXERCISES. 1. Given a triangle whose sides are 2, 3, 4. Construct a triangle having its angles equal respectively to those of the given triangle and having a side 10 corresponding to the given side 2. 2. If each of two triangles is similar to a third triangle, they are similar to each other. 3. If in a right triangle a perpendicular be drawn from the vertex of the right angle to the hypotenuse, show that each of the triangles thus formed is similar to the given tri- angle, and hence that they are similar to each other. 4. In the figure of Ex. 3, make a table showing which angles are equal and which pairs of sides are corresponding in the fol- lowing pairs of triangles: A CD and ACB, CDB and ACB, A CD and CDB. 5. On a given segment as a side show how to construct a triangle similar to a given equilateral triangle. 128 PLANE GEOMETRY. 262. Theorem. The square on the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides. ^ . . ^^"^^^'u \ ^Ia I Given A ABC with a right angle at C. Call the lengths of the sides opposite A A, B, C, respectively, a, b, c. To prove that c^ = a^ -{- h"^. Proof : Let the perpendicular p divide the hypotenuse into tlie two parts m and n so that c = m -\- n. From A ACB and ACB show that ^ = - (1) h c From A CDB and ACB show that - = -• (2) From (1) mc = lP' ^ "^ (3) and from (2) nc = a^ (Why?). (4) From (3) and (4) (m + n)c = 0^ + 1)^ (Why?). That is, c • c = e^ =^ a?" -\- h^. For another proof of this theorem see § 319. Historical Note. The proof given above is supposed to be that given by Pythagoras, who first discovered the theorem. 263. EXERCISES. 1. The radius of a circle is 8. "What is the distance from the center to a chord v^hose length is 6 ? 2. In the same circle, what is the length of a chord whose distance from the center is 5 ? 3. Find the diagonal of a square whose side is 5 ; whose side is a. 4. What is the side of a square whose diagonal is 8 ? whose diagonal isrf? i.„vl V MEASUREMENT OF LINE-SEGMENTS. 129 5. The hypotenuse of a right isosceles triangle is 12 inches. Find the lengths of its sides. 6. The diagonals of a rhombus are 14 and 10 inches respectively. Find the length of its sides. (See § 147, Ex. 4.) C» 7. The square on the hypotenuse of a right triangle is equal to four times the square on the median to the hypotenuse. 8. What is the radius of a circle if a chord 12 inches long is 9 inches from the centre ? 9. Find the altitude of an equilateral triangle whose side is 8 ; whose side is a. 10. If the altitude of an equilateral triangle is h, find its side. (Use the formula obtained under Ex. 9.) ' f 11. Find the altitude of a triangle whose sides are 6, 8, and 10. 12. The oval in the figure is a design used in the construction of sewers. It is constructed as follows : In the O OA let CD, the perpendicular bisector of AB, meet the arc AO'B at 0'. Arcs AM and BN are drawn with the same radius AB and with centers B and A respectively. The lines BO' and A 0' meet these arcs in M and N respectively. The arc MDN has the center 0' and radius O'M. (a) Is arc ACB tangent to AM and BN at A and B respectively ? Why ? Q)) Is arc MDN tangent to AJl and BN at M and N respectively ? Why ? (c) If ^5 = 8 feet, find BO', and hence, O'M, and finally CD. That is, if the sewer is 8 feet wide, what is its depth? (c?) If the width of the sewer is a feet, show that its depth is I (4 - V2). (e) If the depth of the sewer is d feet, show that its width is c? (4 + V2) 7 (/) Compute to two places of decimals the width of a sewer whose depth is 12 feet. 130 PLANE GEOMETRY. 264. Theorem. // in two right triangles the hypote- nuse of the one equals the hypotenuse of the other, and if the sides a, h and a', ¥ are such that a > a\ then h < b\ Proof : Let c be the length of the hypotenuse in each. Then ^2 + 52^ ^2 ^nd a'^+b''^= A (Why?) ... ^2 + 52 ^^^2 ^5^2. (Why?) and a2 _ a'^ = h'^ - h\ (Why ?) «> Since a>a\ the left member of the last equation is positive, and hence the right member is also positive, that is, hA'B'. To prove that AB is nearer the center than A'b'. Proof : Draw the J^ b and b' and the radii c and c'. Then a and a' are halves of AB and A'b' respectively. Now complete the proof, using § 264. 266. Theorem. State and prove the converse of the theorem in § 265, using the same figure. 267. Definitions. A continued proportion is a series of equal ratios connected by signs of equality. ^•^' b-d-f-h The perimeter of a polygon is the sum of its sides. MEASUREMENT OF LINE-SEGMENTS. 131 268. Theoeem. In a continued proportion, the sum of the antecedents and the sum of the consequents form a ratio equal to any one of the given ratios. a. c c Q Given the continued proportion - = - = - = ^ . b d f h To prove that , , ,^ = - • h-Vd+f+h h Proof : Let T = ^• ™, c e q Then, ^ = ., j = r, j = r. ! Hence, a = hr, e = dr^ e=fr^ g = hr^ and a-\-c + e-\-g = hr-^dr -\-fr + hr = (7> + d +/+ h)r, or — ' ■ -^ z=r = —• h+d+f+h b Give all reasons in full. 269. Theorem. The perimeters of two similar poly- gons are in the same ratio as any two corresponding sides. D Proof : By definition of similar polygons ab _ bc _ cd _ be __ ea a'b'~ b'c'~~ c'b'~ d'e' ~~ e'a' Complete the proof. 132 PLANE GEOMETRY. 270. Theorem. // the diagonals drawn from one vertex in each of two polygons divide them into the same number of triangles, similar each to each and similarly placed, then the two polygons are similar, B ^ Given the diagonals drawn from the vertices A and A' in the polygons P and P', forming the same number of triangles in each, such that AI^AI', All ^ All', AIII'^AIII'. To prove that P -^ p'. Outline of Proof : (1) Use the hypothesis to show that Za = Za',^b = Zb\ Zc=Zc', etc. (2) Show that 4^ = 4^ = 44^ etc. ^ ^ a'b' b'c' c'd' TiC CT) Notice that the proportion ——- = ——- follows from b'c' c'd' ¥c'~A^~C^'' ^^'^^^ Give the proof in detail. 271. Theorem. State the converse of the "preceding theorem, and give the proof in full detail. Make an outline of all the steps in the proofs of these two theorems. MEASUREMENT OF LINE-SEGMENTS. 133 272. Theokem. // through a fixed point within a circle any number of chords are drawn, the product of the segments of one chord is equal to the product of the seg- ments of any other. Given O C with any two chords AB and DE intersecting in P. To prove that AP ■ PB = ep • PB. Proof : Draw EB and DA. TheD, Z1 = Z2 andZ3 = Z4. (Why?) Hence, A EPB ~ A PDA, (Why?) Which are corresponding angles and which are corre- sponding sides? Show that — : = EP PA Complete the proof. It follows from this theorem that if a chord AB is made to swing around the fixed point P, the product AP • PB does not change, that is, it is constant. 273. EXERCISES. 1. Which chord through a point is bisected by the diameter through that point? Why? ^y/" 2. Through a given point within a circle which chord is the short- Zest? Why? 3. The product AP- PB is the area of the rectangle whose base and altitude are the segments of AB. (See § 307.) Note that this area is constant as the chord swings about the point P as a pivot. 134 PLANE GEOMETRY. 274. Definition. If a secant of a circle is drawn from a point P without it, meeting the circle in the points A and J3, then PB is called the whole secant and PA the external segment, provided A lies between B and P. 275. Theoeem. If from a fixed point outside a cir- cle any number of secants are drawn, the product of one whole secant and its external segment is the same as that of any whole secant and its external segment, 'P Given secants PB and PE drawn from a point P. To prove that PA- PB — PD- PE. Proof : In the figure show that A PDB ~ A PAE. Complete the proof. 276. EXERCISES. 1. A point P is 8 inches from the center of a circle whose radius is 4. Any secant is drawn from P, cutting the circle. Find the product of the whole secant and its external segment. 2. From the same point without a circle two secants are drawn. If one whole secant and its external segment are 14 and 5 respec- tively and the other external segment is 7, find the other whole secant. 3. Two chords intersect within a circle. The segments of one are m and n and one segment of the other is p. Find the remaining segment. MEASUREMENT OF LINE-SEGMENTS. 135 277. Theorem. If a tangent and a secant meet out- side a circle, the square on the tangent is equal to the product of the whole secant and its external segment. Proof : Show that / / ^^^ A APD ~ A BPD and hence that PB : PB — PD : PA. B^ Complete the proof. 278. EXERCISES. 1. If a square is constructed on PD as a side, and a rectangle with PB as base and PA as altitude, compare their areas as the secant revolves about P as a pivot. 2. Show that the theorem in § 277 may be obtained as a direct consequence of that in § 275 by supposing one secant to swing about P as a pivot till it becomes a tangent. 3. A point P is 10 inches from the center of a circle whose radius is 6 inches. Find the length of the tangent from P to the circle. 4. The length of a tangent from P to a circle is 7 inches, and the external segment of a secant is 4 inches. Find the length of the whole secant. 5. What theorems are included in the following statement : " Froin a point P in a plane a line is drawn cutting a circle in A and B. Then the product PA - PB is the same for all such lines " ? 6. In a circle of radius 10 a point P divides a chord into two seg- ments 4 and 6. How far from the center is P? Suggestion. Use Ex. 5. 7. In two similar polygons two corresponding sides are 3 and 7. If the perimeter of the first polygon is 45, what is the perimeter of the second? 8. The perimeters of two similar polygons are 32 and 84. A side of the first is 11. What is the corresponding side of the second polygon ? 136 PLANE GEOMETRY. 279. Definitions. In a right triangle ABC^ right-angled at C, the ratio — is called the sine of Z A AB and is written sin A, If any other point b' be taken on the hypotenuse or the hypotenuse extended, and a perpendicular b' c' be let fall to AC^ &b' _ CB AB' ~ AB B'/ A^ c c then (Why?) Likewise in A ab"c", in which AB'^ = 1 unit, we have &'b'' C'^B' = c"b^' = sin A. AB" 1 Hence, in a right triangle whose hypotenuse is unity, the length of the side opposite an acute angle is the sine of that angle. 280. Theorem. The ratio of the sides opposite two acute angles of a triangle is equal to the ratio of the sines of these angles. Given A ABC with Z A and Z B both acute angles. _, ,1 , sin A a To prove that ~ = -. sin B Proof: Draw the perpendicular j9. Then sin A =\ and sin B =^. , a Hence, ^HL^=f^^ = f See §241, (3). sin B b a h ^ ' ^ ^ Note. The definitions of § 279 and the theorem of § 280 are given here for acute angles only. In trigonometry, where the subject is studied in full detail, they are extended to apply to any angles whatever. Other ratios called cosines, tangents, etc., are also introduced. MEASUREMENT OF LINE-SEGMENTS. 137 281. The theorem of § 280 is of great importance in finding certain parts of a triangle when other parts are known. By careful measurement (and in other ways) tables may be constructed giving the sine of any angle. 282. EXERCISES. 1. By means of a protractor construct angles of 10°, 20°, 30°, 40°, 50°, 60°, 70°, 80°, and measure the sine of each angle, and so construct a table of these sines. If one decimeter is used as a unit for the hypotenuse, then the length of the side opposite Z vl, expressed in terms of decimeters, is the sine of the angle A. Notice that the values of the sines are the same no matter what unit is used, but in general the larger the unit the more accurately is the sine determined. By means of the table just constructed solve the follow- ing problems, using the notation of the figure : 2. Given ZA= 30°, B = 80°, b = 12, find a, c. B Solution. By the theorem, § 280, - = ^i^ • h sin B Substituting the values h = 12, and sin A = sin 30°, -4~ 6=12 o sin B = sin 80° from the table, we find a. In the same manner find c. 3. Given a=l(y,ZA= 60°, ZC = 70°, find Z B, h, c. 4. A lighthouse L is observed from a ship S to be due northeast. After sailing north 9 miles to S', the lighthouse is observed to be 35° south of east. Find the distance from the ship to the lighthouse at each point of observation. Use § 284. 5. A ladder 25 feet long rests with one end on the ground at a point 12 feet from a wall. At what angle does the ladder meet the ground. 6. If two sides and the included angle of a triangle are known, can the remaining parts be found by means of § 280? 138 PLANE GEOMETRY, 283. In land surveying on an extensive scale, processes similar to that used on the preceding page are constantly employed in finding the sides of triangles. To begin with, a level piece of ground is selected and a line AB measured with great care. ^ e Then a point C is selected, and /-ABC \ ^"^-^^^c^-^"! 1>g and Z.BAC measured very accurately with an instrument. Sides AC and BC may now be computed by means -^ i> of the theorem, § 280, and a table of sines (see page opposite). By measuring Z BAC and Z ACB, CB may be computed. By this process, called triangulating, it is possible to survey over a large territory without directly measuring any line except the first. 284. The saving of labor afforded by this indirect method of measuring is very great, and especially so in a rough and mountainous country, since measuring the straight line distance from one mountain peak to another by means of a measuring chain is impossible. In practice, tables of logarithms are used and the sines are carried out to a larger number of decimal places, but the general process is that used on the preceding page. 285. EXERCISES. Using the table on the next page, solve the following examples: 1. Given A = 53% B =z 65% a = 11.5. Find b, c, and Z C. 2. Given B = 49°, = 71°, a = 19.3. Find h, c, and Z A . 3. Given A = 65% a = U,b= 12. Find ZB, ZC, and c. Solution. §H^ = -^ or sin ^ = sin v4 x - = .91 x — = .78. Sin A a a 14 From the table we find that sin 51° =.78. Hence £ = 5P. ZC and c may now be found as before. MEASUREMENT OF LINE-SEGMENTS. 139 4. Given A =71°, a = 19.5, b = 17. Find ZB, ZC, and c. As be- b fore, sin 5 = sin yl x sin 71° X -^ = .95 X — 19.5 19.5 .828. From the table we find sin 5.5° = .82 and sin 56° = .83. But sin B is nearer .83 than .82, and hence B = 56° is the nearest approximation using a degree as the smallest unit. Angle Sin Angle Sin Angle Sin Angle Sin Angle Sin Angle Sin 0-^ 15° .26 30° .50 45° .71 60° .87 75° .97 1° .02 16° .28 31° ..52 46° .72 61° .87 76° .97 2° .03 17° .29 32° .53 47° .73 62° .88 77° .97 3° .05 18° .31 33° .54 48° .74 63° .89 78° .98 4° .07 19° .33 34" .56 49° .75 64° .90 79° .98 5° .09 20° .34 35° .57 50° .77 65° .91 80° .98 6° .10 21° .36 36° ..59 51° .78 66° .91 81° .99 7° .12 22° .37 37° .60 52° .79 67° .92 82° .99 8° .14 2.3° .39 38° .62 53° .80 68° .93 83° .99 9° .16 24° .41 39° .63 54° .81 69° .93 84° .99 10° .17 25° .42 40° .64 55° .82 70° .94 85° 11° .19 26° .44 41° .66 56° .83 71° .95 86° 12" .21 27° .45 42° .67 57° .84 72° .95 87° 13° .22 28° .47 43° .68 58° .85 73° .96 88° 14° .24 29° .48 44° .69 59° .86 74° .96 89° 286. Definition. An object AB is said to subtend an angle APB from a point P if the lines PA and PB are sup- posed to be drawn from P to the extremities of the object. 287. EXERCISES. 1. A building known to be 150 feet high is seen to subtend an angle of 20°. How far is the observer from the building if he is standing on a level with its base? 2. Show how to find the distance from an accessible to an inacces- ible point by means of § 280 and the table of sines. 3. A flagstaff is 125 feet tall. How far from it must one be in order that the flagstaff shall subtend an angle of 25°? 140 PLANE GEOMETRY. PROBLEMS OF CONSTRUCTION. 288. Instruments. In addition to the ruler, compasses^ and protractor described in § 44, the parallel ruler is convenient for drawing lines through given points parallel to given lines without each time making the construction of § 101. Description. R and R' are two rulers of equal length and width. AB and CD are arms of equal length pivoted at the points A, B, C, and D, making AC = BD. Why do the rulers, when thus constructed, i ^g "TPl ig" remain parallel as they are spread ? 289. Segments of equal length can be laid off with great accuracy on a line-segment by means of the com- passes. The following construction makes use of the compasses and parallel ruler. 290. Problem. To divide a given line-segment into any number of equal parts. Construction. Proceed as in § 159, Ex. 5, using the parallel ruler to draw the parallel lines. Give the construction and proof in full. Historical Note. The idea of similarity of geometric figures, or "sameness of shape," is one of early origin, as is also the simple theory of proportion. It was probably used by Pythagoras to prove the famous theorem known by his name. (See § 262.) But the dis- covery by him of the incommensurable case (§ 230) showed that this theory was inadequate for the rigorous proof of all theorems on simi- lar figures. It remained for Eudoxus, the teacher of Plato, to perfect a rigorous theory of ratio and proportion. MEASUREMENT OF LINE-SEGMENTS. 141 Euclid, following his predecessors, deals with ratios of magnitudes ill general as well as of numbers. Later writers have frequently in- sisted that ratios in general are not numbers. But nothing is gained by this procedure, since they possess all the properties of numbers. In this book a ratio is treated simply as the quotient of two numbers. See §§ 238-242. 291. Problem. Given three line-segments m, n, p, to construct a fourth segment q such that — = - ; that is, to n q find a fourth proportional to m, n, and p. Construction. Draw two indefi- nite straight lines, Ax and Ai/ mak- ing a convenient angle. On Ai/ lay off AB = m, BC = n. On Ax lay off AD=p. Draw BI). ^~^d Through C draw CE II BD. Then DE= q is the required segment. Give the proof in detail. 292. EXERCISES. 1. Apply the method of § 290 to bisect a given line-segment, and compare this with the method of § 51. 2. Divide a line 7| inches long into 11 equal parts by the method of § 290, and compare with the process of measuring by means of an ordinary ruler giving inches and sixteenths. 3. Using the same segments as in § 291, construct a segment q such that —=:- ; also such that - = — . p q p q 4. Using the same segments as in the preceding, construct a seg- ment q such that ^ = - ; also such that ^ = '" . m q n q What is q called in each case ? 5. If the given segments m, n, p are respectively 3| inches, ^f^ inches, 4| inches, compute q such that — = ". Also construct q as in § 291 and compare results. , ^ 142 PLANE GEOMETRY. 293. Definitions. If three numbers «, m, and b are such that a : m = m : b, then m is the mean proportional between a and 5, and b is the third proportional to a and m. 294. Pkoblem. To construct a mean proportional be- tween two given segments m and n. ^y Construction. On an indefinite line xy\2iyo^AB = m and BC = n. On AC as a diameter construct a semicircle. At B erect a perpendicular to AC meeting the semicircle in D. Then BD = jt? is the required segment. Proof. Draw AB and CD and prove AABD r^ ACBD. ^ Complete the proof. 295. EXERCISES. 1. Show that in § 277 DP is a mean proportional between PB and PA. 2. If in the above problem m = 3, and n = 5 show that the con- struction gives Vl5. 3. Show how to construct a segment p = VS, also p = V3, p = V2, .by means of the above process. 4. Show how to construct a square equal in area to a given J rectangle. <» 5. Show how to construct on a given base a rectangle equal in area to a given square. V Suppose AB = m is the given base and BD =p a. side of the given square, the two segments being placed at right angles as in the figure above. The problem is then to find on the line AB the center of a circle which passes through the points A and D. To do this connect A and D and construct a perpendicular bisector of this segment meeting AC in a point which is the center of the required circle. BC = n is then the required side of the rectangle since m-n = p^. MEASUREMENT OF LINE-SEGMENTS. 143 296. Problem. To divide a line-segment into three 'parts proportional to three given segments m, n, p. A D E B Construction. Let AB be the given segment. Con- struct Ax and on it lay off m, w, p as shown in the figure. Complete the figure and give proof in full. ^ 297. EXERCISES. 1. Divide a line-segment 5 inches long into three parts proportional to 2, 3, and 4. 2. Divide a segment 11 inches long into parts proportional to 3, 5, 7, and 9. First compute the lengths of the required segments, then construct them and measure the segments obtained. Compare the results. Which method is more convenient ? Which is more accurate ? 3. Divide a segment 9 inches long into two parts proportional to 1 and V2. Also compute the required segments. Which is more convenient? More accurate? 4. Divide a segment whose length is vTl into two parts propor- tional to V'2 and V5. First construct the segments whose lengths are \/2, a/5, and Vll. Also compute the required segments. 5. Divide a given line-segment into parts proportional to two given segments m and n, (a) if the division point falls on the segment; (ft) if the division point falls on the segment produced. See § 252. ' 6. A triangle is inscribed in another by joining the middle points of the sides. (a) What is the ratio of the perimeters of the original and the inscribed triangles? (6) Is the inscribed triangle similar to the original? 144 PLANE G:E0METRV. 298. Problem. On a given line-segment as a side construct a triangle similar to a given triangle. f Construction. Let ABC be the given triangle and a'b' the given segment, and let it correspond to AB of the given triangle. Construct Z 2 = Z 1 and Z 4 = Z 3 and produce the sides till they meet in &, Then A'b'c' is the required tri- angle (Why?). 299. EXERCISES. 1. In the problem of § 298 construct on A'B' a triangle similar to ABC such that A'B' and BC are corresponding sides. 2. Show that on a given segment three different triangles may be constructed similar to a given triangle. 3. Solve the problem of § 298 by making Z 2 = Z 1 and construct- AB A C ing A'C so that — — = -777-7 * Give the solution and proof in full. A B A Ly 4. On a given segment as a side construct a polygon similar to a given polygon, by first dividing the given polygon into triangles and then constructing triangles in order similar to these. Apply § 270. 5. "On a given segment construct a polygon similar to a given polygon in a manner analogous to the method used in Ex. 3 for con- structing a triangle similar to a given triangle. 6. ABC is a right triangle with the hypotenuse AB, and CD1.AB. Prove AC 2b mean proportional between AB and AD and likewise CB a mean proportional between AB and DB. 7. Use the theorem of § 277 to construct a square equal in area to that of a given rectangle. 8. The tangents to two intersecting circles from any point in their common chord produced are equal. Use § 277. MEASUREMENT OF LINE-SEGMENTS. J 9. Ill the figure AD and BE are tangents at the extremities of a diameter. If BD and AE meet in a point C on the circle prove that AB is a mean pro- portional between AD and BE. 10. The greatest distance to a chord 8 inches long from a point on its intercepted arc (minor arc) is 2 inches. Find the diameter of the circle. Use § 272. 11. At the extremities of a chord AB tangents are drawn. From a point in the arc AB perpen- diculars PC, PD, PF are drawn to the chord and the tangents. Prove that PC is a mean propor- tional between PD and PF. Suggestion.' Draw segments AP and BP and prove AAPF^ABCP and A .1 PC ~ BPD. Then ^.^and^ = ^. PB PC PB PD 145 SUMMARY OF CHAPTER III. 1. Make a list of the definitions in Chapter III. 2. Make a list of the theorems on proportional segments involving triangles. 3. State the various conditions which make triangles similar. 4. State the conclusions which can be drawn when it is known that two triangles are similar. 5. State the theorems on proportional segments involving poly- gons. 6. State the theorems on proportional segments involving straight lines and circles. 7. State ill the form of theorems the various ways in which a proportion may be taken so as to leave the four terms still in proportion. 8. Under what conditions is a segment a mean proportional be- tween two given segments. For instance, see § 277. 9. Make a list of the problems of construction in Chapter III. 10. State some important applications of the theorems on propor- tion. (Return to this question after studying those which follow.) 146 PLANE GEOMETRY, PROBLEMS AND APPLICATIONS. 1. The middle points of adjacent sides of a square are joined. (a) Prove that the inscribed figure is a square. (&) What is the ratio of the inscribed and the original squares. (c) If a side of the original square is a, find a side of the inscribed square. (d) If a side of the inscribed square is ft, find a side of the original square. 2. Two strips intersect at right angles. (a) If the width of each strip is 3 inches, what is the largest square which can be placed on them so that its sides will pass through the corner points as shown in the figure? (The corners will bisect the sides of the square.) (ft) What is the side of this square if the width of each strip is a ? (c) Find the side of the square if the width of one strip is 3 and that of the other 5. (of) Find the side of the square if the width of one strip is a and that of the other ft. 3. In the figure ABCD is a square and EFGH ••• is a regular octagon. (a) Show that A E, EF, FB are proportional to 1, V2, 1. (Assume ^jEJ = 1 an^ find NE.) ^ (ft) Ifyl^^a,showthat — = ^ +^^ = — • (c) Find Z FEN. See § 163, Ex. 2. |, (d) Show that AO = AF, and hence that the regular octagon may be constructed as shown in the figure. (e) Show that ZAOE = 22^° by § 219, and use this to make another proof that EFGH ... is a regular octagon. 4. Find the area of a square whose diagonal is a. Tile Pattern. 17T Tf Tile Pattern. MEASUREMENT OF LINE-SEGMENTS. 147 5. If a side of a regular octagon is a, find its area. 6. A floor is tiled with regular white octagons and black squares as shown in the design. What per cent (approxi- mately) of the floor is black? 7. Divide a circle into eight equal parts and join alternate division points. (a) Prove that LMNPQ ••• is a regular octagon. Cut Glass Design. (?>) If the diameter of the circle is a, show that LE — ~(Vi Suggestion. Find HE and then use Ex. 3, (a). "^ 8. Problem. Inscribe a square in a given semicircle. Construction. Let AB be the diameter of the given semicircle. At B construct a perpendicular to AB, making BD==AB. Connect D with the center (7, meeting tlie circle at E. Let fall EF perpendicular to AB. Then EF is a side of the required square. Complete the figure and make the proof by showing that EF = 2 CF. 9. Problem. Inscribe a square in a given triangle. Construction. Let ABC be the given tri- angle. Draw CD parallel to AB, making CD = CO. Complete the square CDEO. Draw DA meeting CB in F. Draw FG II AB, GH li CO, and FK II CO. Then FGHKr^ CDEO and hence is the required square. 1). -^E 148 PLANE GEOMETRY. t f Touch down ®°'^^ P°«ts ^ , 10. The sides of a triangle are 13, 17, 19. Find the lengtlis of the segments into which the angle bisectors divide the opposite sides. 11. The angles of a triangle are 30°, (J0°, 90°. Find the lengths of the segments into which the angle bisectors divide the opposite sides if the hypotenuse is 10. 12. Prove that the perpendicular bisectors of all the sides of a polygon inscribed in a circle meet in a point. 13. An equilateral polygon inscribed in a circle is equiangular. . 14. The goal posts on a football field are 18| feet apart. If in making a touchdown the ball crosses the goal line 25 ft. from the nearest goal post, how far back should it be carried so that the goal posts shall subtend the greatest possible angle from the place where the ball is 'placed? See Ex. 22, page 111, and § 277. F 15. Solve the preceding problem if the touchdown is made a feet from the nearest goal post, and thus obtain a formula by means of which the distance may be computed in any case. 16. A triangle has a fixed base and a constant vertex angle. Show that the locus of the vertex is two arcs whose end-points are the extremities of the base. See Ex. 20, page 111. Note that the locus also includes an arc on the opposite side of AB from the one shown in the figure. It follows that if two points A and B subtend the same angle from P and from Cf, then a circle may be passed through A,B,P,Q. »| 17. In the figure, A, B, F lie in the same ^traight line, as do also D, C, F, and Z 1 = Z2. (a) If BF =8,CF= 9, and DC = 12, find AB. (b) JiAB = 100, AF = 250, CF = 60, find DC. Note. By holding a small object, say a pencil, at arm's length, and sighting across the ends of it, we may determine approximately whether two given objects subtend the same angle. MEASUREMENT OF LINE-SEGMENTS. 149 f 18. Not having any instruments, an engineer proceeds as follows to obtain approximately the distance from an accessible point B to an inaccessible point A . Walking from B along ^ the line AB he takes 50 steps to F. Then he walks in a convenient direction 50 steps to C, and notes that A and B subtend a certain angle. He then proceeds along the same straight line until he reaches a point D at which A and B again subtend the same angle as at C. He then concludes that DC = AB. Is th;s conclusion correct? Give proof. ^ 19. If the height of a building is known, show how the method of Ex. 18 can be used to determine the height of a flagstaff on it. * 20. A building is 130 feet high, and a flagstaff on the top of it is 60 ft. high; 130 feet from the base of the building in a horizontal plane, the flagstaff subtends a certain angle. How far from the build- ing along the same line is there another point at which the staff sub- tends the same angle? At what distance does it subtend the same angle as it does at 300 feet? ^ 21. If the diagonals of an isosceles trapezoid are drawn, what similar triangles are produced? s' 22. Find the locus of points at a fixed distance from a given tri- angle, ^ways measuring to the nearest point on it. 23. The line bisecting the bases of any trapezoid passes through the point of intersection of its diagonals. Suggestion. Let E bisect DC, and draw EO meeting AB in F. Prove AF= FB. I 24. If two triangles have equal bases on one of two parallel lines, and their vertices on the other, then the sides of these triangles intercept equal segments on a line parallel to these and lying between them. i 25. A segment bisecting the two bases of a D P c ' trapezoid bisects every segment joining its othev two sides and parallel to the bases. (Prove EF^BG and FK^ KH, using Ex. 24.) *^ ""' 26. In a triangle lines are drawn parallel to one side, forming trap- ezoids. Find the locus of the intersection points of their diagonals. E^ 150 PLANE GEOMETRY. 27. In the figure D, E, F are the middle points of the sides of an equilateral A ABC. Arcs are constructed with centers A, B, C as shown in the figure. (a) Prove that these arcs are tangent in pairs at the points D, E, F. (h) Construct a circle tangent to the three arcs, (c) What axes of symmetry has this figure? 28. In the figure ABC is an equilateral triangle. Arcs AB, BC, and CA are constructed with C, A, and B as centers. A F, BE, CD are altitudes of the triangle. Prove QOG tangent to AB, BC, and CA. 29. In the equilateral triangle ABC, AB, BC, and CA are constructed as in the preceding ex- ample. D^, EF, FD are constructed as in Ex. 27. The figure is completed making ADE, DFB, and EEC similar to ABC. (a) Construct a circle tangent to a1), DE, and EA as shown in the figure. (b) Construct a circle tangent to ^D, DF, and FE. •^ 30. In the figure A CB is a semicircle. Arcs DE and DF are constructed with B and A as centers, li AD = 4: feet, find the radius of O OC. Solution. Show that the value of r. is derived from the equation (4 + r)^ = (4 - r)2 + 4^. ^ ^ 31. Construct the accompanying design. Notice that the points A , B, M, N, F, E divide the circle into six equal arcs. See Ex. 5, From Boynton Cathedral, England. MEASUREMENT OF LINE-SEGMENTS. 151 The § 163. The arcs A 0, OB, etc., and the circle with center 0', are constructed as in Ex. 28. The small circle with center at Q is con- structed as in the preceding example. ►. 32. ^5 C is an equilateral arch and AEB is a semicircle. (a) If AB (the span of the arch) is 10 feet, C find the radius of the small circle tangent to the semicircle and the arcs of the arch. (b) Find the radius OE of the circle ii AB = s. (c) If OE = 2, find AB. 5 33. The accompanying design consists of three A semicircles and three circles related as shown in the figure. (a) If AB = 12, find OQ by using the triangle OSD. method is similar to that used in Ex. 30. (b) Using the right triangle DRO', find O'x. Notice that in order to have the circles with the centers and 0' tangent to each other the sum of their radii must be equal to RD. (c) If AB = s show that 0(2=-. and O'x = ~, and construct the figure. 34. Upon a given segment AB construct the design shown in the figure. Notice that it consists of the two preceding figures put together. Compare the radii of the circles in this design. A E H B Ospedale Maggiore, Milau. • 35. Between two parallel lines construct semicircles and circles as shown in the figure. Solution. Suppose the construction made. Let HE = r, and OQ = r'. Using the triangle EHO, show that struct the fig'ure. and then con- 152 PLANE GEOMETRY. \ 36. Between two parallel lines construct circles and equilateral arches as shown in the figure. The Doge's Palace, V'euice. Hint. AQ = 2 PR and AH = 3 PH. Hence HQ^ = WPH^ - YPH^ = TPH \ That is, PH :HQ = 1'. V5. (Why ?) Now divide AC into two segments proportional to 1 and Vo, and so construct the figure. C 37. ABC is an equilateral arch. AD = DB, AF=EB=^AB. ^ Semicircles are constructed with E and F as centers, and radius FB. • (a) If .4J5 = s, find OH and construct the figure. j Solution. Let OH F0 = - + r. r. Then BO = s - r, FB='-, DF = -, 3 6 - i From A ODB, Ojf = Olf - 1)B^ = (.s - r)' From A ODF, Olf = OF^ - DF^ = [t + r From (1) and (2) (. _ r)^ _ (0' = (| + r)'- (01 Hence, show that r — \s. Construct the figure. (6) If in the preceding 0H=^ feet, find AB and AF. (1), (2j MEASUREMENT OF LINE-SEGMENTS. 153 7~^ IR. / m_M: AT saL WQ 38. Between two parallel lines construct circles and semicircles having equal radii as shown in the figure. Prove that the ratio — = — , RLK being HQ RS ^ an equilateral triangle and KS = SL. Divide CA in the ratio of SL : RS^ 39. Through a fixed point on a circle chords are drawn and each extended to twice its length. Find the locus of the end-points of these segments. Compare Ex. 6, § 214. 40. If a quadrilateral is circumscribed about a circle, show that the sums of its pairs of opposite sides are equal. 41. On a diameter produced of a given circle, find a point from which the tangents to the circle are of a given length. Solve this problem by construction, and also algebraically. 42. Compare the perimeters of equilateral triangles circumscribed about and inscribed in the same circle. 43. In a given square construct semicircles each tangent to two sides of the square and terminating on the diameters of the square. Construction. Connect E and F two ex- tremities ol diameters and on EF as a diame- ter construct a semicircle with center at 0'. Draw O'H perpendicular to AB meeting the arc in H. Draw OH meeting AB in K. Draw KO" perpendicular to AB meeting the diagonal ^C in 0". Then 0" is the center of the re- quired circle. Proof. Draw 0"L\\ FE. We need to prove that 0"K =0"L. 00" _ 0"L 00' ~ O'E 00" ^0"K 00' ~ O'H A 00' E ~ A 00"L, and hence Also A 00' H ~ A 00"K, and hence (Why?) (1) (2) V ,^^ A /ON 0"L 0"K From (1) and (2) ^^^ = ^ • But O'E = O'H, and hence 0" L = 0"K. This figure occurs in designs for steel ceilings. (Why?) CHAPTER IV. AREAS OP POLYGONS. AREAS OF RECTANGLES. 300. Heretofore certain properties of plane figures have been studied, such as congruence and similarity^ but no attempt has been made to measure the extent of surface inclosed by such figures. For this purpose we first con- sider the rectangle. 301. The surface inclosed by a rectangle is said to be exactly measured when we find how many times some unit square is contained in it. E.g. if the base of a rectangle is five units ^ long and its altitude three units, its surface con- tains a square one unit on a side fifteen times. 302. The number of times which a unit square is con- tained in the surface of a rectangle is called the numerical measure of the surface, or its area. We distinguish three cases. 303. Case 1. If the sides of the given rectangle are in- tegral multiples of the sides of the unit square, then the area of the rectangle is determined by finding into how many unit squares it can be divided. Thus if the sides of a rectangle are m and n units respectively, then it can be divided into m rows of unit squares, each row containing n squares. Hence the area of such a rectangle is m x n unit squares, that i is, in this case, ^^^^ ^ ^^^^ ^ ^j^.^^^^ ^^ ^ 1 154 --t---i— |— -|--- — i — I — 1.__< — AREAS OF POLYGONS. 155 304. Case 2. If the sides of the rectangle are not inte- gral multiples of the side of the chosen unit square, but if the side of this square can be divided into equal parts such that the sides of the rectangle are integral multiples of one of these parts, then the area of the rectangle may be expressed integrally in terms of this smaller unit square, a,nd fractionally in terms of the original unit. For example, if the base is 3.4 deci- meters and the altitude 2.6 decimeters, then the rectangle cannot be exactly di- vided into square decimeters, but it can be exactly divided into square centi- meters. Each row contains 34 centi- meters and there are 26 such rows. Hence, the area is 34 x 26 = 884 small squares or 8.84 square decimeters. But 3.4 X 2.6 = 8.84. Hence, in this case also, area = base x altitude. (2) 305. Case 3. If a rectangle is such that there exists no common measure whatever of its base and altitude, then there is no surface unit in terms of which its area can be exactly expressed. But by choosing a unit sufficiently small we may determine the area of a rectangle which differs as little as we please from the given rectangle. E.g. if the base is 5 inches and the altitude is v^5 inches, then the rectangle cannot be exactly divided into equal squares, however small. But since V5 = 2.2361 •••, if we take as a unit of area a square whose side is one one-thousandth of an inch, then the rectangle whose base is 5 inches and whose altitude is 2.236 inches can be exactly meas- ured as in cases 1 and 2, and its area is 5 x 2.236= 11.18 square inches. The small strip by which this rectangle differs from the given rec- tangle is less than .0002 of an inch in width, and its area is less than 5 X .0002 = .001 of a square inch. By expressing V5 to further places of decimals and thus using smaller units of area, successive rectangles may be found which differ less and less from the given rectangle. 156 PLANE GEOMETRY. 306. An area thus obtained is called an approximate area of the rectangle. For practical purposes the surface of a rectangle is measured as soon as the width of the remaining strip is less than the width of the smallest unit square available. From the foregoing considerations we are led to the following preliminary theorem ; 307. Theorem. The area of a rectangle is equal to the product of its base and altitude. 308. The argument used above shows that the theorem (§ 307) holds for all rectangles used in the process of approximation, and hence it applies to all practical meas- urements of the areas of rectangles. AREAS OF POLYGONS. 309. From the formula for rectangles, area = base x altitude, we deduce the areas of other rectilinear figures by means of the principle : Two rectilinear figures are equivalent (that is, have the same area) if they are congruent, or if they can be divided into parts which are congruent in pairs. E.g. the two figures here shown are equiv- Aiv / alent since AT and III are congruent respec- y'^\v^y^ /n'/ tively to II and IV. ^ /-— -i/ There are figures of equal area which cannot be thus divided. Thus a rectangle with base 2 and altitude 1 is equal in area to a square whose side is \/2, though these figures cannot be divided in the manner stated above. However, the test specified in the principle is sufficient for our present purposes. The symbol = joining two polygons means equivalent or equal in area. ABEAS OF POLYGONS. 157 310. Theorem. The area of a parallelogram is equal to the product of its base and altitude. III Given the parallelogram ABCD whose base is AB and whose altitude is AE. To prove that area ABCD = AB x AE. Proof : Draw BF ± to DC produced, forming the rec- tangle ABFE, whose base is AB and altitude AE. Then area ABFE = AB x AE (Why ?). If now we prove Al^AlI, then the parallelogram is composed of parts I and III which are congruent respec- tively to parts II and III of the rectangle, and hence the parallelogram and rectangle are equal in area. Give this proof in full. 311. EXERCISES. 1. In the figure CF is perpendicular to AB and AE to CB. Prove that ABxCF = BCxAE. ^^ ^, Suggestion. (Show that A^5J5; ~ AC^F.) It follows that the same result is obtained if either of two adjacent sides of a parallelogram is taken as the base. ^ ^ ^ 2. Construct the figure described in Ex. 1 in such manner that the point E falls on BC extended and prove the theorem in that case. In the following \ and l^ are parallel lines. 3. Two parallelograms have equal bases lying on l^ and l^. Show that they are equivalent. 4. One base of a parallelogram is fixed on l^ and the other moves along I.,. Does the area change? 158 PLANE GEOMETRY. 312. Theorem. The area of a triangle is equal to one half the product of its base and altitude. Given the A ABC whose altitude upon the side AB is CD. To prove that the area of AABC= I (^AB X CD). Proof : Let G be the middle point of AC. Complete the parallelogram ABFG. Prove that A CGH ^ A BFH^ and hence show that A ABC and O ABFG have equal areas. Hence the area of AABC= ED x AB. Hence the required area is J (^ii x CD). But ^D = I CD. 313. EXERCISES. 1. Prove the theorem of § 312 us- ing the accompanying figure. 2. Prove this theorem also by draw- ing a figure in which the given triangle is one half of a parallelogram. 3. If the middle points of two ^i B adjacent sides of a parallelogram are joined, a triangle is formed whose area is equal to one eighth of the area of the parallelogram. In the following exercises l^ and l^ are parallel lines. 4. Show that two triangles are equivalent if their vertices lie in ^2 and their equal bases in ly 5. A triangle has a fixed base in L^. If its vertex moves along Zj, what can you say of its area? €. If the base of a triangle is fixed and if its vertex moves so as to preserve the area constant, what is the locus of the vertex ? 7. A line is drawn from a vertex of a triangle to the middle point of the opposite side. Compare the areas of the triangles thus formed. If a segment is drawn from the vertex to the point P in the base AB, show that the areas of the triangles are in the ratio AP : PB. AREAS OF POLYGONS. 159 314. Theorem. The area of a trapezoid is equal to one half the sum of its bases multiplied by its altitude. Proof : Draw a diagonal, thus forming two triangles having a common altitude. Form the ex- pression for the sum of the areas of these triangles. ^ ^ Give the proof in detail. 315. Theorem. The square erected on the sum of two line-segments as a side is equal to the sum of the squares erected on the two segments separately, plus twice the rec- tangle whose base is one segment and whose altitude is the other segment. Proof : Let a and b represent the numerical measures of the two line-segments. Then, the square erected upon the segment a -{- h may be subdivided as shown in the figure, giving (a -\-by = a'^ + 2ab + b\ Give the construction and proof in full. 316. EXERCISES. 1. Show by a figure that (a-\-b + c)2 = a2 + 5M- c2 + 2 a6 + 2 ac + 2 be. 2. If ABC is any triangle and AD, BE, and CF are its altitudes, show that i^ = 4£, and ^ CF AB AC BE BC Hence show that AD X CB = BE y. CA = CF x AB. 3. The side AB oi CD A BCD is fixed. What is the locus of the points C and D if CD moves so as to leave the area of the parallelogram fixed ? 160 PLANE GEOMETRY. 317. Theorem. The square erected on the difference of two line-segments as a side is equal to the sum of the squares erected on the two segments separately, minus twice the rectangle whose base is one of the segments and whose altitude is the other. Proof : Use the figure to show that (a _ 6)2 = a2 + 52 _ 2 «5. Give the construction and proof in detail. a . f.-' ab y b a-b \ a-b I b 318. EXERCISES. ah •* 1. Show by a figure that *- (a — b)(a — c) = a^ + be 2. Likewise, show that (a -\- b)(a — b) = a^ — b^. 3. Show by counting squares ih the accompany- ing figure that in case of an isosceles right triangle the square constructed on the hypotenuse is equal -4n area to the sum of the squares on the two legs. 4. In case the triangle is scalene, show by counting squares and congruent parts of squares that the square constructed on the hypotenuse is equal in area to the sum of the squares on the other two sides. 5. Using the third figure, show that the same theorem holds for any right triangle. Hint. Call the legs of the given triangle a and b. Describe the construction of the auxiliary lines and give the proof in full. This theorem is proved again in the next paragraph and was also proved in § 262. The proof just given is due to Bhaskara (born 1114 a.d.), who constructed the " Behold ! " 16 8P3 16 -T!\ ^-=^ 6 ^ \ \ 1A A V- W \6 iP_ M ^ ^-^''" T" ± figure simply AREAS OF POLYGONS. 161 319. Theokem. The area of the square described on the hypotenuse of a right triangle is equal to the sum of the areas of the squares described on the other two sides. n Given the right triangle ABC, and the squares I, II, and III con- structed as shown in the figure. To prove that D I = D II 4- □ HI. Proof : Complete the rectangle ABCE. Construct A GFH^ A ACE. Draw EH and produce AE to meet the line FH at K. Prove that (a) AAEC^A HEX. (5) AEHG is a parallelogram whose base AE and alti- tude HK are each equal to a side of D III. (^) ECFH a parallelogram whose base EC and altitude EK are each equal to a side of D II. But these two parallelograms together are equivalent ton I (Why?). The student should give the proof iu detail, making an outline of the steps and showing how each step is needed for the next. For ex- ample, why is it necessary to prove AHEK ^ A A EC. Compare the various proofs of this theorem that are given in this book. The method of counting squares shown on the opposite page is applicable to all cases where the two sides are commensurable. 162 PIANE GEOMETRY. Historical Note. The theorem of § 319 is one of the most impor- tant in all mathematics. It is now fairly certain that the general theorem was first stated and proved by Pythagoras, though the story that he sacrificed 100 oxen to the gods on the occasion may be ques- tioned. Special cases of the theorem were known to the Egyptians as early as 2000 B.C., e.g. that a triangle whose sides are 3, 4, 5 is right-angled. In this connection the Pythagoreans also discovered the irrational number, that is, that there are numbers such as V2 which cannot be expressed exactly as integers or as ordinary fractions. 320. EXERCISES. 1. The bases of a trapezoid are 8 and 12 inches, and its altitude is 8 inches. Find its area. 2. The bases of a trapezoid are 16 and 20 inches, respectively, and the area of the smaller of its component triangles is 80 square inches. Find the area of the trapezoid. See the figure in § 152. 3. State in words the geometric theorem indicated in each of the following, and draw a figure to illustrate each case : (a) h(a +b) = ah + bh. (6) (a + by +(a- by = 2 (a2 + b^). (c) (a -{* by - (a - by :^ i ab. 4. Show that the rectangle whose base is a + J and whose altitude h a — b has the same perimeter as the square whose side is a. By means of Ex. 2, § 318, compare their areas. 5. If two triangles have the same base but their vertices are on opposite sides of it, and if the segment joining their vertices is bi- sected by the common base, extended if necessary, then the two triangles are equivalent. 6. State and prove the converse of the theorem in the preceding exercise. 7. The bases of a parallelogram lie on the parallel lines l^ and I^ A triangle whose base is equal to that of the parallelogram has its vertex in l^ and its base in l^. Compare their areas. 8. Prove that all triangles having the same vertex and equal bases lying in the same straight line are equal in area. AREAS OF POLYGONS. 163 321. Theorem. The areas of similar triangles are in the same ratio as the squares of any two corresponding sides, or as the squares of any two corresponding altitudes. A C D A' C D' Given A ABC and A'BC, with altitudes BD and B'l>. To prove that area A ABC AB^ BC? CA? B^ Proof: But Hence Give the proof for each of the other pairs of corre- sponding sides. 322. EXERCISE. Verify the preceding theorem by counting triangles in the accom- panying figure. kyq^aa'b'c' a'b'^ b'c'^ c'a'^ B'D'"" area, A ABC _ ^(AC x BD) _ AC BD area Aa'b'c' ^a' c' x B^D') A'C' B'D' AC BD A'& B'D' Why? are'dAABC^ A(/ BD^ 'dTesi aa'b'c' a'c'^ b'd'^ Why? 164 PLANE GEOMETRY, 323. Theorem. The areas of two similar polygons are in the same ratio as the squares of any two corre- sponding sides or any two corresponding diagonals, B' E E' Outline of Proof: Let I, II, •••, T, 11', areas of pairs of similar triangles. r ir iir iv I +11 4- III 4- IV ^ I V Show (1) (2) (3) stand for the (§271.) ^- (By §321.) r + ir + iir + iv' ABCDEF _ I AB (By § 268.) Bd^ a'b'c'd'e'f' Give the proof in detail. ¥^'^ 324. EXERCISES. 1. Verify the above theorem by counting the number of equal hexagons into which these two similar hexagons are divided, and also taking the square of the length of one side in each, using a side of a small hexagon as a unit. 2. A certain triangular field containing 2 acres is 10 rods long on one side. Find the area of a similar triangular field whose corresponding side is 50 rods. AREAS OF POLYGONS. 165 3. The areas of two similar triangular flower beds are 24 square feet and 36 square feet respectively. If a side of one bed is 8 feet, find the corresponding side of the other. 4. If similar polygons are constructed on the three sides of a right triangle, show that the one described on the hypotenuse is equivalent to the sum of the other two. Suggestions. Let a, b, c represent the two legs and hypotenuse of the triangle and A^B,C the areas of the corresponding polygons. Then A a2 B &2 — = — and — = — . C c'^ C c^ Hence ^^^^ = ^^^^ = ^ = 1 ov A -\- B = C. Give all reasons in C c2 6-2 full. 5. Find a line-segment such that the equilateral triangle described upon it has four times the area of the equilateral triangle whose side is 3 inches long. 6. Show that the square on the altitude of an equilateral triangle is three fourths the square on a side. 7. If in a right triangle a perpendicular is let fall from the vertex of the right angle to the hypotenuse, show that the areas of the two triangles thus formed are in Uie same ratio as the adjacent segments of the hypotenuse, and also as the squares of the adjacent sides of the triangle. 8. Draw a line from a vertex of a triangle to a point in the oppo- site side which shall divide the triangle into two triangles whose ratio is 2: 5. Also 2:1. 9. Divide a parallelogram into three equivalent parts by lines drawn from one vertex. Use the last construction in Ex. 8. 10. The sides of two equilateral triangles are 8 and 6 respectively. Find the side of an equilateral triangle whose area shall be equivalent to their sum. Use the result in Ex. 4. 11. State and solve a problem like the preceding for the difference of the areas. 12. Corresponding sides of two similar triangles are a and h. Find the side of a third triangle similar to these whose area is equal to the sum of their areas. 13. Likewise for any two similar polygons. > 166 ' PLANE GEOMETRY. CONSTRUCTIONS. The theorems of this chapter lead to numerous construc- tions of practical importance. 325. Problem. To construct a square equivalent to the sum of two or more given squares. Construction. In the case of two given squares, con- struct a right triangle whose legs are sides of the two given squared. How is the desired square obtained ? In the case of three given squares, use the square result- ing from the first construction together with the third square, and so on. Give the construction and proof in full. 326. Problem. To construct a square equivalent to a given rectangle. Construction. If the base and altitude of the rectangle are a and b respectively, we seek the side s of a square such that 8^= ab ; that is, we s^ek a mean proportional between a and b. See Ex. 4, § 295. Give the construction and proof in full here. 327. Problem. To construct a square equivalent to a given triangle. Construction. Show how to modify the preceding con- struction to suit this case. 328. EXERCISES. Show how to construct each of the following : • 1. A square equivalent to the difference of two given squares. 2. A square equivalent to the sum of two given rectangles. 3. A square equivalent to the sum of two given triangles. 4. A square equivalent to the difference of two given rectangles, also to the difference of two given triangles. AREAS OF POLYGONS. 167 329. Problem. To construct a triangle equivalent to a given polygon, c D Given the polygon ABODE. To construct the triangle PCO equivalent to ABODE. Construction. Cut off A ABC by the segment AC. Through B draw BP II AC to meet EA extended at P. Draw OP. In a similar manner draw 00. Then PCO is the required triangle. Proof : PCDE = ABODE since A APO = A ABC (Why ?). Further, PCDE has one less vertex than ABODE. But A PCO = PCDE since A ECD = A ECO (Why?). Hence, A PCO is equivalent to the given polygon. 330. Problem. To construct a rectangle on a given base and equivalent to a given parallelogram. b' Construction. Let b and h be the base and altitude of the given parallelogram, b' the base of the required rectangle, and x the unknown altitude. Then we are to determine x so that b'x = hh, that is^ h' : h = h : x. Hence, x is the fourth proportional to 5', b, and h. (See § 291.) Construct this fourth proportional, showing the complete solution. This construction is attributed to Pythagoras. It represents a much higher achievement than the . discovery of the Pythagorean proposition itself. 168 PLANE GEOMETRY. •% ^ 331. EXERCISES. 1. Show how to modify the last construction in case the given figure is a triangle. Give the construction. 2. Construct a rectangle on a given base equivalent to a given irregular quadrilateral. 3. Construct a rectangle on a given base equivalent to an irregu- lar hexagon. 4. On a side of a regular hexagon as a base construct a rectangle equivalent to the hexagon. 5. Construct a parallelogram on a given base equivalent to a given triangle. Is there more than one solution? 6. Construct a square whose area shall be three times that of a given square; five times. One half the area; one fifth. 7. Construct an isosceles triangle, with a given altitude h, equiva- lent to a given triangle. 8. Draw a line parallel to the base of a triangle and cutting two of its sides. How will the resulting triangle and trapezoid compare in area, (a) If each of the two sides of the triangle is bisected? (h) If each of the two sides of the triangle is three times the length of the corresponding side of the trapezoid ? 9. Construct a triangle whose base and altitude are equal and whose area is equal to that of a given triangle. 10. In a parallelogram ABCD, any point E on the diagonal BD is joined to A and C. Prove that A BE A and BEC are equivalent, and also that A DEA and DEC are equivalent. 11. The sides of a triangle are 6, 8, 9. A line parallel to the longest side divides the triangle into a trapezoid and a triangle of equal areas. Find the ratio in which the line divides the two sides. 12. Draw a line parallel to the base of any triangle, and cutting two of its sides. How do the altitudes of the resulting triangle and trape- zoid compare, (a) If they are equal in area ? (b) If the area of the triangle is three times that of the trapezoid? AREAS OF POLYGONS. 169 13. Through a point on a side of a triangle draw a line dividing it into two equivalent parts. Solution. Let P be the given point. Draw the median BD. Draw BE || PB and draw PE. This is the required line. Prove. Suggestion. Notice that ADPE = ADPB by Ex. 4, § 313, and hence that AEOD = ABOP. 14. Through a given point on a triangle draw a line which divides it into two figures whose areas are in the ratio i. 15. Inscribe a circle in a triangle, touching its sides in the points D, E, F. With the vertices as centers, construct circles passing through these points in pairs. Show that each of these latter circles is tangent to the other two. SUMMARY OP CHAPTER IV. 1. State what is meant by the area of a rectangle. Give the formula. 2. Give formulas for areas of parallelograms and triangles. 3. How is the formula for the area of a trapezoid obtained? 4. What theorems of this chapter can be stated algebraically, as (a + by = a^ + 2ab-\-b^ 5. State the theorem on the ratio of the areas of two similar tri- angles; two similar polygons. Give examples. 6. Tabulate the problems of construction given in this chapter. 7. If two rectangles have the same base, how does the ratio of their areas compare with the ratio of their altitudes? 8. If two triangles have equal altitudes, how does the ratio of their areas compare with the ratio of their bases? 9. State all theorems of this chapter proved *by means of the Pythagorean proposition. 10. State some of the more important applications of the theorems in this chapter. Return to this question after studying the succeeding list of problems. 170 PLANE GEOMETRY. PROBLEMS AND APPLICATIONS. 1. Find the area of a square whose diagonal is 6 inches. 2. Find the area of a square whose diagonal is d inches. 3. ABCD is a square placed at the crossing of two strips of equal width, as shown in the figure. The small black square has two vertices on the sides of the horizontal strips and two on the sides of the vertical strip. (a) Find the area of each square when the width of the strips is 4 inches. (6) Compare the area of the black square and the white border surrounding it. (c) Can squares be placed as in the figure in case the strips are of unequal width? In the two following questions let the small square^ be drawn with two vertices on the sides of the horizontal strip and one diagonal parallel to these sides. (jl) If the horizontal strip is 4 inches wide, what must be the width of the vertical strip in order that the large square may have twice the area of the small one? Hint. The diagonal of the small square is 4 inches. (e) If the horizontal strip is a inches wide, what must be the width of the vertical strip in order that the area of the black square shall be - the area of the larger square ? n 4. Prove that the area of a rhombus is one half the product of its diagonals. 5. Prove that the area of an isosceles right triangle is equal to the square on the altitude let fall upon the hypotenuse. 6. If the diagonals of a quadrilateral intersect at right angles, prove that the sum of the squares on one pair of opposite sides is equal to the sum of the squares on the other two sides. 7. Inscribe a square in a semicircle and in a quadrant of the same circle. Compare their areas. See Ex. 8, page 147. 8. In the triangle ABC, CD is an altitude. E is any point on CD. If DE is one half CD, compare the area of the triangle AEB and the sum of the areas of the triangles AEC and BEC. Also compare these areas if DE is one nth oi DC. AREAS OF POLYGONS, 171 9. A BCD is a square, and E, F, G, H are the middle points of its sides. On EF, points N and Q are taken so that PN = PQ. Similarly KL = KM, also KL = NP, and so on. (a) Prove that .O/OL is a rhombus. (h) If AB=Q inches and if KM = ME, find the sum of the areas of the four rhombuses AMOL, BQON, etc. (c) If HE = 8 inches, what is the area of the whole square ? {(l) What part of KE must KM be in order that the sum of the rhombuses shall be \ the area of the square? (g) If AB a, find KM so that the sum of the rhombuses shall be - the area of the square. Parquet Flooring. (/) Prove that L, 0, R lie in the same straight line. 10. ABCD is a square and E, F, G, Hare the middle points of its sides. SN = PK=QL = RM. (a) li AB = Q inches and ii PK = I inch, find the sum of the areas of the triangles EHN, EFK, FGL, and GHM. * (&) li AB = 6, find PK so that the sum of the four triangles EFK, etc., shall be one fifth of the whole square. , {c) li AB = Q and if the points E, K, Q lie in (a straight line, find the sum of the areas of these triangles. (d) If AB = a and if PK = one nth of PO, /^ find the sum of the areas of the triangles. (g) If AB = a, find PK so that the sum of the triangles shall be one mih of the whole square. What will be the length of PK in case the tri- Parquet Flooring, angles occupy one half of the whole square? 11. The sides of two equilateral triangles are a and h respectively. Find the side of an equilateral triangle whose area is equal to the sum of their areas. PLANE GEOMETRY, lip, / 12. Construct a triangle similar to a given triangle and having 16 times the area. 13. The middle points of the sides of a quadrilateral are con- nected. Show that the area of the parallelogram so formed is half the area of the quadrilateral. 14. ABCD is a square. Each side is divided into four equal parts and the construction com- pleted as shown in the figure. (a) Prove that QNOH, KLOE, etc., are par- allelograms. (b) What part of the area of the square is oc- cupied by these four parallelograms? (c) What part of the area of the square is oc- cupied by the four triangles NEC, LGO, PFO, andilfi^O? {d) If AB = 6, find the lengths of KO and QP. (e) Find the ratio of the segments KO and QP. Does this ratio depend upon the length of ABl {/) In the parquet floor design what fraction is made of the dark wood? Does this depend upon the size of the original square? 15. On a given line-segment AB as a hypotenuse construct a right triangle such that the altitude upon the hypotenuse shall meet it at a given point D. 16. If ABC is a right triangle and CD±AB, Parquet Flooring. prove that AC" CB' AD DB 17. By means of Exs. 15 and 16 construct two segments HK and LM such that the ratio of the squares on these segments shall equal a given ratio. 18. Divide a given segment into two segments such that the areas of the squares constructed upon them shall be in a given ratio. ATB f 19. On a given segment A B find a point D such that ■ = 2. Ajf 20. Construct a line parallel to the base of a triangle such that the resulting triangle and trapezoid shall be equivalent. AREAS OF POLYGONS. 173 21. Construct two lines parallel to the base of a triangle so that the resulting two trapezoids and the triangle shall have equal areas. ^^r ^ D ^ 22. In each of the accompanying de- signs for tile flooring find what fraction of the space between the lines AB and CD is occupied by tiles of eacji color. Study each design with care to see that the character of the figure determines the relative sizes of the various pieces of tile. I 23. ABCD is a square. Points E, F, G, ••• are so taken that AE = AF= GB = BH= -•■ . (a) If AB = 6 and .4F = 1, find the sum of the D_M_ areas of the four triangles EFO, GHO, KLO, j^^ MNO. (b) Find the sum of these areas if AB = a and AF=h. (c) If AB = Q and if the sum of the areas of ^ ^ the triangles is 9 square inches, find A F. (d) li AB = a and if the sum of the areas of the triangles is — , find n \y A F. Interpret the two results. _ 24. Show how to construct a square whose area is n times the area of a given square. 25. Construct a triangle similar to a given triangle and equivalent to n times its area. ^ 26. Construct a hexagon similar to a given hexagon and equivalent '' to n times its area. 27. Show how to construct a polygon similar to a given polygon and equivalent to n times its area. 174 PLANE GEOMETRY. 28. The alternate middle points of the sides of a regular hexagon are joined as shown in the figure. (a) Are the triangles thus formed equilateral ? Prove. (b) Is the star regular (i.e. are its six acute angles equal and its sides equal) ? (c) Compare the three segments into which each triangle divides the sides of the other. (d) Is the inner hexagon regular? Prove. See Ex. 1, p. 76. ^^H I: III MH ~ Tile Pattern. \ (e) If AB = 6 in., find the area of the large hexagon, the star, and the small hexagon. V, 29. A border is to be constructed about a given square with an area equal to one half that of the square. (a) By geometrical construction find the outer side of the border if the side of the square is given. (b) If an outer side of the border is 24 in., find a side of the square, its area being two thirds that of the border. \ . 30. A border is constructed about a given regular octagon, such that its area is equal to that of the octagon. (a) If a side of the given octagon is a , given segment AB, find by geometrical construction a segment equal to an outer side of the border. (b) If a side of the given octagon is 16 in., find an outer side of the border. Ceiling Pattern. Ceiling Pattern. AREAS OF POLYGONS. 175 J n /\c A / r IX X I'x; () ^' \m ■< ' t 31. The accompanying design is based on a set of squares such as ABCD. The small triangles are equal isosceles right triangles constructed as shown. (a) Are the vertical and the horizontal sides of the octagons equal ? (6) Are the octagons regular ? (c) If a side of one of the squares is 6, find the area of one of the octagons. (d) What fraction of the whole tile design is occupied by the light-colored tiles? Does this depend upon the size of the original squares ? 32. Given two lines at right angles to each other. Find the locus of all points such that the sum of the squares of the distances from the lines is 25. 33. Given two concentric circles whose radii are r and r'. Find the length of a chord of the greater which is tangent to the smaller. 34. If two equal circles of radius r intersect so that each passes through the center of the other, find the length of the common chord. 35. The square on the hypotenuse of a right triangle is four times the square on the altitude upon the hypotenuse. Prove it isosceles. 36. In a right triangle the hypotenuse is 10 feet and the difference between the other sides is 2 feet. Find the sides. 37. Two equal circles are tangent to each other and each circle is tangent to one of two lines perpendicular to each other. Find the locus of the points of tangency of the two circles. Suggestion. Note that the point of tangency bisects their line of centers and that the centers move along lines at right angles to each other. 38. The square on a diagonal of a rectangle is equal to half the sum of the squares on the diagonals of the squares constructed on two adjacent sides of the rectangle. 39. Show that the diagonals of a trapezoid form with' the non- parallel sides two triangles having equal areas. CHAPTER V. REGULAR POLYGONS AND CIRCLES. REGULAR POLYGONS. 332. A regular polygon is one which is both equilateral and equiangular. According to this definition, determine whether each of the following polygons is regular or not and state why : An equilateral triangle, an equiangular triangle, a rectangle, a square, a rhombus. Draw a figure to illustrate each. Make a triangle which fulfills neither condition of the definition, also a quadrilateral. 333. The general problem of constructing a regular polygon depends upon the division of a circle into as many equal parts as the polygon has sides. The problem of dividing the circle into equal parts can be solved in some cases by the methods of elementary geometry, and some of these methods will be considered in this chapter. In most cases this problem cannot be solved by elementary methods. E.g. the circle may be divided into 2, 3, 4, 5, 6, 8, 10, 12, 15, equal parts, but not into 7, 9, 11, 13, 14, equal parts. If a circle has already been divided into a certain num- ber of equal parts, it may then be divided into twice, four times, eight times, etc., that number of parts by repeated bisection of the arcs. (See § 204, Ex. 1.) The division of the circle into equal parts depends upon the theorem (§199) that equal central angles intercept 176 REGULAR POLYGONS AND CIRCLES, 177 equal arcs on the circle, and hence it involves the subdi- vision of angles into equal parts. The following exercises depend upon principles already familiar. With each solution give the reasons in full for each step. All constructions are to be made with ruler and com- passes only. 334. EXERCISES. 1. Divide a given angle or arc into four equal parts. 2. Divide a given angle or arc into eight equal parts. 3. If an angle or arc is already divided into a certain number of equal parts, show how to divide it into twice that number of equal parts. 4. Divide a circle into four equal parts. 5. Divide a circle into eight equal parts. 6. If a circle is already divided into a certain number of equal parts, show how to divide it into twice that number of equal parts. 7. Divide a circle into six equal parts. Suggestion. Construct at the center an angle of 60°. 8. Draw the chords connecting in order the four division points in Ex. 4, and show that the figure is a regular quadrilateral. 9. Draw the tangents at the four division points in Ex. 4, and show that a regular quadrilateral is formed. 10. Draw the chords connecting in order the division points in Ex. 7, and show that a regular hexagon is formed. Prove that the side of the hexagon is equal to the radius of the circle. 11. Draw chords connecting alternate division points in Ex. 7, and show that a regular triangle is formed. 12. Construct tangents to the circle at alternate division points in Ex. 7, and show that a regular triangle is formed. 13. Draw tangents to the circle at the division points in Ex. 7, and show that a regular Jiexagon is formed. Note. See also the construction of regular polygons of 3, 4, and 6 sides, § 163, Exs. 3, 4, 5. 178 PLANE GEOMETRY. 335. Theorem. If a circle is divided into any num- her of equal arcs, the chords joini7ig the division points, taken in order, form a regidar polygon. Proof : Show (1) that the chords are equal ; (2) that the angles are inscribed in equal arcs and hence are equal. 336. Theorem. If a circle is divided into any num- ber of equal parts, the tangents at the points of divi- sion, taken in order, form a regular polygon. Given the G OA divided into equal arcs by the points M, N, P, etc., with tangents drawn at these points forming the polygon ABCDEF. To prove that ABCDEF is a regular polygon. Analysis of Proof : (1) To prove that AB = BC=^ CD, etc. We know that BP = BN (Why ?). Hence, if we can show that AN = NB and BP = PC, then it will follow that AB — BC. To prove AN = NB, it must be shown that Z 1 = Z 2, and this is done by showing that Z 1 and Z 2 are halves of the equal angles NOM and NOP. This necessitates proving A ^OiV ^ A AOM. Now state the proof in full. (2) To prove ZABC = Zbcd = Z CDE, etc. From the triangles proved congruent under (1) show REGULAR POLYGONS AND CIRCLES. 179 that Z MAO = Z OAN = Z NHO = ZoBP; and hence, Z MAO-\- Z OAN = Z NBO + Z 07^7^ or Z i^Jl? = Z ABC. In like manner it is proved that Z^7:?C= Z BCD^ etc. Hence the polygon is equilateral and also equiangular, and hence regular. 337. Theorem. If a polygon is regular, a cwcle may he circumscribed about it. D A B Given a regular polygon ABODE. To prove that a point o can be found such that OA = OB — OC — 0B= OE. Outline of Proof : Bisect A A and B and let the bisectors meet in some point O. Then o is the center sought. For we have (1) OA = OB, Why ? (2) A AOB ^ BOC. .', 0A= OC. (S) A BOC ^ A COD. .'.OB=OD. (4) Now prove OC = OE. . •. OA = 0B= OC = OD = OE. Prove each step, showing how it depends upon the pre-' ceding. 338. EXERCISES. 1. Find the locus of the vertices of all regular polygons of the same number of sides which can be circumscribed about the same circle. 180 PLANE GEOMETRY. 2. Find the locus of the middle points of the sides of all regular poly- gons of the same number of sides which are inscribed iu the same circle. 3. Is an equilateral circumscribed polygon regular? Prove. 4. Is an equiangular circumscribed polygon regular? Prove. 339. Theorem. // a ^polygon is regular, a circle may he inscribed in it. Given the regular polygon ABCDE. To prove that a point O may be found such that the per- pendiculars OM, ON, OP, OQ, OR, are equal. /x Outline of Proof : Determine a y< >\^ point O as in the proof of tlie pre- j^^^ \ / ^^^c ceding theorem. Then A AOB, \ '§--'''' / HOC, etc., are equal isosceles trian- /A''' '/ ''\~ ~^~/v gles, and their altitudes are equal. \ /'' j \ / Hence O is the required point. \/^ | '\] Give proof in full. a M B 340. Theorem. An equilateral polygon inscribed in a circle is regular. Suggestion for Proof : Use the fact that equal chords subtend equal arcs, and apply § 335. 341. EXERCISE. Show that the inscribed and circumscribed circles of a regular polygon have the same center. 342» Definitions. The center of a regular polygon is the common center of its inscribed and circumscribed circles. The radius of a regular polygon is the distance from the center to one of its vertices. The apothem of a regular polygon is the perpendicular distance from its center to a side. REGULAR POLYGONS ANU CIRCLES. 181 343. Theorem. The area of a regular polygon is equal to half the product of its apothem and perimeter. Suggestion for Proof : A regular polygon is divided by its radii into a series of congruent triangles, the area of each of which is one half the product of the apothem and a side of the polygon. Complete the proof. 344. Theorem. The area of any polygon circum- scribed about a circle is equal to one half the product of the perimeter of the polygon and the radius of the circle. The proof is left to the student. 345. EXERCISES. 1. Is every equiangular polygon inscribed in a circle regular? Prove. 2. Show that the radius of a regular polygon bisects the angle at the vertex to which it is drawn. 3. Show that the perimeter of a regular polygon of a given num- ber of sides is less than that of one having twice the number of sides, both being inscribed in the same circle. 4. Show that the perimeter of a regular polygon is greater than that of one having twice the number of sides, both being circum- scribed about the same circle. 5. Compare the areas of the two polygons in p]x. 3. 6. Compare the areas of the two polygons in Ex. 4. $ 7. Show that the area of a square inscribed in a circle of radius R is 2 R"^. How does this compare with the area of the circum- scribed square. \ 8. Compute the apothem and area of a regular inscribed hexagon if the radius of the circle is R ; also of the regular circumscribed hexagon. 9. A regular triangle is inscribed in a circle of radius 10. Find the apothem and a side of the triangle. 182 PLANE GEOMETRY, 346. Theorem. Two regular polygons of the same number of sides are similar. Outline of Proof: (1) Show that all pairs of corre- sponding angles are equal. (2) Show that the ratios of pairs of corresponding sides are equal. Hence the polygons are similar (Why?). 347. Theorem. The perimeters of two regular poly- gons having the same number of sides are in the same ratio as their radii or their apothems. Outline of Proof: Show (1) that each triangle formed by a side and two radii in one polygon is similar to the corresponding triangle in the other polygon. A Ti f d (T) That — -- = — = — , where AB and A'B' are the two ^ A^B^ r' a' sides, r and r' the corresponding radii and a and a' the corresponding apothems. And so for the remaining pairs of triangles. ,gN rp, , AB -{- BC -\- CD ^ • • • _ AB _r__o^ Draw the figure and give the proof in full. 348. Theorem. The areas of two regular polygons of the same number of sides are in the same ratio as the squares of the corresponding radii or apothems. Outline of Proof : Divide the polygons into pairs of corresponding triangles as in the preceding proof. AT r^ n^ (1) Show that , =—, = -—, and so for each pair of A r r'^ a'^ AT + AlI-hATTI+ ... r2 triangles. ^ ^ Draw figure and give the proof in full. REGULAR POLYGONS AND CIRCLES. PROBLEMS AND APPLICATIONS. 1. Find the ratio of the perimeters of squares inscribed in and circumscribed about the same circle. 2. Find the ratio of the perimeters of regular hexagons inscribed in and circumscribed about the same circle. 3. Find the ratio between the perimeters of regular triangles inscribed in and circumscribed about the same circle. 4. Find the ratio of the areas of regular triangles inscribed in and circumscribed about the same circle. Also find the ratio of the areas of such squares and of such hexagons. 5. The perimeter of a regular hexagon inscribed in a circle is 24 in6hes. Find the perimeter of a regular hexagon circumscribed about a circle of twice the diameter. 6. The area of a regular triangle circumscribed about a circle is 64 square inches. What is the area of a regular triangle inscribed in a circle of one third the radius? 7. The area of a regular hexagon inscribed in a circle is 48 square inches. What is the area of a regular hexagon circumscribed about a circle whose diameter is If times that of the first? 8. A chord AB bisects the radius perpendicular to it. Find the central angle subtended by the chord. State the result as a theorem. 9. State and prove the converse of the theorem in Ex. 8. 10. Find the area of a regular triangle inscribed in a circle whose radius is 6 inches. mi 11. Find the area of a regular triangle inscribed in a circle whose radius is r inches. 12. One of the acute angles of a right triangle is 60° and the side adjacent to this angle is r inches long. Find the remaining sides of the triangle. , 13. A regular triangle is circumscribed about a circle of radius r. Find its area. Suggestion. First find DC in the figure. 4 14. A regular triangle of area 36 square inches is inscribed in a circle. Find the radius of the circle. 184 PLANE GEOMETRY. / 15. Find the radius of a circle if the area of its regular inscribed triangle is a. 16. Find the radius of the circle if the area of the regular circum- scribed triangle is a. 17. Find the radius of a circle if the difference between the perimeters of the tegular inscribed and circumscribed triangles is 12 inches. 18. Find the radius of a circle if the difference of, the perimeters of the regular inscribed and circumscribed hexagons is 10 inches. 19. If the area of a circumscribed square is 25 square inches greater than that of an inscribed square, what is the diameter of the circle ? 20. Find the radius of a circle if the difference between the areas of the inscribed and circumscribed regular triangles is 25 square inches. 21. Find the radius of a circle if the difference between the areas of the regular inscribed and circumscribed hexagons is 25 square inches. 22. The difference betwieen the areas of the squares circumscribed about two circles is 50 square inches and the differ- ence of their diameters is 4 inches. Find each diameter. 23. If the inscribed and escribed circles and 0' of an equilateral triangle are constructed as shown in the figure, find the ratio of their radii. Does this ratio depend upon the size of the triangle ? 24. Given a triangle ABC and a segment a, show how to construct a segment DE II AB and equal to the segment «, such that the points D and E shall lie on the sides CA and CB respectively or on these sides extended. 25. A triangular plot of ground ABC is to be laid out as a triangular flower bed with a walk of uniform width extending around it. (a) Prove that the flower bed is similar to the original triangle. (ft) Show that the corners of the flower bed lie on the bisectors of the angles of the original triangle. REGULAR POLYGONS AND CIRCLES. 185 (c) Find a segment S so that ''—;;- = 2 and construct A'B' equal to S and parallel to AB, the points A' and B' lying on the bisectors of the angles A and B respectively. See Exs. 16-19, page 172. (rf) Draw A'C II AC and B'C II 5C. ' Prove that the area of the flower bed A' B'C, as thus constructed, is equal to the area of the walk. (e) Construct the figure for the flower bed so that its area is five times that of the walk. 26. Given a rectangular plot of ground. Is it always possible to lay off on it a walk of uniform width running around it so that the plot inside the walk shall be similar to the original figure? Prove. . 27. Is the construction proposed in Ex. 26 always possible in the case of a square ? of a rhombus ? Prove. 28. If a side of a regular hexagon circumscribed about a circle is a, find tlie radius of the circle. 29. On a regular hexagonal plot of ground whose side is 12 feet a walk of uniform width is to be laid oft' around it. Find by algebraic computation the width of the walk if it is to occupy one half the whole plot. 30. Find by geometric construction the width of the walk in Ex. 29. 31. Show that the figure inside the walk in Ex. 29 is a regular hexagon. 32. Given a segment AB, find three points C, D, E on it so that, AC^^l A^^l ^^^ AE' ^ 3 AB^ 4' ab' 2' ab' 4* 33. A regular hexagon ABCDEF is to be divided into four pieces of equal area by segments drawn parallel to its sides forming hexagons as shown in the figure. If AB = 2^ feet, find a side of each of the other hexa- gons and also the apothem of each. 34. Without computing algebraically the apo- them or sides of the inner hexagons in Ex. 33, show how to construct the figure geometrically. See Ex. 6, § 331. Also Ex. 18, page 172. Use §348. C 186 PLANE GEOMETRY, / 35. In a given hexagonal polygon whose side is a find in terms of a. and n the width of a walk around it which will occupy one nth of the area of the whole polygon. 36. By means of hexagons similar to those in Ex. 33, divide a given regular hexagon into'threeparts such that the outside part is \ the whole area and the next ^ of the whole area. 37. Compute the sides and the apothem of the two hexagons con- structed in Ex. 36 if the side of the given hexagon is 12 inches. 38. In the adjoining pattern find two regular hexagons whose areas are in the ratio 1 : 4 and show that this agrees with theorem, § 348. 39. If a polygon is circumscribed about a circle, show that the bisectors of all its angles meet in a point. 40. Given any polygon circumscribed about a circle. draw segments parallel to each of its sides and at the same distance from each side. Show that these segments form a polygon similar to the first. 41. If the bisectors of all the angles of a polygon meet in a point prove that a circle may be inscribed in it (tangent to all its sides). What is the relation of the theorems in Exs. 39 and 41 ? 42. Given a polygonal plot of ground (boundary is a polygon) such that a path of equal width on it around the border leaves a polygonal plot similar to the first. Prove that a circle may be in- scribed in the polygon which forms the boundary of either plot. 43. In the figure, A BCD is a square and EFGHKLMNis a regular octagon. (a) If EF = 4 inches, find AB. d l k C (6) If AB = 12 inches, find EF. (c) Ji EF = a, find AB. (d) It AB = s, find EF. (e) It AB = s, find the apothem. (/) Jt AB = s, find the radius of the octagon. REGULAR POLYGONS AND CIRCLES. 187 44. Find the ratio between the areas of regular octagons inscribed in and circumscribed about the same circle. 45. (a) The apothem of a regular octagon is 10 feet. Find the width of a uniform strip laid off around it which occupies \ its area. (b) Show how to construct this strip geometrically without first computing its width. Prove that the inside figure is a regular octagon. 46. A regular octagon ABCDEFGH is to be divided into four parts of equal area by means of octagons as shown in the figure. (a) li AB = \2 inches find a side of each octagon, (ft) Show how to construct the figure without first computing the sides. (c) Show how to construct such a figure if the four parts I, II, III, IV, are to be in any required ratios. (d) Measure the sides of the two inner parts of the ceiling pattern and hence find the ratio of their areas. 47. The middle points, A,B, C, D, of alternate sides of a regular octagon are joined as shown in the figure. A H is perpendicular to A E and equal to it. BG, CK, and DL are con- structed in the same manner. (a) Prove that ABCD and HGKL are squares. (h) Find the areas of ABCD and HGKL, if EF= 10 inches. (c) What fraction of the whole octagon is occupied by the square HGKL. See the accompanying tile border. ^ 188 PLANE GEOMETRY. MEASUREMENT OF THE CIRCLE. 349. If in a circle a regular polygon is inscribed, its perimeter may be measured. For example, the perimeter of a regular inscribed hexa- gon is 6 r if r is the radius of the circle. See § 334, Ex. 10. If the number of sides of the inscribed polygon be doubled, the resulting perim- eter may be measured or computed in terms of r. See §357. If again the number of sides be doubled, the resulting perimeter may be computed in terms of r, and so on. In a similar manner a regular polygon, say a hexagon, may be circumscribed about a circle and its perimeter expressed in terms of r. If the number of sides of the circum- scribed polygon be doubled, its perim- eter may again be computed, and so on as often as may be desired. 350. By continuing either of these processes it is evident that the inscribed or the circumscribed polygon may be made to lie as close to the circle as we please. 351. The word length has thus far been used in connec- tion with straight line-segments only. Thus, the perime- ter of any polygon is the sum of the lengths of its sides. 352. We now assume that a circle has a definite length and that this can be approximated as nearly as we please by taking the perimeters of the successive inscribed or cir- cumscribed polygons. The length of a circle is often called its perimeter or circumference. REGULAR POLYGONS AND CIRCLES. 189 It is evident that approximate measurement is the only kind possible in the case of the circle, since no straight line unit of measure, how- ever small, can be made to coincide with an arc of a circle. 353. Comparison of the Lengths of Two Circles. In each of two circles, O and o', let regular polygons of 6, 12, 21, 48, 96, 192, etc., sides be inscribed. Call the perimeters of the polygons in O O, Pg, P^2,'> ^24' ^^^-i ^i^d those in O o', P'g, P'i2^ P'24, etc. Let the radii of the circles be R and r' respectively. Then by the theorem of § 347, we have 4 = ^ = ^ = -2.4= etc., /?' p' pi jy n -* 6 12 2-4 however great the number of sides of the inscribed poly- gons. Show that the same relations hold if polygons are circumscribed about the circle. From these considerations we are led to the following : 354. Preliminary Theorem. The lengths of two circles are in the same ratio as their radii. 355. Hence, if C and & are the circumferences of two circles, R and r' their radii, and D and b' their diameters, we have — = —= — ; and, also - = — : • (See § 245) c' r' b' d n' Hence, the ratio of the circumference to the diameter in one circle is the same as this ratio in any other circle. 356. This constant ratio is denoted by the Greek letter tt, pronounced pi. The argument used above shows that a theorem like that of § 354 holds for every pair of polygons used in the approximation process, and hence it is established for all purposes of practical measurement or computation. 190 PLANE GEO MET BY. 357. Problem. To compute the approximate value ofTT, ^ Solution. Suppose a regular polygon of n sides is in- scribed in a circle whose radius is r and let one of the sides AB be called S^- We first obtain in terras of S^ the length AC^ or S^^-* of a side of a regular polygon oi 2n sides inscribed in the same circle. Let AB be a side of the first polygon. Bisect AB at C and draw AC and BC. Then ^C is a side of a regular inscribed polygon of 2 w sides (Why ?). Then in the figure (82.)^=l? = (i^B)2+A2 = (i5j2+^2. (Why?) (1) But hxDE=^h(2r-h) = AD xi)B==A^, (Why?) or lS' = a-s,y=hC2r-h). (2) Solving equation (2) for A, we have 2 r ± V4 r^ - s,,^ 2 Taking only the negative sign, since A < r, and squaring, we have, or, 2r2-rV4r2-s„2_ 1 s^^2^ ^2 4. (1 s^-)^= 2 r2 - rV4 r^ - 8,;\ Hence, from (1), S^„^ = 2r'^ - rVir^ - s,,\ REGULAR POLYGONS AND CIRCLES, 191 and S2^ = ^2r^-rV4:r^- Sr?, Since, by § 355, the value of tt is the same for all circles, we take a circle whose radius is 1. In this case -S2„ == ^2 _ V4 - s„\ (3) If the first polygon is a regular hexagon, then -S^g = 1. Hence, s^^ = V2 _ VI^ = 0.51763809. Denoting the perimeter of a regular inscribed polygon of n sides by P„, we have, P12 = 12(0.51763809)= 6.21165708. In the formula (3) let n = Vl. Then -^24 = V2 _ V4 - (0.51768809)2 = 0.26105238. Hence, P^^ = 24(0,26108238)= 6.26525722. Computing /S^g, P^g, etc., in a similar manner, we have. s,. = ^2- . vri ^ .51763809. .26105238. .13080626. .06543817. .03272346. .01636228. .00818126. -'Pu .-. P,, •••^96 ■^'P^,2 '•'P,S, •••^768 = 6.21165708. Su = V2- . V4~ - (.51763809)2 = P 6.26525722. S4S = V2- . V4- - (.26105238)2 = =. 6.27870041. s^ = V2 - . VI~- - (.13080626)2 = = 6.28206396. •^192 = V2- ■ V4- - (.06543817)2 = = 6.28290510. "^384 = V2 - ■ VT^ - (.03272346)2 = = 6.28311544. •^768 = V2- . vn - (.01636228)2 = = 6.28316941. 358. The Length of the Circle. By continuing this process it is found that the first five figures in the decimal remain unchanged. Hence 6.28317 is an approximation to the circumference of a circle whose radius is 1. 192 PLANE GEOMETRY, 359. By a process similar to the preceding, if circum- scribed polygons of 4, 8, 16, etc., sides are used, the follow- ing results are obtained: Number of Sides Length of Each Side Perimeter 4 2.000000 8.000000 8 0.828428 6.627418 16 0.397824 6.365196 32 0.196984 6.303450 64 0.098254 6.288236 128 0.049078 6.284448 256 0.024544 6.283500 512 0.012272 6.283264 1024 0.006136 6.283205 2048 0.003068 6.283190 360. Since the diameter is 2, the ratio of the circumfer- ence to the diameter is 64^11 = 3.14159, or ^•^^f^'* = 3.14159 according as the inscribed or circumscribed polygons are used. That is, these approximations of tt agree to five decimal places. If so great accuracy is not required, we may use a smaller number of decimal places; such as, 3.1416, 3.14, or3f /-» Since ' =7r, it follows that C= irD = 27rR. J) That is, the circumference is 2 tt times the radius. REGULAR POLYGONS AND CIRCLES. 193 361. The area of the circle. In § 359 the area of each circumscribed polygon is half the product of the perimeter and the apothem, which in this case is the radius of the circle. That is, for every such polygon, -4 = | P • i?, or area equals one half perimeter times radius. We now assume that a circle has a definite area which can be approximated as closely as we please by taking the areas of the successive circumscribed polygons. 362. Since the perimeters of the circumscribed polygons can be made to approximate the length of the circle as nearly as we please, and since C = 2 7rr we have, area of circle = | • 2 irr • r = irr^. The degree of accuracy to which this formula leads depends entirely upon the accuracy with which tt is determined. The old problem of squaring the circle, that is, finding the side of a square whose area equals that of a given circle, involves therefore determining the value of tt. Much time and labor have been expended upon this in the hope that this value could be exactly constructed by means of the ruler and compasses, but it is now known that this is impossible. 363. Since the area of a circle is tt r^, if we have two given circles whose radii are r and r' and whose diameters are d and d' ^ then the ratio of their areas A and A' is A' 7rr'^ r'^ d'^ Theorem. The areas of two circles are in the same ratio as the squares of their radii, or of their diameters. 194 PLANE GEOMETRY. 364. The area of a sector bears the same ratio to the area of the circle as the angle of the sector does to the perigon. E.g. the area of the sector whose arc is a quadrant is one fourth of the area of the circle, that is, ^^^ — . The area of a semicircle is ^^^^. Find the area of a sector whose angle is 60° ; 30° ; 45° ; 72°. 365. The area of a segment of a circle is known if that of the sector having the same arc, and of the central triangle on the same chord, can be determined. Show that the areas of segments of a circle whose arcs are respec- tively 90° and 60° are 4 2 4 "^ ^ 6 4 12 ^ ^ The accompanying figure shows a circle cut into sectors by a series of radii. Each sector ap- ■^ * A A A /> ,^ « ,< A A /» A A proximates the shape of i \ i\ i \ 1 \ j \ < \ I \ / \ < \ I \ ! \ 1 \ / \ a triangle, whose altitude / \' \/ \/ w' \/ \/ \/ \/ \/ \/ \/ \/ \ is the radius of the circle, and whose base is an arc of the circle. Since the sum of the areas of these triangles is the product of the altitude and the sum of the bases, we obtain a verification of the theorem that the area of a circle is one half the product of the circumference and radius. SUMMARY OF CHAPTER V. 1. Make a list of the definitions pertaining to regular polygons. 2. State the theorems concerning regular polygons inscribed in or circumscribed about a circle. 3. State the theorems involving similar regular polygons. 4. Give an outline of the discussion concerning the value of tt and its use in approximating the length and area of the circle. 5. What are some of the more important applications of the theo- rems in Chapter V? (Return to this question after studying the ap- plications which follow.) y ^^ ^j ^ )^-©C )^e( ^ r^^r REGULAR POLYGONS AND CIRCLES. 195 PROBLEMS AND APPLICATIONS ON CHAPTER V. Note. In the following problems, unless otherwise specified, use the value rr = 3f , which differs from 3.1416 by about ^^ of one per cent. 1. The diameter of a circle is 8 inches. Find the circumference and area. 2. The circumference of a circle is 10 feet. Find its radius and area. 3. The area of a circle is 24 square inches. Find its radius and circumference. 4. Centers of circles are arranged at equal distances on a network of lines at right angles to each other as shown in the figure. If r = I AB, what part of the whole area is enclosed by the circles? Suggestion. Consider what part of the square A BCD lies within the circles. 5. If in Ex. 4 the radius of each circle is 3 inches, how far apart must the centers be in order that one half the area shall lie within the circles? 6. If the radius of each circle is r, how far apart must the centers be located in order that - of the whole area may lie within the circles ? 7. If the circles occupy — of the area, and if the centers are d inches apart, find the radii. 8. A square is inscribed in a circle. Find the ratio between the areas of the square and the circle. 9. A square is circumscribed about a circle. Find the ratio be- tween their areas. Do the ratios required in Exs. 8 and 9 depend upon the radius of the circle ? 10. Find the ratio between the area of a circle and its regular in- scribed hexagon. il. Find the ratio between the area of a circle and its regular cir- cumscribed hexagon. Do the ratios required in Exs. 10 and 11 depend upon the radius of the circle ? 196 PLANE GEOMETRY. 12. Find the ratio between the area of a square inscribed in a circle and another circumscribed about a circle having D G C 3 times the radius. 13. Find the ratio of the area of a regular hexagon inscribed in a circle to that of another jj circumscribed about a circle having a radius \ as great. 14. J 7>CZ) is a square whose side is a and the ^ points A^ B, C, D, are the centers of the arcs HE, EF, etc. (a) Find the area of the figure formed by the arcs FLE, EKH, HNG, GMF. {h) Find the area of the figure formed by the arcs ELF2indiEQF. Suggestion. Find the difference between the areas of a circle and its inscribed square. (c) Find the area of the figure formed by the arcs EK, KL, and LE. 15. If the sides of the rectangle A BCD are 8 and 12 inches, respec- tively, find the radii of the circles. (a) AVhat fraction of the area of the rectangle lies within only one circle? (b) Prove that at each vertex of a square two circles are tangent to a diagonal of the square. 16. Two concentric circles are such that one divides the area of the other into two equal parts. (a) Find the ratio of the radii of the circles. (b) Given the outer circle, construct the inner one. 17. Construct three circles concentric with a circle with radius r, which shall divide its area into four equal parts. 18. Prove that six circles of equal radii can be constructed each tangent to two of the others and to a given circle. (a) Show that a circle can be constructed around the six circles tangent to each of them. (b) What fraction of the area of the last circle is occupied by the seven circles within it? 19. A OB is a central angle of 60°. Find the area bounded by the chord AB and AB if the radius of the circle is 3. REGULAE POLYGONS AND CIRCLES, 197 20. ABC is an equilateral gothic arch. (See page 109.) Find the area inclosed by the segment AB and the arcs AC and BC, \iAB = ^ feet. Suggestion. Find the area of the sector with center A, arc BC, and radii AB and AC, aiid add to this the area of the circle-segment whose chord is A C. 21. ABC is an equilateral triangle and A, B, and C are the centers of the arcs. Show that the area of the figure formed by the arcs is three times the area of one • of the sectors minus twice the area of the A ABC. 22. In the figure ABC is an equilateral triangle and D, E, and F are the mid- dle points of its sides. Arcs are constructed as shown. (a) li AB=6 feet, find the area inclosed by CE, EF, FC. (h) Find the area inclosed by .O;, EC, and Cyi. {c) Find the area inclosed by DF, 1^, and El). 23. In the figure ABC is an equi- lateral arch and O is constructed on AB as a diameter. AH and BK are perpendicular to AB. (a) Construct the equilateral arches HED and DFK tangent to the circle as shown in the figure. Suggestion. 0H= OA + DH. (h) Prove that the vertices E and F lie on the circle. SuGGESTiox. What kind of a triangle is HOKl In the following let AB = ^ feet. (c) Find the area bounded by the arcs DF, FE, and ED. (d) Find the area of the rectangle ABKH. (e) Find the area bounded by AH and the arcs HE and EA. (/) Find the area bounded by the upper semicircle AB and the arcs AC Bind BC. (g) Find all the areas required in (c) ••• (/) if AB = a. From the Uniou Park Church, Chicago. C H D K From the First Presbyterian Church, Chicago. 198 PLANE GEOMETRY. 24. In the figure, ABCDEF is a regular hexagon. B is the center of the arc ALC, D the center of CHE, and F the center of EKA. li AB = 1Q inches, find H (a) The circumference and area of the circle, (b) The area bounded by the arc ALC and the segments AB and BC, (c) The area bounded by KA, AL and LK, ^, (d) The area bounded by ALC, CHE and EKA. (e) Find the areas required in (a)-(d) if AB = a inches. 25. ABC is a regular octagon. Arcs are con- structed with the vertices as centers as in the figure. (a) If AB= 10 inches, find the area inclosed by the whole figure. Also if AB = a. 26. ABCDEFGH is a regular octagon. Semi- circles are constructed with the sides as diameters. (a) li AB = 10 inches, find the area of the whole figure. Also \i AB = a. (b) Complete the drawing in the outline figure to make the steel ceiling pattern here shown. 27. In the figure A CB, AFD, DEB, and ECF are semicircles. EF is tangent to two semicircles. (a) Prove that the semicircles AFD, FCE, and DEB are equal, D being given the middle point of AB. If ^5 = 48 inches, find : {b) The area bounded by AC, 7^^ and FA, (c) The area bounded by FD and DE, and the line-segment FE, A D B (d) The area bounded by AF, FCE, EB, and the line-segqient AB. (e) Find the areas required in (b)-(d) if AB = a. 28. (a) If a side of the regular hexagon in the figure is a, find the area inclosed by the arcs (including the area of the hexagon). (b) Show that a circle may be cir- cumscribed about the whole figure. (c) Find the area inside this circle and outside the figure in (a). REGULAR POLYGONS AND CIRCLES, 199 MISCELLANEOUS PROBLEMS AND APPLICATIONS. >•< ►^ ►-< X >< 1 ►^ ^4 ►^ ►-4 ►'^ ►-^ KfefefeM 1. Given a rhombus, two of whose angles are 60°, to divide it into a regular hexagon and two equilateral tri- angles. See Ex. 7, page 78. 2. What fraction of the accompanying design for tile flooring is made up of the black tiles? Show how to construct this design by marking off points along the border and drawing parallel lines. 3. In the accompanying design for a parquet border two strips of wood appear to be intertwined. (a) If the border is 8 inches wide and 3 feet 4 inches long, find the area of one of these strips, including the part which ap- pears to be obscured by the other strip. (b) If the figure consists of squares, find the angle at which the strips meet the sides. Use the table on page 139. (c) If the width of the border is a and its length b, find the com- bined area of these strips. Compare the total area of the obscured part of these strips with the sum of the areas of the small triangles along the edge of the border. 4. A solid board fence 5 feet in vertical height running due north and south is to be built across a valley, 80 rods connecting two points of the same elevation. Find the number of square feet in the fence if the horizontal distance is 80 rods. 5. Are the data given in the preceding problem sufficient to solve it if the fence required is to be an ordinary four- board fence, each board 6 inches wide? 6. Two circles are tangent internally at a point A. Chords AB and ^C of the larger circle are drawn meeting the smaller circle in D and E respectively. Prove that BC and DE are parallel. 7. Two circles, radii r and r', are tangent internally. Find the length of a chord of the larger circle tangent to the smaller if : (a) The chord is parallel to the line of centers. (b) The chord is perpendicular to the line of centers. (c) Meets the larger circle at the same point as the line of centers. 200 PLANE GEOMETRY. A CBE and hence dP^ = ^. BE CE 8. Given a straight line and two points A and B on the same side of it. Find a point C on the line such that the sum of the seg- ments A C and BC shall be the least possible. Solution. In the figure let B' be symmetric to B with respect to the line. Draw AB' meeting the line in C. Then C is the required point. For let C be any other point on the line. Then A C + C'B >AC+ CB. The proof depends upon § 128 and Ax. Ill, § 61. (live it in full detail. 9. If in the figure preceding, A I) is perpendicular to the line, prove that A A DC 10. If in Ex. 8, BE = a, AD = b, BE = c, find CD and CE. 11. Two towns, A and B, are 10 and 6 miles respectively from a river and A is 12 miles farther up the river than B. A pumping station is to be built which shall serve both towns. Where must it be located so that the total length of water main to the two towns shall be the least possible ? 12. Two factories are situated on th^ same side of a railway at different distances from it. A spur ' ^=-^====^' is to be built to each factory and these are to join the railway at the same point. State just what measurements must be made and how to locate the point where these spurs should join the main line in order to permit the shortest length of road to be built. 13. Two equal circles of radius r intersect so that their common chord is equal to r. Find the area of the figure which lies within both circles. 14. In the accompanying design for oak and mahogany parquet flooring the large squares are 6 inches and the small black ones 2| inches on a side. What fraction of the whole is the ma- hogany (the black squares) ? 15. Construct circles on the three sides of a right triangle as diameters. Compare the area of the circle constructed on the hypotenuse with the sum of tlie areas of the other two. Prove. K \ REGULAR POLYGONS AND CIRCLES. 201 16. In the accompanying design for grill work : (a) Find the angles ABC, BAD, and AGC. (&) If the radius of each circle is r, find the distance AF. (c) What fraction of the area of the parallelogram GBCE lies within one circle only? (d) If the radius of each circle is r, find the distance between two horizontal lines. (e) Construct the whole figure. Suggestions. (1) Find AF and lay off points on AB. (2) Find Z. BAD and construct it. (3) Through the points of division on AB draw lines parallel to AD. (4) From A along AD lay off segments equal to AF and through these division points construct lines parallel to AB. (5) Along DC lay off segments equal to AF. Connect points as shown in the figure. (/) From the construction of the figure does it foUow^ that ZDAB = /.EGC'i 17. Prove that the sum of three altitudes of a triangle is less than its perimeter. • 18. In the accompanying design for grill work, the arcs are con- structed from the vertices of the equilateral triangles as centers. (a) Prove that two arcs are tangent to each other at each vertex of a triangle. (b) Find the area bounded by the arcs AB, BC, CD, DA . (c) Find the ratio between the area in (ft) and the area of the triangle DBC. 202 PLANE GEOMETRY. 19. The character of the accompanying design for a window is obvious from the figure. Denote the radius of the large circle by R, of the semicircles by R' , and of the small circles by r. (a) If i2 =: 8 feet find R' and r. (b) Find R' and r in terms of R. (c) What fraction of the area of the large circle lies within the four small circles? {d) What fraction of the area of the large circle lies outside the four semicircles ? (e) If it = 10, find the area inclosed within the four small circles. 20. In the accompanying design for a stained glass window : (a) What part of the square A'B'C'D' lies within ABCD'i (b) If A'B' = 4: feet, find the sum of the areas of the semicircles. (c) Find the area inclosed by the line-seg- ments FB', B'E and the arcs FB and BE, if EB' is 1^ feet. (d) Find the areas required in (b) and (c) if A'B' = a. 21. The accompanying design for tile flooring consists of regular octagons and squares. The design can be constructed by drawing parallel lines as shown in the figure. (a) If a side of the octagon AB is given, find BC, DE, and EF by construction. Find the ratio of any two of these seg- ments. (&) If AB = a, find BC. (c) Ji AB = a find the area of the square xyzw. (d) At what angles do the oblique lines meet the horizontal ? (e) Construct the figure by laying off the required points on the sides, drawing parallel lines in pencil, inking in the sides of the octagons and erasing the remainder of the lines. REGULAR POLYGONS AND CIRCLES. 203 D :,x,x,x,x, :_x_x_x X T T T T 1 T T X X : X X X X r_T_T_r x„x x,x X X X X .X .X 3 ^L -! / / t» I >/ 5 1 ■ ,._ / 7 — < h = 1 (V 'f. ^/ 1 * / / 1 h 2= 10 n / 1 / 1 "^/ /' 1 1 / / 1 / 1 i, -» / ^ A 1 1 -^ ^ -4^ y 1 1 / 1 1 / * 1 / ( 1 1 / ) 1 1 J 3 Alti tudt )- Ax S VARIABLE GEOMETRIC MAGNITUDES. 209 If we suppose that while the base of the rectangle re- mains fixed, the altitude varies continuously through all values from Ag = 10 to Ag = 20 = 2 x 10, then it must take among other values the value 10 V2. Using 10 V2 as an abscissa, the question is, whether CD is the area ordinate corresponding to it. This area ordinate is not less than CD and commen- surable with the base, for in that case the altitude would be less than 10 V2, and for the same reason it is not greater than CD and commensurable with the base. But we can find line-segments less than 10 V2 or greater than 10 V2 and as near to 10 V2 as we please. Hence we conclude that the area ordinate for the rectangle whose altitude is 10 V2 is CD, that is, the point c lies in the line OP^. In like manner, the point determined by any other ab- scissa incommensurable with the base is shown to lie on the line OP^^. Since the abscissa and ordinate of any point on OP are equal, we have for any altitude, 369. The preceding theorem may also be stated : The area of a rectangle ivith a fixed base varies directly as its altitude. This means that if A and h are the varying area and altitude re- spectively, and if ^1 and hi are the area and altitude at any given instant, then -— - = — or ^ = — ^ ' h or A = kh, where k is the Jixed ratio —^' A\ h,\ hi hi The graph representing the relations of two variables when one varies directly as the other is always a straight line. 210 PLANE GEOMETBY, 370. EXERCISES. 1. Make a graph to show that the area of two rectangles having equal altitudes are in the same ratio as their bases. 2. Show by a graph that the area of a triangle having a fixed alti- tude varies as the base, and having a fixed base varies as the altitude. 3. Represent graphically the relation between two line-segments both of which begin at zero, and one of which increases three times as fast as the other. Five times as fast. One half as fast. 4. If areas be represented by the length of a line-segment as in § 368, which question in Ex. 3 applies to the altitude and area of a parallelogram having a fixed base and varying altitude? Which applies to a triangle having a fixed base and varying altitude? 371. Theorem. The area of a rectangle is equal to the product of its base and altitude. Proof : Using the graphic representations of §§ 368, 370, Ex. 1, we have for all cases, ^ = ? and ^=1. a ^1 1 ^, 1 Az b ^1 1 ^^2 1 Hence ^ = ^ x ^ = ? x - = a5. A^ A^ A,^ 1 1 But A^ is the unit of area. Hence ab represents the numerical measure of " ^3 by the area unit. That is, Area = base x altitude. 372. Problem. Make a graphic representation of the theorem : The perimeters of similar polygons are in the same ratio as anyjivo corresponding sides. Solution. First consider the special case of equilat- eral triangles. On the horizontal axis lay off the lengths of one side of several such triangles, and on the vertical VARIABLE GEOMETRIC MAGNITUDES. 211 axis lay off the lengths of the cor- responding perimeters. Show that the points so obtained lie in a straight line. 373. EXERCISES. 1. In the manner above graph the rela- tion between the perimeters and sides of squares. Of regular pentagons. Of regu- lar hexagons. Of rhombuses. 2. If a side of a given regular polygon is a and its perimeter p, graph the relation of the perimeters and corresponding sides of polygons similar to the given polygon. ==="E5" -f^. ::::::7:i; /- l===S:=== |::::^::::: Siiii:::::: 1 :$?:::::: .2 fl «::/::::::: -f f- /::::::::: O Axis ior Sides 374. Theorem, If a line is parallel to one side of a triangle and cuts the other tioo sides, then it divides these two sides in the same ratio. c Graphic representation : Lay off CA along the hori- zontal axis and CB along the vertical axis, thus locating the A point Pj. In like manner find P^ with the coordinates CD and CE. If CD and CA are commensurable, we know that CD _CE CA ~ Cb' Hence o, Pg and Pj are collinear. If CD' and CA are incommensur- able, show, as in § 368, that the corresponding point P lies on the line OP^^ and hence, as in that case, also CD^ _ CE^ CA ~~ cb' CA 212 PLANE GEOMETBY. 375. Pkoblem. To represent graphically the rela- tion hetiveen the area and side of a square as the side varies conthiuously. AO -U? -r ^ - ^ 7 !7_ t 1 t J, t f 1 I A^t Sc ^ 36 -jl 'q tr ^ 1 _i f y T T 7 '^, / 1 7 tt ?: '^1- J uLJ^ ^t «vi ft §z 1 ^/ "I >,' T I T f I J f 16 -H 4 - *2 4 t T y L y o /n P„' / J£V ^^ T ^ T y ' »' /: 4 ^s^ ^2^ i 4 :f2 7"^ i pLy- 1 S^ J^-^ ^^^::3:_--±.__X___J . 5:: i L 12345678 Axir for Sides Solution. On the horizontal axis lay off the segments equal to various values of the side s, and on the vertical axis lay off segments equal to the corresponding areas A, (1) If one horizontal space represents one unit of length of side, and one vertical space one unit of area, then the points Pj, Pg, P3, etc., are found to lie on the steep curve. VARIABLE GEOMETRIC MAGNITUDES. 213 (2) If five horizontal spaces are taken for one unit of length of side and one vertical space for one unit of area, then the points P^\ P^^ Pg', etc., are found and the less steep curve is the result. The student should locate many more points between those here shown and see that a smooth curve can be drawn through them all in each case. The graph of the relation between two variables, one of which varies as the square of the other, is always similar to the one here given. 376. The area of a square is said to vary as the square of one of its sides, that is, A = s^. For example, the theorem : The areas of two similar polygons are in the same ratio as the squares of any two corresponding sides, means that if a given side of a polygon is made to vary continuously while the polygon remains similar to itself, the area of the polygon varies con- tinuously as the square of the side. 377. EXERCISES. 1. From the last graph find approximately the areas of squares whose sides are 3.4 ; 5.25; 6.35. 2. Find approximately from the graph the side of a square whose ^ area is 28 square units ; 21 square units ; 41.5 square units. 3. Construct a graph showing the relation between the areas and sides of equilateral triangles. 4. Given a polygon with area A and a side a. Construct a graph showing the relation between the areas and the sides corresponding to a in polygons similar to the one given. 5. From the graph constructed in Ex. 3, find the area of an equi- lateral triangle whose sides are 4. Also of one whose sides are 6. Compute these areas and compare results. 6. Construct a graph showing the relation between a side and the area of a regular hexagon. By means of it find the area of a regular hexagon whose sides are 6. Compare with the computed area. 214 PLANE GEOMETBY. 378. Problem. To construct a graph showing the relation between the radius and the circumference of a circle, and between the radius and area of a circle^ as the radius varies co7itinuously . Solution. Taking ten horizontal spaces to represent one unit of length of radius, and one vertical space for one unit of circumference in one graph and one unit of area in the other, we find the results as shown in the figure. I ii 2 1 / t 1 t t 2 t s 7 C3 ^ 2 7 ^^0 j_ g30 7 M / O '%9K %t §^ is ^^ ^ .^ a ^ ^ u V .^ I ^^ o ^ ^^ J- ^^ 'n 7 ^^ 2j 4 ^^ - ^ ^'^ 1^^ jL-^ .^/ 10 - ^^-1 oitcl^^ .ZS'p'^ / <.t^ ^ ^ ^^ "" ^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^^ Axis for Radii VABIABLE GEOMETRIC MAGNITUDES. 215 379. EXERCISES. 1. From the graph find approximately the circumference of a circle whose radius is 2.7, also of one whose radius is 3.4. 2. Find the radius of a circle whose circumference is 17, also of one whose circumference is 23. 3. Find the areas of circles whose radii are 1.9, 2.8, 3.6. 4. Find the radii of circles whose areas are 13.5, 25.5, 37, 45. 5. How does the circumference of a circle vary with respect to tHe radius? 6. How does the area of a circle vary with respect to the radius ? 7. Find the radius of that circle whose area in square units equals its circumference in linear units. DEPENDENCE OF VARIABLES. 380. In the preceding pages we have considered certain areas or perimeters of polygons as varying through a series of values. For example, if a rectangle has a fixed base and varying altitude, then the area also varies de- pending on the altitudes. The fixed base is called a con- stant, while the altitude and area are called variables. The altitude which we think of as varying at our pleas- ure is called the independent variable, while the area, being dependent upon the altitude, is called the dependent variable. 381. The dependent variable is sometimes called a func- tion of the independent variable, meaning that the two are connected by a definite relation such that for any definite value of the independent variable, the dependent variable also has a definite value. Thus, in A = s'^ (§ 376), J is a function of s, since giving s any definite value also assigns a definite value to A . Similarly, C is a function of r in C = 2 irr, and ^ is a function of rin A = irr^. 216 PLANE GEOMETRY. 382. EXERCISES. Justify each of the followmg statements, remembering that a variable y varies directly as another variable x ii y = kx, and directly as the square oi x ii y = kx\ where k is some constant. Find k in each case. 1. The area of a rectangle with constant base varies directly as its altitude. Find the value of k. Suggestion. By § 368, 4- = T- ^^ A=^.h. Hence A = kh. Ai ti-y hy In this case k = — ^=b, the constant base. See also § 369. ^i 2. The area of a square varies directly as the square of its side. Suggestion. Since A=s^ and A^ = s^^, we have — = — or 4 A ^' ''' A=^' s^ That is, A = ks^ where k=^=l. See § 376. 3. An angle inscribed in a circle varies directly as the intercepted arc. Show that in this case k = I. 4. A central angle in a circle varies as the intercepted arc. Show that in this case k = 1. 5. In the figure of § 374, show that DE varies directly as CD if DE moves, remaining parallel to AB. 6. An angle formed by two chords intersecting within a circle varies directly as the sum of the two arcs intercepted by the angle and its vertical angle. 7. An angle formed by two secants intersecting outside a circle varies directly as the difference of the two intercepted arcs. 8. If a polygon varies so as to remain similar to a fixed polygon, then its perimeter varies directly as any one of its sides. Show that p k = —I, where P, and s, are the perimeter and side of the fixed h polygon. 9. In the preceding, the area of the polygon varies directly as the square of any one of its sides. 10. The circumference of a circle varies directly as the radius, and its area varies directly as the square of its radius. VARIABLE GEOMETRIC MAGNITUDES. 217 11. Plot y = kx for Jc = 1, h 3, 4. 12. Plot y = kx^ for yfc = 1, i 3, 4. Notice that the graphs in Ex. 11 are all straight lines, while those in Ex. 12 are curves which rise more and more rapidly as the inde- pendent variable increases. See also § 378. LIMIT OP A VARIABLE. 383. If a regular polygon (§ 357) is inscribed in a circle of fixed radius, and if the number of sides of the polygon be continually increased, for instance by repeatedly doubling the number, then the apothem, perimeter, and area are all variables depending upon the number of sides. That is, each of these is a function of the number of sides. Now the greater the number of sides the more nearly does the apothem equal the radius in length. Indeed, it is evident that the difference between the apothem and the radius will ultimately become less than any fixed number, however small. Hence we say that the apothem approaches the radius as a limit as the number of sides increases indefinitely. 384. Similarly by § 352 the perimeters of the polygons considered in the preceding paragraph may be made as nearly equal to the circumference as we please by making the number of sides sufficiently great. - Hence we define the circumference of a circle as the limit of the perimeter of a regular inscribed polygon as the number of sides increases indefinitely. It also follows from § 353 that the circumference of a circle may be defined as the limit of the perimeter of a circumscribed polygon as the number of sides is increased. Likewise we may define the area of a circle as the limit of the area of the inscribed or the circumscribed polygon as the number of sides increases indefinitely. See § 361. 218 PLANE GEOMETRY. 385. The notion of a limit may be used to define the length of a line-segment which is incommensurable with a given unit segment. Thus, the diagonal c? of a square whose side is unity \s d = V2. Hence d may be defined as the limit of the variable line-segment whose successive lengths are 1, 1.4, 1.41, 1.414- •-. See §§ 234, 240. In like manner, the length of any line-segment, whether commensurable or incommensurable with the unit seg- ment, may be defined in terms of a limit. Thus, if a variable segment is increased by successively adding to it one half the length previously added, then the segment will approach a limit. If the initial length is 1, then the successive additions are I, I, \, j\, -^^, etc., and the successive lengths are 1, 1|, If, 1|, li|, lf|, etc. Evidently this segment approaches the limit 2. Hence 2 may be defined as the limit of the variable segment, whose successive lengths are 1, 1^, 1|, 1|, etc., as the number of suc- cessive additions is increased indefinitely. 386. A useful definition of a tangent to a circle, or to any other smooth curve, may be given in terms of a limit. Let a secant cut the curve in a fixed point A and a variable point B, and let the point B move along the curve and approach coincidence with A, thus making the secant continually vary its direction. Then the tangent is defined to be the limiting position of the secant as B approaches A indefinitely. This definition of a tangent is- used in all higher mathematical work. It includes the definition given in the case of the circle in § 183. 387. The functional relation between variables and the idea of a limit as illustrated above are two of the most im- portant concepts in all mathematics. The whole subject is much too difficult for rigorous consideration in this course. VABIABLE GEOMETRIC MAGNITUDES. 219 388. EXERCISES. 1. Find the limit of a variable line-segment whose initial length is 6 inches, and which varies by successive additions each equal to one half the preceding. 2|. Find the limit of a variable line-segment whose initial length is 1 and whose successive additions are .3, .03, .003, etc. 3. Construct a right triangle whose sides are 1 and 2. By approxi- mating a square root, find five successive lengths of a segment which approaches the length of the hypotenuse as a limit. 4. If one tangent to a circle is fixed, and another is made to move so that their intersection point approaches the circle, what is the limiting position of the moving tangent ? What is the limit of the measure of the angle formed by the tangents ? 5. The arc ^^ of 74° is the greater of the two arcs intercepted be- tween two secants meeting at C outside the circle. The points A and B remain fixed while C moves up to the circle. What is the limit of the angle formed by the secants ? The limit of the measure of their angle ? 6. If in the preceding the secants meet within the circle, what is the limit of their angle and also of the measure of this angle ? 7. If in Ex. 5 one secant and the intersection point remain fixed, while the other secant approaches the limiting position of a tangent at the point A, find the limit of the measure of the included angle. 8. If in Ex. 6 one secant and the point of intersection remain fixed while the other secant swings so as to make the included angle approach a straight angle, find the limit of the measure of the angle. 9. If in Exs. 5 and 6 the moving point crosses the circle, state the theorem on the measurement of the angle in question so as to apply equally well whether the point is inside or outside the circle. 10. A fixed segment ^ J5 is divided into equal parts, and equilateral triangles are constructed on each part as a base, as A EEC, CGD, DHA. Then each base is divided into equal parts and equilateral tri- angles are constructed on these parts as bases. What is the limit of the sum of the perimeters of these triangles as the number of them is increased indefinitely ? What is the limit of the sum of their areas ? CHAPTER VII. REVIEW AND FURTHER APPLICATIONS. ON DEFINITIONS AND PROOFS. 389. A definition is a statement that a certain word or phrase is to be used in place of a more complicated expression. Thus the word "triangle" is used instead of "the figure formed by three segments connecting three non-collinear points." In geometry we may distinguish two classes of words : (a) Technical words representing geometric concepts such as line, plane, polygon, circle, etc. (5) Words of ordinary speech not included in the first class. The meaning of words in the second class is taken for granted without any definition whatever. It is not possible to define everi/ word of the first class, for every definition brings in new technical terms which in turn require definition. Thus, one of the many definitions of "point" is "that which separates one part of a line from the adjoining part." In this definition the technical terms " separate," " line," " adjoin- ing " are used. If we try to define these, still other terms are brought in which need to be defined, and so on. The only escape is defining in a circle, which is not permitted in a logical science. Since it is thus evident that some terms must be used without being defined, it is best to state which ones are left undefined. 220 REVIEW AND FUBTHER APPLICATIONS. 221 390. In this book point, straight line, or simply line, plane, size and shape of figures are not defined. Nor do we define the expression, a point is between two other points, or a point lies on a segment. Descriptions of some of these terms which do not, how- ever, constitute definitions in the logical sense are given in §§ 1-5. Other technical terms are defined by means of these simple undefined words with which we start. Thus, "a segment is that part of a line which lies hetioeen two of its points " is a definition of segment in terms of the three undefined words, point, line, and between. 391. A geometric proposition consists of an affirmation that a geometric figure has some property not explicitly specified in its definition. A proposition is said to be proved if it is shown to fol- low from ' other propositions which are admitted to be true. Hence every geometric proposition demands for its proof certain other propositions. 392. It is obvious, therefore, that certain propositions must be admitted without proof. Such unproved propo- sitions are called axioms. In order that a set of axioms for geometry shall be complete it must be possible to prove that every theorem of geometry follows from them. The set of axioms used in this text is not complete. For instance, it is assumed without formal statement that if ^, B, C, are three points on a segment we cannot have at the same time B between A and C and C between A and B ; that the diagonals of a convex quadrilateral intersect each other ; that a ray drawn from the vertex of an angle and included between its sides intersects every segment determined by two points, one on each side of the angle. In like manner many other tacit assumptions are made. 222 PLANE GEOMETRY. 393. In a complete logical treatment every undefined term must occur in one or more axioms, since all knowl- edge of this term in a logical sense comes from the axioms in which it is found. In this text not every undefined term occurs in an axiom, for instance, the word hetiveen. The axioms are, of course, based on our space intuition^ or on our experience with the space in which we live. It is interesting to notice, however, that the axioms tran- scend that experience both as to exactness and extent. For instance, we have had no experience with endless lines, and hence we cannot know directly from experience whether or not there are complete lines which have no point in common. See §§ 89, 96. 394. Proofs are of two kinds, direct and indirect. A direct proof starts with the hypothesis and leads step by step to the conclusion. An indirect proof starts* with the hypothesis and with the assumption that the conclusion does not hold, and shows that this leads to a contradiction with some known proposition. Or it starts simply with the assumption that the conclusion does not hold and shows that this leads to a contradiction with the hypothesis. This kind of proof is based upon the logical assumption that a proposition must either be true or not true. The proof consists in showing that if the proposition were not true, impossible consequences would follow. Hence the only remaining possibility is that it must be true. 395. Every proposition in geometry refers to some figure. See § 12. The essential characteristic of a figure is its description in words and not the drawing that repre- sents it. Each drawing represents just one figure from a class of figures defined by the description. Thus we say BEVIEW AND FURTHER APPLICATIONS. 223 "let ABCD be a convex quadrilateral," and we construct a particular quadrilateral. We must then take care that all we say about it applies to ani/ figure whatever so long as it is a convex quadrilateral. The logic of the proof must be entirely independent of the appearance of the constructed figure. The description of the figure must contain all the con- ditions given by the hypothesis. A good way to show that the description of the figure is what really enters into the proof, is to let one pupil describe the figure in words and each of the others draw a figure of his own to correspond to that description. The proof must then be such as to apply to every one of these figures though there may not be two of them exactly alike. 396. EXERCISES. 1. Every word in the language is defined in the dictionary. How is this possible in view of what has been said about the impossibility of defining every word ? 2. Can we determine experimentally whether or not the space in which we live satisfies the parallel line axiom (§ 96) ? 3. Can we determine experimentally whether or not there can be more than one straight line through two given points? 4. Which theorems of Chapter I are found by direct proof and which by indirect proof ? 5. If two triangles have two angles of the one equal to two angles of the other, and also any pair of corresponding sides equal, the triangles are congruent. 6. If two triangles have two sides of the one equal to two sides of the other, and also any pair of corresponding angles equal, the triangles are congruent in all cases except one. Discuss the various cases according as the given equal angles are greater than, equal to, or less than a right angle, and are, or are not, included between the equal sides, and thus discover the exceptional case. 7. State a theorem on the congruence of right triangles which is included in the preceding theorem. 224 PLANE GEOMETBY, 8. What theorems of Chapter I on parallel lines can be proved without the parallel line axiom ? 9. What theorems of Chapter I on parallelograms can be proved without the parallel line axiom ? 10. What regular figures of the same kind can be used to exactly cover the plane about a point used as a vertex ? 11. What combinations of the same or different regular figures can be used to exactly cover the plane about a point used as a vertex ? 12. Suppose it has been proved that the base angles of an isosceles triangle are equal but that the converse has not been proved. On this basis can it be decided whether or not the base angles are equal by simply measuring the sides ? Can it be decided on the same basis whether or not the sides are equal by simply measuring the base angles? Discuss fully. 13. The sum of the three medians of a triangle is less than the sum of the sides. See Ex. 34, p. 83. 14. The sum of the three altitudes of a triangle is less than the sum of the sides. 15. A triangle is isosceles (1) if an altitude and sm angle-bisector coincide, (2) if an altitude and a median coincide, (3) if a median and an angle-bisector coincide. 16. If two sides of a triangle are unequal, the medians upon these sides are unequal and also the altitudes. 17. An isosceles triangle has two equal altitudes, two equal medians, and two equal angle-bisectors. 18. ABCD is a square and the points E, F, G, H, are so taken that AE = AH = CF = CG. Prove that EFGH is a rectangle of constant perimeter, whatever the length of AE. 19. The bisectors of the exterior angles of a parallelogram form a rectangle the sum of the diagonals of which is the same as the sum of the sides of the parallelogram. 20. Prove that the perpendicular bisectors of the sides of a polygon inscribed in a circle meet in a point. Use this theorem to show that the statement is true of any triangle. REVIEW AND FURTHER APPLICATIONS. 225 21. The bisectors of the exterior angles of any quadrilateral form a quadrilateral whose opposite angles are supplementary. Definition. In a polygon of n sides there are n angles and hence 2 n parts. Parts a, /. A^h^ Z. B, etc., are said to be consecutive if a lies on a side of Z ^, 5 lies on the other side of Z^, and also on a side of ZjB, etc. 22. Is the following proposition true ? If in two polygons each of n sides 2n — 3 consecutive parts of one are equal respectively to 2n — 3 consecutive parts of the other, the polygons are congruent. Suggestion. Try to prove this proposition for n = 3, fhen for n = 4, and finally for the general polygon. 23. What theorems on the congruence of triangles are included in the preceding proposition ? A proposition may be proved not true by giving one ex- ample in which it does not hold. 24. Is the following proposition true? If in two polygons each of n sides 2 n — 3 parts of one are equal respectively to 2 n — 3 correspond- ing parts of the other, the polygons are congruent provided at least one of the equal parts is a side. 25. Given three parallel lines, to construct an equilateral triangle whose vertices shall lie on these lines. Solution. liCt P be any point on the middle line l^. Draw PC and PA, making an angle of 60° with l^. Through the points A, P, C construct a circle meeting /g in -6. Then ABC is the required triangle. Suggestion for Proof. Comparer! A~' BPA and BCA also A BPC and BA C. 26. Show how to modify the construction of the preceding example so as to make ABC similar to any given triangle. 27. If tangents are drawn to a circle at the extremities of a diameter and if another line tangent to the circle at P meets these two tangents in ^ and 5 respectively, show that AP • PB = r% where r is the radius of the circle. G-'- — -. h h i y 1 h 226 PLANE GEOMETRY. LOCI CONSIDERATIONS. 397. Two methods are available to show that a certain geometric figure is the locus of points satisfying a given condition. First method : Prove (a) Every point satisfying the condition lies on the figure. (b) Every point on the figure satisfies the condition. Second method : Prove (a') every point not on the figure fails to satisfy the condition. (6') ^very point on the figure satisfies the condition. The first of the methods is more direct and usually more simple. See § 127. The second is likely to lead to proofs that are not general. PROBLEMS AND APPLICATIONS. 1. ^5 is a fixed segment connecting two parallel lines and per- pendicular to each of them. Find the locus of the vertices of all isosceles triangles whose common B' B _._Q^:1 — - base is AB. Is the middle point of AB a part of — ^C.- -rz>— this locus ? ""-I^^".- A 2. If in Ex. 1 AB is allowed to move always remaining perpendicular to the given lines, and if ABC is any tri- angle remaining fixed in shape, find the locus of the point C. 3. Find the locus of the centers of all parallelo- grams which have the same base and equal altitudes. 4. Find the locus of the centers of parallelograms obtained by cutting two parallel lines by parallel secant lines. 5. Find the locus of the vertices of all triangles which have the same base and equal areas. 6. Find the locus of a point whose distances from two intersect- ing lines are in a fixed ratio. Note that the whole figure is symmetrical with respect to the bisector of the angle formed by the two given lines. BEVIEW AND FURTHER APPLICATIONS. 227 7. Find the locus of a point such, that the difference of the squares of its distances from two fixed points is a constant. Note that the locus must be symmetrical with respect to the perpendicular bisector of the segment connecting the two given points A and B. Show that in the figure AP'^ - FB^ = AC^- CB^- A'^r^rzzi -7*B 1 'i p B ^■o D 8. The lines l^ and l^ meet at right angles in a point A. is any fixed point on l^. Through draw a line meet- ing Zj in B. P is a varying point on this line such that OB - OP is fixed. Find the locus of P as the line swings about as a pivot. Suggestion. Draw PD ± to BP. Show that BO ' OP =A0 • OD. Hence we obtain a set of right triangles whose common hypotenuse is OD. Find the locus of the vertex P. 9. Find the locus of all points the sum of whose distances from two intersecting lines is equal to a fixed segment a. Suggestion. If AB and CB are the given lines, investigate points on the base AC of the isosceles triangle ABC in case the perpendicular from C on AB is equal to a. 10. The figure ABCDEF is a regular hexagon. AG, BH, etc., are equal segments bisecting the angles A, B, etc. (a) Prove GHWAB. (b) Prove that the- inner figure is a regular hexagon. Suggestion. Show ABHG congruent to the other similar parts by superposition. (c) Find how many degrees in ZA GH and Z BHG, as needed by a carpenter in making a hexagonal frame. 11. The figure ABCD is a parallelogram. The points E, F, G, H are taken so that AE = CG and AH = CF. Vrowe A FCG ^ A AEH and A EOH^ A FOG. D G _C This figure is found in an old Roman pavement in Sussex, England. 228 PLANE GEOMETRY. 12. The figure A BCD is a square. AE = BF= CG- DH, and Ey, Fz, Giv, Hx are so drawn that Zl=Z2 = Zo=:Z4. (a) Find A EyF, FzG, etc. Suggestion. Through B draw a line parallel to Fz and make use of the A thus formed. (h) Prove EBFy ^ FCGz = GDHw = HAEx. (c) What kind of figure is :r?/zt<;? Prove. See the accompanying design. 13. In a square ABCD diagonals and diameters are drawn as shown in the figure. (A line connect- ing the middle points of opposite sides of a quadri- lateral is a diameter.) The points K, Z, N are laid off so that MN= MK = ML. The small triangles on the other sides are constructed congruent to KLN. Through the vertices of the triangles lines EF, EG, etc., are drawn parallel to the sides of the given square. \ / ^ s / N Parquet Pattern. D n \H\ / / \ \ / / P / K / \ F / / \ \ A. K M L B Tile Pattern. Prove that (a) AKNE, LBFN, etc., are congruent parallelograms and hence that EF, EG, etc., meet in points on the diagonals. (See Ex. 6, § 123.) (b) ABFE^BCGF. (c) EFGH is a square. (d) KN and QP lie on the same straight line. 14. The figure ^5 CD is a square. PQ and RS are diameters. The points E, F, G, H, :. are so taken that AE = BE = BG = HC = .... Also ON^OK=OL = OM. ^ w^ ^ ^ kS ^ ^ ^ ^ ^ 8^ ^ OF REVIEW AND FURTHER APPLICATIONS. 229 (a) Prove that A ENF ^: A GKH. (b) Prove that TMONEA = FNOKGB. (c) How many axes of symmetry has the figure TAENFBGKH .-."i (d) Has this figure a center of symmetry ? See the four squares in the accompanying design. 15. A cross-over track is constructed as shown in the figure. The rails of the curved track are tangent to those of the main line at A, B, F QXidi G. The curves are tan- gent to each other at D and E. The arcs GE and BD have equal radii as have the arcs A E and DF. C^ is the center of the arcs GE and FD and C2 is the center of the arcs BD and A E. (a) Prove that C^HC^ is a right triangle. (h) If the distance between the tracks is h feet and the distance be- tween the rails in each track (the c^ gauge) is a feet, show that /2 ^^ y (2 x + 3 a + 2 &)2 = ?/2 + (2 a; + 2 a + hy. Since a and h are given, this equation may be solved for x in terms of y or for y in terms of x. Hence if the distance AK \^ known, we may use this equation to compute the radii of the arcs used in constructing the figure. On the other hand, if the radii of the arcs' are known, we may compute the distance A K. This is a very common problem in railway construction. The construction is also used in laying out a curved street to connect two parallel streets. 16. If in the preceding problem a = 4 feet 8| inches, ft = 9 feet, and AK = 200 feet (an actual case), find the radii of the arcs. Using the same values for a and b as in the preceding example, find AK if C^E = 300 feet. 230 PLANE GEOMETBY. FURTHER DATA CONCERNING CIRCLES. 398. In Chapter II numerous theorems were proved concerning the equality of arcs, angles, chords, etc. The three following theorems involve inequalities of these elements. The student should construct figures in each case and give the proof in full. 399. Theorem. In the same circle or in equal circles, of two unequal angles at the center the greater is suh- te7ided hy the greater arc ; and of tivo unequal arcs the greater subtends the greater angle at the center. The proof is made by superposition. 400. Theorem. In the same circle or in equal circleSy of two unequal minor arcs the greater is subtended by the greater chord ; and of two unequal chords • the greater subtends the greater minor arc. The proof depends upon the theorem of § 117. 401. Theorem, hi the same circle or in equal circles, of tioo unequal angles at the center, both less than straight angles, the greater is subtended by the greater chord; and of two unequal chords the greater subtends the greater angle at the center. 402. Problem. On a given line-segment as a chord to construct an arc of a circle in ivhich a given angle may be in- scribed. . „ Construction. Let ^ be the given angle and AB the given segment. REVIEW AND FURTHER APPLICATIONS. 231 It is required to construct a circle in which AB shall be a chord and such that an angle inscribed in the arc AMB shall be equal to the given angle K. At A construct an angle BAC equal to Z K. If now a circle were passed through A and B so as to be tangent to AC at A, then one half the arc AB would be the measure of the angle BAG. See § 219. Then any angle inscribed in the arc AMB as Z APB would be equal to ZbaC—Zk. Hence the problem is to find the center of a circle tan- gent to AC at A and passing through A and B. Let the student complete the construction and proof. 403. EXERCISES. 1. Prove that the problem of § 402 may be solved as follows: With vl, any pomt on one side of the angle K, as center and AB as radius strike an arc meeting the other side of the angle at B. Cir- cumscribe a circle about the triangle ABK. 2. Show that from any point within or outside a circle two equal line-segments can be drawn to meet the circle and that these make equal angles with the line joining the given point to the center. 3. Show that if two opposite angles of any quadrilateral are sup- plementary, it can be inscribed in a circle. Suggestion. Let ABCD be the quadri- p lateral in which /.A -\- ZC = 2 vt.A. Z"^7^N>\ Pass a circle through B, C, and D. To prove / /\ ^^rj that A lies also on the circle, and not at some / /// \ inside or outside point as A' or A" . \k'/Jx 1/ (1) Show that Z A' •> Z A s^ndi /. A" 5^^^>^ (2) Hence Z.4' + Z C>2 rt. J if A' is l^"^^'^ within the circle and Z ^" + Z C<2 rt. A\i A" A" is outside the circle, both of which are contrary to the hypothesis. Hence the fourth vertex must lie on the circle. Show that the condition that the polygon is convex follows from the hypothesis. 232 PLANE GEOMETRY, 404. Problem. To draw a common tangent to two circles ivhich lie wholly outside of each other. Construction. Let the given circles be and 0' of which the radius of the first is the greater. Required to draw a tangent common to both circles. Draw an auxiliary circle with center o and radius equal to the difference of the two given radii in one figure and equal to the sum of these radii in the other figure. In each case draw a tangent to this auxiliary circle from o\ thus fixing the point D. See § 230. Draw the radius OD, thus fixing the point P. Draw O^P^ II DP, thus fixing the point P'. Then PP^ is the required tangent. Proof : Show that, in each case, PP'o'd is a rectangle, thus making PP' perpendicular to the radii OP and o'p', that is, tangent to each circle. Definition. A common tangent to two circles is called direct if it does not cross the segment connecting the centers, and transverse if it does cross it. 405. EXERCISES. 1. Describe the relative positions of two circles if they have two direct common tangents. Also if they have only one. 2. Describe the positions of two circles if they have two transverse common tangents j one ; none. BEVIEW AND FUBTHEB APPLICATIONS, 233 D ^"^^ Tk G ^ A \^^^ A B __4^ D ! yC' \\\/y J PROBLEMS AKD APPLICATIONS. 1. Find the locus of the middle points of all chords of a circle drawn through a fixed point within it. 2. The sides AD and BC of the square ABCD are each divided into four equal parts and a circle inscribed in the square. Lines are drawn as shown in the figure. Show that EFGHKL is a regular hexagon. This is tlie construction by means of which a regular hexagonal tile is cut from a square tile. 3. In the preceding example what fraction of the area of the square is covered by the hexagon. 4. A circle of constant radius passes through a fixed point A. A line tangent to it at the point P remains parallel to a fixed line BD. Find the locus of P. Suggestion. In the figure prove A C'PC a parallelogram. Show that the locus consists of two circles. 5. The same as the preceding except that the circle of constant radius remains tangent to a fixed circle instead of passing through a fixed point. Suggestion. On that diameter of the fixed circle which is perpendicular to the fixed line lay off ^ C = C'P. Prove A C'PC a parallelo- gram. Show that the locus consists of two circles. 6. In the square ABCD, AM = 0B = BU = VC, etc. The point E is the intersection of the segments MV and TU, F is the inter- section of OB and TU, G is the intersection of OW and US, etc. When these segments are partially erased, we have the figure. (a) Prove that THEN, OUGF, etc., are squares. (6) What kind of a figure is MOFE ? Prove, (c) How many axes of symmetry has the figure NEFGH ■• . ? Has it a center of symmetry? 234 PLANE GEOMETRY. 7. ABCD is a square the mid-points of whose sides are joined. On its diagonals points E, K, P, T are taken so that AE = BK = CP = DT. The construction is then completed as shown in the figure. {FG and MN lie in the same straight line.) (a) Prove that E, K, P, T are the vertices of a square. (b) li AB = Q find ^^ so that the area of the square EKPT shall be half that of the square ABCD. (c) If AB = 6, and if CP = PS, find the area of the figure EFGHKLM .... Also find the area of the trapezoid BKLR. (d) If CP = f CS, find the area of the inside figure and also of BKLR. (e) What fraction of CS must CP be in order that the inside figure shall be half the square? (/) If the inside figure is | of the square and if AB find CP. 8 inchesj, 8. In the figure ABCD is a square. Each of its sides is divided into three equal parts by the points E, F, G, H, ■•. The points E, X, Y,H', F, X, W, il/; etc., lie in straight lines. (a) If AB = 6 inches, find the area of that part of the figure which lies outside the shaded band. (ft) If AB = 6 inches, and if the width of the shaded band is J inch, find the area of the band. (c) JiAB = 8 inches and ZP = J inch, find the area of that part of the figure which lies inside the band. (d) If AB = 8 inches, what must be the width of the band in order that it shall occupy 10 % of the area of the whole design ? (e) li AB = a inches find the width of the band if it occupies n per cent of the area of the whole design. Parquet Flooring. REVIEW AND FURTHER APPLICATIONS. 235 ,P N scxx 9. By means of the accompanying figure find the area of a regular dodecagon whose sides are 6 inches. Notice that the dodecagon con- sists of the regular hexagon in the center, the six equilateral triangles, and the six squares. Note how this figure enters into the accom- panying tile design. 10. Find an expression for the area of a regu- lar dodecagon whose sides are a. 11. Find the apothem of a regular dodecagon by dividing its area by half the perimeter. Also find it by finding the apothem of the hexagon and then adding a to this. Compare results. 12. Find the radius of a circle circumscribed about a regular dodecagon whose side is a. 13. If the accompanying design for par- quet border is 10 inches wide, find the width of the strips from which the small squares are made. Also find this result if the width of the border is a. 14. In the figure of the preceding example find the dimensions of the .small triangles along the two edges if the width of the border is a inches. 15. Show that the figure given below may be used as the basis for the design shown beside it. (See Ex. 7, p. 147.) 16. If q,n outside of each of the two squares shown in the figure is 6 inches, find the width of the sti-ips of which they are made in order that each shall fit closely into the corner of the other? Suggestion. Find the altitude on EM in A UEM above if AB = Q inches. This is a common Arabic ornament. 236 PLANE GEOMETRY. 17. In the figure ABCD is a square. On each side a triangle, two of whose sides are parallel to the diagonals of the square, is so constructed that the points E, F, and G lie in the same straight line. (a) Prove that the triangles are right isos- celes triangles. (b) What part of the square lies within the triangles ? (c) If AB = a, find AE so that the tri- angles shall occupy - of the area of the square. n See the accompanying tile pattern. 18. ABCD is a square and the small fig- ures in the corners are squares. (a) Show that if the lines intersect as shown in the figure A E must be \ of AB. (b) li AB — Q inches find the areas of the squares A'B'C'D^ and XYZW. 19. In the tile design show that the figure within the square is the same as that of ^x. 18. What part of the large square is occupied by each shaded part ? 20. ABCD is a square. AE = FB = BK = LC = CP = QD =DR = SA. (a) If AB = a and if AE = b, find the sum j of the areas of the four shaded rectangles. (b) If. AB = 8 inches, find ^^ so that the sum of these rectangles shall be ^ of the whole square. Interpret the two solutions. (c) If AB = a inches, find AE so that the sum of the rectangles shall be - of the square. n 21. (a) Show that in the accompanying de- sign for tile flooring the size of any one piece determines the size of every piece in the figure. (b) What fraction of the figure is occupied by each color? E D /¥ 'X/V/ K\ xf i / 500 J \ uA BEVIEW AND FUBTHEB APPLICATIONS. 237 22. This tile floor design is based on the plane figure and this in turn is based on. the figure of Ex. 17 where AE = AB > < What part of the whole design A E is occupied by tiles of the various colors ? 23. ABCD is a square. Lines are drawn parallel to the sides and intersecting as shown in the figure. , (a) Show how the sides of the square must be divided in order to make the lines intersect as they do. (b) If AB = Q inches, find the areas of the squares SXZQ and RTYP. (c) Find these areas M AB = a. (d) Show that the outline figure is the basis of the tile design here given. 24. ABCD and A'B'C'D' are equal squares, rounded by strips of equal width. (a) liAB = a, find AF, FE, and EBl (b) Find the width of the strips if their outer edge passes through the points A, B, C, D, when AB = S inches! Also when AB = a. (c) If AB = a, find A'H and hence KL. (d) If the width of the strip is 1 inch, find KL. (e) If the width of the strip is b, find KL. (/) If KL = c, find the width of the strip. When the width of the parquet floor border shown in the figure is given, (/) is the problem one needs to solve to know how wide a dark strip to use. Compare Ex. 16, page 235. M L N XX X X X X X X K H A'B'C'D' is sur- 238 PLANE GEOMETRY. THE INCOMMENSURABLE CASES. 406. We have seen, in § 234, that there are segments which are incommensurable ; that is, which have no com- mon unit of measure, — for instance, the side and the diagonal of a square. For practical purposes the lengths of such segments are approximated to any desired degree of accuracy, and their ratios are understood to be the ratios of these approximate numerical measures. See §§ 238-240. All theorems involving the ratios of incommensurable segments, and the lengths and areas of circles, have thus far been proved only for such approximations, and these are quite sufficient for any refinements of measurement which it is possible to make. But for theoretical purposes it is important to consider these incommensurable cases further, just as in algebra we not only approximate such roots as V2, V3, V5, etc., but we also deal with these surds as exact numbers. For instance, in such an operation as (V3 + >/2)( V3 - \^) = 3 - 2 = 1. While the length of the diagonal of a unit square cannot be expressed as an integer or as a rational fraction, that is, the quotient of two integers, we nevertheless think of such a segment as having a definite lengthy or what is the same thing, a definite ratio with the unit segment forming the side of the square. E.g. If d is the diagonal of the square whose side is 1, then d^=l -\-l ov d = V2. Now suppose y/2 = - , a fraction in its lowest 2 ^ terms. Then 2 =^, a fraction also in its lowest terms. Bat a fraction in its lowest terms cannot be equal to 2. Hence V2 is neither an integer nor the quotient of two integers. REVIEW AND FURTHER APPLICATIONS. 239 . 407. The following axiom and that of § 409 -enQ funda- mental in the consideration of incommensurable line-seg- ments. Axiom X. Every line-segment ab has a definite le?igth, which is greater tha7i AC if c lies between A and b. The length of a line-segment is in every case a number, which is rational (an integer or the quotient of two inte- gers) in case the segment is commensurable with the unit segment, but which otherwise is irrational. Under the operations of arithmetic these irrational numbers obey the same laws as the rational numbers. The length of a line-segment is often called its numeri- cal measure. 408. The exact ratio, or simply the ratio, of two line- segments is the quotient of their numerical measures, whether these are rational or irrational. That is, every such ratio is a number. It is obvious that a segment may be constructed whose length is any given rational num|)er. We have also seen how to construct with ruler and compasses segments whose lengths are certain irrational numbers, such as V^, VS, V5, etc. See Ex. 3, § 295. 409. We now assume the following Axiom XI. For any given number k there exists a line segment whose length is K, This does not imply that it is possible by means of the ruler and compasses to construct a segment whose length is any given irrational number. For instance, we cannot thus construct a segment whose length is V2. We now prove the fundamental theorem on the propor- tionality of sides of triangles. 240 PLANE GEOMETRY. 410. Theoeem. a line parallel to the hase of a tri- angle, and meeting the other two sides, divides these sides proportionally. Given A ABC with DE II AB^ /A -, CD CE / V To prove — = — jj/ \q CA CB jy A x. Proof: Whether CD and CA are / \ At — ^5 commensurable or not, we know by §§ 407, 408 that — and — are definite numbers. We prove \yA (yJj that these numbers cannot be different. ^J^. . CD ^ CE t irst, suppose — < CA CB Take F between C and E so that — = Ax. XI (1) CA CB Divide CB into equal parts, each less than EF, Then at least one of the division points, as Gf, lies between E and F. Draw GH II AB. Since CG and CB are commensurable, we have, by § 243, CH^CG 2) CA CB ^ ^ Dividing (1) by (2), we have — = -^. But this can- CH CG not be true, since CD > CH and CF < CG. Hence — cannot be less than — . Why? CA CB CD Secondly, prove in the same manner that — cannot be CE ^^ greater than — . Hence, since the one is neither less than CB nor greater than the other, these ratios must be equal. The following treatment of incommensurable arcs and angles is exactly similar to the above. REVIEW AND FUBTHEB APPLICATIONS. 241 411. Axiom XII. Any given arc ab has a definite ratio ivith a unit arc, which is greater than that of an arc AC, if c lies on the arc ab. 412. Axiom XIII. Any given angle abc has a defi- nite ratio ivith a imit angle, which is greater than that of abb if bd lies within the angle abc. 413. Theorem. In the same or equal circles the ratio of two central angles is the same as the ratio of their intercepted arcs. E Outline of Proof : We show that in the figure ^ arc AB ^^ "^ can neither be less than nor cfreater than . arc CD o „^ Z AOB ^ arc AB suppose < . Z CO'd arc CD rpi .1 „ .-, , Z.AOB arc AE ^^>, Then take E so that — = . (1) Z CO'd arc CD Divide arc CD into equal parts each less than arc EB. Lay oif this unit arc successively on AB reaching a point F between E and B. Then arcs AF and CD are commen- surable and by a proof exactly similar to that of § 243, making use of § 199, we can show that Z AOF _ arc AF Z CO'd ~ arc CD ' Complete the proof as in § 410. (2) 242 PLANE GEOMETRY. 414. Axiom XIV. Any given rectangle with base h and altitude a has a definite area ivhich is greater than that of another rectangle ivith base b' and altitude a' if a^a and b>V or if a> a and b > V. 415. Theorem. Tlie area of a rectangle is the prod- uct of its base and altitude. A B Proof : Denote the base and altitude by b and a, respec- tively, and the area by A. Suppose A < ah, and let a' be a number such that A = a'b. Lay off BE=: a'. Consider first the case where b is commensurable with the unit segment and a is not. Divide the unit segment into equal parts each less than CE and lay off one of these parts successively on BC reach- ing a point F between E and C. Denote the length of BFhj a", and draw FF' II AB. Then by § 307 the area of ABFS^ is a"b. By hypothesis A = a'b, but a'b < a"b smce a' < a" . Hence A < a"b. (1) But by Ax. XIV A > a"b. (2) Hence the assumption that A ab cannot hold. The proof in case both sides are incommensurable with the unit segment is now exactly like the above and is left to the student. REVIEW AND FURTHER APPLICATIONS. 243 416. Axiom XV. A circle has a definite length and incloses a definite area ivhich are greater than those of any inscribed polygon and less than those of any cir- cumscribed polygon. 417. Theoeem. For a giveji circle and for any number k, however small, it is possible to inscribe and to circumscribe similar polygons such that their perimeters or their areas shall differ by less than k. Proof : First, Let p and p' be the perimeters of two similar polygons, the first circnmscribed and the second in- scribed, and let a and a' be their apothems. Then E = ±.,ovP:zPL^1^. §245 Hence Now a— a' can be taken as small as we like. Hence p • — ^-~ can be made as small as we please ; that ^ a' is, p—p' can be made smaller than any given number K. Second, letting P and p' be the areas of the circum- scribed and inscribed polygons respectively, we have P a^ —f = — ^, and the proof proceeds exactly as before. 418. Since the length C of the circle is greater than p' and. less than p, it follows that C is thus made to differ from either p or p' by less than K. And since the area A of the circle is greater than p' and less than P, it follows that A is made to differ from either P or p' by less than K. 244 PLANE GEOMETRY. 419. Theorem. The lengths of two circles are in the same ratio as their radii Proof: Let c and c' and hence p>c' ■—, (2) c r Hence the supposition that — <— leads to the contra- c r diction expressed in (1) and (2) and is untenable. o r • Now prove in same way that — ^ — is untenable. 420. Theorem. The areas of two circles are in the same ratio as the squares of their radii. Using §§ 348 and 418, the proof is exactly similar to that of the preceding theorem. REVIEW AND FURTHER APPLICATIONS. 245 PROBLEMS AND APPLICATIONS. 1. A billiard ball is placed at a point P on a billiard table. In ■what direction must it be shot in order to return to the same point after hitting all four sides ? (The angle at which the ball is reflected from a side is equal to the angle at which it meets the side, that is, Zl = Z2, and Z3 = Z4.) Suggestion, (a) Show that the opposite sides of the quadrilateral along which the ball travels are parallel. (l>) If the ball is started parallel to a diagonal of the table, show that it will return to the starting point. 2. Show that in the preceding problem the length of the path traveled by the ball is equal to the sum of the diagonals of the table. 3. Find the direction in which a billiard ball must be shot from a given point on the table so as to strike another ball at a given point after first striking one side of the table. Suggestion. Construct BE ± to that side of the table which the ball is to strike and make ED = BE. 4. The same as the preceding problem except that the cue ball is to strike two sides of the table before striking the other ball. Suggestion. B'E' = E'D', D'H = HF. 5. Solve Ex. 4, if the cue ball is to strike three sides before strik- ing the other ball, — also if it is to strike all four sides. 6. In the figure ABCDEF is a regular hexagon. Prove that : (a) AD, BE, and CF meet in a point. (6) ABCO is a rhombus. (c) The inner circle with center at and the arcs with centers at A, B, C, etc., have equal radii. (d) The straight line connecting A and C is tangent to the inner circle and to the arc with center at B. (e) The centers of two of the small circles lie ^ on the line connecting A and C. (/) Find by construction the centers of the small circles. F E.^ -.D ^^B 246 PLANE GEOMETRY. 7. In the figure EG and FH are diameters of the square ABCD. On the diagonals, points K, U, V, W are laid off so that AK = BU = CV = DW. Also LM = NP = RS = etc. EN and SF are in the same straight line, and so on around the figure. Prove that : (a) KUVW is a square. (b) AKME and ENUB are equal trapezoids. (c) L, 0, T lie in a straight line. • (d) The four heavy six-sided figures are congruent. AO a,AK and ML (e) JiAB KL, find the areas of the figures KUVW, AKME, LOPNEM. (/) Find the areas required under (e) \tAK = — and LM = m- LK. n AO (g) li AK = -T- , what is the length of ML if the four heavy figures occupy half the square ABCD'i 8. Find a side of a regular octagon of radius r. See Ex. 3, p. 146. 9. Find the area of a regular octagon of radius r. 10. In the figure ABCD-' is a regular octagon inscribed in a circle of radius r. Find the area of the triangle ABC. Suggestion. Find first the altitude on ^ C in the /\ AOC and thus the altitude of A ^5C. 11. In the same fig- ure the lines CH and A D are drawn meeting at K. Prove that A BCK is a parallelogram and find its area if the radius of the circle is r. ® s m F^-~-"- G REVIEW AND FURTHER APPLICATIONS. 247 12. In the parquet floor design given with Ex. 10, the darker parts are parallelograms constructed as under Ex. 11. What part of the area is of white wood ? 13. In the figure ABCD and A'B'C'D' are equal squares, placed as shown. Lines are drawn through A'^ B', C, and D' parallel to AB, BCj CD, and DA, forming a quadrilateral EFGH. (a) Prove EFGH a ^^ square. (&) What part of the large square is inclosed by the out- jj side heavy figure? (c) li AB = a, find the area of the inside heavy figure. E C r D /\ a i V ^ >{ [< ^ K A' F 14. The design opposite consists of white figures constructed like the inner figure preced- ing, together with the remaining black figures. What part of the figure is ^hite? 15. Show that the altitude of an equilateral triangle with side s is - VS. 16. If the angles of a triangle are 30°, 60^ 90°, and if the side opposite the 60° angle is r, show that r - 2 r - the other sides are o V3 and -^ V3. See § 159, Ex. 14. 17. Three equal circles of radius 2 are inscribed in^a circle as shown in the figure. Find the radius of the large pircle. Suggestion. The center of the large circle is at the intersection of the altitudes of the equi- lateral triangle 0'0"0"'. (Why?) Hence O'OD is a triangle with angles 30°, 60°, 90° as in Ex- 16. Solve this problem for any radius r of the 2 r /— small circles. Ans. R = r -\ v 3. o ►♦♦♦♦♦♦ ►♦♦♦♦♦♦ ►♦♦♦♦♦♦ ►♦♦♦#♦♦ ►♦♦♦♦♦♦ Design from the Alhambra. s/^ \ A^r f \ 248 PLANE GEOMETRY, 18. A', B', C, are the middle points of the sides of the equilateral triangle ABC. The sides of the triangle A'B'C are trisected and segments drawn as shown in the figure. A D E B From Church of Or San Michele. If AB = a, (a) Prove A"B"C" an equilateral triangle. (6) Find the area of A"B"C" and of the dotted hexagon, (c) Find the area of the triangle GFC and of the trapezoid DEB" A". A P E B From Westminster Abbey. 19. In the figure ABC is an equilateral triangle. On the equal bases PM, MD, DE equilateral arches are constructed, two of them tangent to the sides of the triangle at L and N respectively. Circles 0' and 0" are each tangent to a side of the triangle and to two of the arches. Circle is tangent to circles 0' and 0" and to both sides of the triangle. (a) liAB = a, find DE. Suggestion. In the right triangle DBL one acute angle is 60°. REVIEW AND FURTHER APPLICATIONS. 249 (h) Find the ratio r:r',r being the base DE of the arch and r^ the radius of the circle 0'. Suggest ion. GD = 2 DL, GO' =2 r>, and O'D = V(r+r'y -r^= y/2 rr' 4- r'^. (c) By what fraction of a will the circles 0' and 0" fail to touch each other? (d) Find the radius of the circle O. 20. In the figure CD = DA = AG = GB, and BE = EA = AF = FG. Semicircles are constructed on the diameters CB, CG, CD, DG, DB, GB. HG = KD= CE. Arcs GL and DM have centers H and K respectively. K C From Church of Or San Michele. Circles are constructed tangent to the various arcs as shown in the figure. Thus O 0" is tangent to semicircles on the diameters CB, CG, and DB. O 0' is tangent to the semicircles on the diameters CG and DB and to the arc DM. Let CB = a. (a) Find the areas of each of the six semicircles. (b) Find the radius r". Suggestion. EA, EP and ^iV are known. (c) Find the radius r'. Suggestion. Enumerate the known parts in A EDO' and FDO'. (d) Fi|id r and r^^. (e) Having determined the radii of the various circles, show how to construct the whole figure. (/) What fraction of the area of the whole figure is occupied by the six circles ? 21. Prove that two segments drawn from vertices of a triangle to points on the opposite sides cannot bisect each other. 250 PLANE GEOMETRY. FURTHER APPLICATIONS OF PROPORTION. 421. Definition. A segment is said to be divided in extreme and mean ratio by a point on it, provided the ratio of the whole segment to the larger part ^ ^ ^ equals the ratio of the larger part to the smaller. E.g. the segment AB is divided in extreme and mean ratio by the point C if ^ = ^. ^ AC CB 422. Problem. To divide a giveri line-segment in extreme and mean ratio. Solution. At one extremity B erect BoA-AB and make bo = \ ^-'''' AB. Draw OA. ^^-' \ On OA take OB = OB and with .--"' ^ as a center and AD as a radius a g b draw the arc DC. Then C is the required point of division. Proof: Let a be the length of AB. Then OB =-. Hence j:o- = i^^ + ^2 = «2 4. |'== | ^2, or ^O = | VsT Then ^C = ^Z) = yio - DO = ^ V5 - ^ =f ("^^ ~ A DC = ^B - ^C = a - -^(^V5 - 1^ = 1^3 - VS)- and Substituting these values in — - and — , and simplifying, AC CB • wo have — and ^"^ ~ _ V5 - 1 3 - V5* By rationalizing denominators these fractions are shown to be equal, and hence — = — • AC CB REVIEW AND FURTHER APPLICATIONS. 251 423. Problem. To consti^uct with ruler and com- passes angles 0/36° and 72°. Solution. On a given segment AB^ determine a point such that i^ = £^. (See § 422.) AC BG Using BD — AG as a base and AB as one leg, construct the isosceles triangle ABD. ^ Then Z 1 = 72° and Z 4 = 36°. Proof : Draw BG. Then A ABD ~ A DBG, (§ 259) since AB BD and Z 1 is common . BC A BD BC AC Hence A ABD and BDG are both isosceles (Why ?), and Z 1 = Z 2, Z 3 = Z 4. Also Z2 = Z3 + Z4 = 2Z4 = Z1. (Why?) Thus in A ABD each base angle is double Z4, making Z4 = | of 2 rt. A = 36°, (Why?) and Z1 = 2Z4 = 72°. 424. EXERCISES. 1. Inscribe a regular decagon in a circle. Suggestion. Construct the central angle A OB = 36", thus determining the side AB oi the decagon. 2. Inscribe a regular pentagon in a circle. Suggestion. Join alternate vertices of a decagon, or construct a central angle equal to 72°. 3. Inscribe a regular polygon of fifteen sides in a circle. Suggestion. Since ^ — J^ = -^■^, it follows that the arc subtended by one side of a hexagon minus that subtended by one side of a deca- gon is the one which subtends one side of a polygon of fifteen sides. 252 PLANE GEOMETRY. 4. The radius of an inscribed regular polygon is a mean propor- tional between its apothem and the radius of a similar circumscribed polygon. 5. A circular pond is surrounded by a gravel walk, such that the area of the walk is equal to the area of the pond. What is the ratio of the radius of the pond to the width of the walk? 6. lla,b,c are three line-segments such as - = - show that - equals he c the ratio of the area of any triangle described on a as a base to the similar triangle described on i as a base. 425. Definitions. A line-segment AB is said to be divided internally in a given ratio — by a point C lying on the segment, li^^Vl. See § 252. ^ ' CB n ^ A line-segment is said to be divided externally in a given ratio _ by a point c' lying c B r" s ^^f on AB produced, if = - • C'B s A line-segment AB is said to be divided harmonically if the points C and C^ lying respectively on AB and on AB AC AC^ produced, are such that — = --— • CB C'B 426. EXERCISES. 1. Show that if AB is divided harmonically by C and C, then CC is divided harmonically by A and B. 2. Show that the base of any triangle is cut harmonically by the bisectors of the internal and external vertical angles. 3. Show how to divide a line-segment externally in extreme and AB C'A mean ratio, that is, in the figure below, so that = . C'A C'B Suggestion. In the figure of § 422 produce BA to a point C such that C'A =A0+ OB. Then use the method there ^ A B used, to show that ^— and -^-^ each reduces to • C'A C'B 2 BEVIEW AND FURTHER APPLICATIONS. 253 427, Theorem. If an angle of one triangle is equal to an angle of the other ^ their areas are to each other as the products of the sides including the equal angles. A B a: b' a Given A ABC and A'BfC in which Z.A = AA\ A ABC AB ' AC To prove that aa'b'c' a'b'-a'g' Proof : Place Aa'b'c' so that Z A' coincides with Z A. Then AABC = Ih- AB and A . A'B'C' = :1A'. A'B' Hence, A ABC aa'b'c' \h' AB 'A'B' h AB A'B' Now show that — = AC A'C' t and hence that '^^^,^ = aa'b'c' AC A'C' AB A'B' AB A'B' 'AC 'A'C' • 428. Theorem. The square on the bisector of an angle of a triangle is equal to the ^ ^\a product of the two adjacent sides minus the product of the segments of the opposite side. Outline of proof: Produce the bisec- tor of the given angle to meet the cir- E cumscribed circle. Since ABDC^ AAEC AC • BC ~ CD - CE. (1) But CD- CE= CDQCD 4- DE) = CD^ + CD • DE (2) and CD- DE=AD' DB. (Why?) (3) Using (1), (2), and (3), complete the proof. 254 PLANE GEOMETRY. 429. Theorem. The product of two sides of a tri- angle is eqmal to the product of the altitude from the vertex in which these sides meet and the diameter of the circumscribed circle. Outline of Proof : Using the figure, show from the similar triangles ACT) and EEC that AC'BC= CE'CB. 430. EXERCISES. 1. The areas of two parallelograms having an angle of the one equal to an angle of the other are in the same ratio as the product of the sides including the equal angles. 2. Three semicircles of equal diameter are arranged as shown in the figure. (a) If ^D = a, find the area bounded by the arcs AB, BEC, and CA. (b) If the area' just found "is 2 square feet, find ylD. 3. Prove that the bisectors of the angles of any- quadrilateral form a quadrilateral whose opposite angles are supplementary. Suggestion. Show that Z3+Z4 + Z5 + Z6 = 2 rt. A, and hence that Z 1 + Z 2 = 2 rt. ^i. 4. On each of two sides of a given triangle ABC a.s chords con- struct arcs in which an angle of 120° may be inscribed. If these arcs meet in a point inside the triangle, show that the three sides of the triangle subtend the same angle from the point 0. 5. If Z 1 + Z 2 + Z 3 = 4 rt. zi, show how to find a point within a given triangle ABC so that Z^0j!3=Z1, ZBOC = Z2, and ZC0A = Z3. REVIEW AND FURTHER APPLICATIONS. 255 6. Given any two segments AB and CD and any two angles a and b, find a point O such that Z A OB = Za and Z COD = Zh. Discuss the various possible cases and the number of points in each case. 7. li ZA OB is a central angle of a circle and if Z CDE is inscribed in an arc of the same circle and if ZAOB = Z CDE = | rt. A, then the chords AB and CE are equal. 431. Definition. A set of lines which all pass through a common point is called a pencil of lines, and the point is called the center of the pencil. 432. EXERCISES. C^ /a:' A \ r /TV /r\ /A' IB' 'c 1. The lines of a pencil intercept proportional segments on parallel transversals. Given three lines meeting in O cut by three parallel transversals. To prove that the corresponding segments are proportional, that is, to show that AB ^ A[R ^ A"B" BC B'C B"C"' 2. Ii two polygons are symmetrical with respect to a point, they are congruent. (See § 170, Ex. 2.) 3. Any two figures symmetrical with respect to a point are congruent. Suggestion. About the point as a pivot swing one of the figures through a straight angle. Then any point P' of the right-side figure will fall on its sym- metrical point P of the left-side figure. 4. By means of the theorem in the preceding example show how to make an accurate copy of a map. Suggestion. Fasten the map to be copied on a drawing board. A long graduated ruler is made to swing freely about a fixed point 0, and by means of it construct a figure symmetrical to the map with respect to the point 0. How are the distances from O measured off ? 256 PLANE GEOMETRY. 433. Theorem. If correspo7iding vertices of kvo polygons lie on the same lines of a pencil and if they cut off proportiotial dis- tances on these lines from the center, then the poly- gons are similar. Prove the theorem first for triangles. OB OC OA -, OA OB OC GiVen. — 7 = — 7 = — 7 and — - = — - = — -. ob' oc' oa' oa" ob" oc" To prove that A ABC, A'b'&, a"b^'c" are similar. Prove the theorem for polygons of any number of sides. 434. Definition. Any two figures are said to have a center of similitude o, if for any two points Pj and P^ the . lines P^O and P^O meet the other figure in points P\ and p'„ such that P^O^P^^ P\0 P\0 Then P\ and P'^ are said to correspond to the points Pj, and Pg. Thus in the figure of § 433 O is called the center of similitude of the two polygons. Any two figures which have a center of similitude are similar. * This affords a ready means of constructing a figure similar to a given figure and having some other required property that is sufficient to determine it. Definition. The ratio of any two corresponding sides of similar polygons is called their ratio of similitude. REVIEW AND FURTHER APPLICATIONS. 257 435. EXERCISES. 1. Construct a polygon similar to a given polygon such that they shall have a given ratio of similitude. Suggestion. Let ABODE F be the given polygon and — the n given ratio of similitude. Select any convenient point O, such that OA =m. Draw lines OA, OB, etc. On OA lay off OA', n units. Lay off points B', C, etc., so that OA ^ OB ^ PC ^ OP ^ OE ^ OF OA' ~ OB' OC OB' OE' OF'' Prove that A'B'C'D'E'F' is the required polygon. 2. Show how the preceding may be used to ehlarge or reduce a map to any required size. Suggestion. Arrange apparatus as under Ex. 4, § 432. 3. In the figure A BCD is a parallelogram whose sides are of con- stant length. The point A is fixed, while the remainder of the figure is free to move. Show that the points P and P' trace out similar figures and that their ratio of simili- tude is — ^^— . AD+ CP This shows the essential parts of an in- strument called the pantograph, which is much used by engravers to transfer figures and to increase or decrease their size. The point P is made to trace out the figure which is to be copied. Hence P' traces a figure similar to it. The scale or ratio of similitude is regulated by adjusting the length of CP. 4. Construct a triangle having given two angles and the median a from one specified angle. Solution. Construct any triangle ABC having the required angles and construct a median AD. Prolong DA to D', making AD' =a. Extend BA and CM and through D' draw B'C'WBC, making AAB'C. Prove that this is the required triangle. (Notice that yl is the center of similitude.) y\ 5. Liscribe a square in a given triangle using the figure given here. Compare this method with that given on page 147. 258 PLANE GEOMETRY. 6. Construct a circle through a given point tangent to two given straight lines. Solution. Let a and h be the given lines and P the given point. Construct any circle tangent to a and t. Draw AP meeting the circle in C and D. Draw CO and OD and through P draw lines parallel to these meeting the bisector of the angle formed by a and h in 0' and 0''. Prove that and 0" are centers of the re- quired circles. Observe that ^ is a center of similitude. This method is used to construct a railway curve through a fixed point connecting two straight stretches of road. 7. On a line find a point which is equidistant from a given point and a given line. The following is another instance of the use of this very important device in constructing figures that resist other methods of attack. It consists essentially in first constructing a figure similar to the one required and then constructing one similar to this and of the proper size. From Westminster Abbey. 8. Given a circle with radius OE. Construct within it the design shown in the figure. That is, the inner semicircles have as diameters the sides of a regular sixteen-sided polygon. Each of the small circles is tangent to two semicircles. The outer arcs have their centers on the given circle and each is tangent to two small circles. All these arcs and circles have equal radii. REVIEW AND FURTHER APPLICATIONS. 259 Solution. Construct an angle equal to one sixteenth of a peri- gon. Through any point D' in the bisector of this angle draw a seg- ment EF perpendicular to the bisector and terminated by the sides of the angle. On' EF as a diameter construct a semi- circle. Make B'D' = EF = B'A' = D'C = C'A' and construct the small figure. E C Now draw radii of the given circle dividing it into sixteen equal parts and bisect one of the central angles by a radius OA. Construct Z. DAB = Z D'A'B'. Then ^ J5 is twice the radius of the required arcs and circles. The whole figure may now be constructed. 9. Given any three non-collinear points A', B', C, to construct an equilateral triangle such that A', B', C shall lie on the sides of the triangle, one point on each side. Suggestion. Through one of the points as A' draw a line such that B' and C lie on the same side of it. 10. Given an equilateral triangle A BC, to construct a triangle sim- ilar to a given triangle A'B'C with its vertices on the sides of ^i^C Suggestion. Construct an equilateral triangle such that A', B', C lie on its sides. Then construct a figure similar to this and of the required size. 11. If jo and/>' are similar polygons inscribed in and circumscribed about the same circle, and if 2.s is a side of the circumscribed polygon p', show that the difference of the areas of p and p' is equal to the area of a polygon similar to these and having a radius s. Suggestion. Let the areas of the polygons whose radii are r, r', and she A, A', A". Prove that A' -A ^ r"^ - r^ ^^^^ _A^ _ ^ _. r"^ - r^ A' ~ r'2 ^" A' ~ r'2 ~ /^ Complete the proof. 12. Two circles are tangent at A. A secant through A meets the circles at B and C respectively. Prove that the tangents at B and C are parallel to each other. 13. Prove that the segments joining one vertex of a regular poly- gon of n sides to the remaining vertices divide the angle at that vertex into n-2 equal parts. 260 PLANE GEOMETBY. FURTHER PROPERTIES OF TRIANGLES. 436. Definition. A segment AB is said to be projected upon a line I if perpendiculars from A and B are drawn to I. If these meet I in points C and D, then CB is the pro- jection of AB upon I. 437. Theokem. The square of a side opposite a7i acute angle of a triangle is equal to the sum of the squares of the other two sides minus twice the product of one of these sides and the projection of the other ujoon it. c m Outline of Proof : In either figure let Z i? be the given acute angle, and in each case BD is the projection of BC upon AB. Call this projection m. We are to prove that 5^ = a^ -{- e^ — 2 cr/i. In the left figure, b^=.h^-{.(c- my. (1) In the right figure, h^ = h^ -\- (^m — c^. (2) In either case W' = c^— m^. (3) Substitute (3) in (1) or in (2), and complete the proof. Modify each figure so as to draw the projection of AB upon BC and call this n. Then give the proof to show that h^ = a^ + c"^ — 2 an. 438. EXERCISE. 1. The area of a polygon may be found by drawing its longest diagonal and letting fall perpendiculars upon this diagonal from each of the remaining vertices. Draw a figure like the one in the margin, only on a much larger scale, measure the necessary lines, compute the areas of the various parts (see § 314), and thus find its total area. REVIEW AND FU ETHER APPLICATIONS. 261 439. Theorem. The square of the side opposite an obtuse angle of a triangle is equal to the sum of the squares of the other two sides plus twice the product of one of these sides and the projection of the other upon it. A ^ B D Outline of Proof : Let Z b he the given obtuse angle and BD the projection of BC upon AB. Call this projection m. As in the preceding theorem show that Also modify the figure so as to show the projection of AB on BC and call this n. Then show that J2 = ^2 + (?2 _p 2 an. 440. Theorem. The sum. of the squares of two sides of any triangle is equal to tivice the square ^ of half the third side plus twice the /^\^"^^^^ square of the median drawn to that side, a d b Suggestion. Make use of the two preceding theorems. 441. EXERCISES. 1. Compute the medians of a triangle in terms of the sides. 2. Show that the difference of the squares of two sides of a tri- angle is equal to twice the product of the third side and the projection of the median upon that side. 3. The base of a triangle is 40 feet and the altitude is 30 feet. Find the area of the triangle cut off by a line parallel to the base and 10 feet from the vertex. 262 PLANE GEOMETRY. 442. Problem. To express the area of a triangle in terms of its three sides. c or Solution. The area of A ABC = | AB x CD = i he. It is first necessary to express h in terms of a, b, c. We have b^= 0^+ e^- 2 cm, (Why?) Also h? = a^ — m^. (Why?) Hence h^= a^ a2^c'i_p \2 _ 4 ^2^2 _ (^2 + g2 _ ^2 )2 ^ (2 gg + a2 4- g2 _ ^2)(2 ^6' - ^2 - g2 _}_ 52) 4^2 ^ [(^ 4- e)2 - p-\ [-62 - (g - 6')2] 4 6'2 ^ (a + e+ h){a -\- c - h){h + a - cfh - a + c) Now call a + b-{- c=2s, or a + c— b = 2s— 2 b. Then a + c-b=2(s-b) b -\- a — c = 2(s - e) b -[- c— a=2{s— a). Hence K^ = 2 ^ -^C^ - ^) ■ 2(. - ^)- 2(a->^)^ 4 ^ — ^ • 5. Use the preceding exercise to find that point in a given line- segment which divides it into two such parts that their product is a maximum. 6. Find the point in a given line-segment such that the sum of the squares on the two parts into which it divides the segment is a minimum. Suggestion. If a and b are the two parts and k the length of the segment, then ^^^^^ ^^^^ a^ + 2 ab + b^ = k% or, a2 + ^,2 ^ ^2 _ 2 ab. Hence, a"^ -\-b^ is least when 2 ab is greatest. Now apply Ex. 5. 7. In the preceding exercise show that a^ + 6^ increases as b grows smaller, that is, as the point of division approaches one end. When is this sum a maximum ? 8. Two points A and B, on opposite sides of a straight stream of uniform width, are to be con- nected by a road and a bridge crossing the stream at right angles. Find by construction the location / of the bridge so as to make the total path from A to B a, minimum. 270 PLANE GEOMETRY, maximum area. 447. Theorem. Of all triangles having equal perim- eters ayid the same base, the isosceles triangle has the a c -ij>J^ A E B Given A ABC isosceles and having the same perimeter as A ABD. To prove that area A ABO area A ABD. Outline of proof: Draw CE JL AB. Construct AAFB having its altitude FE the same as that of A ABD. Pro- long AF making FG = AF. Draw GB and GD. The ob- ject is to prove that EF, the altitude of AABD^ is less than EC, the altitude of A ABC. (1) Show that A AFB, FBG, and GBD are all isosceles, for which purpose it must be shown that GB ± AB and FD II AB. Then AD-^ DB = AD-\- DG>AG. Or AF-h FB a'b'c'd'e'. Proof: In the first figure draw the diameter DK and draw AK and BK. On a'b' construct A a'b'k ^ Aabk and draw d'k'. The circlte whose diameter is b'k' cannot pass through all the points A', b', c\ and E\ else the polygon would be inscribed contrary to hypothesis. If either A' or E' is not on this circle, then AKBE>a'k'd'e'. (Why?) (1) Likewise if either b' or & is not on the circle, then KBCD > k'b'c'b'. (Why ?) (2) In any case, adding (1) and (2), AKBCDE > A'e'b'c'd'e'. By construction, A AKB ^ A A' k' b' . From (3) and (4), ABODE > a'b'c'd'e'. (3) (4) 274 PLANE GEOMETRY. 453. Theokem. Of all regular polygons having equal perimeters, that which has the greatest number of sides is the maximum. F Given T a regular triangle and S a square having the same perimeter. To prove that s > T. Proof : Take any point E in AB. Draw CE and on CE construct A CEF ^ A A EC. Then polygon BCFE has a perimeter equal to that of T and the same area, and has the same number of sides as -s. Hence, s > BCFE. (§ 449) That is, s > T. In like manner it can be shown that a regular polygon of five sides is greater than a square of equal perimeter, and so on for any number of sides. 454, EXERCISE Of the three medians of a scalene C triangle that one is the shortest which is drawn to the longest side. F Suggestion. If ACAE. A D B Jn A A DC and BDC, AD = DB and DC is common. But^C<^C. Hence Z1 .4 and hence that BF>AE. E REVIEW AND FURTHER APPLICATIONS. 275 455. Theorem. Of all regular polygons having a given area, that one which has the greatest number of sides has the least perimeter. Given regular polygons P and Q such that P = Q while Q has the greater number of sides. To prove that perimeter of Q < perimeter of P. Proof : Construct a regular polygon R having the same number of sides as P and the same perimeter as Q. Then r ^^ ^ > 21-100m-12,'46(A2012sl6)4120 # -V" (7442.7 UNIVERSITY OF CALIFORNIA LIBRARY mmmmimmm§ li ill I r^^^^^'^^^^^^ jilj! iiibi|[|llifsili!j| i i