IN MEMORIAM FLOR1AN CAJORI PLANE GEOMETRY BY EDWARD RUTLEDGE ROBBINS, A.B. SENIOR MATHEMATICAL MASTER THE WILLIAM PENN CHARTER SCHOOL NEW YORK .-CINCINNATI --CHIC AGO AMERICAN BOOK COMPANY COPYRIGHT, 1906, BY EDWARD RUTLEDGE ROBBINS. PLANE GEOMETRY. W. P. I FOR THOSE WHOSE PRIVILEGE IT MAY BE TO ACQUIRE A KNOWLEDGE OF GEOMETRY THIS VOLUME HAS BEEN WRITTEN AND TO THE BOYS AND GIRLS WHO LEARN THE ANCIENT SCIENCE FROM THESE PAGES, AND WHO ESTEEM THE POWER OF CORRECT REASONING THE MORE BECAUSE OF THE LOGIC OF PURE GEOMETRY THIS VOLUME IS DEDICATED PREFACE THE motives actuating the author in the preparation of this text in Geometry have been : (a) To present a book that has been written for the pupil. The object sought in the study of Geometry is not solely to train the mind to accept only those statements as truth for which convincing reasons can be provided, but to culti- vate a foresight that will appreciate both the purpose in mak- ing a statement and the process of reasoning by which the ultimate truth is established. Thus, the study of this formal science should develop in the pupil the ability to pursue argument coherently, and to establish one truth by the aid of other known truths, in logical order. The more mature members of a class do not require that the reason for every declaration be given in full in the text ; still, to omit it altogether, wrongs those pupils who do not know and cannot perceive the correct reason. But to ask for the reason and to print the paragraph reference meets the requirements of the various degrees of intellectual capac- ity and maturity in every class. The pupil who knows and knows that he knows need not consult the paragraph cited ; the pupil who does not know may learn for himself the correct reason by the reference. It is obvious that the greater progress an individual makes in assimilating the sub- ject and in entering into its spirit, the less need there will be for the printed reference. 6 PREFACE (6) To stimulate the mental activity of tie pupil. To compel a young student to supply his own demonstra- tions, in other words, to think and reason for himself, fre- quently proves unprofitable as well as unpleasant, and engenders in the learner a distaste for a study he has the right to admire and to delight in. The short-sighted youth absorbs his Geometry by memorizing, only to find that his memory has been an enemy, and while he himself is becom- ing more and more confused, his thoughtful companion is making greater and greater progress. The earlier he dis- covers his error the better, and the plan of this text gives him an opportunity to reestablish himself with his class. It is not calculated to produce accomplished geometricians at the completion of the first book, but to aid the learner in his progress throughout the volume, wherever experience has shown that he is likely to require assistance. It is cal- culated, under good instruction, to develop a clear concep- tion of the geometric idea, and to produce at the end of the course a rational individual and a friend of this particular science. (c) To bring the pupil to the theorems and their demonstra- tions the real subject-matter of Geometry as early in the study as possible. (c?) To explain rather than formally demonstrate the simple fundamental truths. (e) To apply each theorem in the demonstration of other theorems as promptly as possible. (/) To present a text that will be clear, consistent, teach- able, and sound. PRP:FACE 7 % The experienced teacher will observe : (a) The economy of arrangement. Many of the smaller figures are placed at the side of the page rather than at the center. The individual numbers of theorems are omitted. (6) The superior character of the diagrams. (V) The omission of the words " since " and "for." The advance statement is made and the reason asked for and usually cited. The inquiring mind fails to understand the force of preceding and following some statements with the same reason. (d) Originals that are carefully classified* graded* and placed after the natural subdivisions of the subject-matter. (e) The independence of these originals. Every exercise can be solved or demonstrated without the use of any other exercise. Only the truths in the numbered paragraphs are necessary in working originals. (/ ) The setting of every theorem* corollary, and problem of the text proper infullface type. (<7) The consistent use of such terms as " vertical angles" " vertex-angle" " adjacent angles" " angles adjoining a side" and others. (h) The full treatment of measurement and the illustrations of the terms employed. (i) The summaries that precede earlier collections of origi- nal exercises. (/) The emphasis given to the discussion of original oon- structions. 8 PREFACE As in all subjects that are new to a class, the successful teacher will be content with short lessons at the beginning, and will progress slowly until the class is thoroughly familiar with the language and the general method and purpose of the new science. The author sincerely desires to extend his thanks to those friends who, by suggestion and encouragement, have inspired him in the preparation of these pages. EDWARD R. ROBBINS. THE WILLIAM PENN CHARTER SCHOOL, PHILADELPHIA. CONTENTS PAGE INTRODUCTION . . . . " . . . . . . .11 Angles 12 Triangles 14 Symbols ........... 16 Axioms . . .16 Postulates 17 BOOK I. ANGLES, LINES, RECTILINEAR FIGURES . . . 18 Preliminary Theorems ........ 18 Theorems and Demonstrations ... ... 20 Model Demonstrations 24 Quadrilaterals 45 Polygons ........... 54 Symmetry 58 Locus 60 Summary. General Directions for attacking Originals . . 62 Original Exercises 64 BOOK II. THE CIRCLE 70 Preliminary Theorems 81 Theorems and Demonstrations 82 Summary ........... 92 Original Exercises 93 Kinds of Quantities. Measurement ...... 97 Original Exercises 107 Constructions .......... 115 Analysis 1-7 Original Constructions .... ... 128 9 10 CONTENTS PAGE BOOK III. PROPORTION. SIMILAR FIGURES 140 Theorems and Demonstrations ....... 141 Original Exercises (Numerical) ...... 168 Summary. Original Exercises (Theorems) .... 172 Constructions .......... 180 Original Constructions 184 BOOK IV. AREAS 187 Theorems and Demonstrations 187 Original Exercises (Theorems) 197 Formulas 201 Original Exercises (Numerical) ...... 204 Constructions . . . . 207 Original Constructions ........ 213 BOOK V. REGULAR POLYGONS. CIRCLES 217 Theorems and Demonstrations . . . . . . .217 Original Exercises (Theorems) 230 Constructions .......... 233 Formulas ........... 236 Original Exercises (Numerical) 239 Original Constructions 244 Maxima and Minima . . . . . . . . . 245 Original Exercises 249 INDEX OF DEFINITIONS ......... 251 PLANE GEOMETRY INTRODUCTION 1. Geometry is a science which treats of the measurement of magnitudes. 2. A definition is a statement explaining the significance of a word or a phrase. Every definition should be clear, simple, descriptive, and correct ; that is, it should contain the essential qualities or exclude all others, or both. 3. A point is that which has position but not magnitude. 4. A line is that which has length but no other magnitude. 5. A straight line is a line which is determined (fixed in position) by any two of its points. That is, two lines that coincide entirely, if they coincide at any two points, are straight lines. 6. A rectilinear figure is a figure containing straight lines and no others. 7. A surface is that which has length and breadth but no other magnitude. 8. A plane is a surface in which if any two points are taken, the straight line connecting them lies wholly in that surface. 9. Plane Geometry is a science which treats of the proper- ties of magnitudes in a plane. 10. A solid is that which has length, breadth, and thick- ness. A solid is that which occupies space. 11 12 PLANE GEOMETRY 11. Boundaries. The boundaries (or boundary) of a solid are surfaces. The boundaries (or boundary) of a surface are lines. The boundaries of a line are points. These boundaries can be no part of the things they limit. A sur- face is no part of a solid ; a line is no part of a surface ; a point is no part of a line. 12. Motion. If a point moves, its path is a line. Hence, if a point moves, it generates (describes or traces) a line ; if a line moves (except upon itself), it generates a surface ; if a surface moves (except upon itself), it generates a solid. NOTE. Unless otherwise specified the word " line " hereafter means straight line. ANGLES ANGLE ADJACENT VERTICAL ANGLES RIGHT ANGLES ANGLES PERPENDICULAR 13. A plane angle is the amount of divergence of two straight lines that meet. The lines are called tjie sides of the angle. The vertex of an angle is the point at which the lines meet. 14. Adjacent angles are two angles that have the same vertex and a common side between them. 15. Vertical angles are two angles that have the same vertex, the sides of one being prolongations of the sides of the other. 16. If one straight line meets another and makes the ad- jacent angles equal, the angles are right angles. INTRODUCTION 13 17. One line is perpendicular to another if they meet at right angles. Either line is perpendicular to the other. The point at which the lines meet is the foot of the perpendicular. Oblique lines are lines that meet but are not perpendicular. 18. A straight angle is an angle whose sides lie in the same straight line, but extend in opposite directions from the vertex. OBTUSE ANGLE ACUTE COM PLEMENTARY SUPPLEMENTARY ANGLES ANGLE ANGLES 19. Aii obtuse angle is an angle that is greater than a right angle. An acute angle is an angle that is less than a right angle. An oblique angle is any angle that is not a right angle. 20. Two angles are complementary if their sum is equal to one right angle. Two angles are supplementary if their sum is equal to two right angles. Thus the complement of an angle is the difference between one right angle and the given angle. The supplement of an angle is the difference between two right angles and the given angle. 21. A degree is one ninetieth of a right angle. The degree is the familiar unit used in measuring angles. It is evident that there are 90 in a right angle ; 180 in two right angles, or a straight angle ; 360 in four right angles. 22. Notation. A point is usually denoted by a capital letter, placed near it. A line is denoted by two capital letters, placed one at each end, or one at each of two of its points. Its length is sometimes represented advantageously by a small letter written near it. Thus, the line AB; the line RS', the line m. A B 5 _ : 14 PLANE GEOMETRY An angle is usually denoted by three capital letters, placed one at the vertex and one on each side. If only one angle is at a vertex, the capital letter at the vertex is sufficient to designate the angle. Sometimes it is advantageous to name an angle by a small letter placed within the angle. The word " angle " is usually denoted by the symbol " Z " in geometrical x o Z a AND Z Z XMA OR Z M. NOT Z It is important that in naming an angle by the use of three letters, the vertex-letter should be placed between the others. The size of an angle does not depend upon the length of the sides, but only on the amount of their divergence. Thus, Z x = Z P and Z P is the same as Z APR or Z APS or Z BPS, etc. An angle is said to be included by its sides. An angle is bisected by a line drawn through the vertex and dividing the angle into two equal angles. TRIANGLES 23. A triangle is a portion of a plane bounded by three straight lines. These lines are the sides. The vertices of a triangle are the three points at which the sides intersect. The angles of a triangle are the three angles at the three vertices. Each side of a triangle has two angles adjoining it. The symbol for triangle is "A". A ISOSCELES A RIGHT A OBTUSE A ACUTE A SCALENE & EQUILATERAL A EQUIANGULAR A The base of a triangle is the side on which the figure appears to stand. The vertex of a triangle is the vertex opposite the .base. The vertex-angle is the angle opposite the base. 15 24. Kinds of triangles : A scalene triangle is a triangle no two sides of which are equal. An isosceles triangle is a triangle two sides of which are equal. An equilateral triangle is a triangle all sides of which are equal. A right triangle is a triangle one angle of which is a right angle. An obtuse triangle is a triangle one angle of which is an obtuse angle. An acute triangle is a triangle all angles of which are acute angles. An equiangular triangle is a triangle all angles of which are equal. 25. The hypotenuse of a right triangle is the side op- posite the right angle. The sides forming the right angle are called the legs. In an isosceles triangle the equal sides are sometimes called the legs, and the other side, the base. 26. Homologous Parts. If two triangles have the three angles of one equal respectively to the three angles of the other, the pairs of equal angles are homologous. Homologous sides in two triangles are opposite the homologous angles. 27. Homologous parts of equal figures are equal. If the triangles DEF and HIJ are equal in all respects, Z D is homologous to, and = Z H, hence EF is homolo- gous to, and = U. And Z E is homologous to, and = Z /, hence, DF is homologous to, and = flj, and so on. SUPERPOSITION. SYMBOLS 28. Equality and coincidence. Two geometrical figures are equal if they can be made to coincide in all respects. Angles coincide, and are equal, if their vertices are the same point and the sides of one angle are identical with the sides of the other. Superposition is the process of placing one figure upon another. This method of showing the equality of two geometrical figures is employed only in establishing fundamental principles. 16 PLANE GEOMETRY 29. Symbols. The usual symbols and abbreviations em- ployed in geometry are the following: 4- plus. - minus. = equals, or is (or are) equal to. is (or are) equivalent to. > is (or are) greater than. < is (or are) less than. .'. hence, therefore, conse- quently. J_ perpendicular. Js perpendiculars. O circle. Prolong it so that AX = AB. BOBBINS 1 PLANE GEOM. 2 BOOK I ANGLES, LINES, RECTILINEAR FIGURES PRELIMINARY THEORIES 36. A right angle is equal to half a straight angle. Because of the definition of a right angle. (See 16.) 37. A straight angle is equal to two right angles. (See 36.) 38. Two straight lines can intersect in only one point. Because they would coincide entirely if they had two common points. (See 5.) 39. Only one straight line can be drawn between two points. (See 5.) 40. A definite (limited or finite) straight line can have only one midpoint. Because the halves of a line are equal. 41. All straight angles are equal. Because they can be made to coincide. (See 28 and Ax. 13.) 42. All right angles are equal. They are halves of straight angles (36), and hence equal (Ax. 3). 43. Only one perpendicular to a line can be drawn from a point in the line. Because the right angles would not be equal if there were two perpendiculars ; and all right angles are equal. (See 42.) 18 BOOK I 19 44. If two adjacent angles have their exterior sides in a straight line, they are supplementary. Because they together = two rt. A. (See 20.) 45. If two adjacent angles are supplementary, their exterior sides _ are in the same straight line. Because their sum is two rt. A (20); or a straight Z (37). Hence the exterior sides are in the same straight line (18). 46. The sum of all the angles on one side of a straight line at a point equals two right angles. (See Ax. 4 and 37.) 47. The sum of all the angles about a point in a plane is equal to four right angles. (See 46.) 48. Angles that have the same complement are equal. Or, comple- ments of the same angle, or of equal angles, are equal. Because equal angles subtracted from equal right angles leave equals. (See Ax. 2.) 49. Angles that have the same supplement are equal. Or, supple- ments of the same angle, or of equal angles, are equal. ( See Ax. 2.) 50. If two angles are equal and supplementary, they are right angles. Because each is half a straight Z; hence each is a rt. Z. (See 36.) NOTE. A single number, given as a reference, always signifies the truth stated in that paragraph and is usually the statement in full face type only. Tn reciting or writing the demonstrations the pupil should quote the correct reason for each statement, and not give the number of its paragraph. [Consult model demonstrations on page 24.] 20 PLANE GEOMETRY THEOREMS AND DEMONSTRATIONS 51. THEOREM. Vertical angles are equal. Given: A AOM and BOL, a pair of vertical angles. To Prove : Z A OM = Z BOL. Proof : /.AOM is the supple- ment of Z MOB. (Why?) (See 44.) ZBOL is the supplement of Z MOB. (Why?) (See 44.) .-. /LAOM=/.BOL. (Why ?) (See 49.) A A OL and BOM are a pair of vertical angles. These may be proven equal in precisely the same manner. If Z A OL = 80, how many degrees are there in the other zt? 52. THEOREM. Two triangles are equal if two sides and the in- cluded angle of one are equal respectively to two sides and the included angle of the other. c T Given : A ABC&ud'RST; AB = RS\ AC llT\ Z A = Z tf. To Prove : A ABC = A RST. Proof : Place the A ABC upon the A RST so that Z A co- incides with its equal, Z R ; then AB will fall upon RS and point B upon 8. (It is given that AB = RS.) AC will fall upon RT and point C upon T. (It is given that AC = RT.) /. BC will coincide with ST. (Why?) (See. 39.) Hence, the triangles coincide in every respect and are equal (28). BOOK I 21 53. THEOREM. Two right triangles are equal if the two legs of one are equal respectively to the two legs tf the other. Given : Ht. A ABC and DEF', AC DF\ CK = FE. To Prove : A ABC = A DEF. Proof: In the A ABC and DEF, AC = DF (Given) ; CB = FE (Given) ; Z C = Z F. (Why?) (See 42.) /. the A are equal. (Why?) (Theorem of 52.) 54. THEOREM. Two triangles are equal if a side and the two angles adjoining it in the one are equal respectively to a side and the two angles adjoining it in the other. Given: A BCD and JKL ; BC = JK ; Z B = Z J ; Z C = To Prove : A BCD = A JKL. Proof: Place A BCD upon A JKL so that Z B coincides with its equal, Z J, BC falling on JK. Point C will fall on JT. (It is given that BC= JK.) BD will fall on JL. (Because Z I? is given = Z J.) CD will fall on 7TL. (Because Z C is given = Z K.) Then point D which falls on both the lines JL and KL will fall at their intersection, L. (Why ?) (See 38.) .-. the A are =. (Why?) (See 28.) 22 PLANK GEOMETRY 55. THEOREM. The angles opposite the equal sides of an isosceles triangle are equal. A Given : A ABC, AB = AC. To Prove : Z B = Z C. Proof: Suppose AX is drawn dividing Z BAG into two equal angles, and meeting BC at X. In A BAX and CAX, AX=AX (Identical); AB=AC (Given); Z BAX B ~~ * = Z (MX. (Because AX made them= .) .-. A ABX= A (Why?) (52.) .-.ZB = ZC. (Why?) (See 27.) 56. THEOREM. An equilateral triangle is equiangular. (See 55.) 57. THEOREM. The line bisecting the vertex-angle of an isosceles triangle is perpendicular to the base, and bisects the base. Prove A ABX and ACX equal as in 55. Then, Z AXB = Z AXC. (Why?) (27.) /. A AXB and AXGme rt. A (16). /. AX is _L to BC. (Why?) (17.) And, also, BX= CX. (Why?) (27.) 58. THEOREM. Two triangles are equal, if the three sides of one are equal respectively to the three sides of the other. Given : A ABC and RST; AC=ET\ BC= ST. To Prove : A RST= A ABC. BOOK I 23 Proof : Place A ABC in the position of A AST so that the longest equal sides (TiC and ST) coincide and A is opposite ST from B. Draw 724. BS = AS (Given). .*. ASB is an isosceles A. (Def. 24.) /. Z SRA = Z SAB. (Why?) (55.) Likewise TB = AT (?) and Z TEA = Z 7^7?. (Why?) Adding these equals we obtain Z SET = Z SAT (Ax. 2). /. A BST = A AST (52). That is, A BST=A ABC. (Substitution, Ax. 6.) 59. Elements of a theorem. Every theorem contains two parts, the one is assumed to be true and the other results' from this assumption. The one part contains the given con- ditions, the other part states the resulting truth. The assumed part of a theorem is called the hypothesis. The part whose truth is to be proved is the conclusion. Usually the hypothesis is a clause introduced by the word "if." When this conjunction is omitted, the subject of the sentence is known and its qualities, described in the quali- fying words, constitute the "given conditions." Thus, in the theorem of 58, the assumed part follows the word " if," and the truth to be proved is: "Two triangles are equal." 60. Elements of a demonstration. All correct demonstra- tions should consist of certain distinct parts, namely : 1. Full statement of the given conditions as applied to a particular figure. 2. Full statement of the truth which it is required to prove. 3. The Proof. This consists in a series of successive state- ments, for each of which a valid reason should be quoted. (The drawing of auxiliary lines is sometimes essential, but this part is accomplished by imperatives for which no reasons are necessary.) 4. The conclusion declared to be true. The letters "Q.E.D." are often annexed at the end of a demonstration and stand for " quod erat demonstrandum" which means, " which was to be proved." 24 PLANE GEOMETRY MODEL DEMONSTRATIONS The angles opposite the equal sides of an isosceles triangle are equal. Given : A ABC; AB = A C. To Prove : Z B = Z C. Proof : Suppose A X is drawn bisecting B ^.BAC and meeting BC at X. In the & BA X and C^Y ,4Z = .4A' (Identical). AB = AC (Hypothesis). /.BAX = Z CAX (Construction). .'.&ABX = &ACX. (Two & are = if two sides and the included Z of one are = respectively to two sides and the included Z of the other.) Hence, Z B = Z C. (Homologous parts of equal figures are equal.) Q.E.D. Two triangles are equal if the three sides of one are equal respectively to the three sides of the other. Given : A ABC and RST; AB = RS; AC= RT; BC=ST. To Prove : A RST = A ABC. Proof : Place A ABC in the position of A A ST so that the longest equal sides (BC and ST) coincide, and A is opposite 5 T 7 from R. Draw RA. RS = AS (Hypothesis). A A SR is isosceles. (An isosceles A is a A two sides of which are equal.) .'. Z SRA = ZSAR...(\)..(TheA opp. the = sides of an isos. A are = .) Again, TR = A T (Hypothesis). A TRA is isosceles. (Same reason as before.) (Same reason as for (1).) Adding equations (1) and (2). Z SRT = Z SA T. (If = 's are added to = 's the results are = .) Consequently, the A RST = A AST. (Two A are = if two sides and the included Z of one are = respectively to two sides and the included Z. of the other.) That is, A RST = A ABC. (Substitution ; A ABC is the same as &AST.) Q.E.D. TRA = Z TA R . . . (2) BOOK I 25 The preceding form of demonstration will serve to illustrate an excel- lent scheme of writing the proofs. It will be observed that the statements are at the left of the page and their reasons at the right. This arrange- ment will be found of great value in the saving of time, both for the pupil who writes the proofs and for the teacher who reads them. 61. The converse of a theorem is the theorem obtained by interchanging the hypothesis and conclusion of the original theorem. Consult 44 and 45; 79, 80, and others. Every theorem which has a simple hypothesis and a simple conclusion has a converse, but only a few of these converses are actually true theorems. For example : Direct theorem : " Vertical angles are equal." Converse theorem : " If angles are equal, they are verti- cal." This statement cannot be universally true. The theorem of 120 is the converse of that of 55. 62. Auxiliary lines. Often it is impossible to give a simple demonstration without -drawing a line (or lines) not described in the hypothesis. Such lines are usually dotted for no other reason than to aid the learner in distinguishing the lines mentioned in the hypothesis and conclusion from lines whose use is confined to the proof. Hence, lines mentioned in the hypothesis and conclusion should never be dotted. (The figure used in 57 should contain no dotted line.) 63. Superposition. It is worthy of note that demonstration by the method of superposition is never employed unless the hypothesis gives a pair of equal angles. 64. Homologous parts. Triangles are proven equal in order that their homologous sides, or homologous angles, may be proven equal. This is a very common method of proving lines equal and angles equal. 65. The distance from one point to another is the length of the straight line joining the two points. 26 PLANE GEOMETRY 66. THEOREM. If lines be drawn from any point in a perpendicu- lar erected at the midpoint of a straight line to the ends of the line, I. They will be equal. II. They will make equal angles with the perpendicular. III. They will make equal angles with the line. Given : AB _i_ to CD at its midpoint, B ; P any point in AB ; PC and PD. To Prove : I. PC=PD ; II. Z CPB = Z DPB ; and III. Z C = Z D. Proof: In rt. A PBC and c PBD, BC = BD (Hyp.) ; BP = BP (Iden.). = Z DP.B ( Why ?) ; Q.E.D. . (Why?) (53.) /. I. PC=PD(Why?) (27;) II. Z III. Zc = ZD (Why?). 67. THEOREM. Any point in the perpendicular bisector of a line is equally distant from the extremities of the line. (See 66, I.) 68. THEOREM. Any point not in the perpendicular bisector of a line is not equally distant from the extremities of the line. Given : AB _L bisector of CD ; P any point not in AB ; PC and PD. o To Prove : PC not = PD. Proof : Either PC or PD will cut AB. Suppose PC cuts AB at O. Draw OD. DO + OP > PD. (Why ?) (Ax. 12.) But CO = OD (67). .'. co+ OP > PD. (Substitution; Ax. 6.) That is, PC > PD, or PC is not = PD. Q.E.D. BOOK I 27 69. THEOREM. If a point is equally distant from the extremities of a line, it is in the perpendicular bisector of the line. (See 67 and 68.) 70. THEOREM. Two points each equally distant from the extrem- ities of a line determine the perpendicular bisector of the line. Each point is in the _L bisector (69); two points deter- mine a line (5). 71. THEOREM. Only one perpendicular can be drawn to a line from an external point. Given: P7? J_ to AB from P; PD any other line from P to AB. To Prove : PD cannot be J_ to AB ; that is, PR is the only J_ to AB from P. Proof: Extend PR to s, mak- ing RS = PR ; draw DS. In rt. A PDR and SDR, PR = RS (Const.). DR = DR (Iden.). .'.A PDR = A SDR. (Why?) (53.) .-. Z PDR = Z SDR (27). That is, Z PDR = half of Z PDS. Now PRS is a straight line (Const.). .'. PDS is not a straight line (39). .-. Z PDS is not a straight angle (18). .-. Z PDR, the half of Z PDS, is not a right angle (36). .-. PD is not _L (17). .'. PR is the only J_. Q.E.D. Ex. 1. Through how many degrees does the minute hand of a clock move in 15 min. ? in 20 min. ? Through how many degrees does the hour hand move in one hour? in 45 minutes? in 10 minutes? Ex. 2. How many degrees are there in the angle between the hands of a clock at 9 o'clock? at 10 o'clock? at 12 :30? at 2 : 15 ? at 3 : 45? Ex. 3. THEOREM. If two lines be drawn bisecting each other, and their ends be joined in order, the opposite pairs of triangles will be equal. [Use 51 and 52.] PLANE GEOMETRY 72. THEOREM. Two right triangles are equal if the hypotenuse and an adjoining angle of one are equal respectively to the hypote- nuse and an adjoining angle of the other. N r L MR S Given: Rt. A LMN and RST ; LN RT ; and Z L = Zft. To Prove: A LMN = A RST. Proof : Superpose A LMN upon A RST so that Z. L coincides with its equal, Z. R, LM falling along RS. Then LN will fall on RT and point N will fall exactly on T (LN= RT by Hyp.). Now NM and TS will both be _L to RS from T (Rt. A by Hyp.). /. NM will coincide with TS (71). ..-. A LMN = A/?.sr(28). Q.E.D. 73. THEOREM. Two right triangles are equal if the hypotenuse and a leg of one are equal respectively to the hypotenuse and a leg of the other. K R M l"~ J X L Given : Rt. A UK and LMR ; KI = RM ; KJ = RL. To Prove : A UK = A LMR. Proof : Place A UK in the position of A XLE so that the equal sides, KJ and RL, coincide and / is at X, oppo- site RL from M. BOOK I 29 Now, ARLM and RLX are supplementary. (Why ?) (20.) .*. XLM is a str. line (45). Also, A RMX is isosceles. (Why ?) (RX = RM by Hyp.) .-.Z x= Z3/. (Why?) (55.) .-. A XLR = A MLR. (Why?) (72.) That is, A UK = ALMR. (Ax. (J.) Q.E.D. COR. The perpendicular from the vertex of an isosceles triangle to the base bisects the base. Proof : A XLR = A MLR. (Why V) . . XL = ML. (Why ?) 74. THEOREM Two right triangles are equal if a leg and the ad- joining acute angle of one are equal respectively to a leg and the adjoining acute angle of the other. A CD . Given : lit. A ABC and DEF ; AC = DF ; Z^ = Z D. To Prove : A AEC = A DEF. Proof : In the A ABC and DEF, AC = DF. (Why?) (Hyp.) Also Z.A = /LD (Why?) and ZC' = ZF. (Why?) (42.) .'.AABC = ADEF. (Why?) (54.) Q.E.D. Ex. 1. How many pairs of equal parts must two triangles have, in order that they may be proven equal? How many pairs is it necessary to mention in the case of two right triangles? Ex. 2. THEOREM. If a perpendicular be erected at any point in the bisector of an angle, two equal right triangles will be formed. [Use 74.] Ex. 3. Through the midpoint of a line AB any oblique line is drawn : I. The lines JL to it from A and B are equal. [Use 72.] II. The lines _L to AB at A and B, terminated by the oblique line, are equal. [Use 74.] 30 PLANE GEOMETRY 75. THEOREM. The sum of two sides of a triangle is greater than the sum of two lines drawn to the extremities of the third side, from any point within the triangle. B Given : P, any point in A ABC i lines PA and PC. To Prove : AB + BC > AP + PC. Proof : Extend AP to meet BC at X. ~ AB + BX > AP + PX. (Why?) (Ax. 12.) CX + PX > PC. (Why ?) (Ax. 12.) Add : AB + BX + CX + PX > AP + PC + PX(Ax. 8). Subtract PX = PX. .'.AB + BC > AP + PC (Ax. 7). Q.E.D. 76. THEOREM. If from any point in a perpendicular to a line two oblique lines be drawn, I. Oblique lines cutting off equal distances from the foot of the perpendicular will be equal. II. Equal oblique lines will cut off equal distances (converse). III. Oblique lines cutting off unequal distances will be unequal, and that one which cuts off the greater distance will be the greater. B A I. Given : CD -L to AB ; ND = MD ; oblique lines PN and PM. [First figure.] To Prove : PN = PM. BOOK I 31 Proof : PD is J_ bisector of NM (Hyp.). .-. PN = PM (67). II. Given : CD _L to AB ; PN = PM. [First figure.] To Prove : ND = MD. Proof: In the rt. A PND and PMD, PD = PD (Iden.); and PN = PM (Hyp.). .'. A PND = A PMD. (Why?) (73.) .-.ND=MD. (Why?) (27.) Q.E.D. III. Given : CD _L to AB ; oblique lines P.R, PT ; also RD > DT. [Second figure.] To Prove: PR > PT. Proof : Because RD is > DT, we may mark DS (on RD) = DT. Draw P8. Extend PD to X, making DX = PD ; draw RX and SX. AD is _L to PX at its midpoint (Const.). .'.PR = RX and PS = sx (66). Now P.R + RX > PS + sx (75). Hence P# + PR > PS + PS (Ax. 6). That is, 2 PR > 2 PS. .'. PR> PS (Ax. 10). But P5= PT (76, I). .'. P > PT (Ax. 6). Q.E.D. COR. Therefore, from an external point it is not possible to draw three equal lines to a given straight line. 77. THEOREM. The perpendicular is the shortest line that can be drawn from a point to a straight line. Given: PRtoAB; PC not _L. To Prove : Ptf < PC. Proof : Extend PR to X, mak- ing RX=PR. Draw CX. PR + RX < PC+ CX (Ax. 12). But AR is _L to PXat its mid- A ~ point (Const.). .-. PC=CX (66). .'.PR+ PR < PC+PC (Ax. 6). That is, 2 PR < 2 PC. /. PR< PC (Ax. 10). Q.E.D. 32 PLANE GEOMETRY 78. The distance from a point to a line is the length of the perpendicular from the point to the line. Thus " distance from a line " involves the perpendicular. If the perpendicu- lars from a point to two lines are equal, the point is said to be equally distant from the lines. 79. THEOREM. Every point in the bisector of an angle is equally distant from the sides of the angle. Given: Z ACE; bisector CQ; point P in CQ; distances PB and PD. To Prove : Pit = PD. Proof : A PBC and PDC are rt. A (78). In rt. A PBC and PDC, PC =PC (Iden.) ; Z PCS = Z PCD (Hyp.). /.A PJ3C= APDC(?)(72). .-. PB = PD (?). Q.E.D. 80. THEOREM. Every point equally distant from the sides of an angle is in the bisector of the angle. Given : Z ACE; P a point, such that PB = PD (distances); CQ a line from vertex of the angle, and containing P. To Prove : Z ACQ = Z ECQ. Proof: A PJ5C and PDC are right A (?). In rt. A PBC and PDC, PC = PC (?) ; PB == PD (?). ?) (73). /.Z^CQ = Z^CQ(?). Q.E.D. 81. THEOREM. Any point not in the bisector of an angle is not equally distant from the sides of the angle. [Because if it were equally distant, it would be in the bisector (80).] 82. THEOREM. The vertex of an angle and a point equally distant from its sides determine the bisector of the angle. (See 80 and 5.) Ex. Describe the path of a moving point which shall be equally dis- tant from two intersecting lines. BOOK I 33 83. The altitude of a triangle is the perpendicular from any vertex to the opposite side (prolonged if necessary). A triangle has three altitudes. The bisec- tor of an angle of a triangle is the line dividing any angle into equal angles. A triangle has three bisectors of its an- ,, ,, . . i ,1 T THE THREE MEDIANS gles. 1 he median ot a triangle is the line drawn from any vertex to the midpoint of the opposite side. A triangle has three medians. 84. THEOREM. The bisectors of the angles of a triangle meet in a point which is equally distant from the sides. Given: A ABC, AX bisect- ing Z A, B Y and CZ the other bisectors. To Prove: AX, BY, cz meet in a point equally dis- tant from AB, AC, and BC. Proof: Suppose that AX and BY intersect at O. O in AX is equally distant from AB and AC (?) (79). O in BY is equally distant from AB and BC (?). .-. point O is equally distant from AC and BC (Ax. 1).* .-. O is in bisector CZ (?) (80). That is, all three bisectors meet at O, and O is equally dis- tant from the three sides. Q.E.D. * The J_ distances from O to the three sides are the three equals. Ex. 1. Draw the three altitudes of an acute triangle; of an obtuse triangle. Ex. 2. Prove that in an equilateral triangle : (a) An altitude is also a median. [Use 73.] (6) A median is also an altitude. [Use 58, 27, 16, 17.] (c) An altitude is also the bisector of an angle of the triangle. (d) The bisector of an angle is also an altitude. [Use 52.] (e) The bisector of an angle is also a median. BOBBINS' PLANE GEOM. 3 PLANE GEOMETRY 85. THEOREM. The three perpendicular bisectors of the sides of a triangle meet in a point which is equally distant from the vertices. Given : A ABC; LR, MS, NT, the three J_ bisectors. To Prove : LR, MS, NT meet at a point equally distant from A and B and C. Proof : Suppose that LR and MS intersect at O. O in LR is equally distant from^ and 5. (?) (67.) O in MS is equally distant from A and C. (?) ~~N~" .-. point O is equally distant from B and C (Ax. 1).* Hence O is in J_ bisector NT (?) (69). That is, all three _L bisectors meet at O, and O is equally distant from A and B and C. Q.E.D. * The three lines from to the vertices are the three equals. 86. THEOREM. If two triangles have two sides of one equal to two sides of the other, but the included angle in the first greater than the included angle in the second, the third side of the first is greater than the third side of the second. F H Given: A ABC, DEF\ AB = DE; BC=EF; Z ABC > Z E. BOOK 1 35 To Prove : AC > DF. Proof : Superpose A DEF upon A ABC so that line DE coincides with its equal AB, A DEF taking the position of A ABH. There will remain an angle, HBC. (Z ABC is > Z E. ) Suppose EX to be the bisector of Z HBC, meeting AC at X. Draw XH. In A HEX and CBX, EX = EX (?) ; EH = EC (?) ; Z #J3X = ZCJBX(?) (Const.)- .'.&UBX = A CBX (?) (52). .-.#x= XC (?). Now, AX + XH > ^iff (?). .-.AX + XC > ^4fl (Ax. 6). That is, JIC > IXF. Q.E.D. 87. THEOREM. If two triangles have two sides of one equal to two sides of the other but the third side of the first greater than the third side of the second, the included angle of the first is greater than the included angle of the second. [Converse.] s T Given : A ABC and RST; AB = BS ; EC = ST ; uiC> T. To Prove: Z >Zs. Proof : It is evident that Z U < Z 8, or Z B = Z S, or > ZS. 1. If ZB < ZS, AC < ET (86). But AC > #r (Hyp.). .\ZB is not < Z 5. 2. If Z u = Z 8, the A are = (52), and AC is = Rr (27). But AC > ET (Hyp.). /.Z E is wof = Z S. 3. /. the only possibility is that Z E > Zs. Q.E.D. 36 PLANE GEOMETRY 88. The preceding method of demonstration is termed the method of exclusion. It consists in making all possible suppositions, leaving the probable one last, and then proving all these suppositions impossible, except the last, which must necessarily be true. 89. The method of proving the individual steps is called reductio ad absurdum (reduction to an absurd or impos- sible conclusion). This method consists in assuming as false the truth to be proved and then showing that this assumption leads to a conclusion altogether contrary to known truth or the given hypothesis. (Examine 87.) This is sometimes called the indirect method. The theorems of 93 and 94 are demonstrated by a single use of this method. 90. THEOREM. If two unequal oblique lines be drawn from any point in a perpendicular to a line, they will cut off unequal distances from the foot of the perpendicular, and the longer oblique line will cut off the greater distance. [Converse of 76, III.] C Given: CD_LtoAB; PUand PS oblique lines ; PR > PS. To Prove : DR > DS. Proof : It is evident that DR < DS, or DR = DS, or ^ _ DR > DS. A R~~ If DR < DS, PR < PS (76, III). But PR > PS (Hyp.). .-. DR is not < DS. If DR = DS, PR = PS (76, I). But PR > PS (Hyp.). .'. DR is not = DS. Therefore the only possibility is that DR > DS. Q.E.D. 91. Parallel lines are straight lines that lie in the same plane and that never meet, however far extended in either direction. 92. AXIOM. Only one line can be drawn through a point parallel to a given line. BOOK r 37 93. THEOREM. Two lines in the same plane and perpendicular to the same line are parallel. Given : CD and EF in same plane and both J_ to AB. To Prove : CD and EF \\ . E Proof : If CD and EF were not II, they would meet if suffi- ciently prolonged. Then there would be two lines from the point of meeting J_ to AB. (By Hyp. they are J_ to AB.) But this is impossible (?) (71). .-. CD and EF do not meet, and are parallel (91). Q.E.D. 94. THEOREM. Two lines in the same plane and parallel to the same line are parallel. Given : AB \\ to RS and CD || to RS and in the same plane. To Prove : AB \\ to CD. A e Proof : If AB and CD were not ||, they would meet if sufficiently prolonged. Then there would be two lines through the point of meet- ing || to the line RS. (By Hyp. they are || to RS). But this is impossible (92). .-. AB and CD do not meet, and are || (?) (91). Q.E.D. 95. THEOREM. If a line is perpendicular to one of two parallels, it is perpendicular to the other also. Given : LM _L to AB and AB || to CD. To Prove : LM _L to CD. X 7* B Y Proof: Suppose XT is drawn through M J_ to LM. AB is || to XT (?) (93). But AB is || to CD (Hyp.). Now CD and XT both contain M (Const.). .'.CD and XT coincide (92). But LM is _L to XT (?). That is, LM is _L to CD. Q.E.D. 38 PLANE GEOMETRY 19' 96. If one line cuts other lines, it is called a transversal. Angles are formed at the several intersections, and these receive the following names : Interior A are between the lines [b, c, e, h~\. Exterior A are without the lines [a, rf,/, N ^, Another proof: J .... ." ; ' /.Z> Etc. \ Zaissupp.of Z r(?). j 105. THEOREM. If two angles have their sides paraUel each to each, the angles are equal or supplementary. I. Given : Z a and Z b with their sides II each to each and extending in the same directions from their vertices. To Prove : Z a = Z b. Proof: If the non-parallel lines do not intersect, produce BOOK I 41 them till they meet, forming Z o. Now, Z a = Z o. (?) (98). Z o = Z ft (?). /.Z a = Zft (?). Q.E.D. ^ ^ II. Given: Z a and Z AC. To Prove : Z ACE > Z B. Proof : On AB take ^17? = AC. [We may, because AB > AC.'] Draw CR and let Z ARC=x. B Z ARC is an ext. Z of A CBR (?). .-. Z x > Z 5 (109). Also, Z ^iC.R = Z .iflC = Z # (?) (55). Again, Z ^4 OB > Z a; (?) (Ax. 5). .-. Z ^c > Z J5 (Ax. 11). Q.E.D. 123. THEOREM. If two angles of a triangle are unequal, the sides opposite them are unequal and the greater angle is subtended by the greater side. Given: A ABC-, To Prove: AB > AC. Proof : In Z ACB, suppose Z BCR constructed = Z B. Then, CR= BR (?) (120). Also AR + CR > AC (?). /. AR + BR > ^C (Ax. 6). That is, AB > ^O. Q.E.D. 124. THEOREM. The hypotenuse is the longest side of a right tri- angle. (See 123.) BOOK I 45 QUADRILATERALS A quadrilateral is a portion of a plane bounded by four straight lines. These four lines are called the sides. The vertices of a quadrilateral are the four points at which the sides intersect. The angles of a quadrilateral are the four angles at the four vertices. The diagonal of a rectilinear figure is a line joining two vertices, not in the same side. 126. A trapezium is a quadrilateral having no two sides parallel. A trapezoid is a quadrilateral having two and only two sides parallel. A parallelogram is a quadrilateral having its opposite sides parallel (O). TRAPEZOID PARALLELOGRAM SQUARE RECTANGLE RHOMBOID 127. A rectangle is a parallelogram whose angles are right angles. A rhomboid is a parallelogram whose angles are not right angles. 128. A square is an equilateral rectangle. A rhombus is an equilateral rhomboid. 129. The side upon which a figure appears to stand is called its base. A trapezoid and all kinds of parallelograms are said to have two bases, the actual base and the side parallel to it. The non-parallel sides of a trapezoid are some- times called the legs. An isosceles trapezoid is a trapezoid 46 PLANE GEOMETRY whose legs are equal. The median of a trapezoid is the line connecting the midpoints of the legs. The altitude of a trapezoid and of all kinds of parallelograms is the perpendicular distance between the bases. 130. THEOREM. The opposite sides of a parallelogram are equal. Given : O LMOP. PO and To Prove : LM LP = MO. Proof : Draw diagonal PH. In A LMP and OMP, PM = ..--' /. A LMP = A OMP (?) (54). /. LM = PO and LP =MO (?) (27). Q.E.D. 131. COR. Parallel lines included between parallel lines are equal. (See 130.) 132. COR. The diagonal of a parallelogram divides it into two equal triangles. 133. COR. The opposite angles of a parallelogram are equal. (See 27.) 134. THEOREM. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. [Converse of 130.] Given: Quadrilateral ABCD; AB = DC ; AD = BC. To Prove : ABCD is a O. Proof : Draw diagonal BD. In A ABD and CBD, BD = BD (?) ; AB = DC (?), and AD = BC (?). .'. A ABD = A CBD (?) (58). Hence Z a = Z i (?). Therefore AB is || to DC (?) (101). Also, Z y = Z.x (?). Therefore AD is || to BC (?). Hence ABCD is a parallelogram (Def. 126). Q.E.D. BOOK I 47 135. THEOREM If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. Given: Quadrilateral ABCD ; AB = CD and AB \\ to CD. To Prove: A BCD is a O. Proof: Draw diagonal BD. In A ABD and C'fi/), BD = #D(?); ^47* = (?); and Za = Z* (?) (97). .-.A ABD = A CtfD (?) (52). Hence Zy=Z.r (?). .-. AD is || to BC (?) (101). is a parallelogram (?) (126). Q.E.D. 136. COR. Any pair of adjoining angles of a parallelogram are sup- plementary. (See 100.) 137. THEOREM. The diagonals of a parallelogram bisect each other. Given: O EFGH ; diago- F - nals EG and FH intersecting at X. To Prove : FX = X/T and GX = ^E. Proof : In A FXG and .EXtf, E FG = ^H (?) (130) ; Z a = Z o and Z c = Z r (?) (97). .'.A F^G = A #Xff (?) (54). .-. FX = XH and GX = XE (?) (27). Q.E.D. 138. THEOREM. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Given: (?). To Prove: (?). Proof: In A FXG and EXH show three parts of one = etc. Hence certain A are = (?). Then two lines are II (?). Also = (?). Now use 135. Ex. 1. In the figure of 137, if Z a = 20 and Z c = 30, find the four angles at X. Ex. 2. If one angle of a parallelogram is 65, find the other three. If one is 90, find the others. Ex. 3. State and prove the converse of 136. 48 PLANE GEOMETRY 139. THEOREM. Two parallelograms are equal if two sides and the included angle of one are equal respectively to two sides and the in- cluded angle of the other. B _c M N A D L O Given : HJ AC and LN ; AB = LM ; AD = LO ; Z A = Z L. To Prove : The UJ are = . Proof: Superpose O ABCD upon O LMNO, so the equal angles A and L coincide, 4D falling along LO and ^4B along LM. Point D will coincide with point O [^4D = LO (Hyp.)]. Point B will coincide with point Jf [AB = jf (Hyp.)]. BC and MN are both || to LO (?) (126). /. BC falls along JOT (?) (92). CD and 2VO are both || to LM (?). /.CD falls along .zvo (?). Hence C will fall exactly upon N (38). /. the figures coincide, and are equal (?) (28). Q.E.D. 140. THEOREM. Two rectangles are equal if the base and altitude of one are equal respectively to the base and altitude of the other. (See 139.) 141. THEOREM. The diagonals of a rhombus (or of a square) are perpendicular to each other, bisect each other, and bisect the angles of the rhombus (or of the square). Given : Rhombus ABCD ; di- agonals AC and BD. To Prove: AC _L to BD; and BD bisect each other ; they bisect A DAB, ABC, etc. A B Proof : Point A is equally distant from B and D (?) (128). Point C is equally distant from B and D (?). .'.^CistoBD (?) (70). Q.E.D. BOOK T 49 Also AC and BD bisect each other (?) (137). Q.E.D. The A at A are = (?) (66, II). Etc. Q.E.D. The proof if the figure is a square is exactly the same. 142. THEOREM. The line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. Given: A ABC; M, the B midpoint of AB ; P, the midpoint of BC; line MP. To Prove : MP and MP = AC. to AC M Proof: Suppose AR is drawn through A, II to BC *~ and meeting MP produced at B. In A ABM and BPM, AM = BM (Hyp.); Zx = Ze (?) and Z o = Z B (?) (97). .'. A ABM = A BPM (?) (54). Hence, AR = BP (?). But BP = PC (?). .-. AR = PC (?). .-. ACPR is a O (?) (135). Hence RP or MP is II to AC (?). Q.E.D. Also, RP = AC (?) (130). But MP = RM (?) (27). .-. MF = J RP=\ ^C(Ax. 6). Q.E.D. 143. THEOREM. The line bisecting one side of a triangle and parallel to a second side, bisects also the third side. Given: A ABC-, MP bisecting AB and II to AC. To Prove : MP bisects BC also. Proof : Suppose MX is drawn from M, the midpoint of AB to X, the midpoint of BC. MX is II to AC (142) ; but MP A is II to AC (Hyp.). .-. MX and MP coincide (?). That is, MP bisects BC. BOBBINS' PLANE GEOM. 4 Q.E.D. r>0 PLANE GEOMETRY 144. THEOREM. The line bisecting one leg of a trapezoid and parallel to the base bisects the other leg, is the median, and is equal to half the sum of the bases. Given: Trapezoid ABCD\ M, the midpoint of AB; MP II to AD, meeting CD at P. To Prove: I. P is the midpoint of CD. II. MP is the median. III. 3fP= Proof: I. Draw diagonal BD, meeting MP at R. MP is II to BC (94). In A ABD, MR bisects BD (143). In ABDC, RP bisects CD (?) (143). That is, P is the midpoint of CD. II. MP is the median (Def. 129). III. MR = 1 AD (142) and RP = \ BC (?). /. MP = I {AD + J3C) (Ax. 2). Q.E.D. 145. THEOREM. The angles adjoining each base of an isosceles trapezoid are equal. Given : Trapezoid AC ; AB = CD. To Prove : Z A = Z D and Z ABC = Z C. Proof : Suppose BX is drawn through B and II to CD. BX = CD (130); AB = CD (Hyp.). A~ ""x"" ~b .-. AB = BX (?), and Z A = Z a (?), and Z a = Z D (?) (98). /.Z A = Z D (Ax. 1). Again, Z O is supp. of Z D (?). Etc. Q.E.D. 146. THEOREM. If the angles at the base of a trapezoid are equal, the trapezoid is isosceles. Given: (?). To Prove: (?). Proof: Suppose BX is drawn II to CD. Z a = Z D (?); Z A = Z D (?). /. Z A = Z a (?). .'. AB = .BJT (?). Etc. BOOK I 51 NOTE. The verb "to intersect" means merely " to cut." In geometry, the verb ""to intercept " means " to include between" Thus the statement " A B and CD intercept X Y on the line EF" really means, " AB and CD A' intersect EF and include XY, a part of EF, between them." 147. THEOREM. Parallels intercepting equal parts on one trans- versal intercept equal parts on any transversal. Given : Us AB, CD, EF, GH, U intercepting equal parts AC, CE, EG, GI, on the transversal A/ \B AI, and cutting transversal BJ. I \ To Prove : BD = DF = FH c/ \D Proof: The figure ABFE is atrapezoid (?). CD bisects AE and is II to EF (Hyp.). .'. D is midpoint of BF (?). _ That is, BD = DF. Similarly, CDHG is a trapezoid and EF bisects DH (?). That is, DF= FH. Similarly, FH = HJ. . . BD = DF = FH = HJ (Ax. 1). Q.E.D. 148. THEOREM. The midpoint of the hypotenuse of a right triangle is equally distant from the three vertices. Given : Rt. A ABC; M, the midpoint of hypotenuse AB. To Prove : AM = CM = BM. Proof: Suppose MX is drawn II to BC, meeting AC at X. X is midpoint of AC (?) (143). MX is to AC (95). That is, MX is J_ to AC at its midpoint, and AM = MC (?) (67). But AM=BM (Hyp.). .'.AM=CM = BM (Ax. 1). Q.E.D. C 52 PLANE GEOMETRY 149. THEOREM. The median of a trapezoid is parallel to the bases and equal to half their sum. [This is another form of stating the theorem of 144.] 150. THEOREM. The perpendiculars from the vertices of a triangle to the opposite sides meet in a point. Given : A ABC, AX J_ to BC, BY J_ to AC, and CZ _L to AB. To Prove : These three Js meet in a point. Proof : Through A suppose RS drawn II to BC ; through B, TS II to AC ; through C, RT II to AB, forming A RST. The figure ABCR is a O (Const.) and ABTC is a O (?). .'. RC = AB and CT = AB (?) (130). .*. RC = CT (Ax. 1). Now cz is JL to RT (?) (95). That is, CZ is _L to RT at its midpoint. Similarly AX is _L to RS at its midpoint. And BY is _L to TS at its midpoint. Therefore AX, BY, CZ meet at a point (?) (85). Q.E.D. Ex. 1. Draw the three altitudes of an obtuse triangle and prolong them until they meet. Ex. 2. Prove that each of the three outer triangles in the figure of 150 is equal to &ABC. Ex. 3. Prove that any altitude of A RST is double the parallel alti- tude of A ,1 BC. [Use H3 and 130.] BOOK I 53 151. THEOREM The point at which two medians of a triangle inter- sect is two thirds the distance from either vertex of the triangle to the midpoint of the opposite side. Given: A ABC, BD and CE two medians intersecting at O. To Prove : HO = 5 BD and CO = | CE. and 10 =OE (?) (137). .'. BH= HO = OD and C/= 7O= OJ? (Ax. 1). That is, BO = 2 OD = f Z>, and CO=2-EO=%CE. Q.E.D. 152. THEOREM. The three medians of a triangle meet in a point which is two thirds the distance from any vertex to the midpoint of the opposite side. Proof : Suppose AX is the third median of A ABC and meets BD at o'. Then O f = f J3D and B0 = f BD (?) (151). .-. BO' = BO (?). That is O' coincides with O and the three medians meet at o which is J the distance, etc. Q.E.D. Ex. 1. In the figure of 151, prove EH = $ AO = D7,by 142. Ex. 2. In the figure of 151, if BC> A C, prove the angle EEC obtuse. [Use 87.] Ex. 3. If one angle of a rhombus is 30, find all the angles of the four triangles formed by drawing the diagonals. Ex. 4. Show that any trapezoid can be divided into a parallelogram and a triangle by drawing one line. Ex. 5. Prove that every right triangle can be divided into two isos- celes triangles by drawing one line. [Use 148.] 54 PLANE GEOMETRY POLYGONS 153. A polygon is a portion of a plane bounded by straight lines. The lines are called the sides. The points of intersection of the sides are the vertices. The angles of a polygon are the angles at the vertices. 154. The number of sides of a polygon is the same as the number of its vertices or the number of its angles. An exterior angle of a polygon is an angle without the polygon, between one side of the polygon and another side prolonged. 155. An equilateral polygon has all of its sides equal to one another. An equiangular polygon has all of its angles equal to one another. 156. A convex polygon is a polygon no side of which if produced will enter the surface bounded by the sides of the polygon. A concave polygon is a polygon two sides of which if produced will enter the polygon. EQUILATERAL EQUIANGULAR CONCAVE, OR CONVEX POLYGONS RE-ENTRANT NOTE. A polygon may be equilateral and not be equiangular ; or it may be equiangular and not be equilateral. The word " polygon " is usually employed to signify convex figures. 157. Two polygons are mutually equiangular if for every angle of the one there is an equal angle in the other and similarly placed, Two polygons are mutually equilateral, if for every side of the one there is an equal side in the other, and similarly placed. BOOK I 55 158. Homologous angles in two mutually equiangular polygons are the pairs of equal angles. Homologous sides in two polygons are the sides between two pairs of homolo- gous angles. 159. Two polygons are equal if they are mutually equi- angular and their homologous sides are equal ; or if they are composed of triangles, equal each to each and similarly placed. (Because in either case the polygons can be made to coincide.) 160. Two polygons may be mutually equiangular without being mutually equilateral ; also, they may be mutually equilateral without being mutually equiangular except in the case of triangles. The first two figures are mutually equilateral but not mutually equi- angular. The last two figures are mutually equiangular but not mutu- ally equilateral. 161. A 3-sided polygon is a triangle. A 4-sided polygon is a quadrilateral. A 5-sided polygon is a pentagon. A 6-sided polygon is a hexagon. A 7-sided polygon is a heptagon. An 8-sided polygon is an octagon. A 10-sided polygon is a decagon. A 12-sided polygon is a dodecagon. A 15-sided polygon is a pentedecagon. An resided polygon is called an w-gon. Ex. Draw a pentagon and all the possible diagonals from one vertex. How many triangles are formed? Draw a decagon and the diagonals from one vertex. How many triangles are thus formed ? Construct a 20- gon and the diagonals from one vertex. How many triangles are formed? 56 PLANE GEOMETRY 162. THEOREM. The sum of the interior angles of an n-gon is equal to (n-2) times 180. Given : A polygon having n sides. To Prove : The sum of its interior A = (n - 2) 180. Proof : By drawing all pos- sible diagonals from any vertex it is evident that there will be formed (n 2) triangles. The sum of the A of one A = 180 (?) (110). .-. the sum of the A of (rc-2) A = (w-2) 180 (Ax. 3). But the sum of the A of the triangles = the sum of the A of the polygon (Ax. 4). .-. sum of A of the polygon = (n 2) 180 (Ax. 1). Q.E.D. 163. COR. The sum of the interior angles of an n-gon is equal to iSon - 360. 164. COR. Each angle of an equiangular n-gon = 1 * MV 165. COR. The sum of the angles of any quadrilateral is equal to four right angles. 166. COR. If three angles of a quadrilateral are right angles, the figure is a rectangle. Ex. 1. How many degrees are there in each angle of an equiangular pentagon ? of an equiangular pentedecagon ? of a 30-gon ? Ex. 2. If two angles of a quadrilateral are right angles, what is true of the other two ? Ex. 3. How many sides has that polygon the sum of whose interior angles is equal to 20 rt. <4? Ex. 4. How many sides has that equiangular polygon each of whose angles contains 160? Ex. 5. Tf in the figure of 105, Za = 65, how many degrees are there in each of the other angles of the figure ? BOOK 1 57 167. THEOREM. If the sides of a polygon be produced, in order, one at each vertex, the sum of the exterior angles of the polygon will equal four right angles, that is, 360. Given : A polygon with sides prolonged in succession forming the several exterior angles a, two such lines. To Prove : AB = RS and AB II to RS. Proof : Draw AR and BS. AO = OS and BO = OR (Hyp.). .-. ABSR is aO(?) (138). .'. AB = RS (?) and AB is II to RS (?). Q.E.D. 175. THEOREM. If a diagonal of a quadrilateral bisects two of its angles, this diagonal is an axis of symmetry. Given : Quadrilateral ABCD-, AC a diagonal bisect- ing Z BAD and Z BCD. To Prove : ABCD symmet- rical with respect to AC. Proof : In A ABC and ADC, AC= AC (?). Z BAG = Z DAC (?) and Z BCL4 =Z DCM (?). Hence A ABC = A ADC (?). /. ^4(7 is an axis of symmetry (?). Q.E.D. 176. COR. The diagonal of a square or of a rhombus is an axis of symmetry. (Why ?) 177. COR. The diagonal of a rectangle or of a rhomboid is not an axis of symmetry. (Why not?) Ex. 1. Is the altitude of an equilateral triangle an axis of symmetry? Ex. 2. Is the altitude of an isosceles triangle an axis of symmetry? Ex. 3. Are all altitudes of all triangles axes of symmetry? Ex. 4. Has an isosceles trapezoid an axis of symmetry ? 60 PLANE GEOMETRY 178. THEOREM. If a figure is symmetrical with respect to two per- pendicular axes, it is symmetrical with respect to their intersection as a center. Given: Figure MN sym- metrical with respect to the J_ axes XX 1 and YY r which intersect at O. To Prove : Figure MN is x- symmetrical with respect to O as a center. Proof: Take any point P in the boundary. Draw PB -L to FF', intersecting YT 1 at A and meeting the boundary at B. Draw BR J_ to XX 1 , intersecting XX 1 ^ C and meet- ing the boundary at R. Draw AC, OP, OR. [The demonstration is accomplished by proving ROP a straight line, bisected at O.] PB is II to XX 1 and BR is I! to FF ; (?) (93). Hence ABCO is a O (?). .-. BC=AO (?). But BC=CR (?) (172). .'. AO= CR (Ax. 1). .'. ACRO is a O (?) (135). /. RO is = and II to AC (?). Similarly CO is = and II to AP ; hence ACOP is a O (?). /. PO is = and II to AC (?). /. POR is a straight line (?) (92) and PO OR (?). But P is any point, so POR is any line through O. Hence O is a center of symmetry (171). Q.E.D. LOCUS 179. The locus of a point is the series of positions the point must occupy in order that it may satisfy a given condition. It is the path of a point whose positions are limited or defined by a given condition, or given conditions. 180. Explanatory. I. If a point is moving so that it is always one inch from a given indefinite straight line, the BOOK I 61 point may occupy any position in either of two indefinite lines, one inch from the given line, parallel to it, and one on either side of it. And this point cannot occupy any position which is not in these lines- Hence, the locus of points at a given distance from a given line is a pair of parallels to the given line, one on either side of it, and at the given distance from it. II. If a point is moving so that it is always equally distant from two parallels, it must move in a third parallel midway between them. Hence, the locus of points equally distant from two parallels is a third parallel midway between them. III. The method of proving that a certain line or group of lines is the locus of points satisfying a given condition, consists in proving that every point in the line fulfills the given requirement, and that there is no other point that fulfills it. In the above illustrations it is evident that every point in the lines which were called the "locus," did fulfill the conditions of the case. It is also evident that there is no point outside these " loci " which does so fulfill the conditions. That is, these "loci" contain all the points described and no others. IV. THEOREM. The locus of points equally distant from the ex- tremities of a line is the perpendicular bisector of the line. Proof: Every point in the _L bisector of a line is equally distant from its extremities (67). And also, there is no point outside the _L bisector which is equally distant from the extremities of a line (69). Hence it is the locus of points equally distant from the extremities of the line. V. THEOREM. The locus of points equally distant from the sides of an angle is the bisector of the angle. Proof : Like the preceding proof. [Use 79, 80, 81.] VI. The locus of the vertices of all the isosceles triangles that can be constructed on a given base is the perpendicular bisector of the base. (Same as IV.) 62 PLANE GEOMETRY CONCERNING ORIGINAL EXERCISES 181. In the original work which this text contains, the pupil is expected to state the hypothesis and conclusion of each theorem, and apply them to an appropriate figure. He is expected to state completely and logically the proof, giving a correct reason for every declarative statement. In many of these exercises, suggestions are made and such assistance is given as experience has shown average pupils require. This is done in order that the learner may be en- couraged toward definite accomplishment, which is one of the greatest incentives to further effort. To apply the knowledge acquired from the preceding pages is now the student's task. His fascination for this science will depend largely upon the success of his efforts at proving originals. Therefore, many obstacles will be removed or modi- fied, and 110 trouble will be spared in making the mastery of this department of geometry both agreeable and profitable. The student should not draw a special figure for a general proposition. That is, if "triangle " is specified, he should draw a scalene and not an isosceles or a right triangle ; and if "quadrilateral" is mentioned, he should draw a trapezium and not a parallelogram or a square. SUMMARY. GENERAL DIRECTIONS FOR ATTACKING EXERCISES 182. A triangle is proven isosceles by showing that it contains two equal sides, or two equal angles. , 183. A triangle is proven a right triangle by showing that one of its angles is a right angle, or two of its angles are complementary, or one of its angles is equal to the sum of the other two. 184. Right triangles are proven equal, by showing that they have : (1) Hypotenuse and acute angle of one = etc. (2) Hypotenuse and leg of one = etc. (3) The legs of one = etc. (4) Leg and adjoining angle of one = etc. (5) Leg and opposite angle of one = etc. BOOK I 63 185. Oblique triangles are proven equal, by showing that they have : (1) Two sides and the included angle of one = etc. (2) One side and the adjoining angles of one = etc. (3) Three sides of one = etc. 186. Angles are proven equal, by showing that they are : (1) Equal to the same or to equal angles. (2) Halves or doubles of equals. (3) Vertical angles. (4) Complements or supplements of equals. (5) Homologous parts of equal figures. (6) Base angles of an isosceles triangle. (7) Corresponding angles, alternate-interior angles, etc. of parallels. (8) Angles whose sides are respectively parallel or perpendicular. (9) Third angles of triangles which have two angles of one = etc. 187. Lines are proven equal, by showing that they are : (1) Equal to the same or to equal lines. (2) Halves or doubles of equals. (3) Distances to the ends of a line from any point in its perpendicu- lar bisector. (4) Homologous parts of equal figures. (5) Sides of an isosceles triangle. (6) Distances to the sides of an angle from any point in its bisector. (7) Opposite sides of a parallelogram. (8) The parts of one diagonal of a parallelogram made by the other. 188. Two lines are proven perpendicular, by showing that they : (1) Make equal adjacent angles with each other. (2) Are legs of a right triangle. (3) Have two points in one, each equally distant from the ends of the other. 189. Two lines are proven parallel, by : (1) The customary angle-relations of parallel lines. (2) Showing that they are opposite sides of a parallelogram. (3) Showing that they are parallel or perpendicular to a third line. 190. Two lines, or two angles, are proven unequal by the usual axioms and theorems pertaining to inequalities. [See especially, Ax. 5*; Ax. 12; 68; 75; 76, III; 77; 86 ; 87*; 90 109*; 122*; 124.] * These refer to angles. 04 PLANE GEOMETRY ORIGINAL EXERCISES 1. A line cutting the equal sides of an isosceles triangle and parallel to the base forms another isosceles triangle. [Use 186 (7), and 182.] 2. The bisectors of the equal angles of an isosceles triangle form, with the base, another isosceles triangle. [Use 186 (2).] H 3. If the exterior angles at the base of a triangle are equal, the triangle is isosceles. [186 (4).] 4. If from any point in the base of an isosceles tri- angle a line be drawn parallel to one of the equal sides and meeting the other side, an isosceles triangle .< will be formed. To Prove : A DEC isosceles. 5. If the median of a triangle is perpendicular to the base, the triangle is isosceles. [Use 187 (3).] 6. If a line through the vertex of a triangle and parallel to the base, makes equal angles with the sides, the triangle is isosceles. Given : /La = Z x, etc. 7. The median of an isosceles triangle is per- pendicular to the base. [Use 188 (3).] 8. The bisectors of two supplementary-adja- cent angles are perpendicular to each other. Proof: Z.40+ZjB<9C = 180(?); %Z.AOB + =9Q (Ax. 3). \ ^AOB = ZROB; etc. 9. The bisectors of two adjoining- interior angles of two parallels meet at right angles. Proof: /.MAC + /.MCA = 90 as in No. 8. Then use 183. 10. If the bisector of an exterior angle of a triangle is parallel to the base, the triangle is isosceles. 11. If the sum of two angles of a triangle is equal to the third, the triangle is a right triangle. [Use 110.] BOOK I 65 12. If the median of a triangle is equal to half the side to which it is drawn, it is a right triangle. Given : MA = MB = MC. c To Prove : A ABC a rt. A. Proof : /.A = /.ACM (?) ; Z.E = /.BCM (?). /. by adding, etc. (Use Ax. 2 and 183.) A*" 13. If from any point in the bisector of an angle a line be drawn parallel to either side of the angle, an isosceles triangle will be formed. [182.] 14. If the bisector of the vertex-angle of a tri- angle is perpendicular to the base, the triangle is isosceles. Proof: The rt. &are= [184 (4)]. Then use 187 (4); etc. 15. Every isosceles right triangle can be divided by one line into two isosceles right triangles. M 16. The diagonals of a rhombus divide the figure into four equal right triangles. [141.] 17. If a line be drawn perpendicular to the bisec- tor of an angle terminating in the sides, the right triangles formed will be equal. 18. If from each point at which a transversal inter- sects two parallels a perpendicular to the other paral- lel be drawn, two equal right triangles will be formed. 19. If two perpendiculars be drawn to the upper base of a parallelogram from the extremi- R__D T C ties of the lower base, two equal right triangles will be formed. 20. The perpendiculars to the equal sides of an isosceles triangle from the opposite vertices form two pairs of equal right triangles. 21. If two intersecting lines have their extremities in two parallels and their point of intersection bisects one of them, it bisects the other also. Given: AO = EO-, etc. BOBBINS' PLANE GEOM. 5 C 66 PLANE GEOMETRY 22. If two adjacent sides of a quadrilateral are equal and the diagonal bisects their included angle, the other two sides are equal. 23. If a diagonal of a quadrilateral bisects two of its angles, the quadrilateral has two pairs of equal sides. 24. The altitudes of an isosceles triangle upon the legs are equal. 25. If a triangle has two equal altitudes, it is isosceles. 26. The diagonals of a rectangle are equal. 27. The medians drawn from the ends of the base of an isosceles triangle are equal. 28. The three lines joining the midpoints of the sides of a triangle divide the triangle into four equal triangles. [Use 142 and 132.] 29. The bisector of the vertex-angle of an isosceles triangle bisects the base at right angles. [Use 185(1); 27; 50.] 30. The bisectors of the equal angles of an isos- celes triangle (terminating in the equal sides) are equal. [Use 185 (2).] 31. The median to the base of an isosceles triangle bisects the vertex- angle. [Use 185 (3).] 32. The diagonals of an isosceles trapezoid are equal. [Use 145.] 33. The diagonals of an isosceles trapezoid divide the figure into four triangles, of which one pair is isosceles and the other pair is equal. 34. What is the complement of an angle containing 35? 80? 75 25'? 8 18'? 35. What is the supplement of 50? 100? 148? 113 48' ? 36. In a right triangle ABC, if Z. A is 47, find Z B. 37. In an isosceles triangle /. A = Z. B = 80 ; find Z C. 38. In A ABC, if Z A = 25, Z B = 88, find Z C. Find the exterior angle at A . BOOK I 67 39. In A ABC, \i^.A- 40, Z. B = 70 40' ; find C and the exterior angle at B. 40. The vertex-angle of an isosceles triangle is 44. Find each base angle. 41. How many degrees are there in the sum of the angles of a penta- gon ? of a decagon ? of a 9-gon ? 42. How many degrees are there in each angle of an equiangular hexa- gon ? of an equiangular octagon ? 43. How many degrees are there in each exterior angle of an equian- gular pentagon V hexagon ? dodecagon V 16-gon ? 44. If one acute angle of a right triangle is double the other, how many degrees are there in each ? [Denote the less Z by x.~] 45. If the acute angles of a right triangle are equal, how many degrees are there in each ? 46. If one acute angle of a right triangle is five times the other, how many degrees are there in each ? 47. If a base angle of an isosceles triangle is 60, find the vertex-angle. What kind of triangle is this ? 48. If the vertex-angle of an isosceles triangle is 60, find each base angle. What kind of triangle is this ? 49. If the vertex-angle of an isosceles triangle equals twice the sum of the two base angles, how many degrees are there in each angle? 50. If the vertex-angle of an isosceles triangle equals four times the sum of the base angles, find each angle. 51. If the vertex-angle of an isosceles triangle is half each of the base angles, find each angle. 52. If one angle of a parallelogram is 54, how many degrees are there in each of the remaining angles? 53. If a transversal cuts two parallels making one pair of alternate- interior angles each 40, how many degrees are there in each of the other six angles formed ? 54. Find the three angles formed by the bisectors of the angles of a triangle whose angles are 44, 62, and 74. 55. If Z A of A ABC is 33 and the exterior angle at C is 110, find Z B. 68 PLANE GEOMETRY 56. If two angles of a triangle are 80 and 55, how many degrees are there in the angle formed by their bisectors? 57. The vertex-angle of an isosceles triangle is one third of either ex- terior angle at the extremities of the base. Find each angle of the triangle. 58. If two angles of a triangle are 30 and 40, how many degrees are there in the angle formed by the bisector of the third angle and the altitude from the same vertex ? Solution : Z x = Z A BS ABD=\ ABC - comp. of Z A. 59. Theorem. The angle between the altitude of a triangle and the bisector of the angle at the same vertex equals half the difference of the other angles of the triangle. Proof: Z x = Z ABS - Z ABD = \ (180 - /.A - Z C) - (90 - Z A) = etc. 60. Theorem. The exterior angle at the base of an isosceles triangle equals half the vertex- angle plus 90. Proof : Zz = Za + Zr = etc. 61. If in A ABC, Z BAG = 80, Z ABC = 30, find the angle formed by the bisectors of the exte- rior angles at A and B. Solution : Z x = 180 - /.BAD-/. ABD] Z BAD = \(\m- Z.A}; etc. 62. The angle formed by the bisectors of two ex- terior angles of a triangle equals half the sum of the interior angles at the same vertices. 63. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter leg. [Denote the less Z by x. Find the other. Draw the median from vertex of rt. Z. Prove one A formed equilateral.] 64. If one angle of a triangle is double another, the line from the third vertex, making with the longer adjacent side an angle equal to the less given angle, divides the triangle into two isosceles triangles. 65. The bisector .of an angle bisects, if pro- duced, the vertical angle also. Given: Z DON = Z BON. BOOK I 69 66. A line perpendicular to the bisector of an angle at the vertex bisects its supplementary-adjacent angle. _ Given : Z a= /. b and OF to ON. (Use 48.) 67. If the bisectors of two adjacent angles are perpendicular, the angles are supplementary. A" Given: Za=Z6; Zar = Zz; 68. The bisectors of any two adjoining angles of a parallelogram meet at right angles. [Use 136 ; 183.] 69. If from any point in the base of an isosceles triangle perpendiculars to the equal sides be drawn, they will make equal angles with the base. [Use 114; 48.] 70. If a line be drawn through the vertex of an angle and perpendicular to the bisector of the angle, it will make equal angles with the sides. To Prove :. ^ r = z s. 71. The bisector of the exterior angle at the ver- tex of an isosceles triangle is parallel to the base. Proof: Z DCB = 2 /.A (?) and =2 Z. DCR (?). Etc. 72. The line through the vertex of an isosceles tri- angle, parallel to the base, bisects the exterior angle. 73. Parallel lines are everywhere equally distant. Given : ||. A C and BD; AB and CD & to A C. To Prove : AB = CD. (Use 93; 130.) 74. If two lines in a plane are everywhere equally distant, they are parallel. [Use 93 ; 135.] 75. If the diagonals of a parallelogram are equal, the figure is a rectangle. [Use A ABC and DBC; 185 (3); 50.] ' 76. The perpendiculars upon a diagonal of a parallelogram from the opposite vertices are equal. [184 (1).] A' TO PLANE GEOMETRY* 77. The perpendiculars to the legs of an isosceles triangle from the midpoint of the base are equal. 78. State and prove the converse of No. 77. 79. If AB = LM and AL = EM, Z B = Z L audZBAO = ^OML2MdBO=OL. 80. Any line terminated in a pair of oppo- c site sides of a parallelogram and passing " through the midpoint of a diagonal is bisected by this point. To Prove : RO = OS. 81. The midpoint of a diagonal of a paral- lelogram is a center of symmetry. 82. If the base angles of a triangle be bisected and through the intersection of the bisectors a line be drawn parallel to the base and terminating in the sides, this line will be equal to the sum of the parts of the sides it meets, between it and the base. A ""^c 83. In two equal triangles, homologous medians are equal. Homolo- gous altitudes are equal. Homologous bisectors are equal. 84. If two parallel lines are cut by a transversal, the two exterior angles on the same side of the transversal are supplementary. 85. If from a point a perpendicular be drawn to each of two parallels they will be in the same line. [Draw a third II through the point.] 86. One side of a triangle is less than the sum of the other two sides. 87. The sum of the sides of any polygon ABODE is greater than the sum of the sides of triangle A CE. 88. If X is a point in side AB of A ABC, AB + BOAX + XC. 89. In the figure of No. 88 Z AXO Z B. 90. If lines be drawn from any point within a triangle to the ends of the base, they will include an angle which is greater than the vertex angle of the triangle. [Use 109 with figure of 75.] 91. Any point (except the vertex) in either leg of an isosceles triangle is unequally distant from the ends of the base. B 92. If two sides of a triangle are unequal and the median to the third side be drawn, the angles formed with the base will be unequal. [Use 87.] 93. State and prove the converse of No. 92. BOOK I 71 94. If the side LM, of equilateral triangle LMN, be produced 'to />, and PN be drawn, Z PNL > L > Z P. Also PL>PN> LN. 95. If from any point within a triangle lines be drawn to the three vertices: (1) Their sum will be less than the sum of the sides of the triangle. [Use 75 three times.] (2) Their sum will be greater than half the sum of the sides of the triangle. [Use Ax. 12 three times.] 96. The sum of the diagonals of any quadri- lateral is less than the sum of the four sides; but greater than half that sum. 97. The line drawn from any point in the base of an isosceles triangle to the opposite vertex is less than either leg. 98. The bisectors of a pair of corresponding angles are parallel. [Use 98 ; 189, etc.] 99. If two lines are cut by a transversal and the exterior angles on the same side of the transversal are supplementary, the lines are parallel. 100. The bisectors of a pair of vertical angles are in the same straight line. 101. If one angle of a parallelogram is a right angle the figure is a rectangle. 102. The bisectors of the angles of a trapezoid form a quadrilateral two of whose angles are right angles. ^ 103. The bisectors of the four interior - ^/ angles formed by a transversal cutting two parallels form a rectangle. [Prove each Z. of LMPQ a rt. ^.] 104. The bisectors of the angles of a par- allelogram form a rectangle. 105. The bisectors of the angles of a rectangle form a square. [In order to prove EFGH equilat- eral, the & AHB and CDF are proven equal and isosceles; similarly & BG C and AED.] 72 PLANE GEOMETRY 106. The lines joining a pair of opposite vertices of a parallelogram to the midpoints of the opposite sides are . }ual and parallel. [Prove BCEF a O.] 107. If the four midpoints of the four halves of the diagonals of a parallelogram be joined in order, another parallelogram will be formed. 108. If the points at which the bisectors of the equal angles of an isosceles triangle meet the opposite sides, be joined by a line, A it will be parallel to the base. 109. If two angles of a quadrilateral are supple- mentary, the other two are supplementary. [Use 165.] 110. If from any point in the base of an isosceles triangle parallels to the equal sides be drawn, the sum of the sides of the parallelogram formed will be equal to the sum of the legs of the triangle. To Prove: XY + YC + CZ + XZ = AC + EC. 111. If one of the legs of an isosceles triangle be produced through the vertex its own length, and the extremity be joined to the nearer end of the base, this line will be perpendicular to the base. [Use 183.] 112. If the middle point of one side of a triangle is equally distant from the three vertices, the triangle is a right triangle. [Proof and figure same as for No. 111.] 113. If through the vertex of the right angle of a right triangle a line be drawn parallel to the hypotenuse, the legs of the right triangle wiU bisect the angles formed by this parallel and the median drawn to the hypotenuse. [Use 148 ; 97 ; etc.] -A 114. Any two vertices of a triangle are equally distant from the median from the third vertex. 115. If from any point within an angle per- pendiculars to the sides be drawn, they will include an angle which is the supplement of the given angle. M BOOK I 73 116. The lines joining (in order) the midpoints of the sides of a quadrilateral form a parallelogram the sum of whose sides is equal to the sum of the diagonals of the quadrilateral. [Use 142.] 117. The lines joining (in order) the mid- R points of the sides of a rectangle form a rhombus. [Draw the diagonals.] 118. If a perpendicular be erected at any point in the base of an isosceles triangle, meeting one leg, and the other leg produced, another isosceles triangle will be formed. [Z o and Z. S are complements of = A A and B (?). Etc.] 119. The difference between two sides of a triangle is less than the third side. 120. The bisectors of two exterior angles of a triangle and of the interior angle at the third ver- tex meet in a point. 121. The bisectors of the exterior angles of a rectangle form a square. 122. If lines be drawn from a pair of oppo- site vertices of a parallelogram to the mid- points of a pair of opposite sides, they will trisect the diagonal joining the other two ver- tices. [Prove AECF&EJ&nd use 142 in & DYCandABX.] 123. If two medians of a triangle are equal, the triangle is isosceles. [Use 151. AO - OB (Ax. 3). Hence prove & AEO and DBO equal.] 124. How many sides has the polygon the sum of whose interior angles exceeds the sum of its exterior angles by 900 ? 125. If the vertex-angle of an isosceles triangle is twice the sum of the base angles, any line per- pendicular to the base forms with the sides of the given triangle (one .side to be produced) an equilateral triangle. [Use 121.] O B 74 PLANE GEOMETRY 126. The lines bisecting two interior angles that a transversal makes with one of two parallels cut off equal segments on the other parallel from the point at which the transversal meets it. [The & formed are isosceles.] 127. The bisector of the right angle of a right triangle is also the bisector of the angle formed by the median and the altitude drawn from the same vertex. To Prove : Z MCS = LCS. Proof : Z A CS = Z BCS (?) ; Z A CM = Z BCL (?) . Now use Ax. 2. 128. If through the point of intersection of the diagonals of a par- allelogram, two lines be drawn intersecting a pair of opposite sides (produced if necessary), the intercepts on these sides will be equal. 129. If ABC is a triangle, BS is the bisector of Z ABC, and AM is parallel to BS meeting BC produced, at M, the triangle ABMia isosceles. 130. If ABC is a triangle and BS is the bisector of exterior Z ABR and AM is II to BS meeting BC at M, A ABM is isosceles. 131. If A) B, C, and D are points on a straight line and AB = BC, the sum of the perpendiculars from A and C to any other line through D is double the perpendicular to that line from B. [Use 147 ; 144.] 132. If on diagonal BD, of square ABCD, BE be taken equal to a side of the square, and EP be drawn perpendicular to BD meeting AD at P, AP = PE = ED. [Draw BP.~] 133. If in A ABC, Z A is bisected by line meet- ing BC at M, AB>BM and AC> CM. [Use 108; 123.] 134. It is impossible to draw two straight lines from the ends of the base of a triangle terminating in the opposite side, so that they shall bisect each other. [Use 138.] 135. If ABC is an equilateral triangle and D, E, F are points on the sides, such that AD . BE = CF, triangle DEF is also equilateral. [Prove the three small & =.] F C BOOK I 75 136. If A BCD is a square and E, F, <2, H are points on the sides, such that AK = HF - <'<; = /)//, EFGH is a square. [First, prove EFGH equilateral ; then one Za rt. Z. .] 137. If ABC is an equilateral triangle and each side is produced (in order) the same distance, so that A D = BE = CF, the triangle DEF is equilateral. 138. If A BCD is a square and the sides be produced (in order) the same distance, so that AE = BF = CG. = DH, the figure EFGH will be a square. 139. The two lines joining the midpoints of the opposite sides of a quadrilateral bisect each other. [Join the 4 midpoints (in order), etc.] 140. If two adjacent angles of a quadrilateral are right angles, the bisectors of the other angles are perpendicular to each other. 141. If two opposite angles of a quadrilateral are right angles, the bisectors of the other angles are parallel. 142. Two isosceles triangles are equal, if : (1) The base and one of the adjoining angles in the one are equal respectively to the base and one of the adjoining angles in the other. (2) A leg and one of the base angles in the one are equal respectively to a leg and one of the base angles in the other. (3) The base and vertex-angle in one are equal to the same in the other. (4) A leg and vertex-angle in one are equal to the same in the other. (5) A leg and the base in one are equal to the same in the other. 143. If upon the three sides of any triangle equilateral triangles be constructed (externally) and a line be drawn from each vertex of the given triangle to the farthest vertex of the opposite equi- lateral triangle, these three lines will be equal. Proof: Z EAC =Z BA F (?). Add to each R of these, /.CAB. :. Z EAB = Z CAF (?). Then prove & EAB and CAF equal. Similarly, A CA D = A CEB. Etc. 144. If two medians be drawn from two vertices of a triangle and produced their own length beyond the opposite sides and these ex- tremities be joined to the third vertex, these two lines will be equal, and in the same straight line. [Draw MP and use 142.] 7t) PLANK (JKOMKTRY 145. The median to one side of a triangle is less than half the sum of the other two sides. Proof: (Fig. of No. 144.) Produce median BM its own length to 72, draw RA. Prove RA = CB. Prove RR < AB -f BC, etc. 146. The sum of the medians of a triangle is less than the sum of the sides of the triangle. r> C 147. If the diagonals of a trapezoid / ' are equal, it is isosceles. [Draw DR and CS _L to AB ; and prove rt. A ACS and BDR equal, to A " get Z. x = Z ar.] 148. If a perpendicular be drawn from each vertex of a parallelogram to any line outside the parallelogram, the sum of those from one pair of opposite vertices will equal the sum of those from the other pair. [Draw the diagonals ; use 144.] B 149. The sum of the perpendiculars to the legs of an isosceles triangle from any point in the base equals the altitude upon one of the legs. (That is, the sum of the perpendiculars from any point in the base of an isosceles triangle to the equal sides is constant for every point of the base.) [Prove PE = CF by 184 (1).] A P~ ~c 150. The sum of the three perpendiculars drawn from any point within an equilateral triangle, to the three sides, is constant for all positions of the point. [Draw a line through this point II to one side; draw the altitude of the A -L to this line and side ; prove the sum of the three _ls = this altitude and hence, = a constant.] 151. The line joining the midpoints of one pair of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals bisect each other. To Prove : LM and RS bisect each other. 152. If one leg of a trapezoid is perpendicular to the bases, the mid- point of the other leg is equally distant from the ends of the first leg. [Draw the median.] 153. The median of a trapezoid bisects both the diagonals. 154. The line joining the midpoints of the diagonals of a trapezoid is a part of the median, is parallel to the bases, and is equal to half their difference. BOOK I 77 155. If, in isosceles triangle X YZ, A D be drawn from A, the midpoint of YZ, perpendicular to the base XZ, DZ = \ XZ. [Draw alt. from y.] 156. If ABC is an equilateral triangle, the bisectors of angles B and C meet at D, DE be drawn parallel to AB meeting AC at E, and DF, parallel to BC meeting A C at F, then AE = ED = EF = DF= CF. 157. If A is any point in RS of triangle RST, and B is the midpoint of RA, C the midpoint of AS, D the midpoint of ST, and E the mid- point of TR, then BCDE is a parallelogram. 158. In a trapezoid one of whose bases is ^ double the other, the diagonals intersect at a point two thirds of the distance from each end of the longer base to the opposite vertex. Proof: Take M, the midpoint of AO, etc. 159. If lines be drawn from any vertex of a parallelogram to the mid- points of the two opposite sides, they will divide the diagonal which they intersect, into three equal parts. Proof : Draw the other diagonal and use 151. 160. If the interior and exterior angles at two vertices of a triangle be bisected, a quadrilateral will be formed, two of whose angles are right angles and the other two are supplementary. 161. The angle between the bisectors of two angles of a triangle equals half the third angle plus a right angle. 162. If, in triangle ABC, the bisectors of the interior angle at B and of the exterior angle at C, meet at D, the angle BAG equals twice the angle BDC. 163. The four bisectors of the angles of a quadrilateral form a second quadrilateral whose opposite angles are supplementary. Proof: Extend a pair of opposite sides of the given quadrilateral to meet at X. Bisect the base angles of the new A formed, meeting at O. Then show that Z. O equals one of the A between the given bisectors, and Z O is supplementary to the angle opposite. 164. The sum of the angles at the vertices of a five-pointed star (pentagram) is equal to two right angles. Proof: Draw interior pentagon. Find number of degrees in each of its angles. Hence find Z A, etc. 165. The lines joining the midpoints of the op- posite sides of an isosceles trapezoid are perpendicular to each other. 78 PLANE GEOMETRY 166. If the opposite sides of a hexagon are equal and parallel, the three diagonals drawn between opposite vertices meet in a point. 167. In triangle ABC, AD is perpendicular to BC, meeting it at D; E is the midpoint of AB, and F of A C; the angle EDF is equal to the angle EA F. [Use 148 ; 55.] 168. If the diagonals of a quadrilateral are equal, and also one pair of opposite sides, two of the four triangles into which the quadrilateral is divided by the diagonals are isosceles. 169. If angle A of triangle ABC equals three times angle B, there can be drawn a line AD meeting BC in D, such that the triangles ABD and A CD are isosceles. 170. If E is the midpoint of side BC of parallelogram ABCD, AE and BD meet at a point two thirds the distance from A to E and from Dio B. 171. If in triangle ABC, in which AB is not equal to AC, A C' be taken on AB (produced if necessary) equal to A C, and AB' be taken on AC (produced if necessary) equal to AB, and B'C' be drawn meeting BC at D, then AD will bisect angle BA C. Proof: A ABC = A A B'C' (?) (52). /. their homologous parts are equal. Thus prove &BC'D = A B'CD (54). Etc. 172. If a diagonal of a parallelogram bisects one angle, it also bisects the opposite angle. 173. If a diagonal of a parallelogram bisects one angle, the figure is equilateral. 174. Any line drawn through the point of intersection of the diago- nals of a parallelogram divides the figure into two equal trapezoids. [See 159.] 175. If AR bisects angle A of triangle A BC and A T bisects the ex- terior angle at A, any line parallel to AB, having its extremities in AR and A T, is bisected by A C. 176. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. [See 165.] BOOK II THE CIRCLE 191. A curved line is a line no part of which is straight. 192. A circumference is a curved line every point of which is equally distant from a point within, called the center. 193. A circle is a portion of a plane bounded by a cir- cumference. [O.] 194. A radius is a straight line drawn from the center to the circumference. A diameter is a straight line containing the center, and whose extremities are in the circumference. CIRCUMFERENCE SECANT CENTRAL ANGLE SEMI- CIRCLE CHORD INSCRIBED ANGLE CIRCUMFERENCES RADIUS TANGENT ARC SEMICIRCLES DIAMETER POINT OF CONTACT A secant is a straight line cutting the circumference in two points. A chord is a straight line whose extremities are in the cir- cumference. A tangent is a straight line which touches the circumference at only one point, and does not cut it, however far it may be extended. The point at which the line touches the circum- ference is called the point of contact or the point of tangency. 79 80 PLANE GEOMETRY 195. A central angle is an angle formed by two radii. An inscribed angle is an angle whose vertex is on the cir- cumference and whose sides are chords. 196. An arc is any part of a circumference. A semicircumference is an arc equal to half a circum- ference. A quadrant is an arc equal to one fourth of a circumference. Equal circles are circles having equal radii. Concentric circles are circles having the same center. CIRCLES EXTERNALLY TANGENT 197. A sector is the part of a circle bounded by two radii and their included arc. A segment of a circle is the part of a circle bounded by an arc and its chord. A semicircle is a segment bounded by a semicircuraference and its diameter. 198. Two circles are tangent to each other if they are tan- gent to the same line at the same point. Circles may be tangent to each other internally, if the one is within the other, or externally, if each is without the other. 199- POSTULATE. A circumference can be described about any given point as center and with any given line as radius. Explanatory. A circle is named either by its center or by three points on its circumference, as " the O O," or " the O ABC." The verb to subtend is used in the sense of "to cut off." BOOK TI rtl A chord subtends an arc. Hence an arc is subtended by a chord. An angle is said .to intercept the arc between its sides. Hence an arc is intercepted by an angle. The hypothesis is contained in what constitutes the sub- ject of the principal verb of the theorem. (See 59.) PRELIMINARY THEOREMS 200. THEOREM. All radii of the same circle are equal. (See 192. ) 201. THEOREM. All radii of equal circles are equal. (See 196.) 202. THEOREM. The diameter of a circle equals twice the radius. 203. THEOREM. All diameters of the same or equal circles are equal. (Ax. 3.) 204. THEOREM. The diameter of a circle bisects the circle and the circumference. Given : Any O and a diameter. To Prove : The segments formed are equal, that is, the diameter bisects the circle and the circumference. Proof : Suppose one segment folded over upon the other segment, using the diameter as an axis. If the arcs do not coincide, there are points of the circumference unequally distant from the center. But this is impossible (?) (192). /. the segments coincide and are equal (?) (28). Q.E.D. 205. THEOREM. With a given point as center and a given line as radius, it is possible to describe only one circumference. (See 192.) That is, a circumference is determined if its center and radius are fixed. XOTE. The word " circle " is frequently used in the sense of " cir- cumference." Thus one may properly speak of drawing a circle. The established definitions could not admit of such an interpretation save as custom makes it permissible. Ex. Draw two intersecting circles and their common chord. Draw two circles which have no common chord. Draw figures to illustrate all the nouns defined on the two preceding pages. BOBBINS' PLANE GEOM. 6 82 PLANE GEOMETRY THEOREMS AND DEMONSTRATIONS 206. THEOREM. In the same circle (or in equal circles) equal central angles intercept equal arcs. Given : O O = O C ; Z O = Z C. To Prove : Arc AB = arc LM. Proof : Superpose O o upon the equal O <7, making Z o coincide with its equal, Z C. Point A will fall on i, and point B on M (?) (201). Arc AB will coincide with arc LM (?) (192). .'.AB = LM (?) (28). Q.E.D. 207. THEOREM. In the same circle (or in equal circles) equal arcs are intercepted by equal central angles. [Converse.] Given : O O = O C ; arc AB = arc LM. To Prove : Z = Z C. Proof : Superpose O O upon the equal O C, making the centers coincide and point A fall on point L. Then arc AB will coincide with arc LM and point B will fall on point M. (Because the arcs are =.) Hence OA will coincide with C, and OB with CM (?) (39). /.Z = Z c(?) (28). Q.E.D. Ex. 1. Can arcs of unequal circles be made to coincide? Explain. Ex. 2. If two sectors are equal, name the several parts that must be equal. BOOK II 83 208. THEOREM. In the same circle (or in equal circles) : I. If two central angles are unequal, the greater angle intercepts the greater arc. II. If two arcs are unequal, the greater arc is intercepted by the greater central angle. [Converse.] I. Given : O o = O c ; Z LCM > Z o. To Prove : Arc LM > arc AB. Proof : Superpose O O upon O C, making sector AOB fall in position of sector XCM, OB coinciding with CM. GX is within the angle LCM (Z LCM > Z O). Arc AB will fall upon ZJf, in the position XM (192). .-. arc LM > arc XM (Ax. 5). That is, arc LM > arc AB. Q.E.D. II. Given : (?). To Prove : Z LCM > Z o. Proof : The pupil may employ either superposition, as in I, or the method of exclusion, as in 87. NOTE. Unless otherwise specified, the arc of a chord always refers to the lesser of the two arcs. If two arcs (in the same or equal circles) are concerned, it is understood either that each is less than a semicircumfer- ence, or each is greater. Ex. 1. Two sectors are equal if the radii and central angle of one are equal respectively to the radii and central angle of the other. Ex. 2. If in the figure of 206, arcs AB and LM were removed, how would the remaining arcs compare ? Ex. 3. If in the figure of 208, arcs AB and LM were removed, how would the remaining arcs compare ? 84 PLANE GEOMETRY 209. THEOREM. In the same circle (or in equal circles) equal chords subtend equal arcs. Given : O O = O C ; chord AB = chord LM. To Prove : Arc AB = arc LM. Proof : Draw the several radii to the ends of the chords. In A OAB and CLM, OA = CL, OB CM (?) (201). Chord AB = chord LM (Hyp.). .-. A OAB = A CLM (?). .'. arc AB = arc LM (?) (206). Q.E.D. 210. THEOREM. In the same circle (or in equal circles) equal arcs are subtended by equal chords. Given : O = O C ; arc AB = arc LM. To Prove: Chord AB = chord LM. Proof: Draw the several radii to the ends of the chords. In A OAB and CLM, OA = CL, OB = CM (?) (201). ZO=ZC(?) (207). /.A AOB =A CLM (?). .'. chord AB = chord LM (?). Q.E.D. 211. THEOREM. In the same circle (or in equal circles) : I. If two chords are unequal, the greater chord subtends the greater arc. II. If two arcs are unequal, the greater arc is subtended by the greater chord. I. Given: O O = O C ; chord AB > chord BS. To Prove: Arc AB > arc US. BOOK II Proof : Draw the several radii to the ends of the chords. In A AOB and RCS, AO = EC, BO == SC (?) (201). Chord AB > chord RS (Hyp.). .-.Z O > Z C (?) (87). .*. arc AB > arc #s (?) (208, I). Q.E.D. II. Given: O O = O C; arc ^45 > arc E-8. To Prove : Chord AB > chord RS. Proof : Draw the several radii. In A A OB and RCS, AO = RC, BO = SC (?) (201). But Z o > Z C (?) (208, II). .*. chord AB > chord RS (?) (86). Q.E.D. 212. THEOREM. The diameter perpendicular to a chord bisects the chord and both the subtended arcs. Given : Diameter DR J_ to chord AB in O O. To Prove: I. AM=MB-> II. AR = RB and AD = DB. Proof: Draw radii to the ends of the chord. I. In rt. A OAM and OBM, OA = OB (?), OM=OM (?). .-. AOAM= Ao#jf(?). Hence, AM=MB (?). Q.E.D. II. Z.AOM = ZJ503f (27). .'. AR = RB (?) (206). Also Z^OD = Z BOD (?)(49). .'.AD=DB (?)(206). Q.E.D. 86 PLANE GEOMETRY 213. THEOREM. The line from the center of a circle perpendicular to a chord bisects the chord and its arc. Proof : The same as 212. 214. THEOREM. The perpendicular bisector of a chord passes through the center of the circle. [0 is equidistant from A and B (?) (200). .-. it is in the _L bisector of AB (?) (69).] 215. THEOREM. The line perpendicular to a radius at its extremity is tangent to the circle. Given: Radius OA of O O, and RT _L to OA at A. To Prove: RT tangent to the circle. Proof: Take any point P in RT (except A) and draw OP. OP > OA (?) (77). Hence P lies without R the O. (Because OP > radius.) That is, every point (except A) of line RT is without the O. Therefore, RT is a tangent (Def. 194). Q.E.D. 216. THEOREM. If a line is tangent to a circle, the radius drawn to the point of contact is perpendicular to the tangent. Given : RT tangent to O O at A ; radius OA. To Prove : OA JL to RT. Proof: Every, point (except A) in RT is without the O (Def. 194). Therefore a line from O to any point of RT (except A) is > OA. (Because it is > a radius.) That is, OA is the shortest line from o to RT. .*. OA is J_ to RT (?) (77). Q.E.D. 217. COR. The perpendicular to a tangent at the point of contact passes through the center of the circle. (See 43.) BOOK II 87 218. THEOREM If two circles are tangent to each other, the line joining their centers passes through their point of contact. Given : O and c tangent to a line at AI and line OC. To Prove : oc passes through A. Proof: Draw radii OA and CA. OA is _L to the tangent and CA is _L to the tangent (?) (216). .'. OAC is a st. line (?) (43). /. OAC and OC coincide and OC passes through A (39). Q.E.D. Let the pupil apply this proof if the circles are tangent internally. 219. THEOREM. Two tangents drawn to a circle from an external point are equal. NOTE. In this theorem the word " tangent " signifies the distance be- tween the external point and the point of contact. Given : O O ; tangents PA, PB. To Prove : Distance PA = distance PB. Proof : Draw radii to the points of contact, and join OP. A GAP and OBP are rt. A (?) (216). In it. A GAP and OBP, OP = OP (?) ; OA = OB (?). .'.A GAP = A GBP (?). .*. PA = PB (?). Q.E.D. 88 PLANE GEOMETRY 220. THEOREM. If from an external point tangents be drawn to a circle, and radii be drawn to the points of contact, the line joining the center and the external point will bisect : I. The angle formed by the tangents. II. The angle formed by the radii. III. The chord joining the points of contact. IV. The arc intercepted by the tangents. Proof: are rt. A (?). They are = . (Explain.) II. III. o is equidistant from A and B (?). P is also (?) (219). .'. OP is -L to AB at its midpoint (?) (70). IV. Arc AX = arc BX (?) (206). Q.E.D. 221. THEOREM. In the same circle (or in equal circles) equal chords are equally distant from the center. Given : O O ; chord AB = chord Cl>, and distances OE and OF. To Prove : OE = OF. Proof : Draw radii OA and OC. In the rt. A AOE and COF, AE = AB ; CF = | CD (213). But AB = CD (Hyp.). Hence, AE = CF (Ax. 3) ; and AO=CO (?). .'. A AOE=A COF (?). /. (XE= OF (?). Q.E.D. 222. THEOREM. In the same circle (or in equal circles) chords which are equally distant from the center are equal. Given : O O; chords AB and CD ; distance OE= distance OF. BOOK II 89 To Prove : chord AB = chord CD. Proof : Draw radii OA and OC. In rt. A AOE and COF, ^10 = CO (?); OF = OF (Hyp.). /. A AOE = A COF (?). .-. .4F = CF (?). AB is twice J.F and CD is twice CF (?). . . AB = CD (Ax. 3). Q.E.D. 223. THEOREM. In the same circle (or in equal circles) if two chords are unequal, the greater chord is at the less distance from the center. Given : O O ; chord AB > chord CD, and distances OE and OF. To Prove : OE < OF. Proof : Arc AB > arc CD (?) (211, I). Suppose arc AH taken on arc AB = arc CD. Draw chord AH. Draw OK _L to AH cutting AB at I. Chord AH = chord CD (?) (210). Distance OK = distance OF (?) (221). But OE < 01 (?) (77); and OI < OK (?) (Ax. 5). .'. OE < OK (Ax. 11). /. OE < OF (Ax. 6). Q.E.D. 224. THEOREM. In the same circle (or in equal circles) if two chords are unequally distant from the center, the chord at the less dis- tance is the greater. Given : O O ; chords AB and CD ; distance OE < distance OF. To Prove : Chord AB > chord CD. Proof : It is evident that chord AB < chord CD, or = chord CD, or > chord CD. Proceed by the method of exclusion. Another Proof : On OF take OX OE. At X draw a chord RS J_ to OX. Ch. ES is II to ch. CD (?). .-. arc RS > arc CD (Ax. 5). .-. ch. RS > ch. CD (?). But ch. AB = c\\. RS (?). .-. ch. AB > ch. CD (Ax. 6). Q.E.D. 225. COR. The diameter is longer than any other chord. 90 PLANE GEOMETRY 226. THEOREM. Through three points, not in the same straight line, one circumference can be drawn, and only one. Given : Points A and B and C. To Prove: I. (?). II. (?). Proof : I. Draw lines AB, BC, AC. Suppose their J_ bisectors, OZ, OX, OF, be drawn. These Js will meet at a point (?) (85). Using O as center and OA, OB, or OC as radius, a circumference can be described through A, B, C (85). II. These Js can meet at only one point (85) ; that is, there is only one center. The distances from O to A, O to J?, O to C, are all equal (85); that is, there is only one radius.. Therefore there is only one circumference (205). Q.E.D. 227. COR. A circumference can be drawn through the vertices of a triangle, and only one. 228. COR. A circumference is determined by three points. 229. COR. A circumference cannot be drawn through three points which are in the same straight line. [The Js would be II.] 230. COR. A straight line can intersect a circumference in only two points. (229.) 231. COR. Two circumferences can intersect in only two points. 232. THEOREM. If two circumferences intersect, the line joining their centers is the perpendicular bisector of their common chord. Proof: Draw radii in each O to ends of AB. Point O is equally dis- tant from A and B (?). Point C is equally dis- tant from A and B (?). .*. OC is the J_ bisector of AB (?) (70). Q.E.D. HOOK II 233. THEOREM. Parallel lines intercept equal arcs on a circum- ference. P M P Given: A circle and a pair of parallels intercepting two arcs. To Prove : The intercepted arcs are equal. There may be three cases : I. If the Us are a tangent (^45, tangent at P) and a secant (CD, cutting the circle at E and F). Proof : Draw diameter to point of contact, P. This di- ameter is _L to AB (216). PP' is also _L to EF (?) (95). .-. arc EP = arc FP (?) (212). II. If the Us are two tangents (points of contact being M and JV). Proof : Suppose a secant be drawn II to one of the tangents, cutting O at R and S. RS will be II to other tangent (?) (94). /. arc MR = arc MS; arc UN = arc SN (proved in I). Adding, arc MEN = we MSN (Ax. 2). III. If the Us are two secants (one intersecting the O at A and B ; the other at C and D) . Proof: Suppose a tangent be drawn touching O at P, II to AB. This tangent will be II to CD (?). /. arc PC = arc PD; arc PA = arc PB (by I). Subtracting, arc AC = arc BD (Ax. 2). Q.E.D. 234. A polygon is inscribed in a circle, or a circle is cir- cumscribed about a polygon if the vertices of the poly- gon are in the circumference, and its sides are chords. t2 PLANE GEOMETRY A polygon is circumscribed] ;f the gides of ^ about a circle, or a circle is are a ii tangent to the circle, inscribed m a polygon A common tangent to two circles is a line tangent to both of them. The perimeter of a figure is the sum of all its bounding lines. EXERCISES IN DRAWING CIRCLES 1. Draw two unequal intersecting circles. Show that the line joining their centers is less than the sum of their radii. 2. Draw two circles externally (not tangent) and show that the line joining their centers is greater than the sum of their radii. 3. Draw two circles tangent externally. Discuss these lines similarly. 4. Draw two circles tangent internally. Discuss these lines similarly. 5. Draw two circles so that they can have only one common tangent. 6. Draw two circles so that they can have two common tangents. 7. Draw two circles so that they can have three common tangents. 8. Draw two circles so that they can have four common tangents. 9. Draw two circles so that they can have no common tangent. StTMMARY 235. The following summary of the truths relating to magnitudes, which have been already established in Book II, may be helpful in attacking the original work following. I. Arcs are equal if they are : (1) Intercepted by equal central angles. (2) Subtended by equal chords. (3) Intercepted by parallel lines. (4) Halves of the same arc, or of equal arcs. IT. Lines are equal if they are : (1) Radii of the same or equal circles. (2) Diameters of the same or equal circles. (3) Chords which subtend equal arcs. (4) Chords which are equally distant from the center. (5) Tangents to one circle from the same point. III. Unequal arcs and unequal chords have like relations. [See 208; 211; 223; 224.] BOOK II ORIGINAL EXERCISES 1. A diameter bisecting a chord is perpendicular to the chord and bisects the subtended arcs. [Use 70.] 2. A diameter bisecting an arc is the perpendicular bisector of the chord of the arc. [Draw AR and BR.~\ 3. A line bisecting a chord and its arc is perpendicular to the chord. 4. The perpendicular bisectors of the sides of an in- scribed polygon meet at a common point. 5. A line joining the midpoints of two parallel chords passes through the center of the circle. [Suppose diam. drawn _L to AB\ this will be to CD. Etc.] 6. The perpendiculars to the sides of a circumscribed polygon at the points of contact meet at a common point. [Use 217.] 7. The bisector of the angle between two tangents to a circle passes through the center. [Use 80.] 8. The bisectors of the angles of a circumscribed polygon all meet at a common point. 9. Tangents drawn at the extremities of a diameter are parallel. 10. In the figure of 220, prove Z APO-Z. ABO. 11. In the same figure, prove Z PA B = Z. FOB. [Use 48.] 12. If two circles are concentric, all chords of the greater, which are tangent to the less, are equal. [Draw radii to points of contact. Use 216 ; 222.] 13. Prove 225 by drawing radii to the ends of the chord. 14. An inscribed trapezoid is isosceles. [Use 233.] 15. The line joining the points of contact of two parallel tangents passes through the center. [Draw radii to points of contact. Etc.] 16. A chord is parallel to the tangent at the midpoint of its subtended arc. [Draw radii to point of contact and to the ends of the chord. Also draw chords of the halves of the given arc.] 17. The sum of one pair of opposite sides of a cir- cumscribed quadrilateral is equal to the sum of the other pair. [Use 219 four times, keeping R and T on the same side of the equations.] 94 PLANE GEOMETRY 18. A circumscribed parallelogram is equilateral. 19. A circumscribed rectangle is a square. 20. If two circles are concentric and a secant cuts them both, the por- tions of the secant intercepted between the circumferences are equal. [Use 212.] 21. Of all secants that can be drawn to a circumference from a fixed external point, the longest passes through the center. To Prove : PB > PE. rv 22. The shortest line from an external point to a circumference is that which, if produced, would pass through the center. To Prove : PA < PD. Draw CD. 23. If two equal secants be drawn to a circle from an external point, their chord segments will be equal. [Draw OA, OP, OC, OB, OD. Prove & POD and POB c ^ >^ n equal; then A COD and A OB are equal.] ^ "* S ^"--/S> ~ 24. In No. 23 prove the external segments "VVZ \ 25. State and prove the converse of No. 23. 26. If two equal secants be drawn to a circle from an external point, they will be equally distant from the center. 27. If two equal chords intersect on the circumference, the radius drawn to their point of intersection bisects their angle. [Draw radii to the other extremities of the chords.] 28. Any two parallel chords drawn through the ends of a diameter are equal. 29. If a circle be inscribed in a right triangle, the sum of the diameter and hypotenuse will be equal to the sum of the legs. [Draw radii OR, OS; ROSC is a square (?) ; then prove diameter + AB = A C + BC.'] 30. The shortest chord that can be drawn through a given point within a circle is perpendicular to the diameter through the point. Given: P, the point; BOC the diam. ; LS to BC G ' at P; OR any other chord through P. To Prove: (?). Proof : Draw OA _L to GR. Etc. ROOK TT 95 31. What, is the longest chord that can be drawn through a given point within a circle? 32. If the line joining the point of intersection of two chords and the center bisects the angles formed by the chords, they are equal. [Draw .& OE and OF and prove them = . Etc.] 33. AB and AC are two tangents from A ; in the less arc EC a point D is taken and a tangent drawn at Z>, meeting AB at E and AC at F\ AE + EF + AFequals a constant for all positions of D in arc BC. [Prove this sum = AB + BC.] 34. The radius of the circle inscribed in an equi- lateral triangle is half the radius of circle circumscribed about it. [Use 152.] 35. If the inscribed and circumscribed circles of a triangle are concentric, the triangle is equilateral. 36. If two parallel tangents meet a third tangent and lines be drawn from the points of intersection to the center, they will be perpendicular. 37. Tangents drawn to two tangent circles from any point in their common interior tangent are equal. 38. The common interior tangent of two tangent circles bisects their common exterior tangent. 39. Do the theorems of No. 37 and No. 38 apply if the circles are tangent internally ? If so, prove. 40. In the adjoining figure if AE and AD are secants, A E passing through the center, and the ex- ternal part of AD is equal to a radius, the angle DCE = 3Z^. [Draw BC. ZDBC = ext. Z of &ABC = = ZD (explain). Z DCE = an ext. Z, etc.] 41. If perpendiculars be drawn upon a tangent from the ends of any diameter : (1) The point of tangency will bisect the line between the feet of the perpendiculars. [Draw CP. Use 144.] O P (2) The sum of the perpendiculars will equal the diameter. ( 3) The center will be equally distant from the feet of the perpendic- ulars. [Use 67.] 9<) PLANE GEOMETRY 42. The two common interior tangents of two circles are equal. C 43. The common exterior tangents to two circles are equal. [Produce them to intersection.] 44. In the above figure, prove that RH = SF. Proof: AR + RB = CS + SD ; /. AR + (RH + FIF) = (SF + HF) + SD. .'. RH + RH + HF = SF + HF + SF ; /. 2 RH = 2 SF, etc. Give reasons and explain. 45. The common exterior tangents to two circles intercept, on a com- mon interior tangent (produced), a line equal to a common exterior tangent. To Prove : RS = A B. 46. Prove that in the figure of No. 42 the line joining the centers will contain and 0'. 47. Prove that in the figure of No. 42 if. chords AC and BD are drawn, they are parallel. 48. If a circle be described upon the hypotenuse of a right triangle as a diameter, it will contain the vertex of the right angle (142). 49. The median of a trapezoid circumscribed about a circle equals one fourth the perimeter of the trapezoid. 50. If the extremities of two perpendicular diameters be joined (in order), the quadrilateral thus formed will be a square. 51. If any number of parallel chords of a circle be drawn, their mid- points will be in the same straight line. 52. State and prove the converse of No. 35. 53. The line joining the center of a circle to the point of intersection of two equal chords bisects the angle formed by the chords. BOOK IT 97 KINDS OF QUANTITIES MEASUREMENT 236. A ratio is the quotient of one quantity divided by another both being of the same kind. 237. To measure a quantity is to find the number of times it contains another quantity of the same kind, called the unit. This number is the ratio of the quantity to the unit. 238. Two quantities are called commensurable if there exists a common unit of measure which is contained in each a whole (integral) number of times. Two quantities are called incommensurable if there does not exist a common unit of measure which is contained in each a whole number of times. Thus : $17 and $35 are commensurable, but $ 17 and $ V35 are not. Two lines 18^ ft. and 13 yd. are commensurable, but 18 in. and \/T3 mi. are~/io/. 239. A constant quantity is a quantity whose value does not change (during a discussion). A constant may have only one value. A variable is a quantity whose value is changing. A variable may have an unlimited number of values. 240. The limit of a variable is a constant, to which the variable cannot be equal, but from which the variable can be made to differ by less than any mentionable quantity. 241. Illustrative. The ratio of 15 yd. to 25 yd. is written either f or 15-4-25 and is equal to three fifths. If we state that a son is two thirds as old as his father, we mean that the son's age divided by the father's equals two thirds. A ratio is a fraction. The statement that a certain distance is 400 yd. signifies that the unit (the yard), if applied to this distance, will be contained exactly 400 times. Are $7.50 and $3.58 commensurable if the unit is fl? Idime? 1 cent? Are 10 ft. and Vl9 ft. commensurable ? BOBBINS' PLANE GEOM. 7 98 PLANE GEOMETRY The height of a steeple is a constant ; the length of its shadow made by the sun is a variable. Our ages are variables. The length of a standard yard, mile, or meter, etc., is a constant. The height of a grow- ing plant or child is a variable. The limit of a variable may be illustrated by considering a right tri- angle ABC, and supposing the vertex A to move farther and farther from the vertex of the right angle. It is evident that the hypotenuse will become longer, that A C will increase, but EC will re- main the same length. The angle A. must decrease, the angle B must in- crease, but the angle C remains con- stantly a right angle. If we carry vertex A toward the left indefinitely, the Z A will become less and less but cannot become zero. [Because, then there could be no A.] Hence, the limit of the decreasing Z A is zero (240). Likewise, the Z .B will become larger and larger but cannot become equal to a right angle. [Because, then two sides of the triangle would be parallel, which is impossible.] But it may be made as nearly equal to a right angle as we choose. Hence, the limit of Z B is a right angle (240). To these limits we cannot make the variables equal, but from these limits we can make them differ by less than any mentionable angle, how- ever small. The following supplies another illustration of the limit of a variable. The sum of the series l+i + J+| + T V + A + B? + T* + et c- etc., will always be less than 2, no matter how many terms are collected. But by taking more and more terms we can make the actual difference between this sum and 2 less than any conceivable fraction, however small. Hence, 2 is the limit of the sum of the series. The limit is not 3 nor 4, because the difference between the sum and 3 cannot be made less than any assigned fraction. Neither is the limit 1$. (Why not?) Similarly, the limit of the value of .333333 ad infinitum is \. Certain variables actually become equal to a fixed magnitude; but this fixed magnitude is not a limit (240). Thus the length of the shadow of a tower really becomes equal to a fixed distance (at noon). A man's age really attains to a definite number of years and then ceases to vary (at death). Such variables have no limit in the mathematical sense of that word. BOOK II 99 Hence : If a variable approaches a constant, and the difference between the two can be made indefinitely small but they cannot become equal, the constant is the limit of the variable. This is merely another definition of a limit. 242. THEOREM OF LIMITS. If two variables are always equal and each approaches a limit, their limits are equal. Given: Two variables v and v'; v always v f - t also v ap- proaching the limit 1; v' approaching the limit I' . To Prove : 1=1'. Proof: v is always =v r (Hyp.). Hence they may be con- sidered as a single variable. Now a single variable can approach only one limit (240). Hence, 1= I'. Q.E.D. NOTE. In order to make use of this theorem, one must have, first, two variables; second, these must be always equal; third, they must each ap- proach a limit. Then, the limits are equal. 243. (1) Algebraic principles concerning variables. If v is a variable and & is a constant : I. v -f k is a variable. IV. kv is a variable. II. v k is a variable. V. - is a variable. k III. k v is a variable. VI. - is a variable. v These six statements are obvious. (2) Algebraic principles concerning limits. If v is a variable whose limit is Z, and & is a constant : I. v k will approach I k as a limit. II. k v will approach k I as a limit. III. kv will approach Id as a limit. IV. ^ will approach - as a limit. tC /C k k V. - will approach - as a limit. XOTK. In these principles as applied to Plane Geometry, a variable is not added to, nor subtracted from, nor multiplied by, nor divided by another variable. These operations present little difficulty, however. 100 PLANE GEOMETRY Proofs : I. v cannot = / (240). .-. v k cannot = I k. Also, v I approaches zero (240). .-. (v k} (l k) approaches zero. (Because it reduces to v I.) Hence, v k approaches / k (240). II. Demonstrated similarly. III. If ko = kl, then v = l (Ax. 3). But this is impossible (240). .. kv cannot = kl. Also v I approaches zero (240). .-. k(v 1) or kv kl approaches zero. Therefore kv approaches kl (240). IV and V. Demonstrated similarly. 244. THEOREM. In the same circle (or in equal circles) the ratio of two central angles is equal to the ratio of their intercepted arcs. Given : O o = O C ; central A O and C; arcs AB and XY. To Prove: ^= arc ^. Z C arc XT Proof : I. If the arcs are commensurable. There exists a common unit of measure of AB and XY (238). Suppose this unit, when applied to the arcs, is contained 5 times in AB and 7 times in XY. .-. arc AB = j> (Ax. 3). Draw radii HYCXY 7 to the several points of division of the arcs. Z O is divided into 5 parts, Z c into 7 parts ; all of these twelve parts are equal (?) (207). .-. = | (Ax. 3). . c i Z O _ arc AB ,. -,, ' ' Z c ~ arc XY ^ X ' } ' Q-E.D. BOOK II 101 II. If the arcs are incommensurable. There does not exist a common unit (238). Suppose arc AB divided into equal parts (any number of them). Apply one of these as a unit of measure to arc XY. There will be a remainder PY left over. (Because AB and XT are incommensurable.) Draw CP. Now Z = arc AB . (The case of commen- Z XCP arc XP surable arcs.) Indefinitely increase the number of subdivisions of arc AB. Then each part, that is, our unit or divisor, will be indefinitely decreased. Hence PF, the remainder, will be in- definitely decreased. (Because the remainder < the divisor.) That is, arc PY will approach zero as a limit and Z PCY will approach zero as a limit. .'. arc XP will approach arc XY as a limit (240) and Z XCP will approach Z XCY as a limit (240). will approach as a limit (243) Z XCP Z XC 'I-, i_ a will approach a Z O arc AB j arc AB 'I-, i_ arc AB v ., ^o/iox and - - will approach -- as a limit (243). arc XP arc XY Z XCY arc XY Ex. How many degrees are there in a central angle which intercepts I of the circumference? \ of the circumference? ^ of the circumfer- ence ? T 5 y of the circumference ? 102 PLANE GEOMETRY 245. THEOREM. A central angle is measured by its intercepted arc. Given: O o; Z.AOY; arc AY. To Prove: Z AOY is measured by the arc A F, that is, they contain the same number of units. Proof : The sum of all the A about 0=4 rt. A = 360 (?) (47). If the circumference of this O be divided into 360 equal parts and radii be drawn to the several points of division, there will be 360 equal central A (207). Each of these 360 central angles will be a degree of 'angle (21). Suppose we call each of the 360 equal arcs, a degree of arc. Take Z AOT, one of these degrees of angle, and arc AT, one of the degrees of arc. Then, Z ^4OF_arc AY ~arc AT (?) (244). But = Z^ior-5-a unit of angle = the number of Z AOT units in /.AOY (237). And = arc AY -5- a unit of arc = the number of arc AT units in arc AY (237). Hence, the number of units in Z AOY = the number of units in arc AY (Ax. 1). That is, Z AOY is measured by arc AY. Q.E.D. 246. COR. A central right angle intercepts a quadrant of arc. (Because each contains 90 units.) ^ -v 247. COR. A right angle is measured by half a semicircumf erence, that is, by a quadrant. 248. An angle is inscribed in a segment if its vertex is on the arc and its sides are drawn to the ends of the arc of the segment. Thus ABCD is a segment and Z ABD is inscribed in it, BOOK II 103 249. THEOREM. An inscribed angle is measured by half its inter- cepted arc. Given : O O ; inscribed Z A ; arc CD. To Prove : Z A is measured by J arc CD. Proof: I. If one side of the Z is a diameter. Draw radius CO. A AOC is isosceles (?). .-. Z A = Z C (?). Z COD = ZA + Z C (?) (108). .-. Z COD = Z^i-r-^^ = 2Z^i (Ax. 6). That is, Z A = % Z COD (Ax. 3). But Z COD is measured by arc CD (?) (245). .*. J Z COD is measured by ^ arc CD (Ax. 8). Therefore, Z A is measured by J arc CD (Ax. 6). A A X X II. If the center is within the angle. Draw diameter AX. Z.CAX is measured by J arc CX (I). is measured by J arc DX (I). Adding, is measured by arc CD (Ax. 2). III. If the center is without the angle. Draw diameter AX. /.CAX is measured by ^ arc CX (I). ZD-4JT is measured by | arc DJT (I). Subtracting, Z C.4D is measured by -| arc CD (Ax. 2). Q.E.D. NOTE. It is evident that angles measured by | the same arc are equal. Ex. 1 . In the figure of 249, if arc CD is 56, how many degrees are there in angle A ? If arc CD is 108, how many degrees are in angle A V Ex'. 2. If Z.A contains 35, how many degrees are there in arc C/)? 104 PLANE GEOMETRY 250. THEOREM. All angles inscribed in the same segment are equal. Given : The several A A inscribed in segment BAG. To Prove : These angles all equal. Proof : Each Z BAG is measured by 1 arc BG (?) (249). .-. they are equal. (Because they are measured by half the same arc.) Q.E.D. 251. COR. All angles inscribed in a semi- circle are right angles. Proof : Each is measured by half of a semicircumference (?) (249). .-. each is a rt. Z (?) (247). Q.E.D. 252. THEOREM. The angle formed by a tangent and a chord is measured by half the intercepted arc. Given : Tangent TN ; chord PA ; Z TPA ; arc PA. To Prove : Z TPA is measured by ^ arc P^l. Proof : Draw diameter PX to point of contact. Z TPX is a rt. Z (?) (216) ; arc PAX is a semicircum- ference (?) (204). Z TPX is measured by % arc PAX (?) (247). ^ APX is measured by j. arc AX (?) (249). Subtracting, Z TPA is measured by arc PA (Ax. 2). Q.E.D. Similarly, Z NPA is measured by ^ arc PBA. (Use Z NPX, and add.) BOOK II 105 (233). 253. THEOREM. The angle formed by two chords intersecting within the circumference is measured by half the sum of the inter- cepted arcs. (The arcs are those intercepted by the given angle and by its vertical angle.) Given : Chords AB and CD in- tersecting at P; Z APC; arcs AC and DB. To Prove : Z APC is measured by ^ (arc AC+ arc DB). Proof : Suppose CX drawn through C \\ to AB. Now Z c is measured by ^ arc DX (?) (249). That is, Z C is measured by ^ (arc BX + arc But Z C = Z APC (?) (97) and arc BX = arc ^ .-. /.APC is measured by |- (arc .4(7 + arc DB) (?) (Ax. 6). Q.E.D. NOTE. This theorem may be proven by drawing chord A D. Then /.APC is an ext. Z of A ADP and = A + /.D (?). [Use 249.] 254. THEOREM. The angle formed by two tangents is measured by half the difference of the intercepted arcs. Given : The two tan- gents AC and AB\ Z-4; arcs CMB and CNB. To Prove : Z A is measured by J(arc CMB arc CNB). Proof : Suppose drawn || to ^LB. Now Z Z)CX is measured by J arc CX (?) (252). That is, ZDCX is measured by (arc CJtfB arc But Z /)CX =Z ,4 (?) (98) ; arc BX = arc CNB (?). .'. Z ^ is measured by J (arc OJfl? arc CNB) (Ax. 6). Q.E.D. 106 PLANE GEOMETRY 255. THEOREM. The angle formed by two secants which intersect without the circumference is measured by half the difference of the in- tercepted arcs. Given: (?). To Prove: (?). Proof: Suppose BX drawn. Where? How? Z CB x is measured by J arc CX (?). That is, by |(arc CE arc EX). But Z CBX = Z A (?) ; arc EX = arc BD (?). .-. ZA is meas. by J (arc OE arcJ5D) (Ax. 6). Q.E.D. 256. THEOREM. The angle formed by a tangent and a secant which intersect without the circumference, is measured by half the difference of the intercepted arcs. c Given: (?). To Prove : (?). Proof: Suppose BX drawn etc. Z CBX is measured by J arc BX (?) (252). That is, by J (arc B XE A arc EX). Etc. NOTE. The theorem of 254 maybe proven by drawing chord BC. Then Z DCB = Z A + Z OB.4 (?) ; or, Z ^ = Z DOB - Z C.B.4 (Ax. 2). Z Z)<7 is measured by \ arc CATS (?) and Z C5.4 is measured by \ arc CW (?). Hence, Z ^ is measured by | (arc CMB - arc CNB) (?). Ex. 1. Prove the theorem of 253 for angle A PD by drawing chord A C. Ex. 2. Prove the theorem of 255 by drawing chord CD. Again, by drawing chord BE. Ex. 3. Prove the theorem of 256 by drawing chord BD. Again, by chord BE. BOOK II 107 ORIGINAL EXERCISES 1. If an inscribed angle contains 20, how many degrees are there in its intercepted arc ? How many degrees are there in the central angle which intercepts the same arc ? 2. A chord subtends an arc of 74. How many degrees are there in the angle between the chord and a tangent at one of its ends? 3. How many degrees are there in an angle inscribed in a segment whose arc contains 210? in a segment whose arc contains 110? 40? 4. Two intersecting chords intercept opposite arcs of 28 and 80. How many degrees are there in the angle formed by the chords ? 5. The angle between a tangent and a chord contains 27. How many degrees are there in the intercepted arc ? 6. The angle between two chords is 30 ; one of the arcs intercepted is 70 ; find the other arc. [Denote the arc by z.] 7. If in figure of 252, arc AP contains 124, how many degrees are there in A PX ? in Z NPA ? 8. If in figure of 253, arc AC is 85, Z A PC is 47, find arc DB. 9. If the arcs intercepted by two tangents contain 80 and 280, find the angle formed by the tangents. 10. If the arcs intercepted by two secants contain 35 and 185, find the angle formed by the secants. 11. If in figure of 254, arc CB is 135, find the angle A. 12. If in figure of 255, angle A = 42 and arc ED = 70, find arc CE. 13. If in figure of 256, angle A = 18, arc BXE = 190, find arc BD. 14. If the angle between two tangents is 80, find the number of degrees in each intercepted arc. [Denote the arcs by x and 360 x.~] 15. The circumference of a circle is divided into four arcs, 70, 80. 130, and x. Find x and the angles of the quadrilateral formed by the chords of these arcs. 16. Find the angles formed by the diagonals in quadrilateral of No. 15. 17. Three of the intercepted arcs of a circumscribed quadrilateral are 68, 98, 114. Find the angles of the quadrilateral. If the chords are drawn connecting (in order) the four points of contact, find the angles of this inscribed quadrilateral. Also find the angles between its diagonals. 108 PLANE GEOMETRY 18. If the angle between two tangents to a circle is 40, find the other angles of the triangle formed by drawing the chord joining the points of contact. 19. The circumference of a circle is divided into four arcs, three of which are, RS = 62, ST= 142, TU = 98. Find : (1) Arc UR; (2) The three angles at/2; atS; at T\ at [7; (3) The angles A, B, C, D of circumscribed quadrilateral ; (4) The angles between the diagonals R T and SC7; (5) The angle between R U and ST at their point of intersection (if produced) ; (6) The angle between RS and TU at their intersection; (7) The angle between AD and BC at their intersection ; (8) The angle between AB and DC at their intersection ; (9) The angle between RS and DC at their intersection; (10) The angle between AD and ST at their intersection. 20. If in the figure of No. 19, Z A = 96 ; Z B = 112 ; and Z C = 68, find the angles of the quadrilateral RSTU. [Denote arc RU by x. :. in A ARU, 96 + | x + a: = 180. .'. a; =, etc.] 257. It is evident from the theorems relating to the measurement of angles, that : 1 Equal angles are measured by equal arcs (in the same circle). 2. Equal arcs measure equal angles. T 21. Prove theorem of 252 by drawing chord parallel to the tangent. 22. The opposite angles of an inscribed quadri- lateral are supplementary. Proof : Z A + Z C are meas. by | circumference. 23. If two chords intersect within a circle, and at right angles, the sum of one pair of opposite arcs equals the sum of the other pair. [Use 253.] 24. If a tangent and a chord are parallel, and the chords of the two intercepted arcs be drawn they w r ill make equal angles with the tan- gent. [Use 233 ; 252.] 25. The line bisecting an inscribed angle bisects the intercepted arc. BOOK II 109 26. The line joining the vertex of an inscribed angle to the midpoint of its intercepted arc bisects the angle. ... p ^* ^r w 27. The line bisecting the angle between a tangent and a chord bisects the intercepted arc. 28. State and prove the converse of -No. 27. 29. The angle between a tangent and a chord is half the angle between the radii drawn to the ends of the chord. 30. If a triangle be inscribed in a circle and a tan- gent be drawn at one of the vertices, the angles formed between the tangent and the sides will equal the other two angles of the triangle. 31. By the figure of No. 30 prove that the sum of the three angles of a triangle equals two right angles. 32. If one pair of opposite sides of an inscribed quad- rilateral are equal, the other pair are parallel. Proof : Draw Js BX, CF; arc A B = arc CD (?). '. arc ABC = arc BCD (Ax. 2). Hence prove rt. & ABX and CDY equal. 33. If any pair of diameters be drawn, the lines joining their extremi- ties (in order) will form a rectangle. [Use 251.] 34. If two circles are tangent externally and any line through their point of contact intersects the circumferences at B and C, the tangents at B and C are parallel. [Draw common tangent at A. Prove : Z A CT = Z A BS.~\ v / 35. Prove the same theorem if the circles are tangent internally. 36. If two circles are tangent externally and any line be drawn through their point of contact terminating in the circumferences, the two diam- eters drawn to the extremities will be parallel. 37. Prove the same theorem if the circles are tangent internally. 38. If two circles are tangent externally and any two lines be drawn through their point of contact intersecting their circumferences, the chords joining these points of intersection will be parallel. [Draw common tangent at 0. Prove : Z C = Z /).] 39. Prove the same theorem if the circles are tangent internally. 110 PLANE GEOMETRY 40. The circle described on one of the equal sides of an isosceles triangle as a diameter bisects the base. Proof: Draw line BM. The A are rt. & (?) and equal (?). 41. If the circle, described on a side of a triangle as diameter, bisects another side, the triangle is isosceles. 42. All angles that are inscribed in a segment greater than a semicircle are acute, and all angles inscribed in a segment Jess than a semicircle are obtuse. 43. An inscribed parallelogram is a rectangle. Proof: Arc AB= arc CD (?) ; arc EC = arc AD (?). /. arc ABC ' = arc A DC (?); that is, each = a semicir- cumference. Etc. 44. The diagonals of an inscribed rectangle pass through the center and are diameters. 45. The bisectors of all the angles inscribed in the same segment pass through a common point. 46. The tangents at the vertices of an inscribed rectangle form a rhombus. [& A BF and HD C are isosceles (?) and equal ? Etc.] 47. If a parallelogram be circumscribed about a circle, the chords joining (in order) the four points of contact will form a rectangle. [Prove BD a diameter.] 48. A circumference described on the hypotenuse of a right triangle as a diameter passes through the vertices of all the right triangles having the same hypotenuse. 49. If from one end of a diameter a chord be drawn, a perpendicular to it drawn from the other end of the diameter will inter- sect the first chord on the circumference. [Use 148.] 50. If two circles intersect and a diameter be drawn in each circle through one of the points of intersection, the line joining the ends of these diameters will pass through the other point of intersection. [Draw chord AB. Use 251; 43.] 51. If A BCD is an inscribed quadrilateral, AB and DC produced to meet at E, AD and BC produced to meet at F, the bisectors of angles E and F are perpendicular. [Show that the circumference is divided into eight arcs which are equal in pairs ; by correctly adding these, show that four arcs are equal to the others, and hence equal to a semicircumference. The angle be- tween the bisectors is measured by half the sum of these four, and hence it is a right angle.] BOOK II 111 52. If a tangent be drawn at one end of a chord, the midpoint of the intercepted arc will be equally distant from the chord and tangent. [Draw chord A M and prove the rt. & = .] 53. If two circles are tangent at A and a common tangent touches them at B and C, the angle BAG is a right angle. [Draw tangent at A. Use 219; 148.] 54. A circle described on the radius of another circle as diameter bisects all chords of the larger circle drawn from their point of contact. To Prove: AB is bisected at C. Draw chord OC. (Use 251 ; 213.) 55. If the opposite angles of a quadrilateral are supplementary, a circle can be drawn circumscribing it. To Prove : A O can be drawn through A, B, C, P. Proof: A O can be drawn through A, B, C (?). It is required to prove that it will contain point P. Z P + Z B are supp. (?). .*. must be meas. by half the entire circumf. Z B is meas. by arc ADC (V). Hence Z P is meas. by \ arc ABC. If Z P is within or without the circumf. it is not meas. by \ arc ABC. (Why not?) 56. The circle described on the side of a square, or of a rhombus, as a diameter passes through the point of intersection of the diagonals. [Use 141; 148.] 57. The line joining the vertex of the right angle of a right triangle to the point of intersection of the diagonals of the square constructed upon the hypotenuse as a side, bisects the right angle of the triangle. Proof : Describe a O upon the hypotenuse as diameter and use 148 ; 209 ; 249. 58. If two secants, PAB and PCD, meet a circle at A, B, and C, D respectively, the triangles PBC and PAD are mutually equiangular. 59. If PAB is a secant and PM is a tangent to a circle from P, the triangles PAM &udPBM are B mutually equiangular. 60. If two circles intersect and a line be drawn through each point of intersection terminating in the circumferences, the chords join ing these extrem- ities will be parallel. [Draw RS. Z A is supp. of Z RSC (?). Finally use 104.] 112 PLANE GEOMETRY 61. If two equal chords intersect within a circle, (1) One pair of intercepted arcs are equal. (2) Corresponding parts of the chords are equal. c (3) The lines joining their extremities (in order) form an isosceles trapezoid. (4) The radius drawn to their intersection bisects their angle. 62. If a secant intersects a circumference at D and E, PC is a parallel chord, and PR a tangent at P meeting secant at R, the triangles PCD and PRD are mutually equiangular. [Z R and Z CDP are measured by arc RC. (Explain.) Etc.] 63. If a circle be described upon one leg of a right triangle as diameter and a tangent be drawn at the point of its intersection with the hypotenuse, this tangent will bisect the other leg. [Draw OP and OD. CD is tangent (?). OD bisects arc PC (?) (220). Z COD = Z A (?) (257). /. OD is || to AB (V). Etc.] 64. If an equilateral triangle ABC is inscribed in a circle and P is any point of arc AC, AP + PC = BP. [Take PN = PA ; draw AN. &ANP is equilateral. (Explain.) AANB = A A PC (?). Etc.] 65. If two circles are tangent internally at C, and a chord AB of the larger circle is tangent to the less circle at M, the line CM bisects the angle A CB. [Draw tangent CX and chord RS. Z KSC = ^BCX = ZA. .-. AB is || to RS. (Explain.) Then use 233. Etc.] 66. If two circles intersect at A and C and lines be drawn from any point P, in one circum- ference, through A and C terminating in the other at points B and D, chord BD will be of constant p length for all positions of point P. [Draw BC. Prove Z BCD, the e*xt. Z of = a constant. Etc.] 67. The perpendiculars from the vertices of a triangle to the opposite sides are the bisectors of the angles of the triangle formed by joining the feet of these perpendiculars. To Prove: BS bisects ^RST, etc. B HOOK II 11:1 Proof: If a circle be described on AO as diam., it will pass through T and ^ (?) (148). If a circle be described on OC as diam., it will pass through R and 5 (?). .'. ZBAR = ^BST(!) ; and /.BCT = ^BSR (?). But Z BAR = Z BCT. (Each is the comp. of Z ABC. . . Etc.) 68. If ABC is a triangle inscribed in a circle, BD is the bisector of angle ABC, meeting AC at O and the circumference at D, the triangles A OB and COD are mutually equiangular. Also triangles BOC and A OD. Also triangles BOC and ABD. Also triangles A OD and ABD. Also triangles CZ> and COD. 69. If two circles intersect at A and 5, and from P, any point on one of them, lines A P and BP be drawn cutting the other circle again at C and D respectively, CD will be parallel to the tangent at P. 70. If two circles intersect at A and B, and through B a line be drawn meeting the circles at R and S respectively, the angle RAS will be constant for all positions of the line RS. [Prove Z R + Z S is constant. .'. Z. RAS is also constant.] 71. Two circles intersect at A and through A any secant is drawn meeting the circles again at M and N. Prove that the tangents at M and N meet at an angle which remains constant for all positions of the secant. [Prove the angle between these tangents equal to the angle between the tangents to the circles, at A .] 72. Three unequal circles are each externally tangent to the other two. Prove that the three tangents draw r n at the points of contact of these circles meet in a common point. 73. Two equal circles intersect at A and B, and through A any straight line MAN is drawn, meeting the circumferences at M and N respectively. Prove chord BM = chord BN. 74. If the midpoint of the arc subtended by any chord be joined to the extremities of any other chord, (1) The triangles formed will be mutually equiangular. (2) The opposite angles of the quadrilateral thus formed will be sup- plementary. 75. Two circles meet at A and B and a tangent to each circle is drawn at A, meeting the circumferences at R and S respectively; prove that the triangles ABR and ABS are mutually equiangular. BOBBINS' PLANE GEOM. 8 114 PLANE GEOMETRY 76. What is the locus of points at a given distance from a given point? Prove. (Review 179 and 180 now.) 77. What is the locus of the midpoints of all the radii of a given circle ? Prove. 78. What is the locus of the midpoints of a series of parallel chords in a circle ? Prove. 79. What is the locus of the midpoints of all chords of the same length in a given circle ? Prove. 80. What is the locus of all points from which two equal tangents can be drawn to two circles which are tangent to each other? 81. What is the locus of all the points at a given distance from a given circumference? Discuss if the distance is > radius. If it is less. 82. What is the locus of the vertices of the right angles of all the right triangles that can be constructed on a given hypotenuse? Prove. 83. What is the locus of the vertices of all the triangles which have a given acute angle (at that vertex) and have a given base? Prove. 84. A line of given length moves so that its ends are in two perpen- dicular lines ; what is the locus of its midpoint ? Prove. [Suppose AB represents one of the positions of the moving line. Draw OP to its midpoint. In all the positions of AB, OP = \ AB = a constant (148). .'. P is always a fixed distance from 0. Etc.] 85. What is the locus of the midpoints of all the chords that can be drawn through a fixed point on a given circumfer- ence ? Prove. [Suppose AB represents one of the chords from B in circle 0, with radius OB ; and P the midpoint of AB. Draw OP. Z P is a rt.Z (?). That is, where - ever the chord may be drawn, /. P is a rt. Z. .'. locus of P is, etc.] 86. A definite line which is always parallel to a given line moves so that one of its extremities is on a given circumference ; find the locus of the other extremity. [Suppose CP represents one position of the moving line CP. Draw OQ = and II to CP from center 0. Join OC and PQ. Wherever CP is, this figure is a O (?). Its sides are of constant length (?). That is, P is always a fixed distance from Q, etc.] A- O BOOK 11 115 CONSTRUCTIONS 258. Heretofore the figures we have used have been assumed. We have supposed such auxiliary lines to be drawn as conditions required. No methods have been given for drawing any lines, and only our three postulates have been assumed regarding such construction. But the lines that have been used were drawn as aids toward establishing truths, and precise drawings have not been essential. The following simple methods for constructing lines are given that mathematical precision may be em- ployed if necessary in drawing diagrams of a more complex nature. The pupil should be very familiar with the use of the ruler and compasses. 259. A geometrical construction is a diagram made of points and lines. 260. A geometrical problem is the statement of a required construction. Thus: "it is required to bisect a line" is a problem. A problem is sometimes defined as "a question to be solved " and includes other varieties besides those involved in geometry. 261. The word proposition is used to include both theorem and problem. 262. The complete solution of a problem consists of five parts : I. The Given data are to be described. II. The Required construction is to be stated. III. The Construction is to be outlined. This part usually contains the verb only in the imperative. No reasons are necessary because no statements are made. The only limitation in this part of the process is, that every construction demanded shall have been shown possible by previous constructions or postulates. (See 32; 33; 199.) IV. The Statement that the required construction has been completed. V. The Proof of this declaration. Sometimes a discussion of ambiguous or impossible in- stances will be necessary. 116 PLANE GEOMETRY IM 263. NOTES. (1) A straight line is determined by two points. (2) A circle is determined by three points. (3) A circle is determined by its center and its radius. Whenever a circumference, or even an arc, is to be drawn, it is essential that the center and the radius be mentioned. (4) "Q.E.F." = Quod erat faciendum = " which was to be done." These letters are annexed to the statement that the construction which was required has been accomplished. 264. PROBLEM. It is required to bisect a given line. Given : The definite line AB. \ : ..- Required : To bisect AB. Construction : Using A and B as centers and one radius, suffi- ciently long to make the cir- cumferences intersect, describe two arcs meeting at R and r. Draw RT meeting AB at M. Statement: Point M bisects AB. Q.E.F. Proof : R is equally distant from A and B (?) (201). T is equally distant from A and B (?). Hence, RT is the J. bisector of AB (?) (70). That is, M bisects AB. Q.E.D. 265. PROBLEM. To bisect a given arc. Given : Arc AB whose center is O. Required : To bisect arc AB. Construction: Draw chord AB ; using A and B as centers and any sufficient radius, de- scribe arcs meeting at C. Draw OC cutting arc AB at M. O-- BOOK TI 117 Statement: The point M bisects arc AB. Q.E.F. Proof : o and C are each equally distant from A and B (201). .*. OC is the _L bisector of chord AB (?) (70). .-. M bisects arc AB (?) (213). Q.E.D. 266. PROBLEM. To bisect a given angle. Given: Z LON. Required : To bisect Z LON. Construction : Using O as a center and any radius, draw arc AB, cutting LO at A and NO at B. Draw chord AB. Using A and B as centers and any suf- ficient radius, draw two arcs intersecting at S. Draw OS meeting arc AB at M. Statement: OS bisects Z LON. Q.E.F. Proof : O and S are each equally distant from A and B (?). .-. OS is the _L bisector of chord AB (?). .-. M bisects arc AB (?) (213). .-. Z AOM = Z BOM (?) (207). That is, OS bisects Z LON. Q.E.D. 267. PROBLEM. At a fixed point in a straight line to erect a perpen- dicular to that line. Given : Line A B and point P within it. Required: To erect a line _L to AB at P. Construction: Using Pas A a center and any radius, draw arcs meeting AB at C and D. Using C and D as centers and a radius longer than before, draw arcs meeting at S. Draw PS. Statement : PS is _L to AB at P. Q.E.F. Proof : PS is the _L bisector of CD (?) (70). Q.E.D. D B 118 PLANE GEOMETRY Another Construction : Using any point O, without AB, as center, and OP as radius, describe a circumference, cutting AB at P and E. Draw diameter EOS. Join SP. Statement : -SP is JL to AB at P. Q.E.F. Proof : Segment SPE A EX. .xp is a semicircle (?) (204). .-. Z SPE is a rt. Z (?) (251). .-. SP is J. to AB (?) (17). 268. PROBLEM. Through a point without a line to draw a perpen- dicular to that line. P Given : Line AB and point P without it. si/ Required: (?). /f\ Construction : Using P as a center and any suffi- ^ --M cient radius, describe an " arc intersecting AB at M and JV. Using Jf and N as centers and any sufficient radius, describe arcs intersecting each other at C. Draw PC. Statement: PC is J_ to AB from P. Q.E.F. Proof : PC is the J_ bisector of MN (?) (70). Q.E.D. 269. PROBLEM. At a given point in a given line to construct an angle which shall be equal to a given angle. Given: Z.AOB; point P in line CD. Required : To construct at P o an Z = Z AOB. Construction: Using O as a center with any radius, describe an arc cutting OA at E arid OB c BOOK II 119 at F. Draw chord EF. Using P as a center and OE as a radius, describe an arc cutting CD at R. Using B as a center and chord EF as a radius, describe an arc cutting the former arc at X. Draw PX and chord RX. Statement : The Z XPD = Z AOB. Q.E.F. Proof: Chord EF= chord EX (?) (201). .-. arc EF= arc RX (?) (209). .-. Z XPR = Z O (?) (207). That is, Z XPD = Z ^L(XB. Q.E.D. 270. PROBLEM. To draw a line through a given point parallel to a given line. Given : Point P and line AB. \ Required : To draw through p\ X P, a line II to AB. Construction: Draw any line \ PN through P meeting AB at N. ~"N~~ On this line, at P, construct Z NPX=Z. ANP. (By 269.) Statement: PX is II to AB. Q.E.F. Proof: ZNPX=/.ANP (?) (Const.). /. PX is II to AB (?) (101). Q.E.D. 271. PROBLEM. To divide a line into any number of equal parts. Given : Definite line AB. A O N M L B '*. / Required: To divide it "'"^. into five equal parts. Construction: Draw through A any other line AX. On this take any length AC as a unit, and mark off on AX five of these units, AC, CD, DE, EF, FG. Draw GB. Through F, E, D, C, draw II to GB, lines FL, EM, DN, CO. Statement : Then, AO = ON = NM= ML = LB. Q.E.F. Proof : AC = CD = DE = EF = FG (Const.). .'.A0= ON=NM=ML = LB (?) (147). Q.E.D. 120 PLANE GEOMETRY 272. PROBLEM. To draw a tangent to a given circle through a given point: I. If the point is on the circumference. II. .If the point is without the circle. I. Given : O O ; P, a point A" 011 the circumference. Required: To draw a tangent through P. Construction: Draw the radius OP. Draw line AB J_ to OP at P (by 267). Statement : AB is tangent to O O at P. Q.E.F. Proof : AB is _1_ to PO at P (Const.). .-. AB is a tangent (?) (215). Q.E.D. II. Given : O O ; P, a point without it. Required : To draw a tan- . gent through P. Construction: Draw PO; \ bisect it at M (by 264). Using M as a center and PM as a radius, describe a circumference intersecting O O at A and B. Draw PA, PJ5, OA, OB. Statement : PA and PB are tangents through p. Q.E.F. Proof : O M passes through O (P3f = MO by const.). .-. Z P^LO is a rt. Z (?) (251). /. PA is tangent (?) (215). Similarly PB is a tangent. Q.E.D. Ex. 1 . Show by two distinct methods how to bisect a line. Ex. 2. Show how to construct the problem of 270 by use of 268. BOOK II 121 273. PROBLEM. To circumscribe a circle about a given triangle. Given: (?). Required: (?). (See 227.) Construction : Bisect AB, BC, AC. Erect Js at these midpoints, meeting at O. Using O as a center and OA as radius, draw a circle. Statement: This O will pass through vertices -4, fi, and C. Q.E.F. Proof: [Use 85.] 274. PROBLEM To inscribe a circle in a given triangle. Given: (?). Required: (?). Construction: Draw the three bisectors of the A of A ABC, meeting at O (by 266). Draw Js from O to the three sides. Using O as a center and one of these Js as a radius, draw a O. Statement : THis O will be tangent to the three sides of A ABC. Q.E.F. Proof : The bisectors of these angles meet in a point and the Js OL, OM, ON are equal (?) (84). /. the circumference passes through L, M, N (?) (192). Therefore the three sides are tangent to the O (?) (215). That is, the O O is inscribed in A ABC (?) (234). Q.E.D. 122 PLANE GEOMETRY 275. PROBLEM. To construct a parallelogram if two sides and the w s y Q.E.F. included angle are given. Given : The sides a and b and their included angle, x. Required : To construct a O containing these parts. Construction : Take a straight line PQ z= a. At P construct Z P = Z x. g On PW, the side of this Z, ^ take PR = b. At B draw BY \\ to PQ; and at Q draw QZ || to PW. Denote the intersection of these lines by 8. Statement : PQSB is the required parallelogram. Proof : First, it is a parallelogram (?). Second, it is the required parallelogram. (Because it con- tains the given parts.) Q.E.D. 276. PROBLEM. To construct a segment of a circle upon a given line, as chord, which shall contain angles equal to a given angle. Given : Line AB and Zir'. Required : To con- struct a segment upon AB whose inscribed angles shall = Z K*. Construction: Con- struct at A, /-BAC=Z.K'. Bisect AB at M. At M erect OM _L to AB. At A erect OA _L to AC, meeting OM at O. Using O as a center and OA as radius, describe O O. Statement : The A inscribed in AKB = Z K r . Q.E.F. BOOK II 123 Proof: Tin? circumference passes through B (?) (67). /. AB is u chord (?). AC is tangent to the O (?) (215). /. /iBACis -measured by half the arc AB (?) (252). Any inscribed angle AKB is measured by half the arc ^#(?). Therefore, any angle AKB = Z BAG (?) (257, 2). Consequently, any inscribed angle AKB = Z K 1 (Ax. 1). Q.E.D. [If the pupil will draw chords AK and BK, he will under- stand the proposition. These were purposely omitted.] 277. PROBLEM. To construct the third angle of a triangle if two angles are known. Given : A A and .B, two A of Ji A. Required : To construct the third. Construction : At point O in a line ES construct Z a = Z A. R ~~ o" At point O in OT construct Z b = Z B. Statement : The Z VOE = the third Z of the A. Q.E.F. Proof : Za + Z + Z VOE = 2 rt, A (?) (46). /.A + /.B + the unknown Z = 2 rt. ^ (?) (110). .-. Z a + Z b +Z F(XR = Z A + Z + the unknown Z (?). But Z a + Z5 = Z A + Z B (Const, and Ax. 2). .'. Z FO.B = the unknown Z (Ax. 2). That is, ZFOtf = the third Z of the A. Q.E.D. Ex. 1. To circumscribe a triangle about a given circle. Ex. 2. To construct the problem of 276 if the given angle is a right angle ; if it is an obtuse angle. Ex. 3. To construct the problem of 273 if the given triangle is obtuse. Ex.4. Is the problem of 277 ever impossible ? Explain. Ex. 5. In the figure of 274, if Z A = 40 and /.B = 94, how many degrees are there in each of the six acute angles at ? If A LMN is constructed, how many degrees are there in each of its angles ? 124 PLANE GEOMETRY 278. PROBLEM. To construct a triangle if the three sides are known. Given : Sides a, 6, c of a A. . Required: To Ct construct the A. Construction : Draw RS = a. ^ ff Using R as a R center and b as a radius, describe an arc ; using S as a center and c as a radius, describe an arc intersecting the former arc at T. Draw ET and ST. Statement: RST is the required A. Q.E.F. Proof: RST is a A (?) (23). RST is the required A. (It contains a, 5, and their included Z C in a A. Required: To construct the A. Construction : Draw CB a. At C construct Z BCX = given c Z C. On ex take CM = b. Join Statement: (?). Proof: (?). NOTE. The student has probably observed that in constructions cer- tain lines and angles must precede others. In such problems as 266, 267, 269, 272, and 276, the order of the successive steps is exceedingly impor- tant. Problems are not so numerous in geometry as theorems, but it must be apparent that problems are instructive, fascinating, and profitable. Definition. If a circle is described, touching one side of a triangle and the prolongations of the other sides, it is called an escribed circle. BOOK II 125 280. PROBLEM. To construct a triangle if a side and the two angles adjoining it are known. \ Given: (?). Required: ('0- B< Construction : Draw BC = a. At B construct Z CBX Z B 1 ; at C construct Z BCY = Z c'. Denote the point of intersec- tion of BX and CY by A. Statement: (?). Proof: (?). Discussion: (?). 281. PROBLEM To construct a right triangle if the hypotenuse and a leg are known. Given: Hypotenuse*?; leg b. A Required: (?)! Construction: Draw an in- definite line CD and at C erect a _L = b. Using A as a center and c as a radius, describe an arc cutting CD at B. Draw AB. Statement: (?). Proof : (?). NOTE. If it is required to construct a right triangle, having given the hypotenuse and another part, it is often advantageous to describe a semi- circle upon the given hypotenuse as diameter. Every triangle whose base is this diameter and whose vertex is on this semicircumference is a right triangle (251). Hence if the triangle constructed contains the other given part, it is the required triangle. Ex. 1. To construct the problem of 281 by use of the semicircle. Ex. 2. Discuss the constructions of 279, 280, and 281 fully. Ex. 3. To construct a triangle and its three escribed circles. Ex. 4. To construct an isosceles right triangle having given the hypot- enuse. Ex. 5. To construct an isosceles right triangle having given the leg. Ex. 6. To construct a quadrilateral having given the four sides and an angle. Discussion: (?). 126 PLANE GEOMETRY 282. PROBLEM. To construct a triangle if an angle, a side adjoin- ing it, and the side opposite it are known ; that is, if two sides and an angle opposite one of them are known. The known angle may be obtuse, right, or acute. Consider : First, If "side opposite" > "side adjoining." Second, If "side opposite "= "side adjoining." Third, If "side opposite " < "side adjoining." Construction for all of these : Draw an indefinite line, ITX, and at one extremity construct an Z =Z K ; take on the side of this /- a distance from the vertex = "side adjoining." Using the end of this side as a center and the "side opposite " as a radius, describe an arc intersecting KX. Draw radius to the intersection just found. If the known angle is obtuse or right Given : Z K, s.a, and s.o. of a A. v Construction : As above. Discussion: Case I. s.o. >s.#. The A is always possible. Case II. s.o. =s. a. The A is never possible (55 and 112). Case III. s. o. < s. a. The A is never possible (?) (122). If the known angle is acute. Case I. s. o. >s. a. The A is always possible. Case II. s.o. = 8. a. isosceles A. Case III. s. o. < s. a. (1) If s.o. < the _L from A to jBTX, the A is never pos- sible. BOOK II 127 (2) If s.o. = the _L from A to ZX, A is a rt.A (216). (3) If s.o. > the -L from A to KX, there are two A. In this instance the arc described, using A as cen- ter and " s.o" as radius, in- tersects KX twice, at B and B f . Hence, A AKB and A AKB 1 both contain the three given parts. CONCERNING ORIGINAL CONSTRUCTIONS ANALYSIS Many constructions are so simple that their correct solution will readily occur to the pupil. Sometimes, as in the case of complicated con- structions, one requires the ability to put the given parts together, one by one. The following outline may be found helpful if employed intelligently. I. Suppose the construction made, that is, suppose the figure drawn. II. Study this figure in search of truths by which the order of the lines that have been drawn can be determined. This is essential. III. One or more auxiliary lines may be necessary. IV. Finally, construct the figure and prove it correct. EXERCISE. Given the base of a triangle, an adjacent acute angle, and the difference of the other sides, to construct the triangle. Given: Base^ljB; /.A'\ differenced. Required : To construct the A. d [Analysis : Suppose A AE C is the required A. ( It is evident that if CD = CB, they may be sides of an isos. A and AD = d. This isosceles A may be constructed.] Construction: At A on AE construct ZBA\ = ^A' and take on AX, AD = d. Join DB. At Af, midpoint of DB, draw M Y to DB meeting AX at C. Draw CB. Statement : (?) . Proof : (?) . Discussion : (?) . r x 128 PLANE GEOMETRY ORIGINAL CONSTRUCTIONS 1. To construct a right angle. 2. To construct an angle containing 45. 3. To construct the complement of a given angle ; the supplement. 4. To construct an angle of 60. [See 115.] 5. To construct an angle of 30 ; of 15 ; of 120. 6. To construct an angle of 150; of 135 ; of 75 ; of 165. 7. To find the center of a given circle. [See 214.] 8. To construct a tangent to a given circle, parallel to a given line. [Draw a radius _L to the given line.] 9. To construct a tangent to a given circle, perpendicular to a given line. 10. To construct the other acute angle of a right triangle if one is known. 11. To draw through a given point without a given line, another line which shall make a given angle with the line. [Draw a || to the given line through the given point.] 12. To trisect a right angle. 13. To find a point in one side of a triangle equally distant from the other sides. [Use 266.] 14. To construct a chord of a circle if its midpoint is known. [Draw a radius through this point and use 267.] 15. To construct the shortest chord that can be drawn through a given point within a circle. Proof: Draw any other chord through the point, etc. 16. To construct through a given point within a circle two chords each equal to a given chord. 17. To construct in a given circle a chord equal to a second chord and parallel to a third. 18. To construct through a given point a line which shall make equal angles with the sides of a given angle. [Use 266 ; 268.] 19. To construct from a given point in a given circumference a chord which shall be at a given distance from the center. [How many can be drawn from this point?] BOOK ir 129 1. To construct an isosceles triangle, having given: 20. The base and one of the equal sides. 21. The base and one of the equal angles. 22. One of the equal sides and the vertex-angle. 23. One of the equal sides and one of the equal angles. 24 The base and altitude upon it. 25. The base and the radius of the inscribed circle. [Bisect the base; erect a JL = radius; describe 0, etc.] 26. The base and the radius of the circumscribed circle. [First, describe O with given radius and any center.] 27. The altitude and the vertex-angle. [Draw an indefinite line and erect a _L equal the given altitude. Bisect the given Z ; at the end of the altitude construct Z = given Z, etc.] 28. The base and the vertex-angle. [Find the supplement of given Z ; bisect this ; at each end of base construct an Z = this half ; etc.] 29. The perimeter and the altitude. Given: Perimeter = AB; alt. = h. Required: (?). Construction: Bisect AB, erect at M = A; draw AP and BP. Bisect these ; erect _b SC and RE ; etc. II. To construct a right triangle, having given : 30. The two legs. 31. One leg and the adjoining acute angle. 32. One leg and the opposite acute angle. 33. The hypotenuse and an acute angle. 34. The hypotenuse and the altitude upon it. 35. The median and the altitude upon the hypote- nuse. [Same as No. 34.] 36. The radius of the circumscribed circle and a leg. 37. The radius of the inscribed circle and a leg. Given: Radius = r; leg = CM. Required: (?). Analysis : Consider that A BC is the completed figure ; CNOM is a square, whose vertex O is the center of the circle, and side OA r is the given radius. AB is tan- gent from A. Construction : On CA take CN '= r and construct square, CNOM. Prolong CM indefinitely. BOBBINS' PLANE GEOM. 9 R ..- --\ r\\ C H Describe O, etc. 130 PLANE GEOMETRY 38. One leg and the altitude upon the hypotenuse.' 39. An acute angle and the sum of the legs. Given: AD = mm; Z K. Required: (?). Construction : At A construct Z A Z K; at D construct Z D = 45, the sides of these A intersecting at B. Draw BC to AD ; etc. 6 40. The hypotenuse and the sum of the legs. [Use A as center, hypotenuse as radius, etc.] 41. The radius of the circumscribed circle and an acute angle. 42. The radius of the inscribed circle and an acute angle. Construction: Take CS on indefinite line B ZA = r. On CS construct square CSOM. At construct /. MOX = Z K. Draw radius OT 45". JL to OX. Draw tangent at T. Proof : A ABC is a rt. A and it is the rt. A. (Explain.) K III. To construct an equilateral triangle, having given : 43. One side. 44. The altitude. 45. The perimeter. 46. A median. 47. The radius of the inscribed circle. [Draw circle and radius; at center construct Z.ROS = 120 and Z ROT = 120; etc.] 48. The radius of the circumscribed circle. IV. To construct a triangle, having given : 49. The base, an angle adjoining it, the altitude upon it. 50. The midpoints of the three sides. [Draw R S, R T, ST, etc.] 51. One side, altitude upon it, and the radius of the circumscribed circle. / '*'.... / Construction : Draw O with given radius and A ~"C any center. Take chord = given side ; etc. 52. One side, an adjoining angle, and the radius of the circumscribed circle. 53. Two sides and the altitude from the same vertex. Construction: Erect JL altitude, upon an indefinite line. Use the end of this altitude as center and the given sides as radii; etc. BOOK II 131 54 One side, an angle adjoining it, and the D /X sum of the other two sides. - Construction: At A construct Z BA X = given _ / K. On AX take AD = s; draw DB ; bisect DB at M, etc. 55. Two sides and the median to the third side. Given : a, b, m. Construction : Construct A ABR whose three sides, AB = a, BR = b, AR = 2 m. Draw^C II to BR and RC II to AB meeting at C. Drawee. Statement: (?). Proof: (?) 56. A side, the altitude upon it, and the angle opposite it. Given : Side =AB, alt. = h ; opposite Z = Z C'. Construction: Upon AB construct segment ACB which will contain A = Z C' (by 276). At A erect A R JL to AB and = A; etc. 57. A side, the median to it, the angle oppo- site it. [Statement: A ABC is the required A.] 58. One side and the altitude from its extremi ties to the other sides. Given: Side = AB, altitudes x and y. Construction: Bisect AB; describe a semicircle. Using A as center and x as radius, describe arc cutting the semicircle at R; etc. 59. Two sides and the altitude upon one of them. [Given : Sides = A B and BC ; alt. on EC = *.] x 60. One side, and angle adjoining it, and the radius of the inscribed circle. Construction: Describe O with given radius, any center. Construct central Z = given Z. Draw two tangents || to these radii. V. To construct a square, having given : 61. One side. 62. The diagonal. 63. The perimeter. 64. The sum of a diagonal and a side. 132 PLANE GEOMETKY VI. To construct a rhombus, having given: 65. One side and an angle adjoining it. 66. One side and the altitude. 67. The diagonals. 68. One side and one diagonal. [Use 281.] 69. An angle and the diagonal to the same vertex. 70. An angle and the diagonal between two other vertices. 71. One side and the radius of the inscribed circle. VII. To construct a rectangle, having given: 72. Two adjoining sides. 73. A diagonal and a side. 74. One side and the angle formed by the diagonals. 75. A diagonal and the sum of two adjoining sides. [See No. 40.] 76. A diagonal and the perimeter. 77. The perimeter and the angle formed Y \ x by the diagonals. " Construction : Bisect the perimeter and take A B = half it. Bisect Z K. At A con- K struct Z BA X = half Z K. Etc. VIII. To construct a parallelogram, having given: 78. One side and the diagonals. [Use 137 and 278.] 79. The diagonals and the angle between them. 80. One side, an angle, and the diagonal not to the same vertex. 81. One side, an angle, and the diagonal to the same vertex. 82. One side, an angle, and the altitude upon that side. 83. Two adjoining sides and the altitude. IX. To construct an isosceles trapezoid, having given: 84. The bases and an angle adjoining the larger base. 85. The bases and an angle adjoining the less base. 86. The bases and the diagonal. 87. The bases and the altitude. 88. The bases and one of the equal sides. 89. One base, an angle adjoining it, and one of the equal sides. 90. One base, the altitude, and one of the equal sides. BOOK TT 133 91. One base, the radius of the circumscribed circle, and one of the equal sides. [First, describe a 0.] 92. One base, an angle adjoining it, and the radius of the circum- scribed circle. 93. The bases and the radius of the circumscribed circle. 94. One base and the radius of the inscribed circle. Construction : Bisect the base and erect a _L = radius ; etc. X. To construct a trapezoid,* having given: 95. The bases and the angles adjoining one of them. Construction: Take EC = longer base, and on it take ED = less base. Construct A DEC (by 280). 96. The four sides. 97. A base, the altitude, and the non-parallel sides. Construction : Construct a A two sides of which = the given non-11 sides of the trapezoid, and the alt. from same vertex = given alt. (See No. 53.) 98. The bases, an angle, and the altitude. Construction: Construct O on ED, having given altitude and Z. 99. A base, the angles adjoining it, and the altitude. 100. The longer base, an angle adjoining it, and the non-parallel sides. 101. The shorter base, an angle not adjoining it, and the non-parallel sides. XI. To construct the locus of a point which will be : 102. At a given distance from a given point. 103. At a given distance from a given line. 104. At a given distance from a given circumference : (i) If the given radius is < the given distance ; (i) If the given radius is > the given distance. 105. Equally distant from two given points. 106. Equally distant from two intersecting lines. * NOTE. It is evident that every trapezoid may be divided into a parallelogram and a triangle by drawing one A line (as BD) II to one of the non-11 sides. Hence / the construction of a trapezoid is often merely / constructing a triangle and a parallelogram. _' A 134 PLANK GEOMETRY XII. To find (by intersecting loci) * the point P, which will be : 107. At two given distances from two given points. f 108. Equally distant from three given points. 109. In a given line and equally distant from two given points. 110. In a given line and equally distant from two given intersecting lines. 111. In a given circumference and equally distant from two given points, f 112. In a given circumference and equally distant from two inter- secting lines.f 113. Equally distant from two given intersecting lines and equally distant from two given points-! 114. At a given distance from a given line and equally distant from two given points.f 115. At a given distance from a given line and equally distant from two other intersecting lines, f 116. Equally distant from two given points and at a given distance from one of them.f 117. Equally distant from two given intersecting lines and at a given distance from one of them.f 118. At a given distance from a point and equally distant from two other points.f 119. At given distances from two given intersecting lines.f 120. At given distances from a given line and from a given circum- ference.! 121. At given distances from a given line and from a given point.f 122. Equally distant from two parallels and equally distant from two intersecting lines.f 123. At a given distance from a given point and equally distant from two given parallels, f * It is well to draw the loci concerned as dotted lines. (See No. 124.) I In the Discussion, include the answers to questions like these : (1) Is this ever impossible ? (i.e. must there always be such a point ?) (2) Are there ever two such points ? When ? (3) Are there ever more than two ? When ? (4) Is there ever only one ? When ? Etc. BOOK II 135 124. At a given distance from a given point and equally distant from two given intersecting lines. Can C be so taken that there will be no point ? Can C be so taken that there will be only one point? Only two? Only three? More than four ? XIII. To find (by intersecting loci) the center of a circle which will : 125. Pass through three given points.* 126. Pass through two given points and touch a given line at a given point.* 127. Pass through two given points and touch two given intersecting lines.* 128. Have a given radius, touch a given line, and pass through a given point.* 129. Pass through a given point and touch two given parallel lines.* 130. Pass through two given points and touch two given parallels.* 131. Have a given radius and touch two given intersecting lines.* 132. Have a given radius and pass through two given points.* 133. Touch three given indefinite lines, no two of them being parallel.f 134. Touch three given lines, only two of them being parallel. XIV. To construct a circle which will : 135. Pass through two given points and touch a given line at a given point. 136. Pass through two given points and touch two given intersecting lines. 137. Pass through two given points and touch two given parallels. 138. Have a given radius, touch a given line, and pass through a given point. 139. Have its center in one line, touch another line, and have a given radius. * Discussion : Is this ever impossible ? Are there ever two circles and hence two centers ? Are there ever more than two ? Etc. t Four solutions. One is in 274. 136 PLANE GEOMETRY 140. Have a given radius and touch two given intersecting lines. 141. Have a given radius and pass through two given points. 142. Have a given radius and touch a given circumference at a given point. [Draw tangent to the given O at the given point.] 143. Have a given radius and touch two given circumferences. 144. Touch three indefinite intersecting lines.* 145. Touch two given intersecting lines, one of them at a given point. 146. Touch a given line and a given cir- cumference at a given point. Given : Line AB ; O C; point P. Construction : Draw radius CP. Draw tangent at P meeting AB at R. Bisect Z.PRB, meeting CP produced at O; etc. 147. Be inscribed in a given sector. Construction : Produce the radii to meet the tangent at the midpoint of the arc. In this A inscribe a O. 148. Have a given radius and touch two given circles. 149. Have a given radius, touch a given line, and a given circumference. 150. Touch a given line at a given point and touch a given circumference. Given : Line AB; point P; O C. Construction : At P erect PX to A B, and extend it below AB, so PR = radius of O C. Draw CR and bisect it at M. Erect MY to CR at M, meeting PX at 0; etc. 151. What is the locus of the vertices of all right triangles having the same hypotenuse? 152. Through a given point on a given circumference to draw two equal chords perpendicular to each other. 153. To draw a line of given length through a given point and ter- minating in two given parallels. Construction : Use any point of one of , the Us as center and the given length as radius to describe an arc meeting the other ||. Join these two points. Through the given point draw a line II, etc. * Four solutions. One is in 274. BOOK II 137 154. To draw a line, terminating in the sides of an angle, which shall be equal to one line and parallel to another. Statement : RS is = a and || to x. M 155. To draw a line through a given point within an angle, which shall be terminated by the sides of the angle and bisected by the point. Construction: Through P draw PD II to AC. Take on AB, DE = AD. Draw EPF-, etc. ^ 156. To circumscribe a circle about a rectangle. A F ^ 157. To construct three circles having the vertices of a given triangle as centers so that each touches the other two. _^ c Construction : Inscribe a O in the A ; etc. 158. To construct within a circle three equal circles each of which will touch the given circle and the other two. Construction: Draw a radius, OA, and construct /. A OB = 120 and Z A OC = 120. In these sectors inscribe, etc. 159. Through a point without a circle to draw a secant having a given distance from the center. 160. To draw a diameter to a circle at a given distance from a given point. 161. Through two given points within a circle to draw two equal and parallel chords. Construction : Bisect the line joining the given points and draw a diameter, etc. 162. To draw a parallel to side RC of tri- angle ABC, meeting AB in X and ^ICin F, such that XY=YC. 163. Find the locus of the points of contact of the tangents drawn to a series of concentric circles from an external point. 164. Given: Line AB and points C and D on the same side of it; find point X in AB such that Z A XC = Z BXD. \ /" Construction: Draw CEio AB and pro- F duce to F so that EF = CE. Draw FD meeting AB in X. Draw CX. 138 PLANE GEOMETRY 165. To draw r from one given point to another the shortest path which shall have one ^)oint in common with a given line. Statement : CX + XD is < CR + RD. 166. To draw a line parallel to side BC of tri- angle ABC meeting AB at X and AC at F, so tha,tXY=BX + YC. Construction : Draw bisectors of A B and C, meeting at O, etc. 167. To draw in a circle, through a given point of an arc, a chord which will be bisected by the chord of the arc. Construction : Draw radius OP meeting chord at C. Prolong PO to X so CX = CP. Draw XM II to AB meeting O at M. Draw PM cutting AB at Z>; etc. A> Is there any other chord from P bisected by AB7 168. To inscribe in a given circle a triangle whose angles are given. Construction : At the center construct three A, doubles of the given A. 169. To circumscribe about a given circle a triangle whose angles are given. Construction : Inscribe A (like No. 168) first, and draw tangents II to the sides. 170. Three lines meet in a point ; it is required to draw a line terminating in the outer two and bisected by the inner one. Construction : Through any point P, of OB, draw Us to the outer lines. Draw diagonal RS; etc. 171. To draw through a given point, P, a line which will be terminated by a given circumference and a given line and be bisected by P. Construction: Draw any line DX meeting AB at D. Draw PE It to AB meeting DX at E. Take EF=ED; etc. 172. Through a given point without a circle to draw a secant to the circle which shall be bisected by the circumference. Construction : Draw arc at T, using P as center and diam. of O as radius. Using T as center and same radius as before, describe circumference touch- ing O at C and passing through P. Draw PC meeting O O at M. ROOK II 139 173. To inscribe a square in a given rhombus. [Bisect the four A formed by the diagonals.] 174. To bisect the angle formed by two lines without producing them to their point of intersection. Construction : At P, any point in RS, draw PA II to XY-, bisect Z APS by PB. At any point in PB erect ML J_ to PB, meeting the given lines in M and L. Bisect ML at D and erect DC JL to ML, etc. 175. To construct a common external tangent to two circles. Construction : Using as a center and a B radius = difference of the given radii, con- struct (dotted) circle. Draw QA tangent to this O from Q ; draw radius OA and produce it to meet given O at B. Draw radius QC \\ to OB. Join BC. Statement : BC is tangent to both (D. Proof : AB = CQ (Const.). AB is || to CQ (?). .'.ABCQisa. O (?). But Z OA Q is a rt. Z (?) ; etc. 176. To construct a common internal tangent to two circles. Construction : Using as a center and a radius = the sum of the given radii, construct (dotted) circle. Draw QA tan- gent to this O from Q ; draw radius OA meeting given O at B, etc., as above. BOOK III PROPORTION. SIMILAR FIGURES 283. A ratio is the quotient of one quantity divided by another, both being of the same kind. 284. A proportion is the statement that two ratios are equal. 285. The extremes of a proportion are the first and last terms. The means of a proportion are the second and third terms. 286. The antecedents are the first and third terms. The consequents are the second and fourth terms. 287. A mean proportional is the second or third term of a proportion in which the means are identical. A third proportional is the last term of a proportion in which the means are identical. A fourth proportional is the last term of a proportion in which the means are not identical. 288. A series of equal ratios is the equality of more than two ratios. A continued proportion is a series of equal ratios in which the consequent of any ratio is the antecedent of the next following ratio. 289. EXPLANATORY. A ratio is written as a fraction or as an indi- cated division ; -, or a -H 6, or a : b. A proportion is usually written 7- = - or a : b x : y, and is read : " a is to b as x is to y" In this pro- y portion the extremes are a and y ; the means are b and x ; the antece- dents are a and x ; the consequents are b and y ; and y is a fourth proportional to a, 6, x. In the proportion a : m = m : z, the mean pro- portional is m, and the third proportional is z. 140 BOOK III 141 THEOREMS AND DEMONSTRATIONS 290. THEOREM. In a proportion the product of the extremes is equal to the product of the means. Given : ^ = or a : b = x : y. To Prove : ay = bx. b y Proof: = (Hyp.). Multiply by the common denomi- b y nator, by and obtain, ay = bx (Ax. 3). Q.E.D. 291. THEOREM. If the product of two quantities is equal to the product of two others, one pair may be made the extremes of a propor- tion and the other pair the means. Given : ay bx. To Prove : These eight proportions : 1. aib = x:y, 5. xiy = a:b, 2. a:x=b:y, 6. x:a = y:b, 3. b : a = y : x, 7. y : x = b : a, 4. b : y = a : x, 8. y : b x : a. Proof: 1. ay = bx (Hyp.). Divide each member by by, and obtain -^ = -^ (Ax. 3). .-. -^ = -, or a : b = x : y. Q.E.D. by by by 2. Divide by xy\ etc. 3. bx=ay(Hyp.'). Divide by ax; etc. NUMERICAL ILLUSTRATION. Suppose in this paragraph a = 4, b = 14, x = 6, y = 21 ; the truth of the above proportions can be clearly seen by writing these equivalents. 4 x 21 = 14 x 6 (True). 1. 4 : 14 = 6 : 21 (True) ; 2. 4 : 6 = 14 : 21 (True) ; etc. They will all be recognized as true proportions. 292. THEOREM. In any proportion the terms are also in proportion by alternation (that is, the first term is to the third as the second is to the fourth). Given : a : b = x : y. To Prove : a:xb\y. Proof: a:b = x:y (Hyp.). .-. ay = bx (290). Hence, a:x=b: y (291). Q.E.D. 142 PLANE GEOMETRY 293. THEOREM. In any proportion the terms are also in proportion by inversion (that is, the second terra is to the first as the fourth term is to the third). [The proof is similar to the proof of 292.] 294. THEOREM. In any proportion the terms are also in proportion by composition (that is, the sum of the first two terms is to the first, or second, as the sum of the last two terms is to the third, or fourth). ( a + b: a = x-\-y:z, or Given: a: b = x: y. To Prove i\ , , Proof: a: b = x:y (Hyp.). .'. ay = bx (?) (290). Add ax to each, and obtain, ax + ay ax + bx (Ax. 2). That is, a (x + y) = x ( a + ). Hence, a + b : a = x + y : x (?) (291). Similarly, by adding by, a + b : b = x + y : y. Q.E.D. 295. THEOREM. In any proportion the terms are also in proportion by division (that is, the difference between the first two terms is to the first, or second, as the difference between the last two terms is to the third, or fourth). (a b : a = x y '. x, or Given : a : b = x : y. To Prove : \ , , \a-b:b = x-yiy. Proof: a:b = xiy (Hyp.). .-. ay = 5* (?) (290). Subtracting each side from ax, ax ay ax bx (Ax. 2). That is, a(x y) = x(a 6). Hence, a-b : a = x- y : x (?) (291). Likewise, a b :b = x y : y. Q.E.D. NOTE I. The proportions of 294 and 295 may be written in many different forms (292, 293) . Thus, (1) a b:a = xy: x; (2) a b : b = x y : y ; (3) a b : x y = a : x, etc. NOTE II. In any proportion the sum of the antecedents is to the sum of the consequents as either antecedent is to its consequent. (Explain.) Also, in any proportion the difference of the antecedents is to the differ- ence of the consequents as either antecedent is to its consequent. (Explain.) Thus: o + x:b+ y = a:b = x : y. Also, a-x :b y a:b = x:y. BOOK TTT 148 296. THEOREM. In any proportion the terms are also in proportion by composition and division (that is, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference). Given : a : b = x : y. To Prove : a+ , = x "*"^- a b x y ft -I rk / y* I. #/ Proof: a _?. = x _y_ (?) (294). a x a L -^ == x--jt ^ ^ 95 ^ Divide the first by the second, a _ = x ^ (?). Q.E.D. a b x y 297. THEOREM. In any proportion, like powers of the terms are also in proportion, and like roots of the terms are in proportion. Given : a : b = x : y. To Prove : a n : b n = x n : y n ; and Sfa : 3/b = %/x : ^y. Proof : [Write the given proportion in fractional form, etc.] 298. THEOREM. In two or more proportions the products of the cor- responding terms are also in proportion. Given : aib = x:y, and c : d = I : m, and e :/= r : 8. To Prove : ace : bdf= xlr : yms. Proof: [Write in fractional form and multiply.] 299. THEOREM. A mean proportional is equal to the square root of the product of the extremes. Given : a : x x : b. To Prove : x = Va6. Proof : [Use 290 ; etc.] 300. THEOREM. If three terms of one proportion are equal to the corresponding three terms of another proportion, each to each, the re- maining terms are also equal. f a : b = c : m, and ) Given : v , K To Prove : m = r. ( a : b = c : r. } Proof : am = Jc.and ar = be (?) (290). .-. am = ar (Ax. 1). Hence, m = r (Ax. 3). Q.E.D. 144 PLANE (IKOMETRY 301. THEOREM. In a series of equal ratios, the sum of all the ante- cedents is to the sum of all the consequents as any antecedent is to its consequent. Given: f = = Hi- b d f h To Prove: 4= f = l -etc. b + d +f+ h b d Proof : Set each given ratio = m ; thus, a c e a l = m\ -=mi - = w;f = w. b d f h .'. a=bm, c = dm, e=fm, g = hm (Ax. 3). Hence a + c +e+g bm + dm+fm + hm (Substitution) l+d+f+h b+d+f+h = m (Canceling). ' b + d+f+h b d f h Q-B.D. EXERCISES 1. If 3:4 = 6:z, find x. 2. If 8 : 12 = 12 : ar, find x. 3. Find a fourth proportional to 6, 7, and 15. 4. Find a third proportional to 4 and 10. 5. If 11: 15 = x: 25, find x. 6. If 4 : x = x : 25, find x. 7. Find a mean proportional between 8 and 18. 8. If 7: x = 35: 48, find x. 9. Given, that 5:8=15: 24, write seven other true proportions con- taining these same four numbers. 10. If 5x6 = 2x1 5, write eight proportions with these numbers. 11. If 7 : 12 = 21 : 36, write the proportion resulting by alternation ; inversion ; composition ; division ; composition and division. 12. If 6 : 25 = 18 : 75, write the proportions required in No. 11. 13. li x + y:x #=17:7, write the proportions that result by virtue of composition ; division ; composition and division. 14. Apply 301 to the ratios, f = f = f = T \ = ft. BOOK TIT 145 NOTE. We have seen that it is possible to add two lines and subtract one line from another. Now it is essential that we clearly understand the significance implied by indicating the multiplication or the division of one line by another. What is actually done is to multiply or divide the numerical measure of one line by the numerical measure of another. Thus if one line is 8 inches long and another is 18 inches long, we say that the ratio of the first line to the second is T \ or f, meaning that the less line is four ninths of the larger. Also, in referring to the product of two lines we merely understand that the product of their numerical measures is intended. If a line is multiplied by itself, we obtain the square of the numerical measure of the line. The square of the line AB is written AB 2 or (AB)' 2 and the quantity that is squared is the numerical value of the length of A B. In the preceding paragraphs of Book TIT, we have been considering numerical magnitudes. It should be distinctly understood that in the following geometrical propositions and demonstrations, the foregoing- interpretation is implied in multiplication and division involving lines. 302. THEOREM. A line parallel to one side of a triangle divides the other sides into proportional segments. Given : A ABC and line LM A II to BC. To Prove: AL : LB = AM : MC. Proof: I. If the parts AL and LB are commensurable. There exists a common unit _ __ _ _ of measure of AL and LB (238). E Suppose this is contained 9 times in AL and 5 times in LB. Then, ||= | (Ax. 3). LiB O Draw lines through the several points of division II to BC. These will divide AM into 9 parts and MC into 5 parts. All of these 14 parts are equal (?) (147). Hence, = (Ax. 3). .-. = (Ax. 1). MC 5 LB MC Q.E.U. BOBBINS' PLANE GEOM. 10 146 PLANE GEOMETRY II. If the parts AL and LB are incommensurable . There does not exist a common unit (238). Divide AL into sev- eral equal parts (by 271). Ap- ply one of these as a unit of measure to LB. There will be a _ B - C .remainder, #B, left over (238). Draw RS || to BC. Now = (The commensurable case). LR MS Indefinitely increase the number of equal parts of AL. That is, indefinitely decrease each part, the unit or divisor. Hence, the remainder, RB, will be indefinitely decreased. (Because the remainder is < the divisor.) That is, RB will approach zero as a limit, and SC will approach zero as a limit. /. LR will approach LB as a limit (240), and MS will approach MC as a limit (240). .-. will approach as a limit (243), LR LB and - will approach ^ as a limit (243). MS MC Consequently, = (?) (242). Q.E.D. LB MC 303. THEOREM. If a line parallel to one side of a triangle intersects the other sides, these sides and their corresponding segments are proportional. Given : A ABC ; LM \\ to BC. To Prove : I. AB : AC = AL : AM. II. AB: AC= LB : MC. Proof : AL : LB = AM: MC (?) (302). BOOK III 147 /. I. AL + LB : AL = AM + MC : AM (?) (294), and II. AL + LB : LB = AM + M C : M C (?). But AL + LB = AB and AM + MC = AC (Ax. 4). Therefore, I. ^L# : AL = AC: AM (Ax. 6). .Hence, AB : AC = AL : AM (?) (292). II. AB:LB = AC:MC (Ax. 6). Hence, AB : AC = LB : MC (?) (292). Q.E.D. NOTE. Each of these proportions may be written eight ways (291). And they may be combined, thus, = = (Ax. 1). A C A M. Jvl C 304. Two lines are divided proportionally, if the ratio of the lines is equal to the ratios of corresponding segments. 305. THEOREM. If a line parallel to one side of a triangle intersects the other sides, it divides these sides proportionally. (Because the ratio of the sides = the ratio of correspond- ing segments (303). This theorem is the same as 303.) 306. THEOREM. Three or more parallels intercept proportional seg- ments on two transversals. Given: (?). To Prove : AC : BD = CE : DF = EG : FH. Proof: Draw from A, AT II to BH intersecting the 11$, etc. In A AES, / \ \ ^=^ = ^(?) (305). / \ AS AR ES In A AGT, 4E = E^ (?) (305). AS ST . AC _CE_EG , A -,, . . XXA. xi. AR ES ST But, AR = BD, RS=DF, <*> C BC^ Jc~ We Q.E.D. 150 PLANE GEOMETRY 312. Similar polygons are polygons that are mutually equiangular and whose homologous sides are proportional. That is, every pair of homologous angles are equal ; and the ratio of one pair of homologous sides is equal to the ratio of every other pair of homologous sides, a : a' b : V = c : c' = d : d' = etc. Triangles are similar if they are mutually equiangular and their homologous sides are proportional. 313. THEOREM. Two triangles are similar if they are mutually equiangular. Given : A ABC, DEF-, Z ^i = Z D, Z # = Z #, Z c = Z>. To Prove : The A are similar (that is, that their sides are proportional). Proof : Place A ABC upon A DEF so that Z A coincides with its equal, Z D, and A ABC takes the position of A DBS. Then Z DRS = ^E (Hyp.). .-. BS is II to EF (?) (102). Hence, DE : DB = DF : DS (?) (305). That is, DE : AB = DF : AC (Ax. 6). Likewise, by placing Z B upon Z E, we may prove that, DE : AB = EF: EC. .'. DE : AB = DF: AC = EF : BC (Ax. 1). Therefore, the A are similar (?) (312). Q.E.D. BOOK III 151 314. THEOREM. Two triangles are similar if two angles of one are equal to two angles of the other. [See 117 and 313.] 315. THEOREM. Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. [See 314.] 316. THEOREM. If a line parallel to one side of a triangle intersects the other sides, the triangle formed is similar to the original triangle. Given : MN II to BCin A ABC. To Prove : A AMN similar to A ABC. Proof : Z A is common to both; Z^L# = ZJ3; Z ANN = Z C (?) (98). /. A are similar (?) (313). Q.E.D. 317. THEOREM. If two triangles have an angle of one equal to an angle of the other and the sides including these angles proportional, the triangles are similar. B C E F Given : A ABC and DEF\ Z A = Z D; DE:AB = DF:AC. To Prove: The A similar. Proof : Superpose A ABC upon A DEF so that Z A coincides with its equal, Z D, and A ABC takes the position of A DBS. Then DE: DR = DF: DS (Hyp.). /. BS is II to JF(?) (307). /. A DBS is similar to A DEF (?) (316). Q.E.I). 152 PLANE GEOMETRY 318. THEOREM. If two triangles have their homologous sides pro- portional, they are similar. b B C E F Given : A ABC and DEF, and DE : AB = DF: AC = EF : BC. To Prove : A ABC similar to A DEF. Proof: On DE take DKAB\ and on DF take DL = AC. Draw KL. Now, DE : AB = DF: AC (Hyp.)- .'. DE : DK = DF : DL (Ax. 6). .'. KL is I! to EF (?) (307). Therefore, A DKL is similar to A DEF (?) (316). [Now A ABC is to be proven equal to A YX7TL.] DE:DK=EF:KL. (Definition of similar triangles, 312.) That is, DE : AB = EF : KL (Ax. 6). But, DE: AB = EF: BC (Hyp.). .'. BC = KL (300). Hence, A ABC = A~DKL (?) (58). But A DKL has been proven similar to A DEF. Therefore, A ABC is similar to A DEF (Ax. 6). Q.E.D. Ex. 1. Are all equilateral triangles similar? Why? Ex.2. Are all squares similar? Why? Ex. 3. Are all rectangles similar? Why? Ex. 4. The sides of a triangle are 7, 8, and 12, and the longest side in a similar triangle is 30. Find the other sides. Ex. 5. In the figure of 311. if AB = 10, AC = 14, BC = 18, find the four segments of AC made by S and S'. Ex. 0. Prove the theorems of 142 and 143 by proportion. BOOK III 153 319. THEOREM. If two triangles have their homologous sides parallel, they are similar. A D E Given: A ABC and DEF; AB II to DE\ AC II to DF; and BC II to EF. To Prove : A ABC similar to A DEF. Proof: Produce BC of A ABC until it intersects two sides of A DEF at R and 8. Now Z B = Z Z>#S, and Z ZXRS = Z E(?) (98). Likewise, Z.ACB = Z D.S#, and /.DSR = /_ F (?). Therefore, A ABC is similar to A DEF (?) (314). Q.E.D. 320. THEOREM. If two triangles are similar to the same triangle, they are similar to each other. Proof: The three angles of each of the first two triangles are respectively equal to the three angles of the third (312). Hence, the first two A are mutually equiangular (Ax. 1). Therefore they are similar (?) (313). Q.E.D. Ex. 1. Let the pupil prove the theorem of #19 if one triangle entirely surrounds the other. Ex. 2. If one side of one triangle intersects two sides of the other. Ex. 3. If they are so placed that no side of either, when prolonged, intersects any side of the other without being prolonged. [Prolong any side of one and the sides not || to it in the other.] 154 PLANE GEOMETRY 321. THEOREM. If two triangles have their homologous sides per- pendicular, they are similar. c Given: A ABC and DEF-, AB _L to DE-, AC A. to DF-, BC J_ to EF. To Prove : A ABC similar to A DEF. Proof : Through P, any point in EF, construct PR II to AC, meeting DF at M. At R, any point in PM, draw RS II to AB, meeting ED at N. Draw PS II to BC, meeting NR at S, forming the APRS. PJfisJ_toDFand #^is_L toZ)j(?). In quadrilat- eral D3f#, Z D + Z Jf + Z Jffijy + Z 2V = 4 rt. A (?) (165). But, Z 3f +^ JT=2 rt. ^ (Const.). /. Z D + Z J/flJvr = 2 rt. Zs (Ax. 2). But, Z a + Z MRN=2 rt. Zs (?) (46). .-. Z D=Z (?) (49). Similarly, by quadrilateral EPSN, it may be proved that Z^=Z6. .'.A DEF is similar to A PRS (?) (314). But A ABC is similar to A PRS (?) (319). /. A ABC is similar to A DEF (?) (320). Q.E.D. Ex. 1. In the figure of 321, prove that Z F = /. c by using 48. Ex. 2. Draw the figure for the theorem of 321 if P is taken on EF prolonged. Prove the theorem with this figure. Ex. 3. Prove the same theorem if P is taken at a vertex. Ex. 4. Prove, if P is taken within the triangle DEF. HOOK ITT , 155 322. THEOREM. Two homologous altitudes of two similar triangles are proportional to any two homologous sides. Given : (?). BL AB AC BC To Prove: B'L' A'B' A'C' B'C' B L' C 1 A L C A 7 Proof : A ABC is similar to A A' B'C' (?). (312). .-.AABL is similar to A A' B'L' (315). BL AB o m R (312) * But ' AB AC BC l^c 1 Hence, BL B'L' AB _ AC _BC^ (A n A^~"A^~B'C' ^ Ax ' 1 > Q.E.D. 323. It is evident that : In similar figures, 1. Homologous angles are equal. 2. Homologous sides are opposite equal angles (in tri- angles). Thus, shortest sides are homologous. [Opp. smallest A] Medium sides are homologous. [Opposite medium A] Longest sides are homologous. [Opposite largest A] 3. Homologous sides are proportional. The antecedents of this proportion belong to one of the similar figures and the consequents to the other. Ex. 1. Prove the theorem of 322 by use of triangles BLC and B'L'C 1 . Ex. 2. State all the instances under which two triangles are similar. Ex. 3. In the figure of 322, if AB = 13, AC = 15, BL = 9, A'C 1 = 20, find A'B' and B'L'. lf>0 PLANE GKOMKTRY 324. THEOREM. If two parallel lines are cut by three or more trans- versals which meet at a point, the corresponding segments of the par- allels are proportional. Given: (?). To Prove : = = . CG GH HD Proof : In A COG, AE is || to CG (Hyp.). /. A OAE is simi- lar to A OCG (?) (316). Likewise, A OEF is similar to A OGH and A OFB is simi- lar to AOHD (316). OE G H ... f- = ^fl(?) (323, 3); also '- L - = (?). CG OG ^ GH OG ^ *\* lit JtjJy , O s v T *1 jLU I \JJi \ J^ O xOX .-. =--(?). Likewise, = (--) = . (?). CG GH GH \OH/ HD Therefore, = = (?). CG GH HD Q.E.D. 325. THEOREM. If three or more non-parallel transversals intercept proportional segments on two parallels, they meet at a point. Given: Transversals AB, CD, EF; parallels AE and BF ; pro- portion, AC : BD = CE : DF. To Prove : AB, CD, EF meet at a point. Proof : Produce BA and CD until they meet, at O. Draw OF cutting AE at X. D Now, AC : BD = CX : DF (?) (324). But AC : BD = CE : DF (Hyp.). .'. CX = CE (?) (300). Therefore, FE and FX coincide (?) (39). That is, FE produced passes through O. Q.E.D. BOOK III 157 326. THEOREM. The perimeters of two similar polygons are to each other as any two homologous sides. A B A' Given : Polygon R whose perimeter = p and similar polygon S whose perimeter = P r . To Prove : P : P 1 = AB : A' B' = BC : B'c 1 = etc. Proof: AB : A'B' = BC : B f c' = CD : C'D' = etc. (323, 3). .-. AB 4- BC + CD + etc. : A'B' + B'C* + C'D' + etc. = AB : A'B' = BC : B'C' = etc. (?) (301). .-. P : P f = AB : A'B' = BC : B f C f = etc. (Ax. 6). Q.E.D. Ex. 1. In the figure of 324, if AE=EF=FB, prove CO = GH = HD. State this truth in a theorem. Ex. 2. The median of a triangle bisects every line that is parallel to the side to which the median is drawn and has its extremities in the other sides of the triangle. Ex. 3. In the figure of 325, prove that the line bisecting AC and BD will pass through point O. Ex. 4. Prove the theorem of 324 if the point is between the parallels. Ex. 5. If AB and CD are any two parallel lines whose midpoints are R and S respectively, prove that the lines AD, BC, RS meet in a point. Ex. 6. Two homologous sides of two similar polygons are 8 and 15. The perimeter of the less polygon is 60. What is the perimeter of the larger? Ex. 7. The perimeters of two similar polygons are 30 and 125 re- spectively. If the shortest side of the larger is 8$, find the shortest side of the less. Ex. 8. The sides of a polygon are 5, 6, 7. 8, 10 respectively. Find the perimeter of a similar polygon whose medium side is 17J. 158 PLANE GEOMETRY 327. THEOREM. If two polygons are similar, they may be decom- posed into the same number of triangles similar each to each and similarly placed. Given : Similar polygons BE and B'E'. To Prove : A ABC similar to A A'B'C'\ A ACD similar to A A'C'D'; A AED similar to A A'E'D'. Proof : First. AB : A'B' = BC : B'C' (323, 3). Also, Z B = Z'(323, 1). Therefore, A ABC is similar to AA'B'C' (317). Second. In A ABC and A'B'C', -^- = -^- (?) (323, 3). B f C' A'C' In the similar polygons, -? =-. (?) (323, 3). B C CD Consequently, AC CD C'D' (Ax. 1). A'C' In the polygons, Z BCD = /.B'C'D'\ In the A ABC and A'B'C', Z.BCA = ^ B'C'A'\ ' Hence, by subtraction, /.ACD = Z.A'C'D' (Ax. 2). Therefore, A ACD is similar to A A' C'D' (?) (317). Third. A AED is proven similar to A A'E'D' in like manner. Q.E.D. HOOK LIT 159 328. THEOREM. If two polygons are composed of triangles similar each to each and similarly placed, the polygons are similar. K Given : A GUI similar to A G'H'I'; A GIJ similar to A G'I'J' ; A GJK similar to A G r J f K f . To Prove : The polygons HK and H r K f similar. Proof : First. In A HGI and H'G'I', Z H = Z H f (323, 1). Also in these A Z HIG = Z II'I'G' (?) (323, 1). In A GIJ and G'I'J', Z GIJ = Z G'/'j 7 (?). Adding, Z #/.E.D. Ex. 1. If, in 340, .IP = 3, PB = 27, find CP. Ex. 2. If, in 342, AP = 4, PB = 21, find viC and J5C. Ex. 3. If, in 342, AB = 20, A C = 6, find JP, P, CP, and BC. 343. THEOREM. The sum of the squares of the legs of a right tri- angle is equal to the square of the hypotenuse. Given : Rt. A ABC. To Prove : Iff + 7*c 2 = AB 2 . Proof : Draw CP J_ to the hypotenuse AB. Then AC 2 = AB AP (?) (342). And BC 2 = AB - BP (?). Adding, A(+ B! = AB AP + AB- BP (Ax. 2). = AB (AP + BP) = AB-AB = Al? (Ax. 4). That is, AC 2 -f BC 2 = ^1# 2 . Q.E.D. 344. THEOREM. The square of either leg of a right triangle is equal to the square of the hypotenuse minus the square of the other leg. That is, AC? = AB 2 - BC 2 ; and BC 2 = AB 2 - A(? (?) (Ax. 2). Ex. If A C = 28 and BC = 45, find AB. Ex. If AC = 21 and AB = 29, find BC. 345. THEOREM. In an obtuse triangle the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides plus twice the product of one of these two sides and the projection of the other side upon that one. Given : Obtuse A ABC ; etc. ^s^/'- Ta Prove: e 2 = a 2 -f ft 2 + 2 bp. ^>^ / k\ Proof : c 2 = h 2 4- (/> + />) 2 = ^^^ 7 W + p2 + #J + 2 bp (?) (343). ^^ 6 / t> \ But 7< 2 +jt? 2 =rt 2 (?) (343). A .\ side a. To Prove: TT 70 O O Proof: In A ARC and BRC, AE = BR (Hyp.); CR is common; AC > BC (Hyp.). .'. Z ARC > Z BRC (?) (87). That is, Z ARC is obtuse and Z BRC is acute. /. in A ARC, 5 2 = (l(?) 2 + m * + cp (345), and in A BRC, a 2 = ( J c) 2 + m 2 - cp (346). I. Adding, 52 + a z = 2 II. Subtracting, b 2 a 2 = (Ax. 2). (Ax. 2). Q.E.D. BOOK III 167 348. If the vertices of a triangle are denoted by A, B, C, the lengths of the sides opposite are denoted by a, 6, 6 2 ; A = V& 2 - 5 jt> a 2 ; etc. III. Medians. m c = J V2 (a 2 + 6 2 ) - c 2 ; w? a = \ V2(6 2 + c 2 ) -a 2 ; etc. IV. Bisectors. # c = Va6 n c r c * ; a = V6c n a r a *; 6 = Vac n 6 r 6 .* V. Diameter of circumscribed circle = ^; =--! =^' h b h c h a VI. Largest Angle. 1. Z C is right if c 2 = a 2 + & 2 (343). 2. ^ C is obtuse if c 2 > a 2 + 6 2 (345). 3. Z C is acute if c 2 < a 2 + 6 2 (346). Ex. 1. If the sides of a triangle are a = 7, & = 10, c = 12, find the nature of Z C. Ex. 2. In the same triangle find m a . Find m&. Find m c . Ex. 3. In the same triangle find a p b . Find b p a . Find a /> c . Find b p c . Find c /> a . Find c p b . Ex. 4. Find . Find h b . Find A c . Ex. 5. Find the diameter of the circumscribed circle. Ex. 6. Find n a and r a . Find n b and r 6 . Find n c and TV Ex. 7. Find t a . Find ? 5 . Find / c . * The segments n and r can be found by 308 j m, : r b = c : a, etc. 168 PLANE GEOMETRY CONCERNING ORIGINALS 350. We should first determine from the nature of each numerical exercise upon which theorem it depends. By applying the truth of that theorem, the exercise is usually solved without difficulty. ORIGINAL EXERCISES (NUMERICAL) 1. The legs of a right triangle are 12 and 16 inches ; find the hypote- nuse. 2. The side of a square is 6 feet; what is the diagonal? 3. The base of an isosceles triangle is 16 and the altitude is 15; find the equal sides. 4. The tangent to a circle from a point is 12 inches and the radius of the circle is 5 inches ; find the length of the line joining the point to the center. 5. In a circle whose radius is 13 inches, what is the length of a chord 5 inches from the center? [Draw chord, distance, and radius to its extremity.] 6. The length of a chord is 2 feet and its distance from the center is 35 inches ; find the radius of the circle. 7. The hypotenuse of a right triangle is 2 feet 2 inches, and one leg is 10 inches ; find the other. 8. The base of an isosceles triangle is 90 and the equal sides are each 53 ; find the altitude. 9. The radius of a circle is 4 feet 7 inches ; find the length of the tangent drawn from a point 6 feet 1 inch from the center. 10. How long is a chord 21 yards from the -center of a circle whose radius is 35 yards ? 11. Each side of an equilateral triangle is 4 feet ; find the altitude. 12. The altitude of an equilateral triangle is 8 feet ; find the side. [Let x = each side ; \ x = the base of each rt. A.] 13. Each side of an isosceles right triangle is a ; find the hypotenuse. 14. If the length of the common chord of two intersecting circles is 16, and their radii are 10 and 17, what is the distance between their centers ? 15. The diagonal of a rectangle is 82 and one side is 80 ; find the other. 16. The length of a tangent to a circle whose diameter is 20, from an external point, is 26. What is the distance from this point to the center? BOOK III 169 17. The diagonal of a square is 10 ; find each side. 18. Find the length of* a chord '2 feet from the center of a circle whose diameter is 5 feet. 19. A flagpole \va.s broken 16 feet from the ground, and the top struck the ground 63 feet from the foot of the pole. How long was the pole? 20. The top of a ladder 17 feet long reaches a point on a wall 15 feet from the ground. Mow far is the lower end of the ladder from the wall ? 21. A chord 2 feet long is 5 inches from the center of a circle. How far from the center is a chord 10 inches long? [Find the radius.] 22. The diameters of two concentric circles are 1 foot 10 inches and 10 feet '2 inches. Find the length of a chord of the larger which is tan- gent to the less. 23. The lower ends of a post and a flagpole are 42 feet apart ; the post is 8 feet high and the pole, 48 feet. How far is it from the top of one to the top of the other? 24. The radii of two circles are 8 inches and 17 inches, and their cen- ters are 41 inches apart. Find the lengths of their common external tangents; of their common internal tangents. 25. A ladder 65 feet long stands in a street ; if it inclines toward one side, it will touch a house at a point 16 feet above the pavement; if to the other side, it will touch a house at a point 56 feet above the pave- ment. How wide is the street ? 26. Two parallel chords of a circle are 4 feet, and 40 inches long, re- spectively, and the distance between them is 22 inches. Find the radius of the circle. [Draw the radii to ends of chords; these = hypotenuses = 72; the distances from the center = x and 22 x.~\ 27. The legs of an isosceles trapezoid are each 2 feet 1 inch long, and one of the bases is 3 feet 4 inches longer than the other. Find the altitude. 28. One of the non-parallel sides of a trapezoid is perpendicular to both bases, and is 63 feet long ; the bases are 41 feet and 25 feet long. Find the length of the remaining side. 29. If a = 10, h = 6, find />, c, //, ft. 30. If h = 8, />' = 4, find ft, c, p, a. 31. If a = 10, p' = 15, find c, p, h, ft. 32. If = 0, ft = 12, find c, />, />', h. 33. If p = 3, p' = 12, find a, /*, ft. 170 PLANE GEOMETRY 34. The line joining the midpoint of a chord to the midpoint of its arc is 5 inches. If the chord is 2 feet long, what is the diameter? 35. If the chord of an arc is 60 and the chord of its half is 34, what is the diameter ? 36. The line joining the midpoint of a chord to the midpoint of its arc is 6 inches. The chord of half this arc is 18 inches. Find the diameter. Find the length of the original chord. 37. To a circle whose radius is 10 inches, two tangents are drawn from a point, each 2 feet long. Find the length of the chord joining their points of contact. 38. The sides of a triangle are 6, 9, 11. Find the segments of the shortest side made by the bisector of the opposite angle. 39. Find the segments of the longest side made by the bisector of the largest angle in No. 38. 40. The sides of a triangle are 5, 9, 12. Find the segments of the shortest side made by the bisector of the opposite exterior angle. Also of the medium side made by the bisector of its opposite exterior angle. 41. In the figure of 306, if AC = 3, CE = 5, EG = 8, BD = 4; find DF and FH. 42. If the sides of a triangle are 6, 8, 12 and the shortest side of a similar triangle is 15, find its other sides. 43. If the homologous altitudes of two similar triangles are 9 and 15 and the base of the former is 21, what is the base of the latter? 44. In the figure of 324, AE = 4, EF=Q, FB=Q, #=15. Find CG and CD. 45. The sides of a pentagon are 5, 6, 8, 9, 18, and the longest side of a similar pentagon is 78. Find the other sides. 46. A pair of homologous sides of two similar polygons are 9 and 16. If the perimeter of the first is 117, what is the perimeter of the second ? 47. The perimeters of two similar polygons are 72 and 120. The shortest side of the former is 4, what is the shortest side of the latter ? 48. Two similar triangles have homologous bases 20 and 48. If the altitude of the latter is 36, find the altitude of the former. 49. The segments of a chord, made by a second chord, are 4 and 27. One segment of the second chord is 6, find the other. 50. One of two intersecting chords is 19 in. long and the segments of the other are 5 in. and 12 in. Find the segments of the first chord. 51. Two secants are drawn to a circle from a point; their lengths are BOOK Til 171 15 inches and 10| inches. The external segment of the latter is 10; find the external segment of the former. 52. The tangent to a circle is 1 foot long and the secant from the same point is 1 foot 6 inches. Find the chord part of the secant. 53. The internal segment of a secant 25 inches long is 16 inches. Find the tangent from the same point to the same circle. 54. Two secants to a circle from a point are 1 feet and 2 feet long ; the tangent from the same point is 12 inches. Find the external segments of the two secants. 55. The sides of a triangle are 5, 6, 8. Is the angle opposite 8 right, acute, or obtuse ? Same for the triangle 8, 7, 4 ? 56. The sides of a triangle are 8, 9, 12. Is the largest angle right, acute, or obtuse ? Same for the triangle 13, 7, 11 ? 57. The sides of a triangle are z, y, z. If z is the greatest side, when will the angle opposite be right? Obtuse? Acute? 58. The sides of a triangle are 6, 8, 9. Find the length of the projec- tion of side 6 upon side 8 ; of side 8 upon side 9 ; of side 9 upon side 6. 59. The sides of a triangle are 5, 6, 9. Find the length of the pro- jection of side 6 upon side 5 ; of side 9 upon side 6. 60. Find the three altitudes in triangle 9, 10, 17. 61. Find the three altitudes in triangle 11, 13, 20. 62. Find the diameter of circumscribed circle about triangle 17, 25, 26. 63. Find the length of the bisector of the least angle of triangle 7, 8, 20. Also of the largest angle. 64. Find the length of the bisector of the largest angle of triangle 12, 32, 33 ; also of the other angles. 65. Find the three medians in triangle 4, 7, 9. 66. Find the product of the segments of every chord drawn through a point 4 units from the center of a circle whose radius is 10 units. 67. The bases of a trapezoid are 12 and 20, the altitude is 8; the other sides are produced to meet. Find the altitude of the larger tri- angle formed. 68. The shadow of a yardstick perpendicular to the ground is 4 feet. Find the height of a tree whose shadow at the same time is 100 yards. 69. There are two belt-wheels 3 feet 8 inches and 1 foot 2 inches in diameter, respectively. Their centers are 9 feet 5 inches apart. Find the length of the belt suspended between the wheels if the belt does not cross itself. Also the length of the belt if it does cross. 172 PLANE GEOMETRY SUMMARY 351. Triangles are proven similar by showing that they have : (1) Two angles of one equal to two angles of the other. (2) An acute angle of one equal to an acute angle of the other. [In right triangles.] (3) Homologous sides proportional. (4) An angle of one equal to an angle of the other and the including sides proportional. (5) Their sides respectively parallel or perpendicular. 352. Four lines are proven proportional by showing that they are : (1) Homologous sides of similar triangles. (2) Homologous sides of similar polygons. (3) Homologous lines of similar figures. 353. The product of two lines is proven equal to the product of two other lines, by proving these four lines proportional and making the product of the extremes equal to the product of the means. 354. One line is proven a mean proportional between two others by proving that two triangles which contain this line in common are similar, and obtaining the required proportion from their sides. 355. In cases dealing with the square of a line, one uses : (1) Similar triangles having this line in common, or, (2) A right triangle containing this line as a part. ORIGINAL EXERCISES (THEOREMS) 1. If two transversals intersect between tw r o parallels, the triangles formed are similar. [Use 351 (1).] 2. Two isosceles triangles are similar if a base angle of one is equal to a base angle of the other. 3. Two isosceles triangles are similar if the vertex-angle of one is equal to the vertex -angle of the other. 4. The line joining the midpoints of two sides of a triangle forms a triangle similar to the original triangle. 5. The diagonals of a trapezoid form, with the parallel sides, two similar triangles. BOOK III 173 6. Two circles are tangent externally at P; through P three lines are drawn, meeting one circumference in .4, B, C, and the other in A', B', C". The triangles ABC and A'B'C' are similar. . _ 7. Prove the same theorem if the circles are tangent internally. 8. If two circles are tangent externally at P, and BB', CC' be drawn through P, terminating in the circumferences, the triangles PBC and PB'C' will be similar. [Draw the common tangent at P.] 9. Prove the same theorem if the circles are tangent internally. 10. If AD and BE are two altitudes of triangle ABC, the triangles ACD and BCE are similar. 11. Two altitudes of a triangle are reciprocally proportional to the bases to which they are drawn. To Prove : AD : BE = A C : BC. 12. The four segments of the diagonals of a trapezoid are proportional 13. If at the extremities of a right triangle perpen- diculars be erected meeting the legs produced, the new triangles formed will be similar. 14. In the figure of Xo. 13, prove : (1) Triangle ABC similar to each of the triangles A CE and BCD. (2) Triangle ABE similar to triangle ABD. (3) Triangle ACE similar to triangle ABD. (4) Triangle BCD similar to triangle ABE. (5) Triangles ABC, ABD, ABE similar. 15. If AD and BE are two altitudes of triangle ABC (fig. of No. 11), meeting at O, the triangles BOD and A OE are similar. 16. Triangles CED and ABC (fig. of No. 11) are similar. [First show A CA D and CEB similar . /. CA : CB = CD : CE (?). Then use 351 (4).] 17. Triangle ABC is inscribed in a circle and AP is drawn to P, the midpoint of arc BC, meeting chord CB at D. The triangles A BD and A CP are similar. B 174 PLANE GEOMETRY 18. Two homologous medians in two similar triangles are in the same ratio as any two homologous sides. [Prove a pair of the new triangles formed, similar, by 351 (4).] 19. Two homologous bisectors in two similar triangles are in the same ratio as any two homologous sides. 20. The radii of circles inscribed in two similar triangles are in the same ratio as any two homologous sides. [Bisect two pairs of homol. A ; draw the altitudes of these new & ; etc.] 21. The radii of circles circumscribed about two similar triangles are in the same ratio as any two homologous sides. [Erect J_ bisectors ; draw radius in each O.] 22. In any right triangle the product of the hypotenuse and the al- titude upon it is equal to the product of the legs. 23. If two circles intersect at A and B and A C and AD be drawn each a tangent to one circle and a chord of the other, the common chord AB will be a mean proportional between BC and ED. 24. If two circles are tangent externally, the chords formed by a straight line drawn through their point of contact have the same ratio as the diameters of the circles. [Draw com. tang, at point of contact ; draw di- ameters from point of contact ; prove A sim. ; etc.] 25. If AB is a diameter and BC a tangent, and AC meets the circumference at D, the diameter is a mean proportional between A C and AD. [Draw BD. Prove A containing AB similar.] 26. If a tangent be drawn from one extremity of a diameter, meeting secants from the other extremity, these secants and their internal segments will be recip- rocally proportional. To Prove: AC: AD = AS: AR. Proof: Draw RS. In A ARS and ACD, ^.A = Z. A and Z. ARS = Z D. (Explain.) Etc. 27. If AB is a chord and CE, another chord, drawn from C, the midpoint of arc AB, meeting chord AB at D, A C is a mean proportional between CD and CE. Prove the above theorem and deduce that, CE CD is constant for all positions of the point E on arc A EB. BOOK III 175 28. If chord A D be drawn from vertex A of inscribed isosceles triangle ABC, cutting EC at E, AB will be a mean proportional between AD and AE. Prove the above theorem and deduce that, AD AE is constant for all positions of the point D on arc BDC. 29. If a square be inscribed in a right triangle so that one vertex is on each leg of the triangle and the other two vertices on the hypotenuse, the side of the square will be a mean proportional between the other segments of the hypotenuse. To Prove: A D : DE = DE : EB. First prove &ADG and BEF similar. 30. If the sides of two triangles are respectively parallel, the lines joining homologous vertices meet in a point. (These lines to be pro- duced if necessary.) 31. In each, of the following triangles, is the greatest angle right, acute, or obtuse, 7, 24, 25 ? 13, 10, 8 ? 19, 13, 23 ? 32. Prove theorem of 329 by drawing two other auxiliary chords. 33. Prove theorem of 325 if point O is between the parallels. 34. Prove theorem of 336 by drawing A Y and BX. 35. In any triangle the difference of the squares of two sides is equal to the difference of the squares of their projections on the third side. [AB 2 = (?) ; BC 2 = (?) . Subtract, etc.] 36. If the altitudes of triangle ABC meet at 0, AB 2 - AC 2 = B0 2 - CO 2 . [Consult No. 35 and substitute.] 37. The square of the altitude of an equilateral triangle is three fourths the square of a side. [Let side = a, etc.] 38. If one leg of a right triangle is double the other, its projection upon the hypotenuse is four times the projection of the other. Proof: (2a) 2 = c/>; a? = cp' (?). D /. p = - ; p> = (Ax. 3). 176 PLANE GEOMETRY 39. If the bisector of an angle of a triangle bisects the opposite side, the triangle is isosceles. 40. The tangents to two intersecting circles from any point in their common chord produced are equal. [Use 333.] 41. If two circles intersect, their common chord, produced, bisects their common tangents. [Use 333.] 42. If A B and A C are tangents to a circle from A ; CD is perpendicular to diameter BOX from C; then AB . CD = BD . BO. [Use 351 (5).] 43. If the altitude of an equilateral tri- angle is h, find the side. [Denote the side by x and half the base by x.~\ 44. If one side of a triangle be divided by a point into segments which are proportional to the other sides, a line from this point to the opposite angle will bisect that angle. [Converse of 308.] To Prove : /. n =Z m in fig. of 308. Proof: Produce CB to P, making BP = AB; draw AP] etc. 45. State and prove the converse of 310. 46. Two rhombuses are similar if an angle of one is equal to an angle of the other. 47. If two circles are tangent internally and any two chords of the greater be drawn from their point of contact, they will be divided propor- tionally by the circumference of the less. [Draw diameter to point of contact and prove the right & similar.] 48. The non-parallel sides of a trapezoid and the line joining the mid- points of the bases, if produced, meet at a point. [Use method of 325.] 49. The diagonals of a trapezoid and the line joining the midpoint of the bases meet at a point. 50. If one chord bisects another, either segment of the latter is a mean proportional between the segments of the other. 51. Two parallelograms are similar if they have an angle of the one equal to an angle of the other and the including sides proportional. 52. Two rectangles are similar if two adjoining pairs of homologous sides are proportional. 53. If two circles are tangent externally, the common exterior tangent is a mean pro- portional between the diameters. [Draw chords PA, PC, PB, PD. Prove, first, APD and BPC straight lines. Second, &ABC and ABD, similar.] HOOK ill 177 54. In any rhombus the sum of the squares of the diagonals is equal to the square of half the perimeter. 55. If in an angle a series of parallel lines be drawn having their ends in the sides of the angle, their midpoints will lie in one straight line. 56. If ABC is an isosceles triangle and BX is the altitude upon AC (one of the legs), BC 2 = 2 A C - CX. [Use 346.] 57. In an isosceles triangle the square of one leg is equal to the square of the line drawn from the vertex to any point of the base, plus the product of the seg- ments of the base. Proof : Circumscribe a O ; use method of 338. 58. If a line be dra*wn in a trapezoid parallel to the bases, the seg- ments between the diagonals and the non- _ c parallel sides will be equal. Proof : & AHI and ABC are similar (?) ; &DJ# and DOB also (?). /.^L? = ^(?). AB BC A - D T\V V J A ff DJT /c = Jd (?> But ' if = l (?) ' (Use Ax * 1} ; etc> 59. A line through the point of intersection of the diagonals of a trapezoid, and parallel to the bases, is bisected by that point. 60. If M is the midpoint of hypotenuse AB of right triangle ABC, AH? + EC 2 - + AC 2 = 8 CJ/ 2 . 61. The squares of the legs of a right triangle have the same ratio as their projections upon the hypotenuse. 62. If the diagonals of a quadrilateral are perpendicular to each other, the sum of the squares of one pair of opposite sides is equal to the sum of the squares of the other pair. 63. The sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of the diagonals. [Use 347, I.] 64. If DE be drawn parallel to the hypotenuse A B of right triangle ABC, meeting AC at D and CB at E, AE 2 + ~BD 2 = AB 2 + DE 2 . [Use 4 rt. A having vertex C.] 65. If between two parallel tangents a third tangent be drawn, the radius will be a mean propor- tional between the segments of the third tangent. To Prove : BP : OP = OP : PD. Proof : A BOD is a rt. A (?). Etc. BOBBINS* PLANE GEOM. 12 178 PLANE GEOMETRY 66. If A BCD is a parallelogram, BD a diagonal, A G any line from A meeting BD at E, CD at F, and BC (pro- duced) at G, AE is a mean proportional be- tween EF and EG. Proof : & A BE and Z)F are similar (?) ; also &ADE and BEG (?). Obtain two ratios j -BE\ED and then apply Ax. 1. A 67. An interior common tangent of two circles divides the line join- ing their centers into segments proportional to the radii. 68. An exterior common tangent of two circles divides the line join- ing their centers (externally) into segments proportional to the radii. 69. The common internal tangents of two circles and the common external tangents meet on the line deter- mined by the centers of the circles. 70. If from the midpoint P, of an arc subtended by a given chord, chords be drawn cutting the given chord, the product of each whole chord from P and its segment adjacent to P will be constant. Proof: Take two such chords, PA and PC; draw diameter PX ; etc. Rt. & PST and PCX are similar. (Explain.) 71. If from any point within a triangle ABC, perpendiculars to the sides be drawn OR to AB, OS to BC, OT to A C, AR 2 + BS 2 + ~CT 2 = BR 2 + CS 2 + AT 2 . [Draw A O, BO, CO.] 72. If two chords intersect within a circle and at right angles, the sum of the squares of their four seg- ments equals the square of the diameter. To Prove: AP 2 + BP 2 + CP 2 + DP 2 = AR 2 - Proof: Draw BC, AD, RD. Ch. BR is JL to AB (?). /. CD is || to BR (?). .'. arc BC arc RD (?). Hence, ch. BC = ch. RD (?). Now, RD 2 = BC 2 = BP 2 + CP 2 (?). ;fD 2 = etc. (?). Finally, AR 2 = AD 2 + RD 2 = etc. (?). 73. The perpendicular from any point of an arc upon its chord is a mean proportional between the perpendiculars from the same point to the tangents at the ends of the chord. To Prove : PR : PT = PT : PS. Proof : Prove A ARP and BTP are sim., also A APT and PBS~(1). Thus, get two ratios each = PA : PB. BOOK III 179 74. If lines be drawn from any point in a circumfer- ence to the four vertices of an inscribed square, the sum of the squares of these four lines will be equal to twice the square of the diameter. Proof: A A PC, DPB, are rt. A; etc. 75. If lines be drawn from any external point to the vertices of a rectangle ABCD, the sum of the squares of two of them which are drawn to a pair of opposite vertices will be equal to the sum of the squares of the other two. To Prove : PA' 2 + PC 2 = PB* + PD*. Proof : Draw PEF to the base, etc. 76. Is the theorem of No. 75 true if the point is taken within the rectangle ? 77. If each of three circles intersects the other two, the three common chords meet in a point. Given : (?). To Prove : AB, LM, RS meet at 0. Proof : Suppose A B and LM meet at 0. Draw RO and produce it to meet the at X and X'. Prove OX = OX' (by 329). .-. X, X', S are coincident. 78. In an inscribed quadrilateral the sum of the products of the two pairs of opposite sides is equal to the product of the diagonals. Proof: Draw DX making Z CDX = Z ADB', & ADB and CDX are sim. (?) ; also A BCD and ADX (?). Hence, AB - DC = DB XC (?), and AD BC = DB AX (?). Adding ; etc. 79. If AB is a diameter, BC and AD tangents, meeting chords AF and BF (produced) at C and D respectively, AB is a mean proportional between the tangents BC and AD. 80. If ABO is isosceles, AB = CO, and AO : CO - CO : AC, prove: (1) AB = BC = CO ; (2) Z ABO = 2 Z ; (3) Z = 36. Proofs: (1) AO:AB = AB:AC(Ax. 6). /. A ABC is sim. to A A BO (?) (317). /. A A BC is isosceles (?) . Etc. (2) Z ABC = Z (?), and Z CBO = Z (?). /. Z ABO = 2 Z (Ax. 2). (3) A = I the sum of A of A A OB (?). Etc. 81. If from a point A on the circumference of a circle two chords be drawn and a line parallel to the tangent at A meet them, the chords and their segments nearer to A will be inversely proportional. 180 PLANE GEOMETRY CONSTRUCTIONS 356. PROBLEM. To find a fourth proportional to three given lines. Given : Three lines, a, b, c. Q ...-- B Required : To find a fourth b n "" ^ ...-%"' proportional to a, b, c. ...-'' R '"'" Construction : Take two in- ^ ...( definite lines, AB and AC, .,. ; ;:'.'1). Etc. 26. To inscribe in a given semicircle, a rec- tangle similar to a given rectangle. Construction : From the midpoint of the base draw line to one of the opposite vertices. At given center construct an = the Z at the midpoint. Proceed as in No. 25. Proof : First, prove a pair of A similar. Etc. 27. To inscribe a square in a given triangle. ^ Construction : Draw altitude A D ; con- struct the square A DEF upon AD as a side ; draw BF meeting AC at R. Draw RU \\ to AD] RS \\ to BC. Etc. Statement: (?). Proof: &BRU and BFE are similar (V). Also A BRS and BAF (?). Thence show that SR = RU. Etc. draw DC, meeting .0 ou c 186 PLANE GEOMETRY 28. To inscribe in a given triangle a rectangle similar to a given rectangle. Construction : Draw the altitude. On this construct a rectangle similar to the given rectangle. Proceed as in No. 27. 29. To construct a circle which shall pass through two given points and touch a given line. Given : Points A and B ; line CD. Construction: Draw line AB meeting CD ^~~ P R at P. Construct a mean proportional between PA and PB (by 361). On PD take PR = this mean. Erect OR JL to CD at R, meeting bi- sector of A B at 0. Use as center, etc. 30. To construct a line = V2 in. [Diag. of square whose side is 1 in.] 31. To construct a line = VS^in. [Hyp. of a rt. A, whose legs are 1 in. and 2 in. respectively.] 32. To divide a line into segments in the ratio of 1 : V2. 33. To divide a line into segments in the ratio of 1 : \/5. 34. To construct a line as, if x = , and a, b, c are lengths of three given lines. [That is, to construct a;, if c : a = b : x (291).] 35. To construct a line x, if x = . [3c : a = b : x.'] oc 36. To construct a line a:, if a: = Vab. [a : x = x : &.] 37. To construct a line x, if x = . c 38. To construct a line x, if x = V a 2 _ 52 [a + b:x = x:a b.~] 39. To construct a line x, if x = - . c 40. To construct a line y, if ay = 2 b 2 . 41. To construct a line = VlO in. 42. To construct a line = 2\/6 in. 43. To construct a line = Va 2 + & 2 , if a and b are given lines. BOOK IV UNIT OF LENGTH AREAS 366. The unit of surface is a square whose sides are each a unit of length. Familiar units of surface are the square inch, the square foot, the square meter, etc. 367. The area of a surface is the number of units of surface it contains. The area of a sur- face is the ratio of that surface to the unit of surface. Equivalent (=0=) figures are figures having equal areas. NOTE. It is often convenient to speak of "triangle," "rectangle," etc., when one really means " the area of a triangle," or " the area of a rectangle," etc. 368. THEOREM. If two rectangles have equal altitudes, they are to each other as their bases. Given : Rectangles AC and D c H EG having = altitudes, and their bases being AB and EF. To Prove : AC : EG = AB : EF. n Proof: I. If A Band EF are commensurable. There exists a common unit of measure of AB and EF (238). Suppose this unit is contained 3 times in AB and 5 times in EF. Hence, AB : EF= 3 : 5 (Ax. 3). Draw lines through these points of division, J_ to the bases. These will divide rectangle AC into three parts and EG into 5 parts, and all of these eight parts are equal (?) (140). Hence, AC : EG = 3 : 5 (?). .'. AC : EG = AB : EF (Ax. 1). Q.E.D. 187 188 PLANE (JKOMETRY II. If AB and EF are incom- D c H SG mensurable. There does not exist a com- mon unit (?) (238). Divide AB into several equal parts. Apply one of these as a unit of measure to EF. There will be a remainder, RF left over (Hyp.)- Draw RS _L to EF. Now, = (Case I). E8 ER Indefinitely increase the number of equal parts of AB ; that is, indefinitely decrease each part, or the unit or divisor. Hence the remainder, RF, will be indefinitely de- creased (?). That is, RF will approach zero as a limit, and, RFGS will approach zero as a limit. Hence, ER will approach EF as a limit (?), and E8 will approach EG as a limit (?). A C 1 AC 1 Therefore, - - will approach - - as a limit (?), ES EG and will approach as a limit (?). ER EF . u_f , ? . (242). EG EF Q.E.D. 369. THEOREM. Two rectangles having equal bases are to each other as their altitudes. (Explain.) 370. THEOREM. Any two rectangles are to each other as the prod- ucts of their bases by their altitudes. F Given : Rectangles A and B whose altitudes are a and c and bases b and d respectively. BOOK IV 189 ,ful i To Prove : A : B = a - b : c d. Proof : Construct a third rectangle X whose base is b and whose altitude is c. Then A : X= a : c (?) (369). Also, X : B = b : d (?). Multiplying, A : B = a - b : c - d (?) (Ax. 3). Q.E.D. 371. THEOREM. The area of a rectangle is equal to the product of its base by its altitude. Given : Rectangle Zf, whose base is b and altitude, h. To Prove : Area of B = b h. Proof : Draw a square J7, each of whose sides is a unit of length. This square is a unit of surface (366). Now, - = ^ = b - h (370). But - = area of B (367). .-. area of R = b - h (Ax. 1). Q.E.D. 372. THEOREM. The area of a square is equal to the square of its side. (See 371.) 373. THEOREM. The area of a parallelogram is equal to the product of its base by its altitude. Given : O ABCD whose base is b and altitude, h. To Prove : Area of ABCD = b- h. Proof : From A and B, the extremities of the base, draw Js to the upper base meeting F D EC it in F and E respectively. In rt. A ADF and BCE, AF = BE (?), AD = BC (?). .'. A ADF= A BCE (?). Now, from the whole figure A take A ADF and parallelogram ABCD remains ; and from the whole figure take A BCE and rectangle ABEF remains. .'. n ABCD =0 rect. ABEF (Ax. 2). Rect. ABEF = b h (?). .-. O ABCD = b - h (Ax. 1). Q.E.D. 190 PLANE GEOMETRY 374. COR. All parallelograms having equal bases and equal alti- tudes are equivalent. 375. THEOREM. Two parallelograms having equal altitudes are to each other as their bases. Proof: P = b-h: p' = b'-h (?). ?=fcH< A *- 8 >: 376. THEOREM. Two parallelograms having equal bases are to each other as their altitudes. Proof: CO- 377. THEOREM. Any two parallelograms are to each other as the products of their bases by their altitudes. Proof: (?). 378. THEOREM. The area of a triangle is equal to half the product of its base by its altitude. Given : A ABC; base = b ; R altitude = h. To Prove : Area of A ABC = ^ b h. Proof : Through A draw AR II to BC and through C draw CR II to AB, meeting AR at R. Now ABCR is a O (?). Area O ABCR =b-h (?). J O ABCR = l b - h (Ax. 3). .'.AABC=%b h (Ax. 1). Q.E.D. 379. COR. If a parallelogram and a triangle have the same base and altitude, the triangle is equivalent to half the parallelogram. 380. COR. All triangles having equal bases and equal altitudes are equivalent. 381. COR. All triangles having the same base and whose vertices are in a line parallel to the base are equivalent (?). BOOK IV 191 382. THEOREM. Two triangles having equal altitudes are to each other as their bases. Proof : A T = J b - h ; A T 1 = % b'h (?). A T _ ^ b h _ b ^x 11/7 TT v )' 'AT' 383. THEOREM. Two triangles having equal bases are to each other as their altitudes. Proof: (?). 384. THEOREM. Any two triangles are to each other as the products of their bases by their altitudes. 385. THEOREM. The area of a right triangle is equal to half the product of the legs. 386. THEOREM. The area of a trapezoid is equal to half the product of the altitude by the sum of the bases. Given: Trapezoid ABCD; altitude = h ; bases = b and*?. To Prove : / Area ABCD = J h (b + h (Ax. 2). That is, trapezoid ABCD = J h (6 + c~) (Ax. 6). Q.E.D. 387. THEOREM. The area of a trapezoid is equal to the product of the altitude by the median. Proof : Area ABCD = J h (b + c) = h J (5 + #F DE - DF A ABC AB AC . . That is, XT AC AB s<>\ =--x . Now, = --(?). A DEF DE DF DF DE 2 A ABC AB^AB AB , A v ^ .*. - = - X - = - ( A.X. u ). A DEF DE DE j)E But 45 = ^=*Zf?), and = 5 = S DE DF EF DE DF EF (297). A ABC AB A DEF AC_ = BCL (Ax. 1). DF Q.E.D. ANOTHER METHOD. Denote a pair of homologous alti tudes by h and h', and the corresponding bases by b and b 1 '. A DEF Hence, But = That is, A DEF b' b' b' A ABC liC 2 AB 2 (322). Q.E.D. A DEF EF * BOBBINS' PLANE GEOM. 13 DF 194 PLANE GEOMETRY 390. THEOREM. Two similar polygons are to each other as the squares of any two homologous sides. B 1 > B E D E- Given : Similar polygons ABODE and A'B'C'D'E'. To Prove : ABODE : A'B'C'D'E' = AB? : A's' 2 = etc. Proof: Draw from homologous vertices, A and A 1 , all the pairs of homologous diagonals, dividing the polygons into A. These A are similar, in pairs (?) (327). AR A R' A ' B ' and CD AB AS' (?) and A T AT' AB AR __ A.S _ AT , A ~ 7 I /i A. A + A S +AT AB (?) (g()1)- AJZ' v Therefore, Hence, - But . polygon ABODE ' P ol yg n A'B'C'D'E' ~ J 7 /? 2 ~" Wo* 2 AA_._^!_ m AR' ^W 2 ^' } ' Q.E.D. Ex. 1. The base of a triangle is 6. Find the base of a similar tri- angle that is 9 times as large. Five times as large. Ex. 2. The area of a polygon is 104 and its longest side is 12. What is the area of a similar polygon whose longest side is 15? BOOK IV 195 H 391. THEOREM. The square described upon the hypotenuse of a right triangle is equivalent to the sum of the squares described upon the legs. Given: (?). To Prove: (?). Proof : Draw CL _L to AB, meeting AB at K and F< ED at L. Draw BF and CE. Now, AACB, ACG, and BCH&i-e all rt. A (?). Hence, ACH and BCG are straight lines (?) (45). Also, AELK and BDLK are rectangles (?) (127). In A ABF and ACE, AB=AE, AF AC (128), and /_BAF= /_ CAE. (Each = a rt.Z + Z BAG.) \. * s \. J * Also, A4.B.Fand square AG have the same base, AF, and the same altitude, AC. Hence, square AG o 2 A ABF (379). Similarly, rectangle AKLE =c= 2 A ACE (?). Therefore, rectangle AKLE =^ square ACGF (Ax. 1). By drawing ^/ and CD, it is proven in like manner, that rectangle BDLK =c= square BCHI. Then, by adding, square ABDE =c= square ACGF+ square BCHI (Ax. 2). Q.E.D. 392. THEOREM. The square described upon one of the legs of a right triangle is equivalent to the square described upon the hypotenuse minus the square described upon the other leg. (Explain.) Ex. 1. If the legs of a right triangle are 15 and 20, what is the hypote- nuse ? If the legs are i 2 - 1 and 2 m, what is the hypotenuse ? Ex. 2. What is the difference in the wording of the theorems of 343 and 391 ? Which proof is purely algebraic ? Which is geometric ? 196 PLANE GEOMETRY 393. THEOREM. If the three sides of a right triangle are the homolo- gous sides of three similar polygons, the polygon described upon the hypote- nuse is equivalent to the sum of the two polygons described upon the legs. Proof: - == (?). ^ AB 2 And -==-(?). Adding, rt | >TT A /"i* -I- f> C< ~~A~I? ^ = AC 9 B =^=1. (Explain.) .'.R o 8+ r(?). Q.E.D. 394. COR. If the three sides of a right triangle are the homolo- gous sides of three similar polygons, the polygon described upon one of the legs is equivalent to the polygon described upon the hypote- nuse minus the polygon described upon the other leg. 395. THEOREM. The two squares described upon the legs of a right triangle are to each other as the projections of the legs upon the hypotenuse. Proof: s q uare s = Square T AB AP AP , , . = (Explain. AB BP BP 396. THEOREM. If two similar poly- gons are described upon the legs of a right triangle as homologous sides, they are to each other as the projections of the legs upon the hypotenuse. ^2 Proof: -=^- = AP BP (Explain.) P B BOOK IV 197 ORIGINAL EXERCISES (THEOREMS) 1. If one parallelogram has half the base and the same altitude as another, the area of the first is half the area of the second. 2 If one parallelogram has half the base and half the altitude of another, its area is one fourth the area of the second. 3. State and prove two analogous theorems about triangles. 4. If a triangle has half the base and half the altitude of a paral- lelogram, the triangle is one eighth of the parallelogram. 5. The area of a rhombus is equal to half the product of its diagonals. 6. The diagonals of a parallelogram divide it into four equivalent triangles. 7. The diagonals of a trapezoid divide it into four triangles, two of which are similar and the other two are equivalent. 8. If a parallelogram has half the base and half the altitude of a triangle, its area is half the area of the triangle. 9. The line joining the midpoints of two sides of a triangle forms a triangle whose area is one fourth the area of the original triangle. 10. The line joining the midpoints of two adjacent sides of a paral- lelogram cuts off a triangle whose area is one eighth of the area of the parallelogram. 11. If one diagonal of a quadrilateral bisects AT B the other, it also divides the quadrilateral into two equivalent triangles. To Prove : A ABC - A ADC. D 12. Either diagonal of a trapezoid divides the figure into two triangles whose ratio is equal to the ratio of the bases of the trapezoid. Prove two ways. [By 382 and by 388.] 13. If, in triangle ABC, D and E are the midpoints of sides AB and A C respectively, &BCD-&BEC. 14. If the diagonals of quadrilateral AB CD meet at E and A ABE ~ A CDE, the sides A D and BC are parallel. [Prove A ABD ^ A A CD.'] 15. The square described upon the hypotenuse of an isosceles right triangle is equivalent to four times the triangle. 16. The square described upon the diagonal of a square is double the original square. 198 PLANK GEOMETRY 17. Any two sides of a triangle are reciprocally proportional to the altitudes upon them. [Use 378 and 291.] 18. In equivalent triangles the bases and the altitudes upon them are reciprocally proportional. 19. If two isosceles triangles have the legs of one equal to the legs of the other, and the vertex- angle of the one the supplement of the vertex- angle of the other, the triangles are equivalent. Given : & ABC and A CD, etc. 20. Two triangles are equivalent if they have two sides of one equal to two sides of the other and the included angles supplemen- tary. Proof: Z.CAD = Z C'AD' (?) and CA = C'A (V). .% the rt. & are = (?). Etc. C' 21. If two triangles have an angle of one the supplement of an angle of the other, the triangles are to each other as the products of the sides in- cluding these angles. C" Given : & ABD and EEC, A at B supplementary. Proof : Draw DC, use A BCD, and proceed as in 388. 22. The area of a triangle is equal to half the perimeter of the triangle multiplied by the radius of the inscribed circle. Proof: Draw CM, etc. &AOC = $AC r(?); &AOB = \AB - r (?), etc. Add. 23. The area of a polygon circumscribed about a circle is equal to half the product of the perimeter of the polygon by the radius of the circle. 24. The line joining the midpoints of the bases of a trapezoid bisects the area of the trapezoid. 25. Any line drawn through the midpoint of a diagonal of a paral- lelogram, intersecting two sides, bisects the area of the parallelogram. 26. The lines joining (in order) the midpoints of the sides of any quadrilateral form a parallelogram whose area is half the area of the quadrilateral. BOOK IV 199 27. If any point within a parallelogram is joined to the four vertices, the sura of one pair of opposite triangles is equivalent to the sum of the other pair; that is, to half the parallelogram. 28. Is a triangle bisected by an altitude? By the bisector of an angle? By a median? By the perpendicular bisector of a side? Give reasons. 29. If the three medians of a triangle are drawn, there are six pairs of triangles formed, one of each pair being double the other. To Prove: A A OB = 2 A A OE; etc. M V* 30. If the midpoints of two sides of a tri- angle are joined to any point in the base, the quadrilateral formed is equivalent to half the original triangle. 31. If lines are drawn from the midpoint of one leg of a trapezoid to the ends of the other leg, the middle triangle thus formed is equivalent to half the trapezoid. Proof: Draw median EF = m. Then EF is li to the bases (?). Denote the altitude of the trapezoid by h. Then EF bisects h (?). A BFE = | m - \ h (?). A AEF = \ m . \h (?). /.A ABE = \ mh. Consult 387. 32. The area of a trapezoid is equal to the product of one of the non-parallel sides, by the perpendicular upon it from the midpoint of the other. Proof : Prove that A ABE = half the trapezoid, by No. 31. But the A A BE=ABx the _L to AB from E (?). .-. half the trapezoid = \ AB x this J_ (Ax. 1). Etc. 33. If through the midpoint of one of the non -parallel sides of a trapezoid a line is drawn parallel to the other side, the parallelo- gram formed is equivalent to the trapezoid. 34. If two equivalent triangles have an angle of one equal to an angle of the other, the sides including these angles are reciprocally proportional. 35. The sum of the three perpendiculars drawn to the three sides of an equilateral triangle from any point within is constant (being equal to the altitude of the triangle). Proof: Join the point to the vertices. Set the sum of the areas of the three inner & equal to the area of the whole A. Etc. E X 200 PLANE GEOMETRY 36. In the figure of 391, prove : (i) Points /, (7, and F are in a straight line. (it) CE and BF are perpendicular. (in) AG and EH are parallel. 1 3 rn n m m m m On n n n m n [See Ex. 20.] 37. The sum of the squares described upon the four segments of two perpendicular chords in a circle is equivalent to the square described upon the diameter. (Fig. is on page 178.) 38. The square described upon the sum of two lines is equivalent to the sum of the squares de- scribed upon the two lines, plus twice the rectangle of these lines. To Prove : Square A E =0= m 2 + n 2 + 2 mn. 39. The square described upon the difference of two lines is equivalent to the sum of the squares de- scribed upon the two lines minus twice the rectangle of these lines. To Prove : Square AD = m 2 -f n 2 2 mn. 40. A and B are the extremities of a diameter of a circle ; C and D are the points of intersection of any third tangent to this circle, with the tangents at A and B respectively. Prove that the area of ABDC is equal to^AB- CD. '41. If the four points midway between the center and vertices of a parallelogram be joined in order, there will be a parallelogram formed ; it will be similar to the original parallelogram ; its perimeter is half of the perimeter of the original figure ; and its area is one quarter of the area of the original figure. 42. If two equivalent triangles have the same base and lie on opposite sides of it, the line joining their vertices is bisected by the base. 43. What part of a right triangle is the quadrilateral which is cut from the triangle by a line joining the midpoints of the legs? 44. Show by drawing a figure that the square on half a line is one fourth the square on the whole line. 45. From M, a vertex of parallelogram LMNO, a line MPX is drawn meeting NO at P and LO produced, at X. LP and NX are also drawn. Prove triangles LOP and XNP are equivalent. G -in n n C n m-n m-n m-n m-n m-n n n n m BOOK IV 201 FORMULAS 397. PROBLEM. To derive a formula for the area of a triangle in terms of its sides. Given : A ABC, having sides = a, b, c. Required: To derive a for- mula for its area, containing only a, b, and c. C Solution : * Draw altitude AD. Now CD = b p a = a + ~' 6 (349), 2a and AD 2 = Iff - CD 2 (392). Hence, h? = \b + ~\r~ ' \ (^ v f ac t rrn g)^ and _ Q + 6 + g)(a + b c) (c + a b ) ( c a + ~-~ 4a 2 Now, area of A = - a - h = 2 ? . /r + J + c)( + 6 - e)( - 6 + c) ( - a + 6 + e) 2 V- - 4^- /.area = % V( . Q.E.F. To simplify this formula, let us call a + b + c=2s. Then, it is evident that a + b c= 2 (* c) ; a b + c= 2 6) ; a + & + c = 2(* ). 202 PLANE GEOMETRY Substituting, in the formula for area, Area of A = J V2 . 2( - c) - 2( - 5) . 2(s - a). That is, Area of A = Vsfs - a) (s - 6) (s -c) . EXERCISE. Find the area of a triangle whose sides are 17, 25, 28. Here, a = 17, b = 25, c = 28, s = 35, s - a = 18, s - b = 10, s - c = 7. Area = V35 18 . 10 . 7 = Vf*~5*~^W 2 = 210. 398. PROBLEM. To derive formulas for the altitudes of a triangle in terms of the three sides. Solution: Area = | ah a = V* (* a) (* 5) (* 0). _ VsQ- Similarly, ft, = . ft ^ 399. PROBLEM. To derive the formulas for the altitude and the area of an equilateral triangle, in terms of its side. Solution: Let each side = a, and altitude = h. Then, A 2 = a 2 - ^ = (?). Also, Area =i base h = \a - ? V3. 2 2 I Ex. 1. Find the area of the triangle whose sides are 7, 10, 11. Ex. 2. Find the area of the triangle whose sides are 8, 15, 17. Ex. 3. Find the area of the equilateral triangle whose side is 8. Ex. 4. Find the side of the equilateral triangle whose area is 121 V^ Ex. 5. Find the area of the equilateral triangle whose altitude is 10. BOOK IV 203 400. PROBLEM. To derive the formula for the radius of the circle inscribed in a triangle, in terms of the sides of the triangle. Solution : Area of A A OB = | -c - r\ area of A AOC = %b>r (?). area of A BOC = a- Adding, area of AABC=^ (a + b -f c) r = sr. (Because, -| (a + b + c) = s.) Hence, r= areaofA ^ BC . .'. r = Vs (s - a) (8 - ft) (a - c) 401. PROBLEM. To derive the formula for the radius of the circle circumscribed about a triangle, in terms of the sides of the triangle. Solution : 2fl h a = b . c (?) (337). b c R = 2A ..JB = a 4^8(8-0) (*- Ex. 1. Find the radius of the circle inscribed in, and the radius of the circle circumscribed about, the triangle whose sides are 17, 25, 28. Ex. 2. Find for triangle whose sides are 11, 14, 17, the radii of the inscribed and circumscribed circles. '204: PLANK GEOMETRY ORIGINAL EXERCISES (NUMERICAL) 1. The base of a parallelogram is 2 ft. 6 in. and its altitude is 1 ft. 4 in. Find the area. Find the side of an equivalent square. 2. The area of a rectangle is 540 sq. in. and its altitude is 15 m. Find its base and diagonal. 3. The base of a rectangle is 3 ft. 4 in. and its diagonal is 3 ft. 5 in. Find its area. 4. The bases of a trapezoid are 2 ft. 1 in., and 3 ft. 4 in., and the altitude is 1 ft. 2 in. Find the area. 5. The area of a trapezoid is 736 sq. in. and its bases are 3 ft. and 4 ft. 8 in. Find the altitude. 6. The area of a certain triangle whose base is 40 rd., is 3.2 A. Find the area of a similar triangle whose base is 10 rd. Find the altitudes of these triangles. 7. The base of a certain triangle is 20 cm. Find the base of a simi- lar triangle four times as large ; of one five times as large ; twice as large; half as large; one ninth as large. 8. The altitude of a certain triangle is 12 and its area is 100. Find the altitude of a similar triangle three times as large. Find the base of a similar triangle seven times as large. Find the altitude and base of a similar triangle one third as large. 9. The area of a polygon is 216 sq. m. and its shortest side is 8 m. Find the area of a similar polygon whose shortest side is 10 m. Find the shortest side of a similar polygon four times as large; one tenth as large. 10. If the longest side of a polygon whose area is 567 is 14, what is the area of a similar polygon whose longest side is 12 ? of another whose longest side is 21 ? 11. Find the area of an equilateral triangle whose sides are each 6 in. Of another whose sides are each 10 V 3 ft. 12. Find the area of an equilateral triangle whose altitude is 4 in. ; of another whose altitude is 18 dm. 13. The area of an equilateral triangle is 64 V3. Find its side and its altitude. 14. The area of an equilateral triangle is 90 sq. m. Find its altitude. 15. Find the side of an equilateral triangle whose area is equal to a square whose side is 15 ft. BOOK IV 205 16. The equal sides of an isosceles triangle are each 17 in. and the base is 16 in. Find the area. 17. Find the area of an isosceles right triangle whose hypotenuse is 2 ft. 6 in. 18. Find the area of a square whose diagonal is 20 m. 19. There are two equilateral triangles whose sides are 33 and 56 respectively. Find the side of the third, equivalent to their sura. Find the side of the equilateral triangle equivalent to their difference. 20. There are two similar polygons two of whose homologous sides are 24 and 70. Find the side of a third similar polygon equivalent to their sum ; the side of a similar polygon equivalent to their difference. 21. What is the area of the right triangle whose hypotenuse is 29 cm. and whose short leg is 20 cm. ? 22. The base of a triangle is three times the base of an equivalent triangle. What is the ratio of their altitudes? 23. The bases of a trapezoid are 56 ft. and 44 ft. and the non- paral- lel sides are each 10 ft. Find its area. Also find the diagonal of an equivalent square. 24. The base of a triangle is 80 m., and its altitude is 8 m. Find the area of the triangle cut off by a line parallel to the base and at a dis- tance of 3 m. from it. Another, cut off by a line parallel to the base and 6 m. from it. 25. The bases of a trapezoid are 30 and 55, and its altitude is 10. If the non-parallel sides are produced till they meet, find the area of the less triangle formed. [The & are similar. .'.30: 55=ziz + 10. Etc.] 26. The diagonals of a rhombus are 2 ft. and 70 in. Find the area; the perimeter ; the altitude. 27. The altitude (h) of a triangle is increased by n and the base (&) is diminished by x so the area remains unchanged. Find x. 28. The projections of the legs of a right triangle upon the hypote- nuse are 8 and 18. Find the area of the triangle. 29. In triangle ABC, AB is 5, BC is 8, and AB is produced to P, making BP=Q. BC is produced (through B) to Q and PQ drawn so the triangle BPQ is equivalent to triangle ABC. Find the length of BQ. [Use 388.] 206 PLANE GEOMETRY 30. The angle C of triangle ABC is right; AC = 5; EC = 12. BA is produced through A, to Z> making AD = 4; CM is produced through .4, to E so triangle ^4Z) is equivalent to triangle ABC. Find A E. 31. Find the area of a square inscribed in a circle whose radius is 6. 32. Find the side of an equilateral triangle whose area is 25 V?. 33. Two sides of a triangle are 12 and 18. What is the ratio of the two triangles formed by the bisector of the angle between these sides? 34. The perimeter of a rectangle is 28 m. and one side is 5 m. Find the area. 35. The perimeter of a polygon is 5 ft. and the radius of the inscribed circle is 5 in. Find the area of the polygon. In the following triangles, find the area, the three altitudes, radius of inscribed circle, radius of circumscribed circle : 36. a = 13, b = 14, c = 15. 37. a = 15, b = 41, c = 52. 38. 20, 37, 51. 39. 25, 63, 74. 40. 140, 143, 157. 41. The sides of a triangle are 15, 41, 52 ; find the areas of the two triangles into which this triangle is divided by the bisector of the largest angle. 42. Find the area of the quadrilateral ABCD if AB = 78 m., BC = 104m., CD = 50 m., AD = 120 m., and AC = 130 m. 43. One diagonal of a rhombus is j 8 ^ of the other and the difference of the diagonals is 14. Find the area and perimeter of the rhombus. 44. A trapezoid is composed of a rhombus and an equilateral triangle ; each side of each figure is 16 inches. Find the area of the trapezoid. 45. Find the side of an equilateral triangle equivalent to the square whose diagonal is 15 \/2. 46. Which of the'figures in No. 45 has the less perimeter? 47. In a triangle whose base is 20 and whose altitude is 12, a line is drawn parallel to the base, bisecting the area of the triangle. Find the distance from the base to this parallel. 48. Parallel to the base of a triangle whose base is 30 and altitude is 18 are drawn two lines dividing the area of the triangle into three equal parts. Find their distances from the vertex. 49. Around a rectangular lawn 30 yards x 20 yards is a drive 16 feet wide. How many square yards are there in the drive ? BOOK TV CONSTRUCTIONS 207 402. PROBLEM. To construct a square equivalent to the sum of two squares. Y Given : (?). Required : (?). Construction : Construct a rt. Z E, whose sides are EX and EY. On EX take EF = AB, and on EY take EG CD. Draw FG. On FG construct square T. Statement : T=C= R + '+ G'C' = CD(?). Q.E.D. S \F>- -... \ X H F ( 1 ^ r X - U \ \ V E v v E' G 1 II. Given : Square S and line CD. Required : To construct a O =c= s ; base altitude = CD. BOOK IV 211 Construction : On CD as diameter, describe a 0, o. At C erect CEA. to CD and =AB. Draw EFOG meeting O at F and G. Take E'G'^EG and draw XY II to E'G' at a distance from it = EF. On XY take JJI = #G. Draw HE 1 and IG'. Statement: EH E'G' ill ^ s and base alt. = CD. Q.E.F. Proof : #C is tangent to O O (?). .'. EG EF^EC 2 ^?) (333). EG EF= area O E'G'IH (?), and #c 2 = Jfi 2 = area S (?). FG = CD(?. Q.E.D. 412. PROBLEM. To find two lines whose product is given : I. If their sum is also given. | [The game ag m -j II. If their difference is also given, j 413. PROBLEM. To construct a square having a given ratio to a given square. O R / Q / X * \ / E--''' \F\ S / *r.- ""\ \ / .""" *" A L---'' m n ' V X O A B C n.... , Given : Square 12, and lines ra and n. Required : To construct a square such that, The square E : the unknown square = m : n. Construction : On an indefinite line AY take AB = ra, and BC =n. On AC as diameter describe a semicircle. At B erect BD. to AC, meeting arc at _D. Draw AD and DC. On AD take DE=a, and draw ^.F || to AC, meeting DC at F. Using DF =x, as a side, construct square S. Statement : R : S = m : n. Q.E.F. Proof : Z ^DC is a rt.Z(?). /. ^D 2 : Z)C' 2 = m : w (?) (395). DC a 2 = = S (?). /. I? : ,s = m : n (Ax. 6). 212 PLANE GEOMETRY 414. PROBLEM. To construct a polygon similar to a given polygon and having a given ratio to it. D Given: (?). Required: (?). Construction and Statement are the same as in 413. Proof: Z ^DCisart. Z (?); /. AJ? : DC 2 = m : n (?)(395). ff=^5(?) ; .-.!? = 45- = . (Explain.) - = ^ (?) (390). .'. R : S = m : n (Ax. 1). Q.E.D. 415. PROBLEM. To construct a polygon similar to one given poly- gon and equivalent to another. A B a f Given : Polygons R and S. Required : (?). Construction : Construct squares R' oR, and s f =o=8 (by 410). Find a fourth proportional to a, 6, and AB. This is CD. Upon CD, homologous to AB, construct T similar to R. Statement : T ^ s. Q.E.F. Proof : - = T .-.2=?- (Ax. 6). .-.T^s (Ax. 3). (?) (390). ^=^ (Const.). .'.^==; b CD b 2 CD' =c=/s. (Explain.) Q.E.D. BOOK IV 213 ORIGINAL CONSTRUCTIONS 1. To construct a square equivalent to a given right triangle. 2. To construct a right triangle equivalent to a given square. 3. To construct a right triangle equivalent to a given parallelogram. 4. To construct a square equivalent to the sum of two given right triangles. 5. To construct a square equivalent to the difference of two given right triangles. 6. To construct a square equivalent to the sum of two given paral- lelograms. 7. To construct a square equivalent to the difference of two given parallelograms. 8. To construct a square equivalent to the sum of several given right triangles. 9. To construct a square equivalent to the sum of several given paral- lelograms. 10. To construct a square equivalent to the sum of several given triangles. 11. To construct a square equivalent to the sum of several given polygons. 12. To construct a square equivalent to the difference of two given polygons. 13. To construct a square equivalent to three times a given square. To construct a square equivalent to seven times a given square. 14. To construct a right triangle equivalent to the sum of several given triangles. 15. To construct a right triangle equivalent to the difference of any two given triangles; of any two given parallelograms. 16. To construct a square equivalent to a given trapezoid ; equiva- lent to a given trapezium. 17. To construct a square equivalent to a given hexagon. 18. To construct a rectangle equivalent to a given triangle, hav- ing given its perimeter. 19. To construct an isosceles right triangle equivalent to a given triangle. 20. To construct a square equivalent to a given rhombus. 21. To construct a rectangle equivalent to a given trapezium, and having its perimeter given. 214 PLANE GEOMETRY 22. To find a line whose length shall be \/2 units. [See 402.] 23. To find a line whose length shall be V3 units. 24. To find a line whose length shall be v/11 units. 25. To find a line whose length shall be \/7 units. 26. To find a line whose length shall be VlO units. 27. To construct a square which shall be f of a given square. 28. To construct a square which shall be f of a given square. 29. To construct a polygon which shall be f of a given polygon, and similar to it. 30. To construct a square which shall have to a given square the ratio V3 : 4. If the given ratio is 4 : V3. 31. To draw through a given point, within a parallelogram, a line which shall bisect the parallelogram. 32. To construct a rectangle equivalent to a given trapezoid, and having given the difference of its base and altitude. 33. To construct a triangle similar to two given similar triangles and equivalent to their sum. 34. To construct a triangle similar to a given triangle and equivalent to a given square. [See 415.] 35. To construct a triangle similar to a given triangle and equivalent to a given parallelogram. 36. To construct a square having twice the area of a given square. [Two methods.] 37. To construct a square having 3| times the area of a given square. 38. To construct an isosceles triangle equivalent to a given triangle and upon the same base. 39. To construct a triangle equivalent to a given triangle, having the same base, and also having a given angle adjoining this base. 40. To construct a parallelogram equivalent to a given parallelo- gram, having the same base and also having a given angle adjoining the base. 41. To draw a line that shall be perpendicular to the bases of a parallelogram and that shall bisect the parallelogram. 42. To construct an equilateral triangle equivalent to a given tri- angle. [See 415.] 43. To trisect (divide into three equivalent parts) the area of a triangle, by lines drawn from one vertex. BOOK IV 215 44. To construct a square equivalent to of a given pentagon. 45. To construct an isosceles trapezoid equivalent to a given trapezoid. 46. To construct an equilateral triangle equivalent to the sum of two given equilateral triangles. 47. To construct an equilateral triangle equivalent to the difference of two given equilateral triangles. 48. To construct upon a given base a rectangle that shall be equivalent to a given rectangle. Analysis : Let us call the unknown altitude x. Then b . h = b' x (V). Hence, b':b = h:x (?) . That is, the unknown altitude is a fourth pro- portional to the given base, the base of the given rec- tangle, and the altitude of the given rectangle. Construction: Find a fourth proportional, x, to &', b and h. Con- struct a rectangle having base = b' and alt. = x. Statement: This rectangle, B**A. B Proof: b':b = h:x (Const.). .-. b'x = bh (?). But ' b'x = the area of B (?). Etc. 49. To construct a rectangle that shall have a given altitude and be equivalent to a given rectangle. 50. To construct a triangle upon a given base that shall be equiva- lent to a given triangle. 51. To construct a triangle that shall have a given altitude and be equivalent to a given triangle. 52. To construct a rectangle that shall have a given base, and shall be equivalent to a given triangle. 53. To construct a triangle that shall have a given base, and be equivalent to a given rectangle. 54. To construct a triangle that shall have a given base and be equivalent to a given polygon. 55. Construct the problems 49, 50, 51, 53, 54 if the first noun in each problem is the word " parallelogram." 56. To construct upon a given hypotenuse, a right triangle equivalent to a given triangle. 57. To construct upon a given hypotenuse, a right triangle equivalent to a given square. 216 PLANE GEOMETRY 58. To construct a triangle which shall have a given base, a given adjoining angle, and be equivalent to a given triangle. Another, equiva- lent to a given square. Another, equivalent to a given polygon. 59. To construct a parallelogram which shall have a given base, a given adjoining angle, and be equivalent to a given parallelogram. Another, equivalent to a given triangle; to a given polygon. 60. To construct a line, DE, from D in AB of triangle ABC, so that DE bisects the triangle. A Analysis: After DE is drawn, ^ABC='2AADE (Hyp.). Rnt&ABC: & ADE=AB AC : AD > AE (?). Hence, AB . AC = 2 (AD . AE) (Ax. 6). /. 2 AD : AB = A C : x (?). Thus x, (that is, AE) is a fourth proportional to three given lines. 61. To draw a line meeting two sides of a triangle and forming an isosceles triangle equivalent to the given triangle. Analysis: Suppose AX a leg of isosceles A. / . A A B C : A A XX' = A B . A C : A X . A X' . But the A are equivalent and A X = A X' (Hyp.). Hence, AB A C = A X 2 . .'. AX is a mean proper- B \ c tional between AB and AC. 62. To draw a line parallel to the base of a triangle which shall bisect the triangle. [See 389 and use 414.] 63. To draw a line meeting two sides of a triangle forming an isosceles triangle equivalent to half the given triangle. 64. To draw a line parallel to the base of a triangle forming a tri- angle equivalent to one third the original triangle. 65. To draw a line parallel to the base of a O trapezoid so that the area is bisected. /'\ Analysis: A OXX' =*% (A OAD + A OB C) / \ and is similar to A OB C. / \c Construction: [Use 408, 402, 415.] / ^ 66. To construct two lines parallel to the base / \ of a triangle, that shall trisect the area of the A D triangle. 67. To construct a triangle having given its angles and its area. Analysis : The required A is similar to any A containing the given A. The given area may be a square. This reduces the problem to 415. 68. To find two straight lines in the ratio of two given polygons. BOOK V REGULAR POLYGONS. CIRCLES 416. A regular polygon is a polygon which is equilateral and equiangular. 417. THEOREM. An equilateral polygon inscribed in a circle is regular. Given : AG, an equilateral inscribed polygon. To Prove : AG is regular. Proof : Z A is measured by 1 arc BGL (?). Also Z B is measured by Jarc CHA (?), etc. Subtended arcs, AB, BC, CD, etc., are all = (?). Hence, arcs BGL, CHA, DIB, etc., are all = (Ax. 2). /. /.A = ZB = Zc= etc. (?). That is, the polygon is equiangular. Therefore, the polygon is regular (?) (416). Q.E.D. 418. THEOREM. If the circumference of a circle be divided into any number of equal arcs, and the chords of these arcs be drawn, they will form an inscribed regular polygon. Proof: Chords AB, B(\ CD, etc. are all = (?). Therefore, the polygon is regular (?). Q.E.D. Ex. 1. What is the usual name of a regular 3-gon ? of a regular 4-gon? Is an equiangular inscribed polygon necessarily regular? Ex. 2. In the figure of 417, how many degrees are there in each of th^ arcs, AB, BC, etc.? How many degrees are there in each angle? 217 218 PLANE GEOMETRY 419. THEOREM. If the circumference of a circle be divided into any number of equal parts, and tangents be drawn, at the several points of division, they will form a circumscribed regular polygon. Given : (?). To Prove : (?). Proof : Draw chords AB, BC, CD, etc. In A ABH, BCI, CDJ, etc., AB = BC = CD = etc. (?). Z HAB = Z HBA = Z IBC = Z.ICB = Z JCD = etc. (?). .*. these A are isosceles (?) (120), and = (?). .-. Z H = Z 1 = Z j =etc. (?). That is, polygon GJ is equi- angular. Also, AH = HB = BI = 1C = CJ =etc. (?), and HI = IJ = JK = etc. (?) (Ax. 3). That is, polygon Jis equilateral ; .*. it is regular (?). Q.E.D. 420. THEOREM. If the cir- cumference of a circle be divided into any number of equal parts and tangents be drawn at their midpoints, they will form a cir- cumscribed regular polygon. Given: (?). To Prove: (?). G Proof : Arcs AB, BC, CD, etc. are all = (?). Also, arcs A I, IB, BK, KG, R CM, etc. are all = (?) (Ax. 3). .*. arcs IK, KM, MO, etc. are all = (?) (Ax. 3). Therefore, the polygon is regular (?) (419). Q.E.D. 421. THEOREM If chords be drawn joining the alternate vertices of an inscribed regular polygon (having an even number of sides), another inscribed regular polygon will be formed. (See 417.) BOOK V 219 422. THEOREM. If the ver- tices of an inscribed regular polygon be joined to the mid- points of the arcs subtended by the sides, another inscribed regu- lar polygon will be formed (hav- ing double the number of sides). 423. THEOREM. If tangents be drawn at the midpoints of the arcs between adjacent points of contact of the sides of a circum- scribed regular polygon, another circumscribed regular polygon will be formed having double the number of sides (?). 424. THEOREM. The perimeter of an inscribed regular polygon is less than the perimeter of an inscribed regular polygon having twice as many sides, and the perimeter of a circumscribed regular polygon is greater than the perimeter of a circumscribed regular polygon having twice as many sides. (Ax. 12.) 425. THEOREM. Two regular polygons having the same number of sides are similar. Given : Regular w-gons AD and A'D'. To Prove : They are simi- lar. Proof : ?) (164). -2) 180 n Similarly, Z B = Z B f , Z C== Z C f , etc. That is, these polygons are mutually equiangular. Also, ^17? = BC = CD= etc.; A'B' = B'C' = c'z)' = .-. AB : A'B' = BC: B f c' = CD : C f D f = etc. (Ax. 3). That is, the homologous sides are proportional. Therefore, the polygons are similar (?). Q.E.D. 220 PLANK (JKOMKTKY 426. THEOREM. A circle can be circumscribed about, and a circle can be inscribed in, any regular polygon. Given : Regular polygon ABCDEF. To Prove : I. A circle can be circumscribed about the polygon. II. A circle can be in- scribed in the polygon. Proof: I. Through three consecutive vertices, A, B, and C, describe a circumfer- ence, whose center is O. Draw radii OA, OJ5, OC, and draw line OD. In A AOB and COD, AB = CD (?) ; BO = CO (?). Now ^ABC = ZBCD (?) (416). /Lose = Z OCB (?). Subtracting, Z ABO = Z. OCD (Ax. 2). .'.A AOB = A COD (?). .'.AO = OD (?). Hence, the arc passes through D, and in like manner it may be proven that it passes through E and F. That is, a circle can be circumscribed about the polygon. II. AB, BC, CD, DE, etc. are = chords (?) (416). Therefore they are equally distant from the center (?). That is, a circle described, using O as a center and OM as a radius, will touch every side of the polygon. Hence a circle can be inscribed. (Def. 234.) Q.E.D. 427. The radius of a regular polygon is the radius of the circumscribed circle. The radius of the inscribed circle is called the apothem. The center of a regular polygon is the common center of the circumscribed and inscribed circles. 428. The central angle of a regular polygon is the angle included between two radii drawn to the ends of a side. 429. THEOREM. Each central angle of a regular w-gon = BOOK V 221 430. THEOREM. Each exterior angle of a regular /7-gon = * (?) 431. THEOREM. The radius drawn to any vertex of a regular poly- gon bisects the angle at the vertex. (See 80.) 432. THEOREM. The central angles of regular polygons having the same number of sides are equal. (See 429.) 433. THEOREM. If radii be drawn to all the vertices of a circum- scribed regular polygon, and chords be drawn connecting the points of intersection of these lines with the circumference, an inscribed regular polygon of the same number of sides will be formed and the sides of the two polygons will be respectively parallel. Given: (?). To Prove: (?). Proof : Central A at O are all = (?). .-.the intercepted arcs are all = (?). .-. the chords A 1 B'c', etc. are all= (?). .-. the inscribed poly- gon is regular (?). Also, A AOB, A'OB', etc. are isosceles (?). If a line OX be drawn from O, bisecting Z AOB, it is _L to AB and to A'B' (?) (57). .'. AB is || to A'B' (?). Q.E.D. Ex. 1. How many degrees are there in the angle, in the central angle, and in the exterior angle of a regular hexagon? a regular decagon? a regular 15-gon ? Ex. 2. In the figure of 433, prove, (rt) Triangle A OB similar to triangle A'OB'. (it) Triangle XOB similar to triangle X'OB'. 222 PLANE GEOMETRY 434. THEOREM. The perimeters of two regular polygons having the same number of sides are to each other as their radii and also as their apothems. A' B 1 Given : Regular w-gons, EC whose perimeter is P, radius JR, apothem r ; and E f C r whose perimeter is P', radius R 1 , apothem r'. To Prove : P : P' = R : B f = r : r' . Proof: Draw radii OB and O'B'. In ,A AOB and A ! O r B f , Hence, 4^-, = 4^7 (Ax. 3). A'O f B'O f .-. A AOB is similar to A A ! O f B f (?) (317). A 7? 7? T .'. fj = f = (?). Also, the polygons are similar (?). 435. THEOREM. The areas of two regular polygons having the same number of sides are to each other as the squares of their radii and also as the squares of their apothems. Proof : If K and K f denote their areas, we have : ( ? )(390). But TJ K R 2 r 2 , A .<>. Hence, _ = _ = _( Ax. 1). -- (Explain.) Q.E.D. ROOK V 223 436. THEOREM. The area of a regular polygon is equal to half the product of the perimeter by the apothem. Given: (?). To Prove: (?). Proof : Draw radii to all the vertices, forming several isos- celes triangles. Area of A AOB = \AH - r , ? , Area of A BOC = J BC r Area of A COD = \ CD r etc., etc. Area of polygon = 1 (A B -f BC + CD + etc.) r (?), or, = ^ P- r (Ax. 6). Q.E.D. 437. THEOREM. If the number of sides of an inscribed regular poly- gon be increased indefinitely, the apothem will approach the radius as a limit. g^*- ""^p Given : N-gon FC inscribed in O O ; apothem = r ; radius = if. To Prove: That as the number of sides is indefi- nitely increased, r approaches R as a limit. Proof: In the A AOK, or R -r C and C > P 1 and C is constant. Hence, P will approach C and p' will approach C. Q.E.D. ii. f=J|< ? X 486 )- If the number of sides of these polygons be indefinitely increased, OD 2 will approach O.E 2 , and OE thus -will approach unity. OD (The argument continues the same as in I.) BOBBINS' PLANE GEOM. 15 226 PLANE GEOMETRY 441. THEOREM. The circumferences of two circles are to each other as their radii. Given : Two CD whose radii are R and R* and circumfer- ences, C and c r respectively. To Prove : C : C 1 = R : R f . Proof: Circumscribe regu- lar polygons (having the same number of sides) about these (D and let P and p' denote their perimeters. Then, P : P r = R : R f (?) (434). Hence, P R r = P' R (?). Now suppose the number of sides of these polygons to be indefinitely increased, P will approach c (?) (440). P f will approach c' (?). .*. P R r will approach c - -B', and P 1 - R will approach c r R. Hence, C - R f = c' - R (?) (242). Therefore C : C f = R : R f (?) (291). Q.E.D. 442. THEOREM. The ratio of any circumference to its diameter is constant for all circles. That is, any circumference divided by its diameter is the same as any other circumference divided by its diam- eter. Proof: - = ^- (?) (441). But *. = --(?). c' R' R' %R' D' .--! = !, (A*. 1). Hence, ....(?) (292). That is, - = constant. Q.E.D. 443. Definition of TT (pi) . The constant ratio of a circum- c 1 ference to its diameter is called TT. That is, == TT. D BOOK V 227 The numerical value of TT = 3.141592 = 3^, approximately. (This is determined in 470.) 444. FORMULA. Let C = circumference and R = radius. Then, ^ = TT (443). .-. C = 27rR (Ax. 3). 445. THEOREM. The area of a circle is equal to half the product of its circumference by its radius. Given: O whose circumfer- ence = C', area =6', radius = R. To Prove : s = I c R. Proof : Circumscribe a reg- ular polygon about the circle ; denote its area by K and per- imeter by P. Nowir=ip..R(?)(436). Suppose the number of sides of the polygon be in- definitely increased. K will approach s, and P will approach C (?). \ P R will approach \ C R as a limit (?). Hence, s = i C - R (?) (242). Q.E.D. 446. FORMULA. Let S = area of O, C = its circumference, and R = its radius. Then, S = J C R (445). Now C= 2 TT (444). Substituting, S = (2 TTR) fi. .'. S= 7TJR2. Ex. 1. Could 445 be proven by inscribing a regular polygon? Why? Ex. 2. The radius of a circle is 40. Find the circumference and area. Ex. 3. The diameter of a circle is 25. Find the circumference and area. Ex. 4. Prove that the area of a circle equals .7854 D-. 228 PLANE GEOMETRY 447. THEOREM. The areas of two circles are to each other as the squares of their radii, and as the squares of their diameters. To Prove: 8 : S 1 = R 2 : R t2 = D 2 :D f2 . Proof : .'. S:S'=R 2 :R f2 = D 2 : D 12 (Ax. 1). Q.E.D. 448. THEOREM. The area of a sector is the same part of the circle as its central angle is of 360. (Ax. 1.) 449. FORMULA. An arc : circum. = central Z : 360 (244). . . arc : 2 TTR = Z : 360. NOTE. If any two of the three quantities, arc, R, Z, are known, the remaining one can be found by this proportion. 450. FORMULA. Sector : area of O = central Z: 360 (448). . . sector : TrJB 2 = Z : 360. NOTE. If any two of the three quantities, sector, R, Z, are known, the remaining one can be found by this proportion. 451. FORMULA. Sector : area of = arc : circum. (Ax. 1). .-. sector : TT R 2 = arc : 2 TTR (Ax. 6). . . sector = %R arc (290). 452. FORMULA. Area of a segment of a circle = area of the sector minus* area of an isosceles triangle. 453. Similar arcs, similar sectors, and similar segments are those which corre- spond to equal central angles, in unequal circles. Thus, AB, A'B', A"B" are similar arcs; AOB, A' OB 1 , and A" OB" are similar sec- tors ; and the shaded segments are simi- lar segments. * If the segment is greater than a semicircle, the area of the triangle should be added. BOOK V 454. THEOREM. Similar arcs are to each other as their radii. Given : Arcs whose lengths are a and a', radii R and R 1 . To Prove : a : a' R : R f . ..a:a' = 27rfl:27rtf'(292); /.a : a' = R: K 1 . .E.D. 455. THEOREM. Similar sectors are to each other as the squares of their radii. Given : Sectors whose areas are T and T', radii R and B r . To Prove : T : T' = R 2 : R 12 . = _?!_(?). .-. T:T'=R 2 :R' 2 . (Explain.) Q.E.D. 456. THEOREM. Similar segments are to each other as the squares of their radii. Given: (?). To Prove : (?). o Proof: A AOB and A'O'B' are similar (?) (317). A AOB 2 R x 9 x ^QOQX and sector sector O'A C'B' sector OACB sector O'A'C'B' sector OACB - , ' sector o f A'c'B'AA f o'B'~ A A'O'B _ ^ /295 Note f ''~ ''' TT segment ABC A AOB R s\ a\ Hence, =_ -=- = - (Ax. 6). ~- segment A'B'C' A A'O'B' R' 2 Q.E.D. 230 PLANE GEOMETRY ORIGINAL EXERCISES (THEOREMS) 1. The central angle of a regular polygon is the supplement of the angle of the polygon. 2. An equiangular polygon inscribed in. a circle is regular (if the number of its sides is odd). 3. An equiangular polygon circumscribed about a circle is regular. [Draw radii and apothems.] 4. The sides of a circumscribed regular polygon are bisected at the points of contact. 5. The diagonals of a regular pentagon are equal. 6. The diagonals drawn from any vertex of a regular n-gon divide the angle at that vertex into n-2 equal parts. 7. If a regular polygon be inscribed in a circle and another regular polygon having the same number of sides be circumscribed about it, the radius of the circle will be a mean proportional between the apothem of the inner and the radius of the outer polygon. 8. The area of the square inscribed in a sector whose central angle is a right angle is equal to half the square of the radius. [Find z 2 , the area of OEDC.~\ /* 9. The apothem of an equilateral triangle is one O third the altitude of the triangle. 10. The chord which bisects a radius of a circle at right angles is the side of the inscribed equilateral triangle. [Prove the central Z subtended is 120.] 11. If ABCDE is a regular pentagon, and diagonals A C and BD be drawn, meeting at O : (a) AO will = AB. (b) A will be || to ED. (c) ABOC will be similar to A BDC. (e) A C will be divided into mean and ex- treme ratio at O. A 12. The altitude of an equilateral triangle is three fourths the diameter of the circumscribed circle. BOOK V 231 13. The apothem of an inscribed regular hexagon equals half the side of an inscribed equilateral triangle. 14. The area of a circle is four times the area of another circle described upon its radius as a diameter. 15. The area of an inscribed square is half the area of the circum- scribed square. 16. An equilateral polygon circumscribed about a circle is regular (if the number of its sides is odd). 17. The sum of the circles described upon the legs of a right triangle as diameters is equivalent to the circle described upon the hypotenuse as a diameter. 18. A circular ring (the area between two con- centric circles) is equivalent to the circle described upon the chord of the larger circle, which is tan- gent to the less, as a diameter. Proof: Draw radii OB, OC. A OBC isrt.A(?); and OC' 2 - OB* = BC* (?). Etc. 19. If semicircles be described upon the three sides of a right triangle (on the same side of the hypotenuse), the sum of the two crescents thus formed will be equivalentto the areaof the triangle. Proof: f Entire figure =c= },r AB 2 + crescent BDC + crescent AEC (?). \Entire figure- ITT AC 2 + %7rBC 2 + A ABC (?). Now use Ax. 1 ; etc. 20. Show that the theorem of No. 19 is true in the case of a right tri- angle whose legs are 18 and 24. 21. If from any point in a semicircumference a line be drawn perpen- dicular to the diameter and semicircles be de- scribed on the two segments of the hypotenuse as diameters, the area of the surface bounded by these three semicircumferences will equal the area of a circle whose diameter is the perpendicu- lar first drawn. Pttrf : Area = 1 ,('!!) 2 - 4 ,(|) 2 - 4 ,(|) 2 = etc. 22. Show that the theorem of No. 21 is true in the case of a circle whose diameter AB is 25 and AD is 5. 232 PLANE GEOMETRY 23. If the sides of a circumscribed regular polygon are tangent to the circle at the vertices of an inscribed regular polygon, each vertex of the outer lies on the prolongation of the apothems of the inner polygon, drawn perpendicular to the several sides. 24. The sum of the perpendiculars drawn from any point within a regular n-gon to the several sides is constant [= n apothem]. Proof : Draw lines from the point to all vertices. Use 436 and 378. 25. The area of a circumscribed equilateral triangle is four times the area of the inscribed equilateral triangle. 26. If a point be taken dividing the diameter of a circle into two parts and circles be described upon these parts as diameters, the sum of the cir- cumferences of these two circles equals the circumference of the original circle. 27. Show that the theorem of No. 26 is true in the case of a circle the segments of whose diameter are 7 and 12. 28. The area of an inscribed regular octagon is equal to the product of the diameter by the side of the inscribed square. 29. If squares be described on the six sides of a regular hexagon (externally), the twelve exterior vertices of these squares will be the vertices of a regular 12-gon. 30. If the alternate vertices of a regular hexagon be joined by draw- ing diagonals, another regular hexagon will be formed. Also its area will be one third the original hexagon. 31. Show that the theorem of No. 18 is true in the case of two con- centric circles whose radii are 34 and 16. 32. In the same or equal circles two sectors are to each other as their central angles. 33. If the diameter of a circle is 10 in. and a point be taken dividing the diameter into segments whose lengths are 4 in. and 6 in., and on these segments as diameters semicircumferences be described on opposite sides of the diameter, these arcs will form a curved line which will divide the original circle into two parts in the ratio of 2 : 3. 34. If the diameter of a circle is d and a point be taken dividing the diameter into segments whose lengths are a and d a, and on these seg- ments as diameters semicircumferences be described on opposite sides of the diameter, these arcs will form a curved line which will divide the original circle into two parts in the ratio of a : d a. BOOK V 233 CONSTRUCTIONS 457. PROBLEM. To inscribe a square in a given circle. Given : The circle o. Re- quired : To inscribe a square. Construction : Draw any di- ameter, AB, and another diame- ter, CD, -L to AB. Draw AC, BC, BD, AD. Statement: ACBD is an in- scribed square. Q.E.F. Proof: ,4 at Oare= (?). .-. arcs AC, CB, etc. are = (?). .'. ACBD is an inscribed regular polygon (?) (418). .'. ABCD is a square (?). Q.E.D. 458. PROBLEM. To inscribe a regular hexagon in a given circle. Given: (?). Required: (?). Construction : Draw any ra- dius, AO. At A, with radius = AO, describe arc intersecting the given O at B. Draw AB. Statement : AB is the side of an inscribed regular hexagon. Proof : Draw BO. A ABO is equilateral (Const.). .-.A JBOisequiangular(?)(56). .-. Z AOB = 60(?)(H5). ..arc AB =^ of the circumference (J of 360). .'. polygon AD, inscribed, having each side = AB, is an inscribed regular hexagon (?) (418). Q.E.D. Ex. If the radius of a circle is 7 in., find : (a) The circumference and the area. (6) The side and area of the inscribed square, (c) The side and area of the inscribed regular hexagon. 234 PLANE GEOMETRY 459. PROBLEM. To inscribe a regular decagon in a given circle. Given: (?). Required: (?). Construction : Draw any radius AO. Divide it into mean and ex- treme ratio (by 363), having the larger segment next the center. Take A as a center and OB as a radius, draw an arc cutting O at C. Draw AC, BC, OC. Statement : AC is a side of the inscribed regular decagon. Q.E.F. Proof: AO : BO = BO : AB (Const.). That is, AO : AC = AC i AB (Ax. 6). Hence, A ABC and AOC have Z A common and are similar (?) (317). .-. 1st, Z ACB = ZO (?); and 2d, A ABC is isosceles (being similar to A AOC). Hence, AC= BC(?), but AC = BO (?). . '.BC= BO (Ax. 1). Therefore, Z BCO = Z O (?) (55). .-. Z AGO = 2 Z O (Ax. 2). Also, Z A = Z ACO (?) (55). /.Z A = 2ZO (Ax. 1). Z O = 1 Z Q. Adding, A of A^CO = 5Zo(Ax. 2). Hence, 5 y O = 180 (?) (110). .'. Z o = 36 (Ax. 3) ; that is, arc ference Co of 360). Hence, polygon AE, having each side regular decagon (?) (418). C of the circum- is an inscribed Q.E.D. Ex. If the radius of a circle is 20 in., find: (a) The circumference and area. (6) The side and area of the inscribed square, (c) The side and area of the inscribed regular hexagon. (W) The side of the inscribed regular decagon. (See 365.) (e) The area o.f sector AOC (fig. 459). (/) The radius of a circle containing twice the area of this circle. BOOK V 235 460. PROBLEM. To inscribe a regular is-gon (pentedecagon) in a given circle. Given: (?). Required: (?). Construction : Draw AB, the side of an inscribed hexagon, and AC, the side of an inscribed decagon. Draw BC. Statement : BC is the side of an inscribed regular 15-gon. Q.E.F. Proof: Arc BC=arc AB arc AC = i J^ = ^ of the circum- ference. (Const.) Hence, the polygon having each side, a chord, = BC, is an inscribed regular 15-gon (?) (418). Q.E.D. 461. PROBLEM. To inscribe in a given circle: I. A regular 8-gon, a regular i6-gon, a regular 32-gon, etc. II. A regular i2-gon, 24-gon, etc. III. A regular so-gon, 6o-gon, etc. Construction : I. Inscribe a square ; bisect the arcs ; draw chords. Statement: (?). Proof: (?). (See 422.) Etc. II. Inscribe a regular hexagon ; bisect the arcs. Etc. III. Inscribe a regular 15-gon, etc. 462. PROBLEM. To inscribe an equilateral triangle in a circle. Construction: Join the alternate vertices of an inscribed regular hexagon. Proof: (?) (See 421.) 463. PROBLEM. To inscribe a regular pentagon in a given circle. 464. PROBLEM. To circumscribe a regular polygon about a circle. Construction : Inscribe a polygon having the same num- ber of sides. At the several vertices draw tangents. Statement: (?). Proof: (?). (See 419.) 236 PLANE (GEOMETRY FORMULAS Sides of inscribed polygons. 1. Side of inscribed equilat- eral triangle = R VS. Proof : Z ACS is a rt. Z (?). AB = 2 R and CB = R (?) 2. Side of inscribed square Proof: Use fig. of 457. 3. Side of inscribed regular hexagon = R (?). 4. Side of inscribed regular decagon = % R ( V5 1). (365.) 466. Sides of circumscribed polygons. 1. Side of circumscribed equilateral A = 2 R VS. Proof : Z DAB = Z DBA = ZD = 60(?). .'.A ABD is equilateral. AD = AB = R _V3 (?). .'.DF=2R V3. 2. Side of circumscribed square- 2 K (?). 3. Side of circumscribed regular hexagon = R VS. (Explain.) 467. In equilateral triangle, apothem = \ R. Proof: Bisect arc AC at H. Draw OA, OC, ^!JJ, CH. Figure AOCH is a rhombus. (Explain.) ON = OH = R (?) (141). BOOK V 237 468. PROBLEM. In a circle whose radius is /? is inscribed a regu- lar polygon whose side is s ; to find the formula for the side of an inscribed regular polygon having double the number of sides. Given : AB = s, a side of an in- scribed regular polygon in O whose radius is It; C, the mid- point of arc AB; chord AC. Required : To find the value of AC, the side of a regular polygon having double the number of sides and inscribed in the same circle. Construction: Draw radii OA and OC. Computation : OC bisects AB at right A (?) (70). In rt. A AON, O is an acute Z. Hence in A AOC, AC 2 = 5I 2 + OC 2 - 2 - OC - ON (?) (346). But AO = R, OC = R, ON = or AC 469. FORMULA. If R = 1, and given side = s, the side of a regular polygon having twice as many sides = \2-V4 s 2 . Ex. 1. If the radius of a circle is 4, find : (a) The side of the inscribed equilateral triangle. (6) The side of the circumscribed equilateral triangle, (c) The side of the inscribed square. ( A ABD. Proof : Draw altitude DE. Now AD > !)#(?). /. AC > DE (Ax. 6). Multiply each member by ^ AB. Then J AB - AC > J AB DE (?). Now J AB - AC area A ABC (?), and i AB - DE = area A ABD (?). Therefore, A ABC > A ABD (Ax. 6). Q.E.D. This theorem may be stated thus : Of all triangles having two given sides, that triangle whose third side is the diameter of the circle which circumscribes it is the maximum. Therefore, Of all n-gons having n \ sides given, that poly- gon whose n th side is the diameter of a circle which circum- scribes the polygon is the maximum. Ex. 1. Of all parallelograms having two adjacent sides given, the rectangle is the maximum. Ex. 2. Of all lines that can be drawn from an external point to a cir- cumference, which is the maximum ? the minimum V 246 PLANE GEOMETRY ,,E : 473. THEOREM. Of all isoperimetric triangles having the same base the isosceles triangle is the maximum. Given : A ABC and ABD isoperimetric, having the same base, AB, and A ABC isosceles. To Prove : A ABC > A ABD. Proof : Prolong AC to E, making CE = AC, and draw BE. Using D as a center and BD as a radius, describe an arc cutting EB prolonged, at F. Draw CG and DH \\ to AB, meeting EF at G and // respectively. Draw AF. Now, using C as a center and AC or BC or EC as a radius, the circle described will pass through A, B, and E (Hyp. and Const.). .'. /. ABE = rt. Z (?). That is, AB is J_ to EF. Hence, CG and DH are _L to EF (?). AC + CE = AC + CB = AD + DB = AD + DF (Hyp. and Const,). That is, AE = AD + DF (Ax. 1). But AD + DF > AF (?). .'. AE > ^LF (Ax. 6). .-. BE > BF (?) (90), and J BE > BF (?). Now, BG = ^ BE and // = J #F (?) (73, Cor.). .'. BG > 7?ff (Ax. 6). Multiply each member by |- AB. Then, | ^is BG > J ^J5 BII (?). But, | AB - BG = area A^BC (?), and I AB BH = area A ^BD (?). .'.A ABC > A ABD (Ax. 6). Q.E.D. 474. THEOREM. Of isoperimetric triangles the equilateral tiiangle is the maximum. [Any side may be considered the base.] BOOK V 247 475. THEOREM. Of isoperimetric polygons having the same num- ber of sides the maximum is equilateral. Given : Polygon AD, the max- imum of all polygons having / *''\ \-M the same perimeter and the same number of sides. To Prove : AB BC = CD = F< DE = etc. Proof : Draw AC and suppose AB not = BC. On AC as base, construct A ACM isoperimetric with A ABC and isosceles ; that is, make AM=CM. Then A ACM > A ABC (?) (473). Add to each member, the polygon ACDEF. .*. polygon AMCDEF > polygon AD (?). But the polygon AD is maximum (Hyp.). .*. AB cannot be unequal to BC as we supposed (because that results in an impossible conclusion). Hence, AB = BC. Likewise it is proved that BC = CD = etc. Q.E.D. 476. THEOREM. Of isoperimetric polygons having the same num- ber of sides the equilateral polygon is maximum. Proof : Only one such polygon is maximum, and the maxi- mum is equilateral (475). Only one such polygon is equilateral, hence the equilateral polygon and the maximum polygon are the same. Q.E.D. Ex. 1. Of isoperimetric triangles, the maximum is equilateral. Ex. 2. Of all right triangles that can be constructed upon a given hypotenuse, which is maximum? Why ? Ex. 3. Of all triangles having a given base and a given vertex-angle, the isosceles is the maximum. Ex. 4. Of all mutually equilateral polygons, that which can be in- scribed in a circle is the maximum. 248 PLANE GEOMETRY 477. THEOREM, Of isoperimetric regular polygons, the polygon having the greatest number of sides is maximum. Given : Equilateral A ABC and square /S, having the same perimeter. To Prove : Square S>AABC. Proof : Take D, any point in J5C, and draw AD. On AD as base, construct isosceles A ADE, isoperimetric wh\iAABD. Now A AED > A ABD (?) (473). Adding A ADC to each member, AEDC > A ABC (?). AEDC is isoperimetric with A ABC find S (Hyp. and Const.). Hence, S > AEDC (?) (476). Therefore 8 > A ABC (?) (Ax. 11). Similarly we may prove that an isoperimetric regular pen- tagon is greater than 8 ; and an isoperimetric regular hexa- gon is greater than this pentagon, etc. Therefore, the regular polygon having the greatest num- ber of sides is maximum. Q.E.D. 478. THEOREM. Of all isoperimetric plane figures the circle is the maximum, 479. THEOREM. Of equivalent regular polygons the perimeter of the polygon having the greatest number of sides is the minimum. Given : Any two equivalent regular polygons, A and -B, A having the greater number of sides. BOOK V 249 To Prove : The perimeter of A < the perimeter of B. Proof : Construct regular polygon s, similar to B and isoperimetric with A. Then A > S (477), but A =0= B (?). .-. B > s (?) (Ax. 6). Hence, the perimeter of B > perimeter of s (390). But, the perimeter of 8 = perimeter of A (?). .'. perimeter of B > perimeter of A (Ax. 6). That is, the perimeter of A< the perimeter of B. Q.E.D. 480. THEOREM. Of all equivalent plane figures the circle has the minimum perimeter. ORIGINAL EXERCISES 1. Of all equivalent parallelograms having equal bases the rectangle has the minimum perimeter. 2. Of all lines drawn between two given parallels (terminating both ways in the parallels), which is the minimum ? Prove. 3. Of all straight lines that can be drawn on the ceiling of a room 12 feet long and 9 feet wide, what is the length of the maximum? 4. Find the areas of an equilateral triangle, a square, a regular hex- agon, and a circle, the perimeter of each being 264 inches. Which is maximum? What theorem does this exercise illustrate ? 5. Find the perimeters of an equilateral triangle, a square, a regular hexagon, and a circle, if the area of each is 1386 square feet. Which perimeter is the minimum? What theorem does this exercise illustrate? 6. Of isoperimetric rectangles which is maximum? 7. To divide a given line into two parts such that their product (rectangle) is maximum. 8. Of all equivalent triangles having the same base the isosceles triangle has the minimum perimeter. To Prove: The perimeter of &ABC > the perimeter of A AB'C. Proof: AD