E x ^ ^ -~- > < I i VXQ E UNIVERSITY OF GALIFORN BRJBY OF THE UNIVERSITY OF CAUFORK II P UNIVERSITY OF CALIFORNIA. LIBRARY OF THE UNIVERSITY OF CHIFORNI Y y = '- UMIVFPSITY nr rc 1 . OF THE ONIVERSITY OF GHIFORNI IF THE UNIVERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CAL P ~ ^ 1 " V JF THE UNIVERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CA1 1 ^ m ORiDV L TUC I1UIUCDCITV R C PHI PLANE AND SOLID GEOMETRY BY EDWARD RUTLEDGE ROBBINS, A.B. SENIOR MATHEMATICAL MASTER THE WILLIAM PENN CHARTER SCHOOL > NEW YORK : CINCINNATI : CHICAGO AMERICAN BOOK COMPANY COPYRIGHT, 1906, 1907, BY EDWARD RUTLEDGE ROBBINS. PLANE AND 80HD GEOMBTEY. w. P. 3 FOR THOSE WHOSE PRIVILEGE IT MAY BE TO ACQUIRE A KNOWLEDGE OP GEOMETRY THIS VOLUME HAS BEEN WRITTEN AND TO THE BOYS AND GIRLS WHO LEARN THE ANCIENT SCIENCB 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 : 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. 5 6 PREFACE (5) To stimulate the mental activity of the 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. (#) 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. PREFACE T 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. (5) The superior character of the diagrams. ( is (or are) greater than. < is (or are) less than. .*. hence, therefore, conse- quently. _L perpendicular. Js perpendiculars. Q circle. circles. Z angle. A angles. rt. Z right angle, rt. A right angles. A triangle. & triangles, rt. & right triangles. || parallel. || s parallels. O parallelogram. (U parallelograms. ax. axiom, hyp. hypothesis, comp. complementary. supp. supplementary, const, construction, cor. corollary, st. straight, def. definition, alt. alternate, int. interior, ext. exterior. AXIOM, POSTULATE, AND THEOREM 30. An axiom is a truth assumed to be self-evident. It is a truth which is received and assented to immediately. 31. AXIOMS. 1. Magnitudes that are equal to the same thing, or to equals, are equal to each other. 2 . If equals are added to, or subtracted from, equals, the results are equal. 3. If equals are multiplied by, or divided by, equals, the results are equal, [Doubles of equals are equal ; halves of equals are equal.] 4. The whole is equal to the sum of all of its parts. 5. The whole is greater than any of its parts. 6. A magnitude may be displaced by its equal in any process. [Briefly called " substitution."] 7. If equals are added to, or subtracted from, unequals, the results are unequal in the same sense. 8. If unequals are added to unequals in the same sense, the results are unequal in that sense. 9. If unequals are subtracted from equals, the results are unequal in the opposite sense. INTRODUCTION 17 10. Doubles or halves of unequals are unequal in the same sense. 11. If the first of three magnitudes is greater than the second, and the second is greater than the third, the first is greater than the third. 12. A straight line is the shortest line that can be drawn between two points. 13. A geometrical figure may be moved from one position to another without any change in form or magnitude. 32. A postulate is something required to be done, the pos- sibility of which is admitted as evident. 33. POSTULATES. 1. It is possible to draw a straight line from any point to any other point. 2. It is possible to extend (prolong or produce) a straight line in- definitely, or to terminate it at any point. 34. A geometric proof or demonstration is a logical course of reasoning by which a truth becomes evident. 35. A theorem is a statement that requires proof. In the case of the preliminary theorems which follow, the proof is very simple ; but as these theorems are not self- evident they cannot be classified with the axioms. A corollary is a truth immediately evident, or readily estab- lished, from some other truth or truths. EXERCISE 1. Draw an ZABC. In ^ABC draw line BD. What does Z ABD + Z DBC = ? What does Z ABC -ZABD = 1 Ex. 2. In a rt. Z ABC draw line BD. If Z ABD=25, how many degrees are there in Z DBC ? How many degrees are there in the complement of an angle of 38 ? How many degrees are there in the supplement ? Ex. 3. Draw a straight line AB and take a point X on it. What line does AX + BX = ? What line does AB - BX = ? Ex. 4. Draw a straight line AB and prolong it to X so that BX AB. Prolong it so that AX - AB. BOOK I ANGLES, LINES, RECTILINEAR FIGURES PRELIMINARY THEOREMS 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. In 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: . Proof: Z.AOM is the supple- ment of Z. MOB. (Why?) (See 44.) Z.BOL is the supplement of /.MOB. (Why?) (See 44.) /. ^AOM=/: BOL. (Why ?) (See 49.) A A OL and BOM are a pair of vertical angles. These may be proved equal in precisely the same manner. If Z AOL = 80, how many degrees are there in the other A ? 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 B R Given: & ABC and RST; AB = ES\ AC = RT\ Z. A = Z R. 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 s. (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 of the other. Given: Rt. &ABC&ndDEF; AC = DF\ CB = 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 = Z J ; Z C = ZJT. 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 K. (It is given that BC=JK.~) BD will fall on JL. (Because Z B is given = Z ,7.) CD will fall on KL. (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 PLANE GEOMETRY 55. THEOREM. The angles opposite the equal sides of an isosceles triangle are equal. 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 (Because AX made them = .) .-. A ABX=A ACX. (Why?) (52.) .-.ZJ8 = 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 AXC&YQ rt. A (16). .\AX is J_ 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\ AB C=ET; BC=ST. To Prove : A EST=A ABC. 15 BOOK I 23 Proof : Place A ABC in the position of A AST so that the longest equal sides (BC and 8T) coincide and A is opposite ST from R. Draw RA. RS = AS (Given). /. ASR is an isosceles A. (Def. 24.)' . Z 8RA == Z SAR. (Why?) (55.) Likewise TR = AT (?) and Z TEA = Z TAR. (Why?) Adding these equals we obtain Z SRT = /. SAT (Ax. 2). /. ARST=A AST (52). That is, A RST = 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." PLANE GEOMETRY MODEL DEMONSTRATIONS The angles opposite the equal sides of an isosceles triangle are equal. Given: &ABC; AB = AC. To Prove : Z B = Z C. Proof : Suppose A X is drawn bisecting Z.BAC and meeting BC at X. In the A BAX and CAX AX = AX (Identical). AB = AC (Hypothesis). ZBAX = Z CAX (Construction). .\&ABX - &ACX. (Two A 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. B Two triangles are equal if the three sides of one are equal respectively to the three sides of the other. Given : & ABC and RST; AB = RS; AC= RT; BC = ST. To Prove : A RST = &ABC. Proof: Place A ABC in the position of A AST so that the longest equal sides {BC and ST) coincide, and A is opposite ST from R. Draw RA. RS = AS (Hypothesis). A ASR is isosceles. (An isosceles A is a A two sides of which are equal.) .*. Z SRA = Z SAR . . . (1) . . (The A opp. the = sides of an isos. A are = .) Again, TR = A T (Hypothesis). A TRA is isosceles. (Same reason as before.) Z TRA = Z TAR . . . (2) . . (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, &RST = &ABC. (Substitution; A ABC is the same as &AST.) Q.E.D. 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. Converse of definitions. The converse of a definition is true. It is often advantageous to quote the converse of a definition, as a reason, instead of the definition itself. 64. Homologous parts. Triangles are proved equal in order that their homologous sides, or homologous angles, may be proved 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 _L to CD at its midpoint, ; P any point in AB ; PC and PD. To Prove: I. PC=PD; II. Z CPB = Z DPS ; and III. Z C = Z D. Proof: Iii rt. A P-BC and c< PJ3D, BC = BD (Hyp.) ; BP = BP (Iden.). ..APC = APJ3D. (Why?) (53.) .-. I. PC = PD(Why?)(27;) II. Z CPJ3 = Z DPS ( Why ?) ; III. Zc = ZD (Why?). Q.E.D. 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 ; A 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: PR 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 _L to AB from P. Proof: Extend PR to 8, mak- ing RS = PR ; draw DS. In rt. A PDR and SDR, PR = R8 (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). /. /.PDR, the half of Z PDS, is not a right angle (36). .*. PD is not _L (17). /. PR is the only _L. 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.] 28 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 T L MR S Given: Rt. A LMN and EST ; LN = BT ; and Z L = /.E. To Prove: A LMN = A EST. Proof : Superpose A LMN upon A EST so that Z L coincides with its equal, Z E, LM falling along ES. Then LN will fall on ET and point N will fall exactly on r (LN ET by Hyp.). Now jyjf and TS will both be J_ to ES from r (Rt. A by Hyp.). .*. N M will coincide with TS (71). /. A LMN = A KSr (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 I*" J X L Given : Rt. A UK and iJf ; KI = EM ; KJ = EL. To Prove : A UK = A LME. Proof : Place A UK in the position of A XLE so that the equal sides, KJ and EL, coincide and I is at X, oppo- site Ei from M. BOOK I 29 Now, A ELM and ELX are supplementary. (Why ?) (20.) .'. XLM is a str. line (45). Also, A EMX is isosceles. (Why ?) (EX = EM by Hyp.) ,\ZX=ZJf. (Why?) (55.) /. A XLE = A JfLtf. (Why?) (72.) That is, A UK = ALMK. (Ax. 6.) 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 ?) /. 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. Given : Rt. A ABC and DEF ; AC = DF ; To Prove : A ABC = A = Z D. Proof : In the A ABC and DEF, AC = DF. (Why?) (Hyp.) Also Z.A = Z.D (Why?) and ZC = Z.F. (Why?) (42.) .'.A ABC = A DEF. (Why?) (54.) Q.E.D. Ex. 1. How many pairs of equal parts must two triangles have, in order that they may be proved 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.] so PLANE GEOMETRY 75. THEOREM. The sum of two sides of a triangle is greater thaii the sum of two lines drawn to the extremities of the third side, from any point within the triangle. Given: P, any point in A ABC ; lines PA and PC. To Prove : AE + BC > AP + PC. Proof : Extend AP to meet BC at x. A AB + BX > AP + PX. (Why V) (Ax. 12.) CX + PX > PC. (Why?) (Ax. 12.) Add : AB BX + CX + PX > AP + PC + Px(Ax. 8). Subtract PJT = PX. .^LB + SC > 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. 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 JL bisector of NM (Hyp.). .-. PN = PM (67). II. Given: CD _L to AB ; PN = PM. [First figure.] K To Prove : ND = MD. Proof: In the rt. A PND and PM D, PD = PD (Iden.); and PN = PM (Hyp.). .-. APND = APMD. (Why?) (73.) .\ND=MD. (Why?) (27.) Q.B.D. III. Given : CD _L to AB ; oblique lines PR, PT ; also RD > DT. [Second figure.] To Prove: PR > PT. Proof : Because RD is > DT, we may mark z>s (on RD) = Dr. Draw PS. Extend PD to JT, making DX = PD ; draw RX and &Z. ^D is JL to PX at its midpoint (Const.). .'.PR= RX and PS = SX (66). Now PR + RX > PS + sx (75). Hence P# + P# > PS + PS (Ax. 6). That is, 2 PE > 2 PS. /. P# > PS (Ax. 10). But P5= PT (76, I). .*. PU > 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 : PR JL to AB ; PC not _L. To Prove : PR < PC. Proof : Extend PE to X, mak- ing RX s= PR. Draw CX. PR + RX < PC+ CX (Ax. 12). But AR is _L to PXat its mid- A ~~ point (Const.). /. PC=cx (66). .'. PR H- PR < PC+ PC (Ax. 6). That is, 2 PR < 2 PC. .'. PR< PC (Ax. 10). Q.E.D. N 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 To Prove : PB = PD. Proof : A P.BC and PDC are rt. A (78). ~D~ In rt. A PBC and PDC, PC =PC (Iden.) ; Z PCB = Z PCD (Hyp.). /.APBC=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 ^ICQ = Z ECQ. Proof: A P.BC and PDC are right A (?). In rt. A PBC and PDC, PC = PC (?) ; PB = PD (?) . 0(73). .-.Z^4CQ = ZJCQ(?). 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- rrn ,. , T . ,1 T THE THREE MEDIANS gles. Ihe median 01 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 T 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 JL distances from 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. 34 PLANE GEOMETRY M 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 _L 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^i and B. (?) (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. Given: A ABC, DEF; AB = DE; BC = EF; Z ABC> Z E. BOOK I 35 To Prove : AC > DF. Proof : Place the 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 BX to be the bisector of Z HBC, meeting AC at x. Draw XH. In AtfBXand CBX, BX=BX (?); Bff= (?(?); ZCBX(?) (Const.). .'.&HBX = ACBX (?)(52). XC (?). Now, Ar+xfl- > ^ff(?). .-.4-r + xc > ^ifl (Ax. 6). That is, AC > DF. 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.] Given : A ABC and RST; AB = US ; BC = sr ; -AC > #T. To Prove: Z B >Z&'. Proof : It is evident that Z B < Z -8, or Z B = Z 5, or 1. It Z.B < Z.S, AC < BT (86). But ^4(7 > RT (Hyp.)- .*.Z# is W0 < Z S. 2. If Z = Z 5, the A are = (52), and AC is = J2T (27). But AC > RT (Hyp.). /.Z B is no = Z -8. 3. /. the only possibility is that Z B > Z.S. 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 greatei distance. [Converse of 76, III.] C Given: CDj_toAB; PR and. PS oblique lines ; PR > PS. To Prove : DR > DS. Proof: It is evident that DR < DS, or DR = DS, or DR > DS. A ~ 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 I 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 _L 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 _L to AB. (By Hyp. they are 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 B 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 -g Proof: Suppose XY is drawn through M _L to LM. AB is || to XY (?) (93). But AB is || to CD (Hyp.). Now CD and XY both contain M (Const.). .'.CD and XY coincide (92). But LM is _L to XY (?). That is, LM is J_ to CD. Q.E.D. 38 PLANE GEOMETRY 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 [6, c, e, K], Exterior A are without the lines [a, d,f, g~\. Alternate A are on opposite sides of the transversal [b and h ; c and e ; a and g ; etc.]. Alternate-Interior A are b and h; and c and e. Alternate-Exterior A are a and g ; d and/. Corresponding A are a and e ; d and A ; b and/; c and #. Adjoining-Interior A are c and h ; 6 and e. Adjacent-Interior A are & and c ; e and ft. The left-hand transversal makes eight angles similarly related. The "primes" are used only to designate angles which are different from those in the right-hand part of the figure. 97. THEOREM. If a transversal intersects two parallels, the alter- nate interior angles are equal. Given : AB and K. to CD\ transversal EF cutting the Us at H To Prove : Z a = Z i and Proof : Suppose through Jf, the midpoint of HE", ES is drawn J_ to AB. Then ES is _L to CD (95). In rt. A EMH and JOfS, HJf = .Of (Const.); Z EMH=Z. KMS (?) (51). .'.A EMH = A JTAfS (?) (72). Z re is the supp. of Z a (?) (44) ; j < Z v is the supp. of Z i (?). (49). Q.E.D. EXERCISE. If Z. a = 70, in the figure of 97, how many degrees are there in Zx? inZi? Zv? Z.AHE1 Z.EHB1 +CKF'! ZDKF1 BOOK I 39 98. THEOREM. If a transversal intersects two parallels, the corre- sponding angles are equal. Given : AB II to CD ; trans- versal EF cutting the Us and forming the 8 A. A ~~" z To Prove : Z s = Z i ; Z c Proof: Zs = Za (51); Z AC. To Prove : Z ACB > Z B. Proof : On AB take AE = AC. [We may, because AB > AC.~\ Draw CE and let Z AEC= x. c B Z AEC is an ext. Z of A C3S (?). .-. Z z > Z B (109). Also, Z 4 OR = Z ^LE<7 = Z x (?) (55). Again, Z ACB > Z z (?) (Ax. 5). .-. Z 4CB > Z.B (Ax. 11). Q.E.D. 123. THEOREM. If two angles of a triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. Given: AABC\ /_ACB>/-B. To Prove: AB > AC. Proof : In Z ACB, suppose Z BCE constructed = Z B. Then, CE= BE (?) (120). Also AE + CE > AC (?). /. ,4E + BE > ^LC (Ax. 6). That is, AB > AC. Q.E.D. 124. THEOREM. The hypotenuse is the longest side of a right tri- angle. (See 123.) BOOK I QUADRILATERALS 45 125. 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 RHOMBOID SQUARE RECTANGLE 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. i 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. L To Prove : LM = PO and LP = MO. Proof : Draw diagonal PM. In A LMP and OMP, PM = .'. ALMP = 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- 9 A B AB = DC l AD = BC. To Prove : A BCD is a O. Proof : Draw diagonal BD. / ...""" In &ABD and CBD, BD = BD (?) ; AB = DC (?), and tf AD = BC (?). .'. A ABD = A CBD (?) (58). Hence Z a = Z i (?). Therefore AB is || to DC (?) (101). Also, Z y = Zz (?). Therefore ^D 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 A BCD ; AB = CD and AB \\ to CD. To Prove: A BCD is a O. Proof : Draw diagonal BD. In A ABD and CDD, DD = DD(?); ^D = CD (?); and Za = Z.i (?) (97). .-. A.1DD = ACDD (?) (52). Hence /_y=/.x (?j. .-. ^4D is || to DC (?) (101). .'.ABCD 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 = XH and GX XE. Proof : In A FXG and EXH, E FG = ^JT (?) (130) ; Z a = Z and Z c = Z / (?) (97). .-. A FXG = A EXH (?) (54). /. FX = XH and GX = XE (?) (27). Q.B.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. c M N A D Given : UJAC and LN; AB = LM; AD = LO ; Z A = /_ L. To Prove : The UJ are = . Proof: Superpose UJ ABCD upon n LMNO, so the equal angles A and L coincide, AD falling along LO and AB along LM. Point D will coincide with point o [AD LO (Hyp.)]. Point B will coincide with point M [AB LM (Hyp.)]. BC and MN are both || to LO (?) (126). .-. BC falls along MN (?) (92). CD and JVO are both || to LM (?). .-.CD falls along JVO (?). Hence C will fall exactly upon N (38). .*. the figures coincide, and are equal (?) (28). Q.B.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). o Given : Rhombus ABCD ; di- agonals AC and BD. To Prove: AC J_ to BD; AC and BD bisect each other ; and 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 (?). .'. AC is _L to BD (?) (70). Q.E.D. BOOK I 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 midpoint of AB\ P, the midpoint of BC> line MP. To Prove : MP II to AC and MP = 1 AC. Proof: Suppose AE is drawn through A, II to BC A and meeting MP produced at E. In A AEM and BPM, AM = BM (Hyp.); ^x = Ze (?) and Z o = Z B (?) (97). .'. A AEM = A BPM (?) (54). Hence, AR = BP (?). But BP = PC (?). /. AE = PC (?). .-. ACPE is a O (?) (135). Hence EP or JfP is II to AC (?). Q.E.D. Also, EP = AC (?) (130). But MP = E3f (?) (27). .-. MP=\ RP = \ AC (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 jf, the midpoint of AB to X, the midpoint of BC. MX is II to AC (142) ; but MP A ~~O is II to AC (Hyp.). .*. MX and MP coincide (?). That is, MP bisects BC. Q.E.D. 50 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. M Z. ^3 \P III. MP = | (AD + BC). 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 = 1 (AD + BC) (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 (7. Proof : Suppose BXis drawn through B and II to CD. BX = CD (130); AB = CD (Hyp.). A~~ "T> /. AB = BX (?), and Z. A = Z a (?), and Z. a = Z. D (?) (98). .'. Z. A = Z. D (Ax. 1). Again, Z C 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^ = Z a (?\ .\AB = BX(?}. Etc. BOOK I NOTE. The verb "to intersect" means merely " to cut." In geometry, the verb " to intercept " means " to include between" Thus the statement " AB and CD intercept XY on the line EF" really means, "AB and CD A intersect 'EF and include XY, a part of EF, between them." 51 7 i L 147. THEOREM. Parallels intercepting equal parts on one trans- versal intercept equal parts on any transversal. Given : Us AB, CD, EF. GH, IJ intercepting equal parts -4C, CE t EG, Gl, on the transversal A/ AI, and cutting transversal BJ. To Prove : BD = DF = FH HJ. 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 4.8(7; Jf, the midpoint of hypotenuse AB. To Prove : AM = CM Proof: Suppose MX is drawn II to BC, meeting AC at X. X is midpoint of AC (?) (143). MX is JL to AC (95). That is, MX is J_ to AC at its midpoint, SUL&AM=MC (?) (67). But AM=BM (Hyp.). .'. AM = CM = BM (Ax. 1). Q.E.D. 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. A'-. f Given : A ABC, AX _L to BC, BY _L 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 JS, T8 II to AC ; through C, ET II to AB, forming A EST. The figure ABCE is a O (Const.) and ABTC is a O (?). /. EC = AB and CT = AB (?) (130). .-. EC = CT (Ax. 1). Now CZ is to ET (?) (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 A AB C. Ex. 3. Prove that any altitude of A EST is double the parallel alti- tude of A ABC. [Use 143 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 : BO = f BD and CO = f CE. Proof : Suppose H is the midpoint of BO and J is the mid- point of CO. Draw ED, DI, IH, HE. In A ABC, DE is II to BC and = J BC (?) (142). In A OBC, HI is II to BC and = JBC(?). .-. ED = HI (Ax. 1), and ED is II to HI (?) (94). _ B c /. HO = OD and 10 =OE (?) (137). .'. BH HO = OD and CI= IO = OE (Ax. 1). That is, BO 2 OD = f BD, and CO = 2 #O = f 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 1 . Then BO r = f BD and BO = f BD (?) (151) . .-. BO f = BO (?). That is o' coincides with O and the three medians meet at O which is the distance, etc. Q.E.D. Ex. 1. In the figure of 151, prove EH = \A = DI, by 142. Ex. 2. In the figure of 151, if BO AC, prove the angle BEG 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, OB 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 n-sided polygon is called an n-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 (n-2) A = (n-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 polygons (w 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 = \ n ~ 2 ) * . 'n/ 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. A ? Ex. 4. How many sides has that equiangular polygon each of whose angles contains 160? Ex. 5. If 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 , ; Z.x Zz ; Z NO Y = rt. Z. 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 : Z 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 Z 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 ; A B and CD _fc 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 & ABC and XXRC; 185 (3); 50.] 76. The perpendiculars upon a diagonal of a parallelogram from the opposite vertices are equal. [184 (1).] A* t 70 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 = BM, ZB = ZL and Z BA = Z. OML and BO=OL. 80. Any line terminated in a pair of oppo- 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, A B -f BO A X + 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. Q 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 P, and PN be drawn, Z PNL > Z 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. Z.] 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 & A HE and CDF are proved equal and isosceles; similarly &BGC 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 equal and parallel. [Prove , BCEF'a.a.'] 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, 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 + BC. 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 will 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. A 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- points of the sides of a rectangle form a rhombus. [Draw the diagonals.] A 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 AECFa, 7 and use 143 in DYCandABX.] 123. If two medians of a triangle are equal, the triangle is isosceles. [Use 151. AO = OB (Ax. 3). Hence prove A AEO and DEO 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.] 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. A Proof: /.ACS= ^BCS (?); ZACM = 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 /. ABC, and A M is parallel to BS meeting EC produced, at M, the triangle ABM is isosceles. 130. If AB C 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 A BCD, 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 109; 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. t [Prove the three small A = .] A BOOK I 75 136. If ABCD is a square and E, F, G, H are points on the sides, such that AE = BF = CG = DH, EFGH is a square. [First, prove EFGH equilateral ; then one Z a rt. Z .] 137. If ABC is an equilateral triangle and each side is produced (in order) the same distance, so that AD = BE=CF, the triangle DEF is equilateral. 138. If ABCD 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 twp 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 otjher 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: ZEAC = Z BAF (?). Add to each R of these, Z CAB. :. Z EAB = Z CA F (?). Then prore & 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.] 76 PLANE GEOMETRY 145. The median to one side of a triangle is less than half the sum ot the other two sides. Proof: (Fig. of No. 144.) Produce median EM its own length to #, draw RA. Prove RA - CB. Prove BR < AB + EC, 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 get Z. x = 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.] 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 J. to this line and side ; prove the sum of the three Js = 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, AD be drawn from A, the midpoint of YZ, perpendicular to the base XZ, DZ = $ XZ. [Draw alt. from 7.] 156. If ABC is an equilateral triangle, the bisectors of angles B and C meet at Z>, DE be drawn parallel to AB meeting AC at E, and DF, parallel to EC meeting A C at F, then AE = ED = EF= DF= CF. 157. If A is any point in RS of triangle RST, and 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 Z>, the angle BA C 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 0. 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 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 t6 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 Z>; E is the midpoint of AB, and F of A C; the angle EDF is equal to the angle EAF. [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 A BCD. 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^nd AB' be taken on AC (produced if necessary) equal to AB, and B'C' be drawn meeting BC&tD, then AD will bisect angle BA C. Proof: A ABC = A A B'C' (?) (52). /. their homologous parts are equal. Thus prove ABC'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 ABC 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 INTERNALLY CIRCLES EXTERNALLY TANGENT 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 semicircumference 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 II 81 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. NOTE. 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. 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 G. To Prove : Arc AB = arc LM. Proof : Superpose O O upon the equal O O, making Z O coincide with its equal, Z C. Point A will fall on L, and point B on M (?) (201). Arc AB will coincide with arc LM (?) (192). .-. AB arc = arc 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 o = 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 Ci, and OB with CM (?) (39). .-.Zo = ZC(?) (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: OO = OC; Z LCM > Z o. To Prove : Arc LM > arc AB. Proof : Superpose O O upon O O, making sector AOB fall in position of sector XCM, OB coinciding with CM. CX is within the angle LCM (Z LCM > Z o). Arc AB will fall upon LM, 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. C 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 (?). Hence Z O = Z (7 (?). .'. 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 = 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 RS. To Prove : Arc AB > arc RS. BOOK II 85 X R Proof: Draw the several radii to the ends of the chords. In &AOB and RCS, AO = RC, BO = SC (?) (201). Chord AB > chord RS (Hyp.). /.Z O > Z C (?) (87). /. arc AB > arc RS (?) (208, I). Q.E.D. II. Given: O O = O C; arc AB > arc R8. To Prove : Chord AB > chord RS. Proof : Draw the several radii. In A AOB 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 (?). .'. A OAM= A OBJf (?). Hence, ^Jf= J/B (?). Q.E.D. II. Z ^OJf = Z BOJf (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 JL 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 ET _L to OA at A. To Prove: RT tangent to the circle. Proof: Take any point P in ET (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 ET is without the O. Therefore, ET 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 J_ to ET. Proof: Every point (except A) in RT is without the O (Def. 194). Therefore a line from O to any point of RT (except .4) is > OA. (Because it is > a radius.) That is, OA is t]ae shortest line from o to RT. .-. OA is _L 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 A, and line OC. To Prove : OC passes through A. Proof: Draw radii OA and CA. OA is JL 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: OO; tangents P^L, PB. To Prove : Distance PA = distance PB. Proof : Draw radii to the points of contact, and join OP. A OAP and OBP are rt. A (?) (216). In rt, A O^P and OBP, OP = OP (?) ; OA = OB (?). .'. A O^IP = A OBP (?). .' 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: AoAPandOBP are rt. A (?). They are = . (Explain.) II. Z.IOP=Z J BOP(?). 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 EX (?) (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 CD, and distances OE and OF. To Prove : OE = OF. Proof : Draw radii OA and OC. In the rt. A AOE and OOF, AE = \ AB ; CF = CD (213). But AB = CD (Hyp.). Hence, AE= CF (Ax. 3) ; and OE= 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 A B 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, AO = CO (?); OE = OF (Hyp.). .'. A AOE = A COF (?). .-. AE CF (?). AB is twice AE and CD is twice CF (?). .'. ^1J5 = 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 CZ>, 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 J_ to AH cutting AB at I. Chord AH = chord CD (?) (210). Distance OK = distance OF (?) (221). But OE < 01 (?) (77); and Ol < OK (?) (Ax. 5). .*. O.E < OK (Ax. 11). .'. OJ < 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 jilS JL to OX. Ch. BS is II to ch. CD (?). .'. arc RS > arc CD (Ax. 5). .-. ch. ES > ch. CD (?). But ch. AB = ch. 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 _L bisectors, OZ, OX, OF, be drawn. These Js will meet at a point (?) (85). Using O as center and OA, O.B, 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 B, O to (7, 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 (?). .*. OCis the J_ bisector of AB (?) (70). Q.E.D. BOOK II 91 233. 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 (AB, 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 J_ to EF (?) (95). .-. arc EP = arc FP (?) (212). II. If the Us are two tangents (points of contact being M and N*). Proof : Suppose a secant be drawn II to one of the tangents, cutting O at E and S. RS will be II to other tangent (?) (94). /, arc MR = arc MS; arc RN = arc SN (proved in I). Adding, arc MRN=nrv 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 J if the vertices of the poly- gon are in the circumference, and its sides are chords. 92 PLANE GEOMETRY A polygon is circumscribed] . f ,, .-, . ,-, J . . it the sides of the polygon about a circle, or a circle is n , are all tangent to the circle. inscribed in 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. SUMMARY 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. II. 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 93 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. [Supposediam.drawn_Lto^4J5; this will be _L 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 ZPAB=ZPOB. [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. 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 A POD and POB equal; then & COD and A OB are equal.] 24. In No. 23 prove the external segments equal. 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. Of all chords that can be drawn through a given point within a circle, the chord perpendicular to the diameter through the given point is the shortest. Given: P, the point; BOG the diam.; LS _L to BC at P; GR any other chord through P. To Prove: (?). Proof: Draw OA to GR. Etc. BOOK II 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 center bisects the angle formed by the chords, they are equal. [Draw Js 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 + .4Fequals a constant for all positions of D in arc BC. [Prove this sum = AB + A C.] 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. jj^ 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 = 3ZA. [Draw BC. Z DEC = ext. /. of A ABC = = Z D (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 e~~ (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.] 96 PLANE GEOMETRY 42. The two common interior tangents of two circles are equal. A 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 + HF) = (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 = AB. 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 A C 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 (148). 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 II 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 $ \/35 are not. Two lines 18 ft. and 13 yd. are commensurable, but 18 in. and Vl3 mi. are not. 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 $1? Idime? 1 cent? Are 10 ft. and Vl9 ft. commensurable? 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 B to move farther and farther from the vertex of the right angle. It is evident that the hypotenuse will become longer, that AC 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- A * A A 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 /. 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 /. 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+4 + | + | + T * + *V + & + T * + e *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 1 '; v always =v'- 9 also v ap- proaching the limit I; v 1 approaching the limit V . To Prove : ll l . Proof: v is always = v f (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, jirst, 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 k is a constant : I. v 4- 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 ?, and k is a constant : I. v Jc will approach I k as a limit. II. k v will approach k I as a limit. III. kv will approach kl as a limit. IV. ^ will approach - as a limit. K K k k V. - will approach - as a limit. v I NOTE. 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). .-. vk cannot = I k. Also, v I approaches zero (240). .. (v k) (I k) approaches zero. (Because it reduces to v I.) Hence, v k approaches / k (240). II. Demonstrated similarly. III. If kv = kl, then v = I (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. Y Given : O o = O C ; central A O and C; arcs AB and XY. To Prove : = ^-2 Z. C arc XY 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 - g (Ax. 3). Draw radii 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 A B f ., ^ ''"7~C~l^^^ >' Q.K.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 XT. There will be a remainder PFleft 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 PF will approach zero as a limit and Z PCF will approach zero as a limit. .'. arc XP will approach arc XT as a limit (240) and Z XCP will approach Z XCT as a limit (240). Z will approach -^- as a limit (243) Z XCP Z XCT j arc AB -n T arc AB v , and will approach - - as a limit arc XP arc XT Z o arc AB (?) (242) Z XCF arc XF Ex. How many degrees are there in a central angle which intercepts of the circumference? of the circumference? ^ of the circumfer- ence? of the circumference? 102 PLANE GEOMETRY 245. THEOREM. A central anglers measured by its intercepted arc. Given: O 0; /. AOY \ arc AY. To Prove: Z AOT is measured by the arc AY, that is, they contain the same number of units. Proof : The sum of all the A about 0=4 rt. Zs= 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 are. Take Z AOT, one of these degrees of angle, and arc AT, one of the degrees of arc. Then, Z^tOF^arc AY Z AOT arc AT (?) (244). But = Z4OF-*-a unit of angle = the number of Z AOT units in 2.AOY (237). And = arc AY -4- 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.) "/C ^s. 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 ^.OC is isosceles (?). .-. ZA = ZC (?). Z COD = ZA + Z C (?) (108). /. Z COD = Z^l+Z^ = 2Zj. (Ax. 6). That is, Z A = Z COD (Ax. 3). But Z COD is measured by arc CD (?) (245). .*. \ Z COD is measured by J arc CD (Ax. 3). Therefore, Z A is measured by J arc CD (Ax. 6). 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. is measured by J arc CX (I). T is measured by J arc DX (I). Subtracting, Z C^ID is measured by J arc CD (Ax. 2). Q.E.D. NOTE. It is evident that angles measured by $ the same arc are equal. are Ex. 1. In the figure of 249, if arc CD is 56, how many degrees there in angle A ? If arc CD is 108, how many degrees are in angle A ? Ex. 2, If ^ A contains 35, how many degrees are there in arc CZ>? 104 PLANE GEOMETRY 250. THEOREM. All angles inscribed in the same segment are equal. Given : The several A A inscribed in A segment BAC. To Prove : These angles all equal. Proof : Each Z BAC is measured by 1 arc BC (?) (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 (?) (24T). 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 PA. 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 J arc PAX (?) (247). Z APX is measured by -|- 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 -flTP-*, and add.) BOOK II 105 253. 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 A PC is measured by ^ (arc AC+ arc DB). Proof : Suppose CX drawn through C \\ to AB. Now Z C is measured by J arc DX (?) (249). That is, Z C is measured by J (arc BX+ arc But Z C = Z ^LPC (?) (97) and arc J?x = arc -4C(?) (233). .*. Z.APC is measured by | (arc -4(7 + arc DB) (?) (Ax. 6). Q.E.D. NOTE. This theorem may be proved by drawing chord AD. Then ZAPC is an ext. ^ of &ADP and = /.A + Zi> (?). [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.A\ arcs CMB and CNB. To Prove : Z A is measured by |(arc OB arc CNB). Proof : Suppose CX drawn || to AB. Now Z DCJT is measured by J arc CX (?) (252). That is, Z DCX is measured by J (arc CMB arc But Z DCX =Z ^1 (?) (98) ; arc BX = arc CNB (?). /.Z.4 is measured by ^ (arc CMB arc CJVB) (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 CBX is measured by 1 arc CX (?). That is, by J(arc CE arc EX) . But Z CBX = ZA (?); arc #JT=arc BD (?). .'. ZA is meas. by J (arc CJ arc^D) (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 ?) (252). That is, by^ (arc BXE A arc EX). Etc. NOTE. The theorem of 254 may be proved by drawing chord BC. Then Z DCB = Z A + Z CBA (?) ; or, Z A = Z DCB - /. CBA (Ax. 2). Z DOB is measured by \ arc CMB (?) and Z CBA is measured by (?). Hence, Z A is measured by $ (arc CMB - arc CW5) (?). 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 40 ; find the other arc. [Denote the arc by x.~\ 7. If in figure of 252, arc AP contains 124, how many degrees are there in Z APX ? in Z NPA ? 8. If in figure of 253, arc AC is 85, Z APC 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 BD = 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 a; 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 R-, at S] at T 7 ; at U; (3) The angles A, B, C, D of circumscribed quadrilateral ; (4) The angles between the diagonals RT and SU; (5) The angle between RU 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. 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 : ^ A + Z C are meas. by 1 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 will 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. T p 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, CY; arc AB = arc CD (?). .-. arc A BC = arc BCD (Ax. 2) . A\x y/D 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 ABS.] 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 less than a semicircle are obtuse. 43. An inscribed parallelogram is a rectangle. Proof: Arc AB = arc CD (?) ; arc BC = arc AD (?). .'. arc ABC = arc ADC (?) ; that is, each = a semicir- cumference. Etc. 44. The diagonals of an inscribed rectangle pass through the center and are diameters. e 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 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 A * 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, .C v the line joining the ends of these diameters will pass through the other point of intersection. [Draw chord AB. Use 251 ; 43.] 51. If ABCD 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. [The difference of one pair of arcs = difference of a second pair ; the difference of a third pair = difference of a fourth pair. (Explain.) Transpose negative terms and add correctly, noting that the sum of 4 arcs = sum of 4 others, and hence = 180. Half the sum of these 4 arcs measures the angle between the bisectors. (Explain.) Etc.] ff 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; 251.] 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 (?). 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 andP BM are 3 mutually equiangular. 60. If two circles intersect and a line be drawn through each point of intersection terminating in the circumferences, the chords joining 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. (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 PC. (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). ZCOD = ^A (?) (257). /. OD is || to AB (?). 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.) &ANB = A APC (?). 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 RSC = Z BCX = Z A. .: 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 ext. 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 Z RST, etc. BOOK II 113 Proof: If a circle be described on AO as diam., it will pass through T and 5 (?) (148). If a circle be described on OC as diam., it will pass through R and S (?). /. /.BAR = ^BST (?) ; and ZBCT = ZBSR (?). 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 and the circumference at D, the triangles A OB and C0Z) are mutually equiangular. Also triangles BOC and AOD. Also triangles BOC and ^Z>. Also triangles A OD and Also triangles BCD and COD. 69. If two circles intersect at A and B, 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 drawn 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. 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 V 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 = 1 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. A [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 /" xP 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 B BOOK ii 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 jive 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 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 E and T. Draw ET meeting AB at M. \ I / Statement: Point M bisects /T AB. Q.E.F. Proof : E is equally distant from A and B (?) (201). T is equally distant from A and B (?). Hence, ET 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 C and any sufficient radius, de- scribe arcs meeting at C. Draw 0(7 cutting arc AB at M. BOOK II 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 J_ 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 Jkf. Statement : OS bisects Z LON. \/s Q.E.F. Proof : O and 8 are each equally distant from A and B (?). .*. OS is the J_ 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 AB and point P within it. Required: To erect a line _L to AB at P. Construction : Using P as A f 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 J_ to AB at P. Q.E.F. Proof : PS is the _L bisector of CD (?) (70). Q.E.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 J. to AB at p. Q.E.F. Proof: Segment SPE A EX. ...- p 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. Required : (?). Construction : Using P as a center and any suffi- A . cient radius, describe an .M \ N.. arc intersecting AB at M and N. Using M and N as centers and any sufficient radius, describe arcs intersecting each other at C. Draw PC. Statement: PC is _L to AB from P. Q.E.F. Proof : PC is the _L 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. ^ A Given : Z A OB ; point P in line CD. Required : To construct at P o- an Z=Z^405. Construction: Using O as a center with any radius, describe an arc cutting OA at E and 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 E 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 ^LOU. 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 ^L^P. (By 269.) Statement: PX is II to AB. Q.E.F. Proof: Z.NPX=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 ">,. / / w %^ through A any other line AX. ""p... / On this take any length AC '"g... 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.). / , AO = 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 on 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 J_ 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, PB, OA, OB. Statement : PA and PB are tangents through P. Q.E.F. Proof : O M passes through O (PJf = MO by const). .-. Z P40 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, EC, 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, B, and (7. Q.E.F. Proof: [Use 85.] 274. PROBLEM. To inscribe a circle in a given triangle. B Given: (?). Required: (?). Construction: Draw the three bisectors of the A of A ABC, meeting at O (by 266). Draw J 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 &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 i, 3f, 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 Q.E.F. included angle are given. w 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 = a. At P construct Z P = Z x. u On PTF, the side of this Z, b take PR = b. At R draw RT \\ to PQ; and at Q draw QZ || to PW. Denote the intersection of these lines by S. Statement : PQSR 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 1 . Construction: Con- struct at A, Z.BAC=/.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 1 . Q.E.F, BOOK II 123 Proof: The circumference passes through B (?) (67). .-. AB is a chord (?). AC is tangent to the O (?) (215). .-. ZBACis measured by half the arc AB (?) (252). Any inscribed angle AKB is measured by half the arc A B (?). Therefore, any angle AKB = Z BAG (?) (257, 2). Consequently, any inscribed angle AKB = Z E f (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 J5, two A of a A. A- Required : To construct the third. Construction : At point O in a line Its construct Z a =Z A. R ~" o' At point O in or construct Z b = Z B. Statement : The Z VOB = the third Z of the A. Q.B.F. Proof : Z.a + Z& + Z VOE = 2 rt. A (?) (46). Z^l + Z + the unknown Z = 2 rt. A (?) (HO). .-. Z a + Z 6 + Z VOB = Z.A + /.B + the unknown Z (?). But Ztf + Z5 = Z A + Z B (Const, and Ax. 2). .*. Z VOB = the unknown Z (Ax. 2). That is, ZFOS = the third Z of the A. Q.E.D. Ex. 1. To circumscri.be 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 Z = 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, 5, c of a A. Required : To tf construct the A. Construction : Draw ES = a. Using E as a R ' center and 6 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 : EST is the required A. Q.B.F. Proof: EST is a A (?) (23). EST is the required A. (It contains a, 5, c.) Q.E.D. Discussion : Is this problem ever impossible ? When ? 279. PROBLEM. To construct a triangle if two sides and the included angle are known. Given : The sides a and 5, and their included Z C in a A. Required: To construct the A. Construction : Draw CB =a. At C construct /. BCX = given Z C. On CX take CA b. Join AB. 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. a -^-- -.-- r-r-.-- cf Given: (?). Required: (?). a Construction : Draw BC = a. At B construct Z CB X Z B 1 ; at C construct Z BCY = Z C r . 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: Hypotenuses; leg b. A Required: (?). Construction: Draw an in- definite line CD and at C erect a J_ = b. Using A as a center and c as a radius, describe an arc cutting CD at B. Draw AB. Statement: (?). Proof: (?). Discussion: (?). 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. 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, EX, and at one extremity construct an Z =Z E ; take on the side of this Z 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 EX. Draw radius to the intersection just found. If the known angle is obtuse or right Given : Z E, s.a, and s.o. of a A. \ Construction : As above. Discussion: Case I. s.o. >s.a. The A is always possible. Case II. 8.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. An isosceles A. Case III. s. o. < s. a. (1) If s.o. < the J_ from A to EX, the A is never pos- sible. BOOK II 12? A X (2) If 8.0. = the _L from A to KX, A is a rt.A (216). (3) If s.o. > the J_ from to JBTJr, 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 A KB 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 AB ; Z A' ; difference d. Required : To construct the A. [Analysis : Suppose A ABC 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 AB construct Z.EAX. /.A' and take on AX, AD = d. Join DB. At M, midpoint of DB, draw MY JL to DB meeting AX at C. Draw CB. Statement : (?). Proof : (?). Discussion : (?). A- 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 o circumference a chord which shall be at a given distance from the center. [How many can be drawn from this point?] BOOK II 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 O, 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 /. \ 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 A B, erect at M _L = h; draw AP and BP. Bisect these ; erect Js 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 EC is the completed figure ; CNOM is a square, whose vertex O is the center of the circle, and side ON is the given radius. AB is tan- gent from A. Construction : On CA take CN= r and construct square, CNOM. Prolong CM indefinitely. 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 = sum; 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 EC to AD ; etc. 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 ZA = r. On CS construct square CSOM. At O construct Z MOX = /.K. Draw radius OT -L to OX. Draw tangent at T. Proof: A ABC is a rt. A and it is the rt. A. (Explain.) 2""~ 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 RS, 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 ~~r~~ "^ ! fc 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 _L = altitude, upon an indefinite line. Use the end of this altitude as center and the given sides as radii; etc. BOOK IJ 131 54. One side, an angle adjoining it, and the sum of the other two sides. Construction: At A construct Z BA X = given ZK. 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 A BR whose three sides, AB = a, BR &, A R = 2 m. Draw^C II ioBR and RC II to AB meeting at C. Draw^C. Statement: (?). Proof: (V). 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 J.R JL to ^4 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. m Given: Side = AB, altitudes x and y. Construction: Bisect A B; 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. A M ""B [Given : Sides = A B and BC ; alt. on EC = r.] x 60. One side, an 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 GEOMETRY 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 278.] 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 by the diagonals. Construction : Bisect the perimeter and take AB = half it. Bisect Z K. MA con- K struct Z BAX = 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 II 133 91. One base, the radius of the circumscribed circle, and one of the equal sides. [First, describe a O.] 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 n on- II 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 d 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 : ({) If the given radius is < the given distance; (ti) 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 parallelogram and a triangle by drawing one line (as BD) II to one of the non-|| sides. Hence the construction of a trapezoid is often merely constructing a triangle and a parallelogram. 134 PLANE 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, f 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, f 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.! * It is well to draw the loci concerned as dotted lines. (See No. 124.) t 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 ? V I XIII. To find (by intersecting loci) the center of a circle which will : 125. Pass through three given points.* 126. Pass through a given point and touch a given line at a given point.* 127. Have a given radius and be tangent to a given line at a given point.* 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. Touch two given parallels, one of them at a given point.* 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 a given point and touch a given line at a given point. 136. Touch two given parallel lines, one of them at a given point. 137. Pass through a given point 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 (7; point P. Construction : Draw radius CP. Draw tangent at P meeting AB at R. Bisect PRB, meeting CP produced at ; 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 intersecting 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 A B; point P; O C. Construction : At P erect PX toAB, and extend it below AB, so PR = radius of O C. Draw CR and bisect it at M. Erect MY _L to CR at AT, meeting PX at 0; etc. B 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 It 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 II to x. 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. 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 Z. 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 BC of tri- angle ABC, meeting AB in X and AC in. Y, 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. j / Construction: Draw CEto 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 from one given point to another the shortest path which shall have one point 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 Y, so Y = 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 Oat M. Draw PM cutting AB at D; etc. A> Is there any other chord from P bisected by AB1 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 JR S; 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 II 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 at M. BOOK 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 to PB, meeting the given lines in M and L. Bisect ML at D and erect DC _L 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 EC. Statement : BC is tangent to both (D. Proof : AB=CQ (Const.). AB is || to CQ (?). /. ABCQisz a (V). 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 -r- b, or a:b. A proportion is usually written d x r = - or a : b = x : y, and is read : " a is to & as a; is to y." In this pro- o 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 , 6, x. In the proportion a : m m : z, the mean pro- portional is m, and the third proportional is z. 140 BOOK m 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, #y 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. a : b = x : y, 5. x : y = a : b, 2. a : x = b : y, 6. x : a = y : 5, 3. b : a = y : x, 1. y : x b : a, 4. b : y = a : x, 8. y : b = x : a. Proof: 1. ay = bx (Hyp.). Divide each member by by, j i , . ay bx , A ON ax -, and obtain - = - (Ax. 3). /. = -, or a : b = x : y. Q.E.D. by by by 2. Divide by xy\ etc. 3. &=?/ (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 : x= b : y. Proof: a\l> 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 term 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). Given: a:b-*:,. To Prove: f + *' \a + b:b 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 -f y) x ( a + 5) . Hence, a + b: a = x + y : x (?) (291). Similarly, by adding by, a -f- 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). ~. , fa b:a = x y:x, or Given : a : b = x : y. To Prove : \ \a-b:b = x-y.y. Proof: a:b = x:y (Hyp.). .-. ay = bx (?) (290). Subtracting each side from ax, ax ay = ax bx (Ax. 2). That is, a (x y) = x (a 6) . Hence, a-bi a = x- y : x (1) (291). Likewise, a b:b x :. 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) ab:b = xy: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: a + x:b+ y = a:b = x : y. Also, a x:b y = a:b = x:. BOOK m 143 296. 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: ail = x:y. To Prove : 2| a o x Proof: il = l(?)(294). Ct X ) (295) . a x Divide the first by the second, ^ = x -& (?). Q . E . D . a o 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 3/a : 1/b = tyx : tyy. 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 : a:b = x:y, and c : d = I : m, and e :/= r : 8. To Prove : ace : bd/= 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 = Voi. 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. ( a: b = c: m, and ) Given : I [.To Prove : m = r. ( a : b = c : r. } Proof : am = be and ar = be (?) (290). /. am ar (Ax. 1). Hence, m = r (Ax. 3). Q.E.D. 144 PLANE GEOMETRY 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. r,. a c e g Given: -~ To Prov e: +* + *+f = g = b + d +/+ hid Proof : Set each given ratio = m ; thus, a c e a - = side a. To Prove: II. b*- Proof: In A ARC and BBC, AB = BE (Hyp.); CB is common; AC > BC (Hyp.). .-. Z ABC > Z BBC (?) (87). That is, Z. ABC is obtuse and Z BBC is acute. .'. in A ABC, 5 2 = (l a 2 + & 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, b = 10, c = 12, find the nature of Z C. Ex. 2. In the same triangle find m a . Find ?n 5 . Find m c . Ex. 3. In the same triangle find a p b . Find b p a . Find a j9 c . Find t />c. Find c p a . Find C jo 5 . Ex. 4. Find fc. Find A s . Find A c . Ex. 5. Find the diameter of the circumscribed circle. Ex. 6. Find n a and r a . Find n 6 and r&. Find n c and r,.. Ex. 7. Find * a . Find t b . Find f c . * The segments n and r can be found by 308 ; TI& : r& = 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 24. 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 was 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. How 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 = R ; the distances from the center = x and 22 a;.] 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 p, c, p', b. 30. If h = S,p' = 4, find 6, c, p, a. 31. If a = 10, p' = 15, find c, p, h, b. 32. If a = 9, b = 12, find c, p, p', h. 33. lip = 3,/ = 12, find a, h,b. 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 =6, FB = 9, GH=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 III 171 15 inches and 10| inches. The external segment of the latter is 10; find the external segment 01 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 x, 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, 15, 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 proved 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 proved 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 proved 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 proved 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 two 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 A, B, C] and the other in A', B', C 1 . 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 A ' P, and BB', CC' be drawn through P, terminating in the circumferences, the triangles PEC 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 the hypotenuse of a right triangle perpendiculars be erected meeting the legs pro- duced, the new triangles formed will be similar. 14. In the figure of No. 13, prove : (1) Triangle A BC similar to each of the triangles ACE 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 0, the triangles BOD and A OE are similar. 16. Triangles CED and ABC (fig. of No. 11) are similar. [First show A CAD and CEB similar. /. CA : CB = CD : CE (?). Then use 351 (4).] 17. Triangle ABC is inscribed in a circle and ^4P is drawn to P, the midpoint of arc BC, meeting chord CB at D. The triangles ABD and A CP are similar. 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 BD. 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 & 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 AC and AD. [Draw BD. Prove & 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 = ASiAR. Proof: Draw RS. In & ARS and A CD, ZA = Z A and /. ARS = /. 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 AEB. BOOK in 175 28. If chord A D be drawn from vertex A of inscribed isosceles triangle ABC, cutting BC 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. &ADG and BEF similar. First prove C F 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 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 A BC meet at 0,AB 2 -AC 2 = B0 2 -C0 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 = cjo; a* = cp' (?). 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 <7; 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^lP; 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 A 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 Ax. 3 and 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, A PD and BPC straight lines. Second, & ABC and ABD, similar.] BOOK III 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 drawn in a trapezoid parallel to the bases, the seg- ments between the diagonals and the non- parallel sides will be equal. Proof : & A HI and AB C are similar (?); H / J^^J \K A DJK and DCB also (?). :. = (?). AB BC j- Q We = c (?) " But 'if=ff (?) ' (Use As ' 1)! ete - 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, AB* + EC* + AC* = 8 CM 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 AB of right triangle ABC, meeting A C at D and CB at E, AE 2 -f ~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. 178 PLANE GEOMETRY B. 66. If ABCD is a parallelogram, BD a diagonal, A G any line from A meeting BD at E, CD at F, and BC (pro- duced) at 6r, ^1,E is a mean proportional be- tween EF and EG. Proof : & ABE and EZXF are similar (?) ; also A ^ DE and .BEG (?). Obtain two ratios = 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 0, 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* + CP 2 + DP 2 = AR*> Proof : Draw BC, AD, RD. Ch. BR is ioAB (?). .*. CD is || to BR (?). /. arc BC = arc RD (?). Hence, ch. BC = ch. RD (?). Now, RD 2 = BC 2 = BP 2 + CP 2 (?). AD 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 aim., also A APT and PBS (?). Thus, get two ratios each = P4 : 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 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* + PC* = PB 2 + PD 2 . Proof : Draw PEF _L 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 AB and LM meet at 0. DrawJ?0 and produce it to meet the A D^F DE - DF Q.E.D. Ex. 1. Prove theorem of 386 by drawing through C a line parallel to ^4Z), dividing the trapezoid into a parallelogram and a triangle. Ex.2. Which includes the other, the word "equal" or the word "equivalent"? Which of these words conveys no idea of shape? Ex. 3. What is the area of a parallelogram whose base is 8 inches and altitude is 5 inches ? What is the area of a triangle having the same base and altitude ? Ex. 4. Is the area of a triangle equal to half the base multiplied by the whole altitude? Or half the altitude multiplied by the whole base? Or half the base multiplied by half the altitude? Or half the product of the base by the altitude ? Ex. 5. If, in the figure of 388, one triangle is four times as large as the other, AB = 10, A C = 6, DE = 16, find DF. Ex. 6. The base of a triangle is 20 and its altitude is 15. The bases of an equivalent trapezoid are 13 and 11; find its altitude. BOOK IV 193 389. THEOREM. Two similar triangles are to each other as the squares of any two homologous sides. Given : Similar A ABC and DEF. TO Prove : A DEF Proof: ONE METHOD. Z BAC = /.EDF (?) (323,1). i\ ABC AB ' AC/ /-o-v /^OQQN (r) (doo). = 4* x ^. Now, ^ = ^ (?). A DEF DE DF DF DE A DEF DE DF That is, x = - (Ax. 6). A DEF DE DE DE DF EF * AABC_ AJ? __ AC* _ BC 2 -^- a- 2 = = (297). up* EF Q.B.D. ANOTHER METHOD. Denote a pair of homologous alti tudes by h and A', and the corresponding bases by b and 6'. But - That is, A DEF (822) Q . E . D . 194 PLANE GEOMETRY 390. THEOREM. Two similar polygons are to each other as the squares of any two homologous sides. B' E D E' Given : Similar polygons ABODE and A'B'C'D'E'. To Prove : ABODE : A'B'C'D'E' = Ze 2 : A'B'* = etc. Proof: Draw from homologous vertices, A and A f , all the pairs of homologous diagonals, dividing the polygons into A. These A are similar, in pairs (?) (327). AR AB 2 , - AS ~CD 2 AB 2 'A*' A 75* v A.' W Z5" A r _ ^ 2 _ is 2 , 9 . AT' JxP A > B >^ Therefore, Hence AR_ A 8 _ AT x^ x ^ A R f ~ A $' ~ A T f ? +AS +AT AR ( ^ r3()n But . polygon t' + As' + AT* AR' ABODE _ AB 2 __ BO 2 polygon A'B'C'D'E' jr#* ^,2 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 195 391. 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 J_ to AB, meeting AB at K and ED at L. Draw BF and CE. Now, AACB, ACG, and BCH are 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, AFAC (128), and Z BAF = /_ CAE. (Each = a rt.Z + /.BAG.) .'. A ABF = A ACE (?) (52). Also, A ABF and square AG have the same base, AF, and the same altitude, AC. Hence, square AG ^=2 AABF (379). Similarly, rectangle AKLE o2AACE(?). Therefore, rectangle AKLE =0= square ACGF (Ax. 1). By drawing AI and CD, it is proved in like manner, that rectangle BDLK =c= square BCHI. Then, by adding, square ABDE =0= 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 ?n 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, ff 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 Adding, 8 + T AC + BC = " "" = gg- = l. (Explain.) And .'. E =c= S + T (?). 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 : Sc l uare 8 = ^L Square T AB AB AP AP sv ^ - = -- (Explain. 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: ,4Lii4f. (Explain.) T ~~( BP P 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 the other, it also divides the quadrilateral into two equivalent triangles. To Prove : A ABC - A ADC. 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 A BCD meet at E and A ABE =0= A CDE, the sides A D and EC 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 PLANE 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 tfiangles are equivalent. Given : A AB C 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: ZCAD = ^C'AD' (?) and CA = L C'A (V). /, the rt. A are = (?). Etc. 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- \ eluding these angles. C Given : A 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 OA, etc. AAOC = &AOB = \AE 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 sum 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 AOB = 2 A A OE-, etc. ** E ** 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 h to the bases (?). Denote the altitude of the trapezoid by h. Then EF bisects h (?). A BFE = \ m . \ h (?). A AEF = I m - %h (?). .'.&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 &ABE = \ABx the to AB from E (?). .-. half the trapezoid = \ AB x this _L (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 A equal to the area of the whole A. Etc, 200 PLANE GEOMETRY 36. In the figure of 391, prove : (t) Points /, C, and F are in a straight line. (ft) CE and BF are perpendicular. (fff) A G and BH are parallel. 3 TO n m m m m On n n n m n 1 - [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 =c= 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* + 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 C -n-l n n n m-n m-n m-n m-n m-n n [ n m D n 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 Required: To derive a for- mula for its area, containing only a, 6, and e. c Solution : Draw altitude AD. Now CD = b p a = <**+&-^ jx\ are produced till they meet, find the area of the >"^~ \ less triangle formed. / 6g " \ [The A are similar. .'.30 : 55= x : x + 10. Etc.] 26. The diagonals of a rhombus are 2 ft. and 70 in. Find the area; the perimeter ; the altitude. 27. The altitude (Ji) 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 5) 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 ; A C = 5; EC = 12. BA is produced through A, to D making AD = 4; CA is produced through A, to E so triangle AED is equivalent to triangle ABC. Find ^4 #. 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 V3". 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 A BCD if AB = 78 m., EC - 104m., CD = 50 m., AD = 120 m., and AC = 130 m. 43. One diagonal of a rhombus is ^ 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 V2. 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 ? 4A2 Pan BOOK IV CONSTRUCTIONS 207 402. PROBLEM. To construct a square equivalent to the sum of two squares. A B C D E F Given : (?). Required : (?). Construction : Construct a rt. Z E, whose sides are EX and ET. On EX take EF = AB, and on ET take EG= CD. Draw FG. On FG construct square T. Statement: T^B+S. Q.B.F. Proof : ~GF 2 = EF* + 2?(? (?). But GF 2 = T ; EF 2 = ; EG 2 = 5 (?) (372). Hence, r =c= u +s (Ax. 6). Q.E.D. 403. PROBLEM. To construct a square equivalent to the sum of several squares. Given : Squares whose sides are a, b, ). From the vertex between these (C), draw CG II to BD, meeting AB prolonged, at G. Draw DG. Repeat (2d figure) by drawing EG, then DH II to EG, and EH. Repeat again by drawing AE, FI II to AE, and El. Statement: A IEH o= polygon ABCDEF. Q.E.F. Proof : In 1st fig., A BGD =e= A BCD (381 ; BD is the base). Add polygon ABDEF = polygon ABDEF. .'.polygon AGDEF =c= polygon ABCDEF (Ax. 2). Likewise, AHEF=o= AGDEF -, and A lEH^AHEF. (Explain.) .'. A IEH =0= polygon ABCDEF (Ax. 1). Q.E.D. 410. PROBLEM. To construct a square equivalent to a given poly- gon. Tllse 409 and 408.] 210 PLANE GEOMETRY 411. PROBLEM. To construct a parallelogram (or a rectangle) equivalent to a given square, and having : I. The sum of its base and altitude equal to a given line. II. The difference of its base and altitude equal to a given line. c 1 E : F >:: ... \ V .. \ \ s \ c ; D' A B C G I. Given : Square S and line CD. Required : To construct a O o S, whose base + altitude shall = CD. Construction : On CD as a diameter describe a semicircle. At C erect CE _L to CD and = AB. Through E draw EF II to CD, meeting the circumference at F. Draw FG _L to CD. Take G'D'= GD and draw XY II to G r D r at the distance from it = CG. On XY take HI = GD. Draw HG f and ID' . Statement: O G'D'IH ^ S and base + alt. = CD. Q.E.F. Proof : G'D'lHis a a (?) (135)._ GDXCG=F(? (?) (341). But GD x C=area G'D'IH (?). FG 2 =^c 2 = area S. (Explain.) /. /U G'D r IHoS(Ax. 6). Also (?'!>' +(?'(/ = CD (?). Q.E.D. X H Cr X.. H r !o^ : E' II. Given : Square 5 and line CD. Required : To construct a O =^ s ; base altitude = C-D. BOOK IV 211 Construction : On CD as diameter, describe a O, o. At C erect CE A- to CD and AB. Draw EFOG meeting O at F and G. Take E'G f = EG and draw XY II to E'G' at a distance from it = EF. On XY take HI = .EG. Draw HE' and 1C?'. Statement: 7 E'G'IH^ s and base alt. = CD. Q.E.F. Proof: M7 is tangent to Oo(?). .'.EG EF=EC 2 (?) (333). EG EF= area O E'G'IH (?), and #C 2 = JJ? = area S (?). FG = CD(?. Q.E.D. 412. PROBLEM. To find two lines whose product is given : I. If their sum is also given. e game ag II. If their difference is also given, 1 [Th , j 413. PROBLEM. To construct a square having a given ratio to a given square. D "" -.. R .--' S \ *'*'% a ''' \X \ \ \ / E./ .*F\ S * !"*' m \ n '-I v X Q A B C i7? _/ Given : Square E, and lines m and n. Required : To construct a square such that, The square R : the unknown square = m : n. Construction : On an indefinite line A Y take AB = m, and BC =n. On AC as diameter describe a semicircle. At B erect #D_L to J.C, meeting arc at D. Draw AD and DC. On ^1Z> take DE=a, and draw .F H to AC, meeting DC at F. Using DF =#, as a side, construct square S. Statement : R : S = m : n. Q.E.F. Proof : Z. ADC is a rt.Z(?). .-. ii? : DC 2 = m : n (?) (395). -:=^ ?); p=^<^)- (Ax. 6). Q.E.D. PLANE GEOMETRY 414. PROBLEM. To construct a polygon similar to a given polygon and having a given ratio to it. Given : (?). Required : (?). Construction and Statement are the same as in 413. Proof : ^ ADCis a rt. Z (?). .-.Al?:~J)C 2 = m:n (?)(395). = ^ (?); ...^ = ^ 2 = ^. (Explain.) * = ^ (?)(390). x DC^ & w* n ^ S x* ^ .'. E : 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 3 Given : Polygons B and s. Required : (?). Construction : Construct squares R f =o=R, and S r =o=S (by 410). Find a fourth proportional to a, b, and AB. This is CD, Upon CD, homologous to AB, construct T similar to R. Statement: r=o& Q.E.F. CD' ( . | =^5 (Const.). /- = - b CD b* CD . Now, a 2 = .-.5 = 5 (Ax. 6). .\T^8 (Ax. 3). ='s'=o=/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 V2 units. [See 402.] 23. To find a line whose length shall be V3 units. 24. To find a line whose length shall be v/II units. 25. To find a line whose length shall be Vf 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 : VF. 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 f 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(t). A b 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, #, to &', b and h. Con- struct a rectangle having base = b' and alt. = x. Statement: This rectangle, B^A. I n Proof: b':b = h:x (Const.). .-. Vx = bh (?). But L I 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=2&ADE Hence, AB A C = 2 (AD . A E) (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 & leg of isosceles A. .:AABC:&AXX'=AB.AC:AX.AX'. But the A are equivalent and AX = AX' (Hyp.). Hence, AB - AC A X 2 . :.AX is a mean propor- B~ tional between AB and A C. 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' - i (A OAD + A OBC) / \ and is similar to A OBC. &Z. \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 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 potygon. To Prove : AG is regular. Proof : /. A is measured by I arc BGL (?). Also Z B is measured by \ arc CHA (?), etc. Subtended arcs, AB, BC, CD, etc., are all = (?). Hence, arcs BGL, CHA, DIB, etc., are all = (Ax. 2). /. Z.A = /.B = ZO= 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, BC, 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 the 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., G AB = BC = CD = etc. (?). Z HAB = Z HBA = ZlC = Z.ICB = Z JCD = etc. (?). .*. these A are isosceles (?) (120), and = (?). /. ZH= Z/= Z J=etc. (?). That is, polygon GJ is equi- angular. Also, AH = HB = BI = 1C = CJ =etc. (?), and HI = u = JK = etc. (?) (Ax. 3). That is, polygon GJis 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: (?). Proof : Arcs AB, BC, CD, etc. are all = (?). Also, arcs AI, 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 r . To Prove : They are simi- lar. rc-2^ 180 Proof: /.A Z.A' (n-2) 180 n (?). .-. Z A = Z A' (?). Similarly, Z B = /. B f , Z. C= Z c', etc. That is, these polygons are mutually equiangular. Also, AB = BC = CD= etc.; A f B f = B f C f = C'D' = /. AB : A'B' = BC : b'c' = CD : C'D' = etc. (Ax. 3). That is, the homologous sides are proportional. Therefore, the polygons are similar (?). (?). .E.D. 220 PLANE GEOMETRY 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, OB, OC, and draw line OD. In A AOB and COD, AB = CD (?) ; BO = CO (?). Now ZABC = ^BCD (?) (416). ZotfC = 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 proved 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 n-gon _ 3 6o c 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'B', B'C', etc. are all= (?). /. the inscribed poly- gon is regular (?). Also, A AOB, A f OB f , etc. are isosceles (?). If a line OX be drawn from O, bisecting /. AOB, it is J_ 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, (a) Triangle A OB similar to triangle A' OB'. (b) 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. B 1 Given : Regular w-gons, EG whose perimeter is P, radius B, apothem r\ and E'C' whose perimeter is P', radius u', apothem r r . To Prove : P : P' = E : E f = r : r f . Proof: Draw radii OB and o r B f . In A AOB and A'O'B', ; AO = BO and ^L'o' = J3' Hence, ^ (Ax. 3). A'O 1 B'0 f .'. A AOB is similar to A A r O r B r (?) (317). A ~f> "D M *. f f = = (?). Also, the polygons are similar (?). A B R T Q.E.D. P' A'B' p' 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 1 denote their areas, we have : A (?)(390). But AB A 7 ^' '- (Explain.) Q.E.D. BOOK 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 = %AB -r Area of A BOC = ^BC Area of A COD = CD etc., etc. Area of poly gon=%(AB or, area = J P-r (Ax. 6). A B. CD -f- etc.) >r (?), 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. E^-* ^^D Given : -ZV-gon FC inscribed in O ; apothem = r ; radius = R. To Prove: That as the number of sides is indefi- nitely increased, r approaches B as a limit. Proof : In the A AOK, or R Now as the number of sides of the polygon is indefinitely increased, AB will be indefinitely decreased. Hence, ^ AB, or AK, will approach zero as a limit. .'. R r will approach zero (because R r < AK). That is, r will approach R as a limit (240). Q.E.D. NOTE. It is evident that if the difference between two variables approaches zero, either (1) one is approaching the other as a limit ; or (2) both are approaching some third quantity as their limit. 224 PLANE GEOMETRY Ex. 1. The apothem of a regular polygon perpendicular to a side bisects the side and the central angle of the polygon. Ex. 2. Are all pairs of squares similar? Why? Are all pairs of rhombuses similar ? Why ? Is a rhombus a regular polygon ? Is a rectangle a regular polygon ? Ex. 3. What is the area of a square whose perimeter is 40 inches ? Ex. 4. What is the area of a regular hexagon whose side is 8 and apothem is 4 V3 ? Ex. 5. What is the area of a regular dodecagon whose side is 10 ^2 - V3 and whose apothem is 5 \2 + \/3 ? Ex. 6. Prove that the lines joining the midpoints of the sides of a regular polygon, in order, form a regular polygon of the same number of sides. Ex. 7. Prove that a polygon is regular if the inscribed and circum- scribed circles are concentric. Ex. 8. Draw a figure showing an inscribed equiangular polygon that is not regular. Ex. 9. Draw a figure showing a circumscribed equilateral polygon that is not regular. Ex. 10. Prove that the circumference of a circle is greater than the perimeter of any inscribed polygon. 438. The theorems of 439 and 440 are considered so evident, and rigorous proofs are so difficult for young students to comprehend (like the demonstrations for many fundamental principles in mathematics) that it is advisable to omit the profound demonstrations and insert only simple explanations. 439. THEOREM. The circumference of a circle is less than the per- imeter of any circumscribed polygon. By drawing tangents at the midpoints of the included arcs another circumscribed polygon is formed; the perim- eter of this polygon is less than the perimeter of the given polygon (?). This can be continued indefinitely, decreas- ing the perimeter of the polygons. Hence, there can be no circumscribed polygon whose perimeter can be the least of all such polygons ; because, by increasing the number of sides, the perimeter is lessened. Hence, the circumference must be less than the perimeter of any circumscribed polygon. BOOK V 225 440. THEOREM. If the number of sides of an inscribed regular poly- gon and of a circumscribed regular polygon be indefinitely increased, I. The perimeter of each polygon will approach the circumference of the circle as a limit. II. The area of each polygon will approach the area of the circle as a limit. Given: A circle O, whose cir- cumference is C and area is 8 ; AB and A r B r , sides of regular cir- cumscribed and inscribed poly- gons, having the same number of sides ; P and P 7 , their perimeters ; K and E!, their areas. To Prove : That if the number of sides be indefinitely increased : I. P will approach C and P 1 will approach C as limit. II. K will approach 8 and K 1 will approach s as limit. Proof: I. The polygons are similar (?) (425). ... = .2? (?) (434). Now, if the number of sides of P 1 OD these polygons be indefinitely increased, OD will approach OE (?) (437). /~l 77* 7-* Hence, will approach 1. That is, - will approach 1, or P OD P' and P 1 will approach equality ; that is, they will approach the same constant as a limit. But P > C and C > P' and C is constant. Hence, P will approach C and P' will approach C. Q.E.D. III. .fL-^L (?) (435). If the number of sides of these ** OD 2 _ polygons be indefinitely increased, OD 2 will approach (XE 2 , and thus OD approach unity. (The argument continues the same as in I.) 226 PLANE GEOMETRY 441. THEOREM. The circumferences of two circles are to each other as their radii. Given : Two whose radii are R and R r and circumfer- ences, C and C 1 respectively. .../ To Prove : C : C f = R : R f . Proof: Circumscribe regu- lar polygons (having the same number of sides) about these and let P and P f denote their perimeters. Then, P : P f R : R' (?) (434). Hence, P R f = P f R (?). Now suppose the number of sides of these polygons to be indefinitely increased, P will approach C (?) (440). P' will approach c' (?). .'. P - R f will approach C R f , and P' R will approach C 1 - R. Hence, C-R' = c r -R (?) (242). Therefore C : C r = R : R f (?) (291). Q.E.D. 442. THEOREM. The ratio of any circumference to its diameter i& 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 ^ = |^ = ^(?). JLV R Zt R D .-.=*, (Ax. 1). d n f Hence, - = ~ (?) (292). That is, - = constant. Q.E.D. 443. Definition of TT (pi) . The constant ratio of a circum- ference to its diameter is called IT. That is, - = TT. Z) BOOK V The numerical value of TT = 3.141592 (This is determined in 470.) 227 , approximately. 444. FORMULA. Let C = circumference and B = radius. Then, ^-=TT (443). .'. C = 27rR (Ax. 3). 2* 445. THEOREM. The area of a circle is equal to half the product of its circumference by its radius. Given : O whose circumfer- ence = (7, area =6', radius =1?. To Prove : 8 = % c B. Proof : Circumscribe a reg- ular polygon about the circle; denote its area by K and per- imeter by P. NowJT=Jp..B(?)(436). Suppose the number of sides of the polygon be in- definitely increased. K will approach S, and P will approach C (?). |- P B will approach ^ C B as a limit (?). Hence, 8 = C - B (?) (242). .B.D. 446. FORMULA- Let 8 = area of 0, C = its circumference, and B = its radius. Then, 8 = J C B (445). Now C= 2 TT U (444). Substituting, s = % (2 TTR) B. .*. S Ex. 1. Could 445 be proved 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 2 . 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 : 8 1 = R 2 : R r2 = D 2 :D f2 . And = 8' (|Z/) 2 l D .'. S:S' = R 2 ": It' 2 = I} 2 : D' 2 (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 : 7rR z = 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 O = 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 n B n are similar arcs; AOB, A f OB r , and A"OB rf 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 229 454. THEOREM. Similar arcs are to each other as their radii. Given : Arcs whose lengths are a and a', radii B and R f . To Prove : a : a' = R : R f . ; .'.a : a = B: E. Q.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 R'. To Prove: T : T' = R 2 : R 1 ' 2 . ... _JL = JE (?). ... T : T 1 = R 2 : R 12 . (Explain.) Q.E.D. 456. THEOREM. Similar segments are to each other as the squares of their radii. Given: (?). To Prove: (?). o Proof: &AOB and A'O'B' are similar (?) (317). and sector , = ~ (?) (455). ' 1 ' 2 ^ sector O'A'C'B' R' 2 sector PACE _ AAOB ,^ x sector O'A'C'B' A A'O'B' sector - A n)< sector O'A'C'B' - A vl'o's' A 4'o'B Hence, segment ^nc = ^AOB_ = & (Ax< 6> segment A'B'C' A ^'o'^' .B /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 ABODE is a regular pentagon, and diagonals AC and ED be drawn, meeting at 0: (a) AO will = AB. (b) A O will be || to ED. (c) ABOC will be similar to A BDC. (O ^ACB will = 36. (e) A C will be divided into mean and ex- treme ratio at 0. 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 OB C is rt. A (?) ; and OC' 2 -OB 2 = 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. f Entire figure ~ \irAl? + crescent B DC + crescent A EC (?). :: + $7rBC 2 + A ABC (?). Entire figure- ^ir 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. Proof: Area = * ,() The side and area of the inscribed square. (c) The side and area of the inscribed regular hexagon. (d) The side of the inscribed regular decagon. (See 365.) (e) The area of 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 i5-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 : B C is the side of an inscribed regular 15-gon. Q.E.F. Proof: Arc JB(7=arc AB arc AC = 1 -^ = -g 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 465. Sides of inscribed polygons. 1. Side of inscribed equilat- eral triangle = R V3. Proof : Z ACS is a rt. Z (?). AB = 2 R and CB = R (?) 2. Side of inscribed square = tfV2. 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 rv B e- equilateral A = 2 R V3. Proof : Z. DAB = Z. DBA = ZD = 60(?). /.A ABD is equilateral. AD = AB = R_V3 (?). .'.DF=2R V3. 2. Side of circumscribed square= 2 B (?). 3. Side of circumscribed regular hexagon = | R V3. (Explain.) 467. In equilateral triangle, apothem = | P. Proof: Bisect arc AC at H. Draw OA, OC, Figure AOCH is a rhombus. (Explain.) 0^=i OH = IE (?) (141). BOOK V 237 468. PROBLEM. In a circle whose radius is R 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 R; 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, is an acute Z. Hence in A AOC, AC* = ol 2 + oc 2 - 2 . oc ON (?) (346). [0 = R, 00= R, ON = .'. AC 2 = 2 # 2 - 2 R 1 V4 R 2 - s 2 (Ax. 6) ; AC = or Q.E.F. 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. (b) The side of the circumscribed equilateral triangle. (c) The side of the inscribed square. (d) The side of the circumscribed square. (e) The side of the inscribed regular hexagon. (y) The side of the circumscribed regular hexagon. (g) The apothem of the inscribed equilateral triangle. (Ji) The apothem of the inscribed regular hexagon. (i) The side of the inscribed regular dodecagon (468). (/) The side of the inscribed regular octagon. 238 PLANE GEOMETRY 470. PROBLEM. To find the approximate numerical value of ir. Given : A circle whose diameter = D and circumference = c. Required : The value of TT, that is, the value of C -j- D. Method: 1. We may select a O of any diameter (442). Hence, for simplicity, we take the O in which D = 2 ; /. R = 1. 2. We compute the perimeter of some inscribed regular polygon (by 465). 3. We compute the length of a side of the inscribed regular polygon having double the number of sides, by the formula = \2 V4 s 2 (469). From this we can find the perimeter of this polygon. 4. Using the side of this polygon as known, we compute, by the same formula, a side of the inscribed regular polygon having still double the number of sides. Hence its perim- eter can be found. 5. By continuing this process we may approximate the value of the circumference (440, I). 6. Thus we can find the value of C -J- D, or TT. Computation : 1. Assume E = 1. 2. Consider the regular hexagon and let s 6 represent its side and P 6 its perimeter. Then s 6 = 1 and P 6 = 6 (?) . 3. Then, a side of the inscribed regular 12-gon is /2-V4^1= 0.5176381, and P 12 = 6.2116572. '12 4. Thus we may find s 24 = 0.2610524 and P 24 = 6.2652576. 5. By continuing, s 3072 = 0.002045, and P 3072 = 6.283184. 6. /. it is evident that C= 6.283184, approximately. But, ir=- (?). 7T = 6.283184 = 3.141592 +. Q.E.F. This calculation is tabulated for reference. * 6 =1, .-./>= 8. s 12 = 0.517638, .-. P 12 = 6.211657. s 24 = 0.261052, .-. P 24 = 6.265257. $48 = 0.130806, .-. P 48 = 6.278700. fe = 0.065438, .-. P 96 = 6.282063. s m = 0.032723, .-. P s 384 = 0.016362, .-. P s m = 0.008181, .-. P s ls36 = 0.004091, .-. P 1 $3072= 0.002045, .-. P 3 = 6.282904. = 6.283115. = 6.283169. = 6.283180. = 6.283184. BOOK V 239 ORIGINAL EXERCISES (NUMERICAL) MENSURATION OF REGULAR POLYGONS AND THE CIRCLE 1. Find the angle and the central angle of: (i) a regular pentagon ; (ii) a regular octagon ; (iii) a regular do- decagon; (iv) a regular 20-gon. 2. Find the area of a regular hexagon whose side is 8. 3. Find the area of a regular hexagon whose apothem is 4. 4. In a circle whose radius is 10 are inscribed an equilateral tri- angle, a square, and a regular hexagon. Find the perimeter, apothem, and area of each. 5. About a circle whose radius is 10 are circumscribed an equilat- eral triangle, a square, and a regular hexagon. Find the perimeter and area of each. 6. Find the circumference and area of a circle whose radius is 5 inches. [Use TT = 3^.] 7. Find the circumference and area of a circle whose diameter is 42 centimeters. 8. The radius of a certain circle is 9 meters. What is the radius of a second circle whose circumference is twice as long as the first? Of a third circle whose area is twice as great as the first ? 9. If the circumference of a circle is 55 yards, what is its diameter ? 10. If the area of a circle is 113 square meters, what is its radius? 11. In a circle whose radius is 35 there is a sector whose angle is 40. Find the length of the arc and the area of the sector. 12. The area of a circle is 6^ times the area of another. If the radius of the smaller circle is 12, what is the radius of the larger circle ? 13. If the angle of a sector is 72 and its arc is 44 inches, what is the radius of the circle ? What is the area of the sector ? 14. In a circle whose radius is 7 find the area of the segment whose central angle is 120. [See 451.] 15. If the radius of a circle is 4 feet, what is the area of a segment whose arc is 60 ? of a segment whose arc is a quadrant? 16. Find the area of the circle inscribed in a square whose area is 75. 240 PLANE GEOMETRY 17. Find the area of an equilateral triangle inscribed in a circle whose area is 441 TT square meters. 18. If the length of a quadrant is 8 inches, what is the radius ? 19. Find the length of an arc subtended by the side of an inscribed regular 15-gon if the radius is 4f inches. 20. The side of an equilateral triangle is 10. Find the areas of its in- scribed and circumscribed circles. 21. Find the perimeter and area of a segment whose^ chord is the side of an inscribed regular hexagon, if the radius of a circle is 5. 22. A circular lake 9 rods in diameter is surrounded by a walk \ rod wide. What is the area of the walk ? 23. A locomotive driving wheel is 7 feet in diameter. How many revolutions will it make in running a mile? 24. What is the number of degrees in the central angle whose arc is as long as the radius ? 25. Find the side of the square equivalent to a circle whose diameter is 4.2 meters. 26. Find the radius of that circle equivalent to a square whose side is 5.5 inches. 27. Find the radius of the circumference which divides a given circle whose radius is 10^ into two equal parts. 28. Three equal circles are each tangent to the other two and the diameter of each is 40 feet. Find the area between these circles. [Required area = area of an eq. A minus area of three sectors.] 29. Find the area of the three segments of a circle whose radius is 5\/3, formed by the sides of the inscribed equilateral triangle. 30. If a cistern can be emptied in 5 hours by a 2-inch pipe, how long will be required to empty it by a 1-inch pipe ? 31. Find the side, apothem, and area of a regular decagon inscribed in a circle whose radius is 6 feet. 32. What is the area of the circle circumscribed about an equilateral triangle whose area is 48 V3? 33. The circumferences of two concentric circles are 40 inches and 50 inches. Find the area of the circular ring between them. 34. A circle has an area of 80 square feet. Find the length of an arc of 80. BOOK V 241 35. Find the angle of a sector whose perimeter equals the circum- ference. 36. Find the angle of a sector whose area is equal to the square of the radius. 37. Find the area of a regular octagon inscribed in a circle whose radius is 20. [Inscribe square, then octagon. Draw radii of octagon. Find area of one isosceles A formed, whose altitude is half the side of the square.] 38. A rectangle whose length is double its width, a square, an equi- lateral triangle, and a circle all have the same perimeter, namely 132 meters. Which has the greatest area? the least? 39. Through a point without a circle whose radius is 35 inches two tangents are drawn, forming an angle of 60. Find the perimeter and area of the figure bounded by the tangents and their smaller intercepted arc. 40. In a circle whose radius is 12 are two parallel chords which sub- tend arcs of 60 and 90 respectively. Find the perimeter and area of the figure bounded by these chords and their intercepted arcs. 41. A quarter mile race track is to be laid out, having parallel sides but semicircular ends whose radius is 105 feet. Find the length of the parallel sides. 42. If the diameter of the earth is 7920 miles, how far at sea can the light from a lighthouse 150 feet high be seen ? 43. The diameter of a circle is 18 inches. Find the area of the figure between this circle and the circumscribed equilateral triangle. 44. How far does the end of the minute hand of a clock move in 20 minutes, if the hand is 3 1 inches long ? 45. The diameter of a circle is 16 inches. What is the area of that portion of the circle outside the inscribed regular hexagon ? 46. Using the vertices of a square whose side is 12, as centers, and radii equal to 4, four quad- rants are described within the square. Find the perimeter and area of the figure thus formed. 47. Using the four vertices of a square, whose side is 12, as centers and radii equal to 6, four arcs are described without the square (see figure). Find the perimeter and area of the figure bounded by these four 242 PLANE GEOMETRY 48. Using the vertices of an equilateral triangle, whose side is 16, as centers and radii equal to 8, three arcs are described within the tri- angle. Find the perimeter and area of the figure bounded by these arcs. Do the same if the three arcs are described without the triangle (terminat- ing in the sides, in each case). 49. Using the vertices of a regular hexagon, whose side is 20, as centers and radii equal to 10, six arcs are described within the hexagon. Find the perimeter and area of the figure bounded by these arcs. Do the same if the six arcs are described without the hexagon (terminating in the sides, in each case). 50. If semicircumferences be described within a square, whose side is 8 inches, upon the four sides as diameters, find the areas of the four lobes bounded by the eight quadrants. Find the area of any one. In the following exercises let n = the number of sides of the regular polygon ; s = the length of its side ; r = its apothem ; R = its radius ; K = its area. 51. If n = 3, show that s = tf V3 ; r=*R; K = 52. If n = 4, show that s = R V2 = 2r; K = 2 2r V3 53. If n = 6, show that s = R = 3 54. If n = -8, show that s = R^2 - \/2 = 2r (V2- 1) ; r =^ J2 + V2 ; R = r^4:-2 V2; ^= 2 .R 2 \/2 ^ 8r 2 (V2- 1). 55. If n = 10, show that s = (\/5-l); r=-JlO + 2V5. 56. If n = 5, show that s = 57. If n = 12 show that s = R*\J2-V3 = 2 r(2-V3); R = 2 r BOOK V 243 58. The apothem of a regular hexagon is 18 V3 inches. Find its side and area. .Find the area of the circle circumscribed about it. 59. What is the radius of a circle whose area is doubled by increas- ing the radius 10 feet V 60. If an 8-inch pipe will fill a cistern in 3 hours 20 minutes, how long will it require a 2-inch pipe to fill it ? 61. The radius of a circle is 12 meters. Find : (a) The area of the inscribed square. (i) The area of the inscribed equilateral triangle. (c) The area of the inscribed regular hexagon. (rf) The area of the inscribed regular dodecagon. (e) The area of the circumscribed square. (/) The area of the circumscribed equilateral triangle. (<7) The area of the circumscribed regular hexagon. Qi) The area of the circumscribed regular dodecagon. 62. The radius of a circle is 18. Find : (a) The side and apothem of the inscribed square. (b) The side and apothem of the inscribed equilateral triangle. (c) The side and apothem of the inscribed regular hexagon, (e?) The area of the inscribed square. (e) The area of the inscribed equilateral triangle. (/) The area of the inscribed regular hexagon. (g) The area of the inscribed regular octagon. (h) The area of the circumscribed regular hexagon. 63. Prove that the area of an inscribed regular hexagon is a mean pro- portional between the areas of the inscribed and the circumscribed equilateral triangles. [Find the three areas in terms of jR.] 64. A B is one side of an inscribed equilateral triangle, and C is the midpoint of AB. If AB be prolonged to making BO equal to BC, and OT\)& drawn tangent to the circle at T 7 , OT will be f the radius. 65. A square, an equilateral triangle, a regular hexagon, and a circle all have the same area, namely 5544 sq. ft. Which figure has the least perimeter ? the greatest ? 66. A square, an equilateral triangle, a regular hexagon, and a circle all have the same perimeter, namely 396 in. Find their areas and compare. 67. The circumferences of two concentric circles are 330 and 440 in. respectively. Find the radius of another circle equivalent to the ring between these two circumferences. 244 PLANE GEOMETRY ORIGINAL CONSTRUCTIONS 1. To circumscribe a regular hexagon about a given circle. 2. To circumscribe an equilateral triangle about a given circle. 3. To circumscribe a regular decagon about a given circle; a regular 16-gon ; a regular 24-gon ; a square. 4. To construct an angle of 36; of 18; of 72; of 24; of 6; of 48; of 96. 5. To construct a regular hexagon upon a given line as a side. 6. To construct a regular decagon upon a given line as a side. 7. To construct a regular octagon upon a given line as a side. 8. To construct a regular pentagon upon a given line as a side. 9. To construct a square which shall have double the area of a given square. 10. To inscribe in a given circle a regular polygon simi- lar to a given regular polygon. Construction : From the center of the polygon draw radii. At the center of the circle construct A- = these cen- tral A of the polygon. Draw chords. Etc. 11. To construct a regular pentagon which shall have double the area of a given regular pentagon. 12. To construct a circumference equal to the sum of two given cir- cumferences. 13. To construct a circumference which shall be three times a given circumference. 14. To construct a circumference equal to the difference of two given circumferences. 15. To construct a circle whose area shall be five times a given circle. 16. To construct a circle equivalent to the sum of two given circles ; another, equivalent to their difference. 17. To construct a circle whose area shall be half a given circle. 18. To bisect the area of a given circle by a concentric circumference. 19. To divide a given circumference into two parts which shall be in the ratio of 3:7; into two other parts which shall be in the ratio of 5:7; into still two other parts, in the ratio of 8 : 7. BOOK V 245 MAXIMA AND MINIMA 471. Of geometrical magnitudes which satisfy a given condition (or given conditions) the greatest is maximum, and the least is minimum. Thus, of all chords that can be drawn through a given point within a circle, the diameter is the maximum, and the chord perpendicular to the diameter at the point is the minimum. Isoperimetric figures are figures having equal perimeters. 472. THEOREM. Of all triangles having two given sides, that in which these sides form a right angle is the maximum. Given: A ABC and A ABD having AB common, and D AC = AD, but Z CAB a rt. Z and Z DAB not a right Z. To Prove : A ABC > A ABD. Proof : Draw altitude DE. Now AD > DE (?). .'. AC > DE (Ax. 6). Multiply each member by ^ AB. Then J AB AC > J AB DE (?). Now AB - AC = area A ABC (?), and l 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 ~L 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 ? 246 PLANE GEOMETRY 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 ABG isosceles. To Prove : A ABC > A ABD. Proof : Prolong AC to E 9 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 H 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.). /. Z ABE = rt. Z (?). That is, AB is _L to EF. Hence, CG and DH are J_ 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 > AF (Ax. 6). .-. BE > BF (?) (90), and BE > BF (?). Now, BG = BE and BH = BF (?) (73, Cor.). /. BG > # (Ax. 6). Multiply each member by \ AB. Then, AB BG> AB BH(?). But, % AB BG = area A ABC (?), and J ^# - BH = area A ^4D (?). .'. A ABC > A ABD (Ax. 6). 474. THEOREM. is the maximum. [Any side may be considered the base.] Q.E.D. Of isoperimetric triangles the equilateral triangle 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 the same perimeter and the same number of sides. To Prove : AB = EC = 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). 1 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.B.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 BC, and draw AD. On AD as base, construct isosceles A ADE, isoperimetric with A ABD. Now A AED > A ABD (?) (473). Adding A ADC to each member, AEDC > A ABC (?). AEDC is isoperimetric with A J.BCand 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 5, 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 Ifeoperirnetric with A. Then A > 8 (477), but A =c= B (?). .-. B > s (?) (Ax. 6). Hence, the perimeter of B > perimeter of S (390). But, the perimeter of /S = 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 D 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 A ABC < the perimeter of A A B' C. 4 Proof: AD = DX (?). .-. D^ is J_ to ^X (?) (70). That is, AF is J_ to all lines in MN through F. Consequently, AF is J. to plane MN (?) (483). Q.E.D. 502. THEOREM. All straight lines perpendicular to a line at one point are in one plane, which is per- pendicular to this line at this point. Given : AB _L to BC, BD, BE, etc. ; plane MN containing BC and BD. To Prove: BE is in the plane MN and MN is to AB at B. Proof : Pass plane AE contain- ing AB and BE, and intersecting MN in line BX. L - = - 1 N Now, AB is J_ to MN (?) (501). BOOK VI 257 That is, plane MN is J- to AB. .-. AB is J_ to BX (?) (489). But AB is _L to BE (Hyp.). That is, BX and BE are both -i. to AB, in the plane AE, at B. Hence, BE and BX coincide (?) (43). That is, BE is in the plane MN. Q.E.D. 503. THEOREM. Through a point in a straight line one plane can be passed perpendicular to the line, and only one. (502.) 504. THEOREM. Through an external point one plane can be passed perpendicular to a given straight line, and only one. Given : The line AB and point P out- side of AB. To Prove : Through P, one plane can be passed JL to AB, and only one. Proof : I. Draw from P, PC _L to AB, and at C draw CX, another line _L to AB. PC and CX determine a plane MN (?). Plane MN contains P and is _L to AB (?) (501). II. Only one line -L to AB can be drawn from P (?) (71). And only one plane J_ to AB can be passed at C (?) (503). That is, MN is the only plane J- to AB, that can be passed through P. Q.E.D. N IB Ex. 1. If two lines are each parallel to a plane, are the lines neces- sarily parallel ? Ex. 2. If two planes intersect, how can a line be drawn through a given point that shall be parallel to both planes ? Ex. 3. How many planes are determined by two intersecting lines (a and 6) and two points (R and S) not in the plane of the lines? 258 SOLID GEOMETRY B 505. THEOREM. Two planes perpendicular to the same straight line are parallel. Given : Planes MN and OP 1_ to AB. To Prove : MN II to OP. Proof : If the planes MN and OP are not II, they will meet when sufficiently extended (?) (Def. 484). Then there would be two planes from the same point _L to AB^JL by hyp.). / / But this is impossible (?) (504). ~~]~~ P .-. the planes never meet and therefore they are parallel (Def. 484). Q.E.D. 506. THEOREM. At a given point in a plane one line can be drawn perpendicular to the given plane, and only one. Given: Plane MN and point P within it. To Prove : One line can be drawn _L to MN at P, and only one. Proof : I. In MN draw any line AB, through P. Suppose plane CD be passed J_ to AB at P, meeting the plane MN in CE. In plane CD draw PR _L to CE, from P. Now, AB is _L to plane CD (Const.). .-. AB is -L to PR (?) (489). PR is _L to CE (Const.). .-. PR is _L to plane MN (?) (501). II. Suppose another line PX to be _L to plane MN at P. Then PX and PR would determine a plane CD, intersecting plane MN in CE (?) (495 and 497). PR and PX would then both be J_ to CE at P (?) (489). But this is impossible (?) (43). That is, PR and PX coincide, and PR is the only line _L to plane MN at P. Q.E.D. BOOK vi 259 .1 507. THEOREM. Through a given external point one line can be drawn perpendicular to a given plane, and only one. Given: Plane MN and point P outside of it. To Prove : One line can be drawn through P _L to MN, and only one. Proof : I. In plane MN draw any line AB. Suppose a plane GH be passed through P J_ to AB, meet- ing plane 'MN in EC, and AB at C. In plane GH draw PR _L to KG and prolong PR to X, making RX=PR. Draw RD to any point in AB, except C. Draw PC, PD, CX, DX. Now RC is -L to PX at its midpoint (Const.). Also AB is _L to plane GH (Const.). Hence, A DCP and DCX are rt. A (?) (489). In rt. A DCP and DCX, DC = DC (?), PC= CX (?) (67). .-. A DCP = A DCX (?) (53). Hence, PD = XD (?). .-. RD is _L to PX (?) (70). That is, PR is J_ to RC and RD, in plane MN. Consequently, PR is _L to plane MN from P (?) (501). II. Suppose there is another line PL _L to plane MN from p. Then PR and PL determine a plane GH (?). This plane intersects plane MN in EC (?). PR and PL would then both be J_ to EC (?) (489). But this is impossible (?) (71). That is, PR and PL coincide and therefore PR is the only line J. to plane MN from P. Q.E.D. Ex. In the figure of 507 name the six right angles at C. 260 SOLID GEOMETRY 508. THEOREM. If a plane is perpendicular to a line in another plane, any line in the first plane perpendicular to the intersection of the planes is perpendicular to the second plane. Proof: Identical with the proof of 507, I. 509. THEOREM. If a plane is perpendicular to one of two parallel lines, it is perpendicular to the other also. Given: Plane MN JL to line AB, and AB II to CP. To Prove : CP J_ to plane MN. Proof : Lines AB and CP determine a plane (?) (494). Pass this plane BC, intersecting plane MN in line BP. Draw BX _L to BP, in plane MN. AB is _L to BX (?) (489). .-. BX is J_ to plane BC (?) (501). BP is _L to AB (?) (489). .-. BP is _L to CP (?) (95). That is, plane BC is J_ to BX and CP is J_ to the intersec- tion BP. .'. CP is _L to plane MN (?) (508). Q.E.D. 510. THEOREM. Two lines perpendicular to the same plane are parallel. Given : Lines AB and CD J_ to plane MN. To Prove : AB II to CD. Proof: Through D, the foot of CD, draw DX II to AB. Then DX is _L to plane MN (?) (509). ~* N But CD is J_ to plane MN at D ( Hyp. ) . /. DX and DC coincide (?) (506). That is, AB is II to CD. Q.E.D. M :X BOOK VI 261 511. THEOREM. Two straight lines that are parallel to a third straight line are parallel to each other. Given : Lines CD and EF each II to AB. To Prove : CD II to EF. Proof : Suppose plane MN be passed to AB. .-. MNis _LtoCZ>andto^F(?) (509). .-. CD is II to EF (?) (510). Q.E.D. M. B M :E N 512. THEOREM. A line perpendicular to one of two parallel planes is perpendicular to the other also. Given : Plane MN II to plane E8\ AP _L to plane ES. To Prove : AP _L to MN. Proof : Through AP pass any two planes, AB and AC, intersecting MN in AD and AE, and intersecting US in PB and PC. AD is II to PB, and AE is II to PC(?) (500). AP is _L to PB and PC (?) (489). /. ^4P is _L to AD and AE (?) (95). .-. ^iP is _L to plane JOT (?) (501). Q.E.D. 513. THEOREM. If two planes are each parallel to a third plane, they are parallel to each other. Proof : Draw a line _L to the third plane. This line is _L to each of the other planes (?). .-. the planes are II (?) (505). a Ex. Why are not AB, CD, EFiu figure of 511 all parallel? 262 SOLID GEOMETRY 514. THEOREM. If two intersecting lines are each parallel to a plane, the plane of these lines is parallel to the given plane. Given : Intersecting lines AS and AC in plane JOT; each line II to plane PQ. To Prove: Plane MN II to plane PQ. Proof : Draw AE JL to MN at A, meeting PQ at R. Through AE and AB pass plane AS, and through AE and AC pass plane AT, intersecting plane PQ in E8 and ET, respectively. AB is II to ES, and AC is II to ET (?) (499). But, AE is _L to AB and AC (?) (489). /. AE is -L to ES and ET (?) (95). Hence, AE is _L to plane PQ (?) (501). .-. plane MN is II to plane PQ (?) (505). Q.E.D. 515. THEOREM. If two angles, not in the same plane, have theii sides parallel each to each, and ex- tending in the same directions from their vertices, the angles are equal and the planes are parallel. Given : Z BAG in plane MN and Z EDF in plane PQ ; AB II to DE; AC II to DF, and extend- ing in the same directions. To Prove: I. Z BAC= Z EDF. II. Plane MN II to plane PQ. Proof : I. Take DE and AB equal, and DF and AC equal, Draw AD, BE, OF, BC, EF. The figure ABED is a O (?) (135). .'. AD = BE (?), BOOK VI 263 Also ACFD is a O (?). .'. AD = CF (?). .'. BE= CF (?). Also AD is II to BE and AD is II to CF (?). .-. BE is II to CF (?) (511). /. BCFE is a O (?) (135). Now, in A ABC and DEF, AB = DE (?) ; AC = DF (?), and BC = ^F (?) (130). /. A ABC = A DEF (?) (58). II. AB is II to plane PQ and AC is II to plane PQ (?) (498). .-. plane MN is II to plane PQ (?) (514). Q.E.D. 516. THEOREM. If three parallel planes intersect two straight lines, the corresponding intercepts are proportional. Given : Parallel planes, LM, NP, QR, intersecting line AB at A, E, B, and CD at C, F, D, respectively. To Prove : AE : EB = CF : FD. | f_ Proof: Draw .BC, meeting plane / U-"""G ' ^P at G. Through AB and BC pass a plane cutting LM in AC and NP in EG. Through BC and CD pass a plane cutting NP in GF and QR in BD. Now, EG is II to AC and GF is II to BD (?) (500), Consequently, AE : EB= CF : FD (Ax. 1). Q.E.D Also, AB and CD are intercepts between planes LM and QR And ^ = (\ = (?) (305). AB \CBj CD Ex. Prove, in figure of 516, AB: CD = AE : CF = EB: FD. 264 SOLID GEOMETRY R J 517. THEOREM. The projection of a straight line upon a plane is a straight line.* Given: Line AB and plane air. To Prove : The projection of AB on MN is a straight line. Proof : Draw PJ JL to plane MN from any point P, in AB. AB and PJ determine a plane (?) (495). This plane AD intersects plane MN in a straight line CD (?). Now in plane AD draw XR I! to PJ from JT, any other point in AB. XR is -L to MN (?) (509). Now R is the projection of X (?) (485). .-. CD is the projection of AB (?) (485). That is, the projection of AB upon the plane MN is a straight line. Q.E.D. 518. COR. A line and its projection upon a plane are in the same plane. 519. THEOREM. A line not parallel to a plane is longer than its projection upon the plane. Given: A plane and line LN not II to the plane, and DE the projection of LN upon the plane. To Prove: LN>DE. Proof: Drawxz) and NE. Draw LX _L to NE from i, in the plane LE. LD and NE are -L to the plane (Def. of projection, 485). LXED is a rectangle (?) (166). Now LN > LX (?) (77). But LX = DE (?) (130). Hence, LN > DE (Ax. 6). Q.E.D. * Except only if the given straight line is a normal to the given plane. BOOK VI 265 520. THEOREM. Of all lines that can be drawn to a plane from a point : I. The perpendicular is the shortest. II. Oblique lines having equal projections are equal. Ill , Equal oblique lines have equal projections. IV. Oblique lines having unequal projections are unequal, and the line having the greater projection is the longer. V. Unequal oblique lines have unequal projections, and the longer line has the greater projection. I. Given: Plane MN; point P ; PR _L to MN ; any other line from P to plane MN, as PA. To Prove : PR < PA. Proof: Draw AR. Now PR (?) ; PA is not A. to AR N II. Given: Oblique lines PA and PB, whose projections A R and BR are equal. To Prove : PA = PB. Proof : The rt. A PEA and PRB are =. (Explain.) III. Given : Equal oblique lines PA and PB. To Prove : Their projections, AR and BR, are equal. Proof: The rt. A PRA and PRB are = . (Explain.) IV. Given : Oblique lines PC and PA\ proj. EC > proj. RA To Prove : PC > PA. Proof: In A PCR, take on RC, RX = RA, and drawPJT. Now PC > PX (?) (76, III). But PA = PX (?) (520, II) .-. PC> PA (Ax. 6). Y. Given : Unequal oblique lines, PC > PA. To Prove : Projection EC > projection RA. Proof : By reductio ad absurdum. (See 88, 89). 266 SOLID GEOMETRY 521. THEOREM. The acute angle that a line makes with its own projection upon a plane is the least angle that the line makes with any line of the plane. Given : AB, any line meeting plane MN at B ; BP, its projection upon MN ; BD, any other line in MN, through B. To Prove : Z ABP < Z ABD. Proof: On BD take BX=BP and draw AX. In A ^4PJ5 and ABX, AB = AB (?) ; BP = BX (?) (Const.). But AP < AX (?) (520, I). .'. Z ^IBP < Z ^BD (?) (87). Q.E.D. 522. THEOREM. Through a given point one plane can be passed parallel to any two given lines in space, and only one. Given : Point P; two lines, AB and CD. To Prove : Through P one plane can be passed II to AB and CD, and only one. Proof : I. Through P draw a line II to AB and another II to CD. Pass a plane , containing these lines. MN is II to both AB and CD (?) (498). II. Only one line can be drawn through P II to AB, and only one, II to CD (?) (92). .-. there is only one plane (?) (495). Q.E.D. Ex. 1. If a line meets a plane, with what line in the plane does this line make the greatest angle ? Ex. 2. In the figure of 515, prove A D parallel to the plane CE. Also BE parallel to the plane A F. Ex. 3. In the figure of 516, if AE = Q,EB = 5, CD = 16J, find DF. BOOK VI 267 523. THEOREM. Through a given point one plane can be passed parallel to a given plane, and only one. x Given: (?). To Prove: (?). l~ ~ V Proof : I. Suppose PR be drawn J_ to / i \y AB-, and XY be passed _L to PR at P. Then XY is II to AB (?) (505). II. Only one line _L to AB can be ' drawn from P (?) (507). Only one plane _L to PR can be passed at P (503). /. only one plane can contain P and be II to AB. Q.E.D. 524. THEOREM. Parallel lines included between parallel planes are equal. Given : (?). To Prove: (?). Proof : AB and CD determine a plane (?). This plane intersects RS and PQ in lines AC and BD, which are II (?) (500). .'.ABDC is a O (?). Hence, AB = CD (?). Q.E.D. v 525. THEOREM. The plane perpendicular to a line at its midpoint is the locus of points in space, equally distant from the extremities of the line. Given : RS _L to AB at its midpoint, M. To Prove: (?). Proof : (1) Take P, any point in RS. Draw PM, PA, PB. PM is J_ to AB (?) (489). .-. PA = PB (?) (67). That is, any point in RS is equally distant from A and B. (2) Take P ; , any point outside of RS. Draw P f M. P ! M is not _L to AB (?) (502). .'. P', any point outside of plane RS, is not equally distant from A and B (?) (68). Hence, plane RS is the locus of points in space equally dis- tant from A and B (?) (179). Q.E.D. M /P- 268 SOLID GEOMETRY 526. THEOREM. The locus of points in space equally distant from all the points in the circumference of a circle is the line perpen- dicular to the plane of the circle at its center. Given : (?). To Prove : (?). Proof : I. Any point in AC is equally distant from all the points in the cir- cumference of the circle (?) (520, II). II. Any point equally distant from all points of the circumference of the circle is in AC (?) (520, III). M /. AC is the required locus (179). Q.E.D. Ex. 1. What is the locus of points equally distant from two given points ? Ex. 2. What is the locus of points equally distant from three given points ? 527. The distance from a point to a plane is the length of the perpendicular from the point to the plane. Thus, the word "distance," referring to the shortest line from a point to a plane, implies the perpendicular. The inclination of a line to a plane is the angle between the line and its projection upon the plane. ORIGINAL EXERCISES 1. Through one straight line a plane can be passed parallel to any other straight line in space, and only one. Through a point of the first line draw a line || to the second. 2. Two parallel planes are everywhere equally distant. 3. If a line and a plane are parallel, another line parallel to the given line and through any point in the given plane lies wholly in the given plane. Through the given line and the point P pass a plane cutting the given plane in PX. Use 499. BOOK VI 269 4. A straight line parallel to the intersection of two planes, but in neither, is parallel to both planes. 5. If two straight lines are parallel and two intersecting planes are passed, each containing one of the lines, the intersection of these planes is parallel to each of the given lines. 6. If three straight lines through a point meet the same straight line, these four lines all lie in the same plane. y 7. If a straight line meets two parallel planes, its inclinations to the planes are equal. ^8. Two parallel planes can be passed, each containing one of two given lines in space. Is this ever impossible? 9. If each of three straight lines intersects the other two, the three lines all lie in a plane. 10. The projections of two parallel lines on a ^rx^fv plane are parallel. | J D Proof: AB is || to CD (?). AE is || to CG (?). / E Ljp H\ .-. planes AF and CH are || (?) ; etc. 11. If two lines in space are equal and parallel, their projections on a plane are equal and parallel. 12. If a plane is parallel to one of two parallel lines, it is parallel to the other. 13. If a straight line and a plane are perpendicular to the same straight line, they are parallel. 14. Equal oblique lines drawn to a plane from one point have equal inclinations with the plane. 15. If a line and a plane are both parallel to the same line, they are parallel to each other. 16. Four points in space, A, B, C, A are joined, and these four lines are bisected. Prove that the four lines joining (in order) the four midpoints of the first lines form a parallelogram. Proof: Pass plane DP through points A, D, B, and plane DX through points B, C, D, these planes intersecting in BD. ST is || to BD and = I BD (?) ; etc. 270 SOLID GEOMETRY 17. If a plane is passed containing a diagonal of a parallelogram and perpendiculars be drawn to the plane from the other vertices of the parallelogram, they are equal. To Prove : A E = CF. Proof : Draw diagonal AC. Draw EO and OF in plane MN. EO, OF, and EOF are projections; etc. 18. If from the foot of a perpendicular to a plane, a line be drawn at right angles to any line in the plane, the line connecting this point of inter- section with any point in the perpendicular is per- pendicular to the line in the plane. Given: AB J_ to plane RS; EC JL to DE in the plane; PC drawn from C to P, in AB. To Prove: PC is JL to DE. Proof : Take CD = CE, draw PD, PE, BD, BE. EC is JL to DE at its midpoint (?). .-. BD = BE (?). PD = PE (?) (520, II). .-. PC is J_ to DE (?) (70). 19. A line PB is perpendicular to a plane at B, and a line is drawn from B meeting any line DE, of the plane, at C. If PC is perpen- dicular to DE, BC is perpendicular to DE. 20. Are two planes that are parallel to the same straight line necessarily parallel ? 21. If each of two parallel lines is parallel to a plane, is the plane of these lines also parallel to the given plane? 22. Is a three-legged chair always stable on the floor? Why? Is a four-legged chair always stable ? Why ? 23. What is the locus in space of points equally distant from two parallel planes ? From two parallel lines ? 24. What is the locus of points in space at a given distance from a given plane ? 25. What is the locus of points in a plane at a given distance from an external point? 26. What is the locus of points in space equally distant from two points and equally distant from two parallel planes? BOOK VI 271 27. What is the locus of points ir. space, equally distant from the vertices of a given triangle ? 28. What is the locus of all straight lines perpendicular to a given straight line at a given point? 29. What is the locus of all lines parallel to a given plane and drawn through a given point? 30. If the points in a line satisfy one condition and the points in a plane satisfy another condition, what will be true of their intersection? What will be true if they do not intersect? 31. If the points in one plane satisfy one condition and the points in another plane satisfy another condition, what is true of their inter- section ? What is true if the planes are parallel ? 32. To construct a plane perpendicular to a given line at a given point in the line. 33. To construct a plane perpendicular to a given line through a given external point. 34. To construct a line perpendicular to a given plane, through a given point in the plane. 35. To construct a line perpendicular to a given plane, through a given external point. 36. To construct a plane parallel to a given plane, through a given point. 37. To construct a number of equal oblique lines to a plane from a given external point. 38. To construct a line through a given point parallel to a given plane. 39. To construct a line through a given point and parallel to each of two given intersecting planes. 40. To construct a plane containing one given line and parallel to another. 41. To construct a plane through a given point parallel to any two given lines in space. 272 SOLID GEOMETRY 42. To construct a line through a given point in space which will intersect two given lines not in the same plane. When is there no such line? Is there ever more than one? 43. To find a point in a plane such that the sum of the two lines joining it to two fixed points on one side of the plane shall be the least possible. ,g : rw o ne - long it to X, making CX = AC. Draw BX, meet- r j\ ing plane at P. Draw A P. Take any other point / v, R in plane MN. J Statement: AP + PB PC. Then prove AD>AC, etc. 19. PA is perpendicular to plane 72S; AB and PC are perpendicular to plane MR. Prove BC perpendicular to RT. 20. If two parallel planes are cut by a third plane, the alternate-interior dihedral angles are equal; the corresponding dihedral angles are equal ; the alternate- exterior dihedral angles are equal ; the adjoining in- terior dihedral angles are supplementary. 21. State and prove the converse theorems of those in No. 20. 22. To construct a plane perpendicular to a given plane and contain- ing a given line in that plane. 23. To construct a plane perpendicular to a given plane and contain- ing a given line without that plane. 24. To construct through a given point a line which will intersect any two given lines in space. Construction : Pass a plane through the point and one of the lines. This plane intersects the other line at a point. This point and the given point determine the required line. Explain. Discuss. 25. To bisect a given dihedral angle. Construction : Pass a plane JL to the edge. ' Bisect the plane Z formed, pass the plane determined by this bisector and the given edge. 284 SOLID GEOMETRY 26. To construct a line whose points shall be equally distant from the ends of a given line and also equally distant from the faces of a dihedral angle. 27. To find the locus of points equally distant from two points and also equally distant from two intersecting planes. Discuss. 28. To find a point equally distant from three given points and equally distant from two intersecting planes. Is this problem ever impossible? Will there ever be two points? When will there be only one ? 29. To find a point equally distant from three given points and equally distant from two intersecting lines. Discuss fully. POLYHEDRAL ANGLES 554. If three or more planes meet at a point, they form a polyhedral angle. The opening partially surrounded by the planes is the polyhedral angle. The point common to all the planes is the vertex. The planes are the faces. The intersections of adjacent faces are the edges. The angles formed at the vertex, by adjacent edges, are the face angles. Thus, V ABODE is a polyhedral angle ; V is the vertex ; A V, BV, etc., are edges; planes AVB, BVC, etc., are faces; A A VB, BVC, etc., are face angles. 555. A plane section of a polyhedral angle is the plane A figure bounded by the intersec- tions of all the faces by a plane. Polygon LMNOP is a plane section of polyhedral angle V ABODE. A convex polyhedral angle is one whose plane sections are all convex. EQUAL POLYHEDRAL ANGLES VERTICAL POLYHEDRAL ANGLES VERTICAL DIHEDRAL ANGLES SYMMETRICAL POLYHEDRAL ANGLES 556. Two polyhedral angles are equal if they can be made to coincide in all particulars. That is, if two polyhe- dral angles are equal, their homologous dihedral angles are equal; their homologous face angles are equal, and they are arranged in the same order. The length of the edges or the extent of the faces does not affect the size of the poly- hedral angle. Two polyhedral angles are vertical if the edges of one are the prolongations of the edges of the other. Two polyhedral angles are symmetrical if all the parts of one are equal to the corresponding parts of the other, but arranged in opposite order. NOTE. It is apparent from the definitions that equal polyhedral angles are mutually equiangular, as to the face angles and as to the dihe- dral angles. Vertical polyhedral angles are mutually equiangular, as to their face angles and as to their dihedral angles, but the order is reversed. Symmetrical polyhedral angles are also mutually equiangular as to their face angles and as to their dihedral angles, but the order is reversed. Thus, if one follows around the polygon A'D' in alphabetical order, he is moving as the hands of a clock if the eye is at the vertex 0'; but if he follows around AD alphabetically, he is moving in a direction opposite to the motion of the hands of a clock if the eye is at the Yertex 0. Hence, it is apparent that, in general, symmetrical polyhedral angles cannot be made to coincide. 286 SOLID GEOMETRY 557. A trihedral angle is a polyhedral angle having three and only three faces. A trihedral angle is rectangular if it contains a right dihe- dral angle; birectangular, if it contains two right dihedral angles; trirectangular, if it contains three right dihedral angles. A trihedral angle is isosceles if two of its face angles are equal. 558. THEOREM. Two vertical polyhedral angles are symmetrical. Proof : Their homologous face angles are equal and arranged in reverse order, and their homologous dihedral angles are equal and arranged in reverse order. /.they are symmetrical (Def. 556). 559. THEOREM. If two polyhedral angles are symmetrical, the vertical polyhedral angle of the one is equal to the other. Because the corresponding parts are equal and they are arranged in the same order. 560. THEOREM. Provided two trihedral angles have their parts arranged in the same order, they are equal : I. If two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other. II. If a face angle and the two dihedral angles adjoining it of the one are equal respectively to a face angle and the two dihedral angles adjoining it, of the other. Proof : By method of superposition, as in plane A. BOOK VI 287 561. THEOREM. Provided two trihedral angles have their parts arranged in the same order, they are equal, if the three face angles of one are equal respectively to the three face angles of the other. Given: Trih. A O and o'; . / Atn'R' O O' C'O'A'. To Prove: Trih. Z O = trih. Z o', that is, dih. Z OA = dih. Z O'A', etc. Proof : Take OA = OB = oc = O'A' = O'B' = o'c'. Draw AB, BC, AC, A'B', B'C', A'c'. Take, on edges AO and A'O', AP = A'P' and in face AOB draw PD _L to AO. Z OAB is acute (A A OB is isosceles). .-. PD will meet AB. In face AOC draw PE _L to AO, meeting .4 Cat E. Draw DE. Similarly draw P'D', P' E', D'E' . . Now A EPD and E'P'D' are the plane A of the dihedral A AO and. A 1 o' (?) (530). To prove that these A are equal requires the proof that eight pairs of A are equal. AB=A'B', BC=B'C', (1) A OAB = A O'A'B'. (Explain.) (2) A OBC=A O'B'C'. (Explain.) (3) A OAC = A o'A'c'. (Explain.) AC = A'C', Z OAB = Z O'A'B', Z OAC = Z O'A'C' (?) (27). . AD = A'D', AE = !'#'; PD = P'l/,etc.(?). /r' A' R' fv\ Z.C/ A K {. J. = E'D' (?). = Z^'P'D'(?). Hence, dih. Z ^4O = dih. Z A'O' (?) (535). Similarly one may prove the other pairs of homologous dihedral angles equal. .'. trihedral Z O = trihedral Z o' (?) (556). Q.E.D. (4) A^PD = A^L'P'D'. (Explain.) (5) A!P^=A^ / P'^ / . (Explain.) (6) A ABC = A A' B'C'. (Explain.) (7) A AED = A A'E'D'. (Explain.) (8) A PED= A P'E'D'. (Explain.) 288 SOLID GEOMETRY 562. THEOREM. Provided two trihedral angles have their parts arranged in reverse order, they are symmetrical : I. If two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other. II . If a face angle and the two dihedral angles adjoining it of the one are equal respectively to a face angle and the two dihedral angles adjoining it of the other. III. If the three face angles of one are equal respectively to the three face angles of the other. Proof : In each case construct a third trihedral Z symmet- rical to the first. This third figure will have its parts = to the parts of the second, and arranged in the same order (Def . 556). .-.the third = the second (?) (560 and 561). /. the first is symmetrical to the second (Ax. 6). Q.E.D. 563. THEOREM. The sum of any two face angles of a trihedral angle is greater than the third face angle. Given: Trih. Z.O-RST\\\ which face angle ROT is the greatest. To Prove: Z EO-S + Z SOT > Z ROT. Proof : Construct, in face KOT, Z ROD = Z ROS. Take OD=OB\ draw ADC, meet- ing OT at C. Draw AB and BC. A AOD = A AOB. (Explain.) .:AB = AD(?). Now AB + BC > AD + DC (?) (Ax. 12). But, AB = AD (?). Subtracting, BC > DC (Ax. 7). Now, OB = OD (?), OC = OC (?) and BC > DC (Just proved). .-. Z BOG > Z DOC (?) (87). But Z AOB = Z AOD (?). Adding, Z AOB + BOC > Z AOC (Ax. 7). That is, Z ROS +Z SOT > Z ROT (?) (Ax. 6). Q.E.D. BOOK VI 289 564. THEOREM. The sum of the face angles of any polyhedral angle is less than four right angles, or 360. Given: Polyhedral Z O, having n faces. To Prove : The sum of the face A at O < 4rt. 4 or 360. Proof : Pass a plane AD, intersecting all the faces, and the edges at A, B, C, etc. In this section take any point X and join X to all the vertices of the polygon. |B (1) There are n face A having their vertices at O (Hyp.). (2) There are n base A having their vertices at X (Const.). (3) The sum of the A of the face A = 2 n rt. A (110). (4) The sum of the A of the base A = 2 n rt. A .(110). (5) .-. the sum of the A of the face A = the sum of the A of the base A (Ax. 1.). Now, Z OAE + Z OAB > Z.EAB (?) (563). And Z OBA + Z OBC > /-ABC (?), etc., etc. Adding, the sum of the base A of the face A > the sum of the base A of the base A (Ax. 8). Subtracting this inequality from equation (5) above, the sum of the face A at O < the sum of the A at X (Ax. 9). But the sum of all the A at X= 4 rt. A (?) (47). .-.the sum of the face A at O < 4 rt. A, or 360 (Ax. 6). Q.E.D. Ex. 1. Prove theorem of 563 for the case of an isosceles trihedral angle. Ex. 2. In the figure of 564, as the vertex O approaches the base, does the sum of the face angles at O increase or decrease? What limit does this sum apDroach? Does the sum ever become equal to this limit? \ \ 290 SOLID GEOMETRY ORIGINAL EXERCISES 1. The three planes bisecting the three dihedral angles of a trihedral angle intersect in a straight line. 2. The three planes containing the three bisectors of the three face angles of a trihedral angle and perpendicular to those faces intersect in a straight line. 3. If two face angles of a trihedral angle are equal, the dihedral angles opposite them are equal. Given: /.RVS=SVT. To Prove: Dih. Z R V= dih. Z TV. Proof: Pass plane SVX bisecting dih. Z SV. Prove trih. A V-RSX and V-TSX are sym. by 562, 1. 4. An isosceles trihedral angle and its symmet- rical trihedral angle are equal. 5. Find the locus of points equally distant from the three faces of a trihedral angle. 6. Find the locus of points equally distant from the three edges of a trihedral angle. 7. If the three face angles of a trihedral angle are equal, the three dihedral angles also are equal. 8. If the three face angles of a trihedral angle are right angles, the three dihedral angles also are right angles. 9. In any trihedral angle the greatest dihedral angle has the greatest face angle opposite it. 10. If the edges of one trihedral angle are perpendicular to the faces of a second trihedral angle, then the edges of the second are perpen- dicular to the faces of the first. 11. To construct the plane angles of the three dihedral angles of a trihedral angle, if the three face angles are known. [To be accomplished by constructions in a plane.] 12. To construct, through a given point, a plane which shall make, with the faces of a polyhedral angle having four faces, a section that is a parallelogram. Construction: Extend one pair of opp. faces to obtain their line of intersection. Similarly extend the other pair. Any plane section || to these lines will be a 7. (Explain.) BOOK VII POLYHEDRONS 565. A polyhedron is a solid bounded by planes. The edges of a polyhedron are the intersections of the bounding planes. The faces are the portions of the bounding planes included by the edges. The vertices are the intersections of the edges. The diagonal of a polyhedron is a straight line joining two vertices not in the same face. POLY- TETKA- HEXAHEDRON OCTA- DODECA- ICOSA- HEDRON HEBRON CUBE HEDRON HEDRON HEBRON 566. A tetrahedron is a polyhedron having four faces. A hexahedron is a polyhedron having six faces. An octahedron is a polyhedron having eight faces. A dodecahedron is a polyhedron having twelve- faces. An icosahedron is a polyhedron having twenty faces. 567. A polyhedron is convex if the section made by every plane is a convex polygon. Only convex polyhedrons are considered in this book. 291 292 SOLID GEOMETRY PRISMS 568. A prism is a polyhedron two of whose opposite faces are equal polygons in parallel planes, and whose other faces are all parallelograms. The bases of a prism are the equal, parallel polygons. The lateral faces of a prism are the parallelograms. The lateral edges of a prism are the intersections of the lateral faces. The lateral area of a prism is the sum of the areas of the lateral faces. The total area of a prism is the sum of the lateral area and the areas of the bases. The altitude of a prism is the perpendicular distance be- tween the planes of the bases. A triangular prism is a prism whose bases are triangles. PRISM TRIANGULAR REGULAR OBLIQUE PRISMS RIGHT SECTION PRISM PRISM TRUNCATED PRISMS 569. A right prism is a prism whose lateral edges are per- pendicular to the planes of the bases. A regular prism is a right prism whose bases are regular polygons. An oblique prism is a prism whose lateral edges are not per- pendicular to the planes of the bases. A truncated prism is the portion of a prism included between the base and a plane not parallel to the base. BOOK VII 293 A right section of a prism is the section made by a plane perpendicular to the lateral edges of the prism. 570. A parallelepiped is a prism whose bases are parallelo- grams. A right parallelepiped is a parallelepiped whose lateral edges are perpendicular to the planes of the bases. A rectangular parallelepiped is a right parallelepiped whose bases are rectangles. PARALLELEPIPED RIGHT RECTANGULAR CUBE PARALLELEPIPED PARALLELEPIPED An oblique parallelepiped is a parallelepiped whose lateral edges are not perpendicular to the planes of the bases. A cube is a rectangular parallelepiped whose six faces are squares. 571. The unit of volume is a cube whose edges are each a unit of length. The volume of a solid is the number of units of volume it contains. The volume of a solid is the ratio of that solid to the unit of volume. The three edges of a rectangular parallelepiped meeting at any vertex are the dimensions of the parallelepiped. Equivalent solids are solids that have equal volumes. Equal solids are solids that can be made to coincide. Ex. What is the base of a rectangular parallelepiped? Of a right parallelepiped ? Of an oblique parallelepiped ? 294 SOLID GEOMETRY NOTE. The space that is bounded by the surfaces of a solid, independ- ent of the solid, is called a geometrical solid. That is, if a material or physical body occupy a certain position and then be removed elsewhere, there is a definite portion of space that is the exact shape and size as the solid, and can be conceived as bounded by exactly the same surfaces as bounded the solid when in that original position. In order that we may pass planes and draw lines through solids, and superpose one solid upon another, it is convenient in studying the properties of solids to consider them usually as geometric solids, the material body being removed for the time. The three kinds of parallelepipeds can be illustrated by removing the cover and bottom of an ordinary cardboard box, and distorting the shape of the frame that remains. PRELIMINARY THEOREMS 572. THEOREM. The lateral edges of a prism are all equal (?). 573. THEOREM. Any two lateral edges of a prism are parallel. (See 511.) 574. THEOREM. Any lateral edge of a right prism equals the altitude. (See 524.) 575. THEOREM. The lateral faces of a right prism are perpen- dicular to the bases. (See 540.) 576. THEOREM. The lateral faces of a right prism are rectangles. (Def. 569.) 577. THEOREM. The faces and bases of a rectangular parallele- piped are rectangles. (Def. 569.) 578. THEOREM. All the faces of any parallelepiped are paral- lelograms (?). 579. AXIOM. A polyhedron cannot have fewer than four faces. 580. AXIOM. A polyhedron cannot have fewer than three faces at each vertex. BOOK VII 295 THEOREMS AND DEMONSTRATIONS 581. THEOREM. The sections of a prism made by parallel planes cutting all the lateral edges are equal polygons. Given : Prism AB\ II sections CF and C'F'. To Prove: Polygon (?>'= polygon Proof : CD is II to C'D', DE is II to D'E', etc. (?) (500). CD r , DE r , Erf, etc., are HJ (?). .-. CD = C'D', DE=D'E', EF = E'F', etc. (?) (130). Z GCD = Z. G'C'D', Z. CDE =^ C'D'E', etc. (?) (515). .'. polygon OT=polygon C f F f (?) (159). J.E.D. 582. THEOREM. The opposite faces of a parallelepiped are equal and parallel. Given: (?). To Prove: H G Face AF = and II to face DG. Proof: Faces AF and DG are 17 (?). AB= DC, AE = D#(130). AB is II to DC and AE is II to DH (?). .'. /- EAB = Z. HDC (?) (515). .'. face AF = face DG (?) (139). Also face AF is II to face DG (?) (515). Q.E.D. Ex. 1. All right sections of a prism are equal. Ex. 2. Any section of a parallelepiped made by a plane cutting two pairs of opposite faces is a parallelogram. 296 SOLID GEOMETRY 583. THEOREM. The lateral area of a prism is equal to the prod- uct of a lateral edge by the perimeter of a right section. Given: Prism EU 1 ; edge = E) right section AD. To Prove: Lateral area of EU 1 = E x perimeter of AD. Proof : AB is _L to EE 1 , BG is i to SS 1 , etc. (?) (489). Area O RS f = E- AB (373). Area O ST r = E BC (?). Area O TU' = ^ CD (?). etc. etc. Adding, The lateral area = E- (AB +BC + CD + etc.) (Ax. 2). = jE?' perimeter of rt. sect. (Ax. 6). Q.E.D. Ex. 1. Any section of a parallelepiped made by a plane parallel to any edge is a parallelogram. Ex. 2. The sum of the face angles at all the vertices of any parallele- piped is equal to 24 right angles. Ex. 3. The sum of the plane angles of all the dihedral angles of any parallelepiped is equal to 12 right angles. Proof : Pass three planes _L to three intersecting edges. Prove these sections [U whose A are the plane angles of the dihedral angles, etc. Ex. 4. Enunciate a theorem for the lateral area of a right prism. Ex. 5. Find the lateral area of a right prism whose altitude is 8 feet and each side of whose triangular base is 5 feet. I V fc *- Ex. 6. Find the total area of a regular prism whose base is a regular hexagon, 10 inches on a side, if the altitude of the prism is 15 inches. l > Ex. 7. Find the lateral area of a prism whose edge is 12 inches and whose right section is a pentagon, the sides of which are 3, 5, 6, 9, and L\ < 11 inches. BOOK VII 297 584. THEOREM. Two prisms are equal if three faces including a trihedral angle of one are equal, respectively, to three faces including a trihedral angle of the other, and similarly placed. Given : Prisms AO and A r O f ; face AM face A'M* '; face AP = face A'P'-, face AD = face A'D'. To Prove : Prism AO = prism Proof : The three face A at A are respectively = to the three face A at A 1 (?) (27). .-. trih. Z A = trih. Z A 1 (?) (561). Superpose prism AO upon prism A?o', making the equal trihedral A A and A' coincide. Face AD will coincide with face A'D', face A M with A f M f , face ^1P with A'P'. (They are = by hyp.) That is, point L will fall on I/; Jf on M'I and P on P f . .-. the plane iO will fall upon the plane L'O' (?) (493). Polygon LO polygon L 1 o' (Ax. 1). /. these bases coin- cide (?). Similarly face BN coincides with B'N'., CO with c'o'. Etc. Consequently, the prisms are equal (?) (571). Q.E.D. 585. THEOREM. Two right prisms are equal if they have equal bases and equal altitudes. (Explain.) 586. THEOREM. Two truncated prisms are equal if three faces in- cluding a trihedral angle of one are equal, respectively, to three faces including a trihedral angle of the other, and are similarly placed. (Explain.) 298 SOLID GEOMETRY 587. THEOREM. An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to the lateral edge of the oblique prism. p 1 O r Given : Oblique prism right prism PN 1 whose base is PN, a right section of AC 1 , and whose altitude PP f = edge EE 1 . To Prove : Oblique prism AC 1 =0= right prism PN r . Proof: Edge EE' = PP'( Hyp.). Subtract PE' from each, and Likewise AL = A'L',BM=B'M', CN=C r N r , etc. (1) Face AC = face A'C f (?) (581). (2) In faces AP and A'P', EP = EP', AL = A'L' (Ax. 2), AE=A'E', PL = P'L 1 (?)(130). That is, face AP and face A'P' are mutually equilateral. Also Z EAL = Z E'A'L', Z PEA = Z p'E*A\ ZA T "D - / A T ~f^ s ~IPT>T - / Jv 1 ' T^ T ^ _/\_ /y r ^\_ JLt Jr *- JL-L-Lj -^ Hf -L *J That is, face AP and A'P' are mutually equiangular. /. face AP = face A r P f (?) (159). (3) Similarly, face AM = face A'M'. .'. truncated prism A N truncated prism A'N' (?) (586). Now, add, solid PC 1 = solid PC 1 (Iden.). Oblique prism AC'o= right prism PN' (Ax. 2). Q.E.D. (?) (98). Ex. Prove, in the figure of 587, that truncated prism A N is equal to truncated prism A'N' by the method of superposition. Proof: Polygon PN = polygon P'N 1 (V). Superpose solid AN upon solid A'N' so that base PN will coincide with its equal P'N'. Etc. BOOK VII 299 588. THEOREM. The plane containing two diagonally opposite edges of a parallelepiped divides the parallelepiped into two equiva- lent triangular prisms. Given : Parallelepiped EH and plane AG containing the opposite edges AE and CG. To Prove: Prism ABC -F =c= prism ADC-H. Proof: Pass a right sec- tion BSTV intersecting the given plane in RT. Face AF is II to DG (?). .-. RS is II to VT (?) (500). Likewise, RVis II to ST (?). .-.RSTVis a O (?). /. A RST= A RVT (?) (132). Prism ABC-F =0= a right prism whose base is R8T and whose altitude = EA (?) (587). Prism ADC-H =o= a right prism whose base is RVT and whose altitude = EA (?). But these imaginary right prisms are equal (?) (585). .'. prism ABC-F =e= prism ADC-H (Ax. 1). Q.E.D. 589. THEOREM. Two rectangular parallelepipeds having equal bases are to each other as their altitudes. Given : Rectangular parallele- pipeds P and Q, having = bases, and their altitudes AB and CD, respectively. To Prove : P : Q = AB : CD. Proof : I . If the altitudes are commensurable. Consult 244, 302, 368, 539. B / D / / R / '/ :/ / I 300 SOLID GEOMETRY B D X C / P / R 7 \ ^ - r - " / " v II. If the altitudes are incommensurable. There does not exist a common unit (238). Suppose AB divided into equal parts. Apply one of these as a unit of measure to CD. There will be a remainder, DX (?). Pass plane XF, through X and II to base. Now, = (?). CY CX Indefinitely increase, etc., as in 244, 302, 368, 539. 590. THEOREM. Two rectangular parallelepipeds having two dimen- sions of the one equal respectively to two dimensions of the other, are to each other as their third dimension. The faces having the sides of one equal to the sides of the other, re- spectively, may be considered the bases and the third dimensions the altitudes. Thus this statement is the same as 589. 591. THEOREM Two rectangular parallelepipeds having equal altitudes are to each other as their bases. / R h / / I / Given : Rect. parallelepipeds R and S, having the same alti- tude h ; and other dimensions a, 6, and c, d, respectively. To Prove: ?L = a LL^. S c-d Proof : Construct a third rectangular parallelepiped, r, having altitude = h, another dimension = a, a third = d. H and H< 59 > f-^C-Az.'S). Q-E.D. ^ BOOK VII 301 592. COR. Two rectangular parallelepipeds having one dimension in common, are to each other as the products of the other dimensions. 593. THEOREM. Any two rectangular parallelepipeds are to each other as the products of their three dimensions. HI VL I T7 b 1 Given : Rectangular parallelepipeds L and J/, whose dimen- sions are a, 6, A, and a', >', 7i', respectively. L a-b-h To Prove : = ~T~T7~~T7* Proof: Construct 2V, whose dimensions are a, 5, h f . Then, | = p (?) (590), and J = ~ (?) (592). I/ a-b-h . A o-\ Multiplying, x = a f -b f -h'^ '' Q.E.D. 594. THEOREM. The volume of a rectangular parallelepiped is equal to the product of its three dimensions. Given: (?) To Prove: (?). Proof : Let U be a unit of vol. But_= vol. of P (?) (571). .'.vol. of P = a - b h (Ax. 1). / / f=> / -m-? w / / 302 SOLID GEOMETRY 595. THEOREM. The volume of a rectangular parallelepiped is equal to the product of its base by its altitude. (See 594.) 596. COR. The volume of a cube is equal to the cube of its edge. 597. THEOREM. The volume of any parallelepiped is equal to the product of its base by its altitude. Given: Parallelepiped #, whose base = B and alt. = h. To Prove : Volume of E = B - h. Proof : Prolong the edge AD and all edges II to AD.' On the prolongation of AD, take EF = AD. Through E and F pass planes EG and FH, JL to EF, forming the right paral- lelepiped 8. Again, prolong FI and all the edges II to FT. On the pro- longation of FI, take KL = FI. Through K and L pass planes KM and LN, _Lto KL, forming the rectangular parallelepiped T. Consider EG the base of s, and EF its altitude, then B =c= S (?) (587). Also B =0= B ' (?) (374). Consider EP the base of S, and KM the base of T, and KL its altitude, then S =0= T (?) (587). Also B'= C (?)(140). Hence, R =c= r (Ax. 1); and B =0= c (Ax. 1); and altitude of T = h (?) (524). But volume of T = C - h (?) (595). /.volume of E = B - h (Ax. 6). Q.E.D. BOOK VII 303 598. THEOREM. Two parallelepipeds having equal altitudes and equivalent bases are equivalent (Ax. i). 599. THEOREM. Two parallelepipeds having equal altitudes are to each other as their bases. Proof : Q = B . h and R = B f . h (?) (597). .. by dividing, Q = f (Ax. 3). Q.E.D. 600. THEOREM. Two parallelepipeds having equivalent bases are to each other as their altitudes (?). 601. THEOREM. Any two parallelepipeds are to each other as the products of their bases by their altitudes (?). 602. THEOREM. The volume of a triangular prism is equal to the product of its base by its altitude. Given : Triangular prism ACD-F; base = ; alt. =A. To Prove: Volume of ACD-F=B-h. Proof: Construct parallele- piped AS having as three of its lateral edges AE, CF, DG. Hence, | volume of AS ^ ACRD h (Ax. 3). But i volume of AS= volume of prism ACD-F (?) (588) andj ACRD = B (?)(132). .'. volume of ACD-F = B - h (Ax. 6). Q.E.D. Ex. 1. Which rectangular parallelepiped contains the greater volume, one whose edges are 5, 7, 9, or one whose edges are 4, 6, 13 ? Ex. 2. The base of a prism is a right triangle whose legs are 8 and 12, and the altitude of the prism is 20. Find its volume. 304 SOLID GEOMETRY 603. THEOREM. The volume of any prism is equal to the product of its base by its altitude. Given : Prism AD ; base = B ; altitude = h. To Prove: Vol. of AD=B-h. Proof : Through any lateral edge, AC, and other lateral edges not adjoining AC, pass^ planes cutting the prism into triangular prisms I, II, III, having bases E, s, T, respectively. Vol. of prism I = E - h Vol. of prism II = 8 - h f (?) (602). Vol. of prism III = T- h J Adding, Vol. of prism AD = (# -f s + T)h = B h (Ax. 2). Q.E.D. 604. THEOREM. Two prisms having equal altitudes and equiva- lent bases are equivalent. 605. THEOREM. Two prisms having equal altitudes are to each other as their bases. 606. THEOREM. Two prisms having equivalent bases are to each other as their altitudes. 607. THEOREM. Any two prisms are to each other as the prod- ucts of their bases by their altitudes. ORIGINAL EXERCISES 1. How many faces has a parallelepiped? Edges? Vertices? How many faces has a hexagonal prism? Edges? Vertices? 2. Every lateral face of a prism is parallel to the lateral edges not in that face. 3. Every lateral edge of a prism is parallel to the faces that do not contain it. 4. Every plane containing one and only one lateral edge of a prism is parallel to all the other lateral edges. BOOK VII 305 5. Any lateral face of a prism is less than the sum of the other lateral faces. [Use fig. of 583.] 6. The diagonals of a rectangular parallelepiped are equal. Proof: Pass the plane A CGE. This is a rectangle (?), etc. 7. The four diagonals of a parallelepiped bisect each other. [First prove that one pair bisect each other, thus prove that any pair bisect each other, etc.] 8. Two triangular prisms are equal if their lateral faces are equal each to each. B 9. Any prism is equivalent to the parallelepiped having the same altitude and an equivalent base. 10. The square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions. To Prove: ZC 2 = AE' 2 + ED 2 + DC*. Proof : A D is the hypotenuse of rt. A A ED, and AC, of rt.&ACD. 11. The diagonal of a cube is equal to the edge multiplied by \/3. 12. The volume of a triangular prism is equal to half the product of the area of any lateral face by the perpendicular drawn to that face from any point in the opposite edge. [Use the fig. of 602.] 13. Every section of a prism made by a plane parallel to a lateral edge is a parallelogram. To Prove : LMRN a O. Proof : LM is \\ to NR (?). LN and MR are each \\ to any edge. (Explain.) 14. Every polyhedron has an even number of face angles. Proof: Consider the faces as separate polygons, sides of these polygons = double the number of edges of the polyhedron. (Explain.) But the number of sides of these polygons = the number of their angles, that is, the number of face angles. .-. the number of face angles = double the number of edges = an even number (?). 15. There is no polyhedron having fewer than 6 edges. The number or 306 SOLID GEOMETRY 16. A room is 7 m. long, 5 m. wide, 3 m. high. Find its contents and its total area. 17. Find the volume, lateral area, and total area of an 8-in. cube. 18. A right prism whose height is 12 ft. has for its base a right triangle whose legs are 6 ft. and 8 ft. Find the volume, lateral area, and total area of the prism. 19. Find the altitude of a rectangular parallelepiped whose base is 21 in. x 30 in., equivalent to a rectangular parallelepiped whose dimen- sions are 27 in. x 28 in. x 35 in. ; 20. A cube and a rectangular parallelepiped whose edges are 6, 16, and 18, have the same volumes. Find the edge of the cube, f i/ ^j 21. Find the volume of a rectangular parallelepiped whose total area is 620 and whose base is 14 x 9. V "^ ^ ^( "^ \ A $ /w , vwUAfl^ 22. How many bricks each 8 x 2| x 2 in. will be required to build a wall 22 x 3 x 2 ft. (not allowing for mortar) ? 8j (Ax. 5). and s t = s t .-. s c v < s e Si (Ax. 9). .*. F s t < s c 8 t (Ax. 7). That is, s c V and F S< are each less than s c S& which itself approaches zero. Hence, 8 e F approaches zero and F S { approaches zero. .-. S c approaches F as a limit, and s t approaches F as a limit (?) (240). (See note on p. 223.) Q.E.D. Ex. 1. If a plane is passed parallel to the base of a pyramid, cutting the lateral edges, the section is to the base as the square of the altitude of the pyramid cut away by this plane is to the square of the altitude of the original pyramid. (See proof of 620.) Ex. 2. If two pyramids have equal altitudes and are cut by planes parallel to the bases and at equal distances from the vertices, the sections formed will be to each other as the bases of the pyramids. Ex. 3. In the figure of 622 prove the planes of the faces of the prisms, that are opposite OC, are parallel to OC and to AB. Ex. 4. State the theorems leading up to the theorem of 588. Ex. 5. State the theorems leading up to the theorem of 597. Ex. 6. State the theorems leading up to the theorem of 603. Ex. 7. The base of a pyramid is 180 sq. in. and its altitude is 15 in. What is the area of the section made by a plane parallel to the base, and 5 in. from the vertex ? Iff I I # I Jt > 7^- T*0 Ex. 8. The base of a pyramid is 200 sq. in., and its altitude is 12 in. At what distance from the vertex must a plane be passed so that the section shall contain half the area of the base ? { V ' 7U* . V ,' f O* O 314 SOLID GEOMETRY 623. THEOREM. Two triangular pyramids having equal altitudes and equivalent bases are equivalent. o or Given: Triangular pyramids O-ABC and O f -A f B r C r having equal altitudes and base ABC =0= base A'B'C ! . To Prove: O-ABC =c= O'-A'B'C'. Proof : Divide the altitude of each pyramid into any num- ber of equal parts. Through these points of division pass planes II to the bases, forming triangular sections. Upon these sections as bases construct inscribed prisms. Denote the volumes of the pyramids by V and F ; , and the sums of the volumes of these series of prisms by 8 and S r . The corresponding sections are =0= (?) (620). .*.. the corresponding prisms are =c= (?) (604). Hence, s=s f (Ax. 2). By indefinitely increasing the number of equal parts into which the altitudes are divided, the number of prisms be- comes indefinitely great. .*. 8 approaches V as a limit, (?) (622), and S r approaches V 1 as a limit (?). .-. v -* v' (?) (242). That is, O-ABC =c= O r -A f B f C f . Q.B.D. NOTE. As in plane geometry, A ABC is the same as A BA C, so in solid geometry the pyramid O-ABC is the same as the pyramid A -B CO or B-A CO or C-ABO. BOOK VII 315 624. THEOREM. The volume of a triangular pyramid is equal to one third the product of its base by its altitude. Given: Triangular pyramid O-AMC, whose base = B and alti- tude = h. To Prove : Volume O-AMC = ^ B - h. Proof: Construct a prism AMC- DOE, having AMC as its base, and , OM as one of its lateral edges. Pass a plane through DO and OC, cutting the face AE in line CD. The prism is now divided into three triangular pyramids. In pyramids O-AMC and C-ODE, the altitudes are=(?)(524). The bases AMC and ODE are equal (?) (568). .-. pyramid O-AMC =c= pyramid C-ODE (?) (623). In the pyramids C-AMO and C-AOD, the altitudes are the same line from c _L to plane DM (?) (507). The bases AMO and AOD are = (?) (132). .*. pyramid C-AMO =0= pyramid C-AOD (?) (623). Hence, O-AMC =0= C-ODE =0= C-AOD (Ax. 1). That is, O-AMC = ^ the prism. But the volume of the prism = B h (?). /. volume of pyramid o-AMC= J B- h (Ax. 6). Q.E.D. Ex. 1. In the figure of 624, prove pyramid 0-A CD - 0-CDE. __ Ex. 2. The area of the base of a triangular pyramid is 30 sq. in., and its altitude is 20 in. Find the volume. Find the volume of the prism having the same base and altitude. j*4M^ <2^. '*++, ^ (ffotf xc, Ex. 3. A pyramid whose base is b and altitude is h is equivalent to another pyramid whose base is d and altitude is x. Find x. Ex. 4. If in the figure of 602, a plane is passed through E and CD, what part of the whole parallelepiped is the pyramid E-A CD? 316 SOLID GEOMETRY . 625. THEOREM. The volume of any pyramid is equal to one third the product of its base by its altitude. Given: Pyramid o-CDEFG, whose base = B and altitude = h. To Prove : Volume of O-CDEFG == J B - h. Proof : Through any lateral edge, OC, and lateral edges not adjoining C, pass planes dividing the pyra- mid into triangular pyramids. Vol. of 0-CDE=CDE-h] Vol. of O-CEF=ICEF- h Vol. of = CFG-h Adding, Vol. of o-CDEFG=^B-h (Ax. 2 and Ax. 4). Q.E.D. 626. THEOREM. Any two pyramids having equal altitudes and equivalent bases are equivalent. (Ax. 1.) 627. THEOREM. Two pyramids having equal altitudes are to each other as their bases. (Prove.) 628. THEOREM. Two pyramids having equivalent bases are to each other as their altitudes. (Prove.) 629. THEOREM. Any two pyramids are to each other as the prod- ucts of their bases by their altitudes. (Prove.) {630. THEOREM. The volume of the frustum of a triangular pyra- mid is equal to one third the altitude multiplied by the sum of the lower base, the upper base, and a mean proportional between the bases of the frustum. Given : The frustum BD of a triangular pyramid whose lower base = B ; upper base= b ; altitude = h. To Prove: Volume of BD = $h [B + b + V \\ BOOK VII 317 Proof : Pass a plane through edge CE and vertex s, and another through edge RS and vertex E, dividing the frustum into three triangular pyramids, S-CDE, E-RST, E-CRS. II. III. We shall now prove E-CRS = h- VjT~b. E-CSD _ A CSD E-CRS A CRS (?) (627). = RS ^ T .-, . Likewise, E-CRS US S-CER A - =- S-ERT A = (Ax. 1). A XON -, (?) and A ERT ET S-CER _ CE ~ RT But A CDE and RST are similar (?) (619, II). CD_CE , ? . ' RS RT TT E-CSD S-CER or E-CRS , A ' 1N Hence, - = - (Ax. 1). E-CRS S-ERT ~h 7? 7? 1 That is, ^ - = -!- (Substituting from I and II). ECRS -k fl /. E-CRS (?) (299). .*. volume of the frustum = J h \_B + 6 + V^ b~\ (Ax. 2). Q.E.D. NOTE. Theorem 630 is sometimes stated thus : The frustum of a triangular pyramid is equivalent to the sum of three pyramids whose altitudes are the same as the altitude of the frus- tum and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum. 318 SOLID GEOMETRY 631. THEOREM. The volume of the frustum of any pyramid is equal to one third the altitude multiplied by the sum of the lower base, the upper base, and a mean proportional between the bases of the frustum. Given: Pyr. O-ADEFG', frustum A'F, whose lower base = J5, upper base = 6, altitude = h. To Prove: Vol. of frus- tum = ^h[B + b + Proof : Construct a A QRS =0= polygon AF (by 409). Upon A QRS as a base, construct a pyramid whose altitude = the altitude of O-ADEFG. Pass a plane Q'R'S' II to QRS and at a distance from QRS = h. Vol. of Q' J R = lA[A QRS + A Q'R'S' + VA QRS A Q'R's 1 ] (630). The alt. of P-Q'R'S' = alt. of O-A'D'E'F'G' (?) (Ax. 2). Also, QRS^B (Const.); and Q'R'S' ^b (?) (620). .'. vol. of O-ADEFG = vol. of P-QRS (?) (626), and vol. of 0-A'D'E'F'G' = vol. of P-Q'R'S' (?). Subtracting, vol. of frustum A'F= vol. of frustum Q'R (Ax. 2). Vol. of frustum A'F= J h[B+b + VB b] (Ax. 6). Q.E.D. Ex. 1. Find the volume of a pyramid whose altitude is 18 in. and whose base is 10 in. square. bU*A/vx. '^~ Ex. 2. Find the volume of the frustum of a pyramid whose altitude is 20 and the areas of whose bases are 18 and 32. 7,-d, ^ c^J, Ex. 3. The bases of the frustum of J a pyramid are regular hexagons whose sides are 10 in. and 6 in., respectively. The altitude of the frus- tum is 2 ft. Find its volume. % % ^ v ^__^ * BOOK VII 319 632. THEOREM. A truncated triangular prism is equivalent to three triangular pyramids whose bases are the base of the prism and whose vertices are the three vertices of the face opposite the base (the inclined section). Given : The truncated triangular prism ABC-RST, whose base is ABC and whose opposite vertices are R, S, T. Let it be divided by the planes ACS, ABT, BCR. To Prove : ABC-RST 0= R-ABC + S-ABC + T-ABC (III). Proof : In Fig. I, S-ABCis obviously one of these pyramids. In Fig. II, A-CST =0= A-BCT (?) (626). That is, A-C8T =0= T-ABC. In Fig. Ill, T-ARS =0= T-ABR (?) (626). T-ABR =0= C-ABR (?) (626). .*. T-ARS =c= R-ABC (Ax. 1). Now, ABC-RST =c= T-ARS + S-ABC + A-CST (Ax. 4). Hence, ABC-RST R-ABC + S-ABC + T-AB C (Ax. 6). Q.E.D. 633. COR. The volume of a truncated triangular prism is equal to the product of the base by one third the sum of the three altitudes drawn to the base from the three vertices opposite the base, 634. COR. The volume of a truncated right tri- angular prism is equal to the product of the base by one third the sum of its lateral edges. 320 SOLID GEOMETKY 635. THEOREM. The volume of any truncated triangular prism is equal to the product of its right section by one third the sum of its lateral edges. Proof: The right section divides the solid into two truncated right prisms. Hence, volume = right section x ^ sum of lateral edges (634). Q.E.D. 636. THEOREM. Two triangular pyramids (tetrahedrons) having a trihedral angle of one equal to a trihedral angle of the other are to each other as the products of the three edges including the equal trihedral angles. Given: Triangular pyra- mids S-ABC, S-PQR-, having the trih. A at 8 equal ; and their volumes F and V 1 . To Prove : , = V 1 SP-SQ-SR Proof: Place the pyra- mids so that the equal trihedral A coincide. Draw the alti- tudes AX and PY and the projection SXY in plane SQR. A SAX is similar to A SPY (?) (315). V A SBC -AX A SBC AX But, = 8*L* (?) (388) ; and = (?) (323, 3) A SQR SQ-SR^ PY SP^ TT V SB SC SA SA SB - SC , . Hence, , = - (Ax. 6), V S SR SP SP - S SR ^ Q.E.D, BOOK VII 321 637. THEOREIVL In any polyhedion the number of edges increased by two is equal to the number of vertices increased by the number of faces. Given : A polyhedron ; E = number of edges ; F= number of faces ; F = number of vertices. To Prove : = V+F. Proof: Suppose the surface of the polyhedron is put to- gether, face by face. For one face, E V (154). (Begin with the base.) By attaching an adjoining face, the number of edges is one greater than the number of vertices. That is, for two faces, E= F + 1. Similarly, for three faces, E = V + 2, and, for four faces, E= F+ 3, and, for five faces, E = F -{- 4, and, for n faces, E = V + (n 1), and, for F 1 faces, E= F+ (F- 2). By attaching the last face, neither the number of edges nor the number of vertices is increased. That is, for F faces, # = F + F - 2. .. for the complete solid, E+ 2 = F-f F (Ax. 2). Q.E.D. 322 SOLID GEOMETRY 638. THEOREM. In any polyhedron the difference between the num- ber of edges and the number of faces, is two less than the number of vertices, that is, E - F = V- 2. (See 637.) 639. THEOREM. The sum of all the face angles of any polyhedron is equal to 4 right angles multiplied by two less than the number of vertices, that is, 8 ^ = ( F- 2) 4 rt. A = ( V- 2) 360. t J Given : A polyhedron ; E = number of edges ; F= number of faces ; F = number of vertices. . To Prove : Sum of all the face A = (F 2) 4 rt. A, ors 4 = (F-2) 360. Proof: Considering the faces as separate polygons, it is obvious that each edge is a side of two polygons, that is, the number of sides of the several faces = 2 E. .. the number of vertices of all the polygons = 2 E (154). Suppose an exterior Z formed at each of these 2 E vertices. Then the sum of the exterior A of each face = 4 rt. A (?). Sum of int. and ext. A at each vertex = 2 rt. A (?). .-. the int. A 4- ext. A at all the 2 E vertices = 4 E rt. A. But, the ext. A of all the F faces = 4 F rt. A (?). Hence, the int. A of these polygons = 4^ rt. A-4FTt. ^(?). = (#-*') 4 rt. A But E-F= F-2 (?) (638). .-. s 4 = (F-2) 4 rt. z = (F-2) 360 (Ax. 6). Q.E.D. BOOK VII 323 REGULAR AND SIMILAR POLYHEDRONS 640. A regular polyhedron is a polyhedron whose faces are equal regular polygons and whose polyhedral angles are all equal. Similar polyhedrons are polyhedrons which have the same number of faces similar each to each and similarly placed, and which have their homologous polyhedral angles equal. 641. THEOREM. There cannot exist more than five kinds of regular polyhedrons. Proof : The faces must be equilateral A, squares, regular pentagons, or some other regular polygons (?) (640). There must be at least three faces at each vertex (?) (580). Sum of the face A at each vertex must be < 360 (?) (564). Let us now discuss the possible regular polyhedrons. I. Each Z of an equilateral A = 60 (?). Hence, we may form a polyhedral Z by placing 3 equilateral A at a vertex, or 4 at a vertex, or 5 at a vertex. But not 6 at a vertex (?). That is, only three regular polyhedrons can be formed having equilateral triangles for fac*es. II. Each Z of a square = 90 (?). Hence, we may form a polyhedral Z by placing 3 squares at a vertex. But not 4 at a vertex (?). That is, only one regular polyhedron can be formed having squares for faces. III. Each Z of a regular pentagon = 108. (?) (164). Hence, we may form a polyhedral Z by placing 3 regular pentagons at a vertex. But not 4 at a vertex (?). That is, only one regular polyhedron can be formed having regular pentagons for faces. IV. Each Z of a regular hexagon = 120 (?). Hence, no polyhedral Z can be formed by hexagons (?). Consequently, there can be no more than five kinds of regular polyhedrons, three kinds bounded by triangles, one kind by squares, and one by pentagons. Q.E.D. 324 SOLID GEOMETRY 642. The names of the regular polyhedrons. NAMES TOTAL NUMBER OF FACES NUMBER OF FACES AT EACH VERTEX KINDS OF FACES Regular tetrahedron 4 3 Equilateral triangles Regular hexahedron (cube) 6 3 Squares Regular octahedron 8 4 Equilateral triangles Regular dodecahedron 12 3 Regular pentagons Regular icosahedron 20 5 Equilateral triangles A REGULAR TETRAHEDRON REGULAR REGULAR REGULAR OCTAHEDRON DODECAHEDRON ICOSAHEDRON BOOK VII 325 DIRECTIONS. The regular polyhedrons may be constructed as fol- lows : Mark on cardboard figures similar to the drawings on page 324, but much larger; with a knife cut the dotted lines half through and the solid lines entirely through ; fold along the dotted lines, closing the solids up and forming the figures as illustrated on page 324; paste strips of paper along the edges to hold them in position. 643. THEOREM. In two similar polyhedrons : I. Homologous edges are propor- tional. II. Homologous faces are to each other as the squares of any two ho- mologous edges. III. Total surfaces are to each other as the squares of any two ho- mologous edges. Proof: I. Homologous faces are similar (?) (640). AB BC CD DH AE BF A'B' B'C' C'D' Face DG DH D'H' 2 II. III. A'E' B'F' face AH AE 2 = etc. (?) (823, 3). Face D'G' Face DG ITJ! j)' H ' 2 face A'H face AH face GE = etc. (?) (390). feach being Face D'G' Total surface of AG = etc. J lace G'E' equal to face DG DH' Total surface of A'o' face D f G 1 TV* -^-=etc. (301). A'E' 2 644. THEOREM. If a pyramid is cut by a plane parallel to the base, the pyramid cut away is similar to the original pyramid. (Def . of 640.) Ex. 1. Show that the theorem of 637 holds true in the case of a cube. Ex. 2. Show that the theorem of 639 holds true in the case of a cube. Ex. 3. Show that the theorems of 637 and 639 are true in the case of a regular octahedron. 326 SOLID GEOMETRY O' 645. THEOREM. Two similar tetrahedrons are to each other as the cubes of any two homologous edges. Given: Similar tetrahedrons O-ABC and o'-A's'c'-, whose volumes = F and F'. To Prove: F : F' = AB* = J 7 F 3 = etc. Proof : Trihedral Z A = tri- hedral Z A' (?) (640). V_ AB AC " F'~ AO A'B'. A' C'- A'O 1 AB AC AO (636). But AC T^C'" v_ '"' F' .!'' ^jg I 7 !* 7 ^45 ^I'O' ^10 A'O'' AB AB A'B' A'B' A'B' (643,1). x. 6), That is, F : F' = etc. Q.E.D. Ex. 1. If, in the figure of 645, A = 4 and A'O'- 1, what is the ratio of the total surfaces of the tetrahedrons ? Of their volumes ? Ex. 2. Two homologous edges of two similar polyhedrons are 2 and 3. What is the ratio of their total surfaces? Ex. 3. Two homologous edges of two similar tetrahedrons are 2 and 5. The area of the total surface of the less is 28, and its volume is 40. Find the area of the total surface and the volume of the other, ^-ji^- \i Ex. 4. Show that the theorems of 637 and 639 are true in the cases of regular dodecahedrons and regular icosahedrons. Ex. 5. A pyramid whose volume is F and altitude is h is bisected by a plane parallel to the base. Find the distance of this plane from the vertex. Ex 6. Find the total surface of a regular tetrahedron whose edges are each equal to 4 in. ^XO-\/3 - l(o\/^ ^ V] % T- Ex. 7. Find the total surface of a regular octahedron whose edges are each equal to 18 in. f I \fe X f , . (^f ff \g .=. n BOOK VII 327 646. THEOREM. Two similar polyhedrons can be decomposed into the same number of tetrahedrons similar each to each and similarly placed. Given: (?). To Prove: (?). Proof: Suppose diagonals I I" I EF/I B' drawn in every face of AT, except the faces containing vertex A, dividing the faces into A. (The figure shows only SV.) Suppose lines drawn from A to the several vertices of these A. (The figure shows only AS, Obviously, this process divides the solid (by planes) into tetrahedrons, each of which has a vertex at A. Then construct homologous lines in solid A r T r . There will evidently be as many lines in A f T f as in AT and as many tetrahedrons, and these will be similarly placed. Now, in the tetrahedrons A-SVR and A'-S'V'R^ A AVR is similar to A A'v'lt' i A ARS is similar to A A r R's' ; A SVR is similar to A s'v'R 1 (?) (327). AV VR VS RS AS Also, ,000 (3 8 ' AV VS AS ' A r V l " V'S 1 A f S Hence, A ASV is similar to A A r s'v f (?) (318). Also, the trihedral A R and R f are = ; 8 and S r are = ; F and v' are = , etc. (?) (561). .*. the two tetrahedrons are similar. Furthermore, after removing these tetrahedrons, the re- maining polyhedrons are similar (Def. 640). By the same process other tetrahedrons may be removed and proved similar, and the process continued until the polyhedrons are completely decomposed into tetrahedrons similar each to each and similarly placed. Q.E.D. 328 SOLID GEOMETRY 647. THEOREM. The volumes of two similar polyhedrons are to each other as the cubes of any two homologous edges. Given: Similar polyhe- drons.^ and A'T'I volumes F and F'; AR and A f R f , any two homologous edges. To Prove: (?) Proof: These solids may be decomposed, etc. (646). Denote vol. of tetrahedrons of AT by w, x, y, z, etc. ; of A'T' by w', ', y', z r , etc. B f W rp, Then, ; & ' V 1 A'R'* ; etc. (?) (645). Hence, , = - t - = = etc. (Ax. 1). Therefore y*i' S**f /til w' etc. " (gel). That is - = w' + x r + y 1 + etc. w 1 V 1 AR (Ax. 6). Q.E.D. 648. THEOREM. The volumes of two similar pyramids are to each other as the cubes of their altitudes. (Explain.) FORMULAS OF BOOK VII Let B = area of base. b = area of upper base. E = number of edges. e, e' = homologous edges. F number of faces. h = altitude. L = lateral area. Parallelepiped . . Prism Regular pyramid . Pyramid .... Frustum of pyramid Polyhedron . . . Similar polyhedrons T T' = e' 2 P = perimeter of base. P r = perimeter of right section. p = perimeter of upper base. s = slant height. T = total area. V = volume or number of vertices. V=B h. = !*,. -edge; T = L + 2 B ; V=B-h. Li=\P-8\ T= L + B. ; Sum of face A =(F- 2)360 ; V-.V = e* :e' s . BOOK VII 329 ORIGINAL EXERCISES 1. What plane through the vertex of a given tetrahedron will divide it into two equivalent parts? Prove. 2. The area of the base of any pyramid is less than the sum of the lateral faces. [Draw the altitudes of the lateral faces and the projections of the alti- tudes upon the base.] 3. Three of the edges of a parallelepiped that meet in a point are also the lateral edges of a pyramid. What part of the parallelepiped is this pyramid? 4. A plane is passed containing one vertex of a parallelepiped and a diagonal of a face not con- taining that vertex. What part of the volume of the parallelepiped is the pyramid thus cut off? 5. Any section of a tetrahedron made by a plane parallel to two opposite edges, is a parallelogram. Given : Section DEFG II to OA and EC. To Prove : DEFG is a O . Proof: EF is II to OA, DG is to OA (?). Also DE is II to BC and GF is || to BC (?) etc. 6. The three lines that join the midpoints of the opposite edges of a tetrahedron meet in a point and bisect one another. Given : LM, PQ, RS, the three lines, etc. To Prove : (?). Proof : Join PS, SQ, QR, PR. PS is II to and = %BC; RQ is I! to and = \ BC. (Explain.) A* Similarly, discuss LM and SR. 7. A pyramid having one of the faces of a cube for its base and the center of the cube for its vertex, contains one sixth of the volume of the cube. O 8. A plane containing an edge of a regular tetra- hedron and the midpoint of the opposite edge, (a) contains the medians of two faces ; (5) is perpendicular to the opposite edge ; (c) is perpendicular to these two faces; (d) contains two altitudes of the tetrahedron. 330 SOLID GEOMETRY 9. The altitude of a regular tetrahedron meets the base at the point of intersection of the medians of the base. 10. The altitude of a regular tetrahedron = ^V6 times the edge. 11. The altitudes of a regular tetrahedron meet at a point. 12. The lines joining the vertices of any tetrahedron to the point of intersection of the medians of the opposite face meet in a point that divides each line into segments in the ratio 3 : 1. Given: OM, CR, two such lines. To Prove : The four such lines meet, etc. Proof : OM and CR lie in the plane determined by OC and point D, the midpoint of AB. :. OAT and CR intersect. Draw RM. &4 ' "DM :. RM is II to OC (?). /. A DMR and DCO are similar (?) ; and DR:DO = RM: OC (?). Thus RM = i OC. (Explain.) Also & PRM and OPC are similar (?). Hence, OP : PM= OC : RM =3:1.) and CP : PR = OC : RM = 3 : 1. | ' O x P lam O Q . E . D . NOTE. This point P is called the center of gravity of the tetrahedron. 13. There can be no polyhedron having seven V edges and only seven. 14. The planes bisecting the dihedral angles of any tetrahedron meet in a point that is equally distant from the faces. 15. The lines perpendicular to the faces of any tetrahedron, at the centers of the circles circumscribed about the faces, meet in a point that is equally distant from the vertices. Proof : RX and SY are loci of points, etc. (526). Plane MN, JL to AB at M, the midpoint of AB, is the locus of points, etc. .'. RX and SY lie in MN and intersect at 0, etc. 16. If a plane be passed through the mid- points of the three edges of a parallelepiped that meet at a vertex, what part of the whole solid is the pyramid thus cut off ? BOOK VII 331 17. The plane bisecting a dihedral angle of a tetrahedron divides the opposite edge into two segments proportional to the areas of the faces that form the dihedral angle. 18. Two tetrahedrons are similar if a dihedral angle of one equals a dihedral angle of the other and the faces forming these dihedral angles are respectively similar. 19. If from any point within a regular tetrahedron perpendiculars to the four faces are drawn, their sum is constant and equal to the altitude of the tetrahedron. : 20. To construct a regular tetrahedron upon a given edge. Construction : Upon AB, construct an equilateral A ABC. Erect ED JL to plane of A ABC, at D > the center of circumscribed O. Take V on ED such that A V=BV = CV= AB, etc. 21. To construct a regular hexahedron upon a given edge. Construction : Upon A B, construct a square ABCD. At the vertices erect Js = AB and join the extremities, etc. 22. To construct a regular octahedron upon a given edge. Construction : Upon A B, construct a square ABCD. At M, the center of the square, erect XX' JL to plane of ABCD. On XX' take MV.=MV = MD. Draw the edges from V and V . Statement : VV is a regular octahedron. Proof: The right A DMV, DMC, DMV, are equal. (Explain.) Thus the 12 edges are equal and the 8 faces are equal. (Explain.) Figures AVCV, DVBV, ABCD are equal squares. (Explain.) Then, pyramids V-ABCD, D-A FCF',etc., are equal and the 6 poly- hedral angles are equal. (Explain.) .'. Etc. 23. To pass a plane through a cube so that the section will be a regular hexagon. 332 SOLID GEOMETRY 24. To pass planes through three given lines in space, no two of which are parallel, which shall inclose a parallelepiped. 25. Find the lateral area and the total area of a regular pyramid whose slant height is 20 in. and whose base is a square, 1 ft. on a side. 26. Find the volume of a pyramid whose altitude is 18 in. and whose base is an equilateral triangle each side of which is 8 in. 27. A regular hexagonal pyramid has an altitude of 9 ft. and each edge of the base is 6 ft. Find the volume. ' 28. The base of a pyramid is an isosceles triangle whose sides are 14, 25, 25, and the altitude of the pyramid is 12. Find its volume. 29. The altitude of the frustum of a pyramid is 25, and the bases are squares whose sides are 4 and 10, respectively. Find the volume of the frustum. 30. The frustum of a regular pyramid has hexagons for bases whose sides are 5 and 9, respectively. The slant height of the frustum is 14. Find its lateral area. Find its total area. 31. The altitude of a regular pyramid is 15, and each side of its square base is 16. Find the slant height, the lateral edge, the total area, and the volume. OA 2 = OD 2 + DA 2 = (15)2 + ( 8 )2 = 2 89. /. A O = 17. OC 2 = OA 2 + AC 2 =289 + 64 = 353. :. OC= V353 = 18.78+. ' 32. The slant height of a regular pyramid is 39, the altitude is 36, and the base is a square. Find the lateral area and volume. 33. The lateral edge of a regular pyramid is 37 and each side of the hexagonal base is 12. Find the slant height, the lateral area, total area, and volume. AD = 6 V3. In rt. A A CD, CD = 12, A C = 6, In rt. A A CO, CO = 37, A C = 6, .'. AO- VHJa. In rt. A CDO, CO = 37, CD = 12, .'. OD = 35, etc. BOOK VII 333 34. Find the total area and volume of a regular tetrahedron whose edge is six. The four faces are equal equilateral &. .'. AO = AC = 3V3; /.^Z> = V3~and CD = 2V3. Hence, OD = 2 v'o. Area of any face = 9 \/3, etc. 35. Find the total area and volume of a regular tetrahedron whose edge is 10. 36. Find the total area and volume of a regular hexahedron whose edge is 8. 37. Find the total area and volume of a regular octahedron whose edge is 16. The 8 faces are equal equilateral A. A = 8 V3. In &ADO, one finds OD = SV2. The volume of the octahedron = volumes of two pyramids, etc. 38. Find the total area and volume of a regular octahedron whose edge is 18. 39. The altitude of a regular pyramid is 16 and each side of the square base is 24. Find the lateral area and volume. 40. The slant height of a regular pyramid is 26 and its base is an equilateral triangle whose side is 20\/37 Find the total area and volume. 41. The altitude of a regular pyramid is 29 and its base is a regular hexagon whose side is 10. Find the total area and volume. 42. Find the total area and volume of a regular tetrahedron whose edge is 18. 43. Find the total area and volume of a regular octahedron whose edge is 20. 44. If the edge of a regular tetrahedron is e, show that the total area ise 2 v/3 and the volume is j 1 ^ e s V2 . 45. If the edge of a regular octahedron is e, show that the total area is 2 e 2 V3 and the volume is e s V2 . 46. A pyramid whose base is a square 9 in. on a side, contains 360 cu. in. Find its height. 47. A pyramid has for its base a hexagon whose side is 7| units and the pyramid contains 675 cu. units. Find the altitude. 48 The volume of a regular tetrahedron is 144 A/2 ; find its edge. 49. The volume of a regular octahedron is 243 V^; find its edge. 334 SOLID GEOMETRY 50. The volume of a square pyramid is 676 cu. in. and the altitude is a foot. Find the side of the base. Find the lateral area. 51. The altitude of the Great Pyramid is 480 ft. and its base is 764 ft. square. It is said to have cost $ 10 a cu. yd. and $ 3 more for each sq. yd. of lateral surface (considered as planes). What was the cost? 52. The total surface of a regular tetrahedron is 324 V3 sq. in. ; find its volume. 53. The base of a pyramid is a rhombus whose diagonals are 7 m. and 10 m. Find the volume if the altitude is 15 m. 54. The areas of the bases of the frustum of a pyramid are 3 sq. m. and 27 sq. m. The volume is 104 cu. m. Find the altitude. 55. The base of a pyramid is an isosceles right triangle whose hypotenuse is 8. The altitude of the pyramid is 15. Find the volume. 56. The altitude of a square pyramid, each side of whose base is 6 ft., is 10 ft. Parallel to the base and 2 ft. from the vertex a plane is passed. Find the area of the section. Find the volumes of the two pyramids concerned, and hence find the volume of the frustum. 57. Find the area of the section of a triangular pyramid each side of whose base is 8 in. and whose altitude is 18 in., made by a plane parallel to the base and a foot from the vertex. 58. The altitude of a frustum of a pyramid is 6, and the areas of the bases are 20 sq. in. and 45 sq. in. Find the altitude of the complete pyramid. Find the volume of this frustum by two distinct methods. 59. A granite monument in the form of a frustum of a pyramid, having rectangular bases one of which is 8 ft. wide and 12 ft long, and the other 6 ft. wide, is 30 ft. high. It is surmounted by a granite pyramid having the same base as the less base of the frustum, and 10 ft. in height. Find the entire volume. If one cu. ft. of water weighs 62| Ib. and granite is three times as heavy as water, what is the weight of the entire monument? 60. If a square pyramid contains 40 cu. in. and its altitude is 15 in., find the side of its base. 61. A church spire in the form of a regular hexagonal pyramid whose base edge is 8 ft. and whose altitude is 75 ft. is to be painted at the rate of 18^ per square yard. Find the cost. 62. Find the edge of a cube whose volume is equal to the volumes of two cubes whose edges are 4 and 6. BOOK vil 335 i 63. The base of a certain pyramid is an isosceles trapezoid whose parallel sides are 20 ft. and 30 ft. and the equal sides are each 13 ft. Find the volume of the pyramid if its altitude is 12 yards. 64. The lateral edge of the frustum of a regular square pyramid is 53 and the sides of the bases are 10 and 66. Find the altitude, the slant height, the lateral area, and the volume. 65. The sides of the base of a triangular pyramid are 33, 34, 65, and the altitude of the pyramid is 80. Find its volume. 66. The sides of the base of a tetrahedron are 17, 25, 26, and its altitude is 90. Find its volume. 67. If there are l^- cu. ft. in a bushel, what is the capacity (in bushels) of a hopper in the shape of an inverted pyramid, 12 ft. deep and 8 ft. square at the top ? 68. In the corner of a cellar is a pyramidal heap of coal. The base of the heap is an isosceles right triangle whose hypotenuse is 20 ft. and the altitude of the heap is 7 ft. If there are 35 cu. ft. in a ton of coal, how many tons are there in this heap ? 69. How many cubic yards of earth must be removed in digging an artificial lake 15 ft. deep, whose base is a rectangle 180 x 20 ft. and whose top is a rectangle 216 x 24 ft. ? [The frustum of a pyramid.] 70. One pair of homologous edges of two similar tetrahedrons are 3 ft. and 5 ft. Find the ratio of their surfaces. Of their volumes. 71. A pair of homologous edges of two similar polyhedrons are 5 in. and 7 in. Find the ratio of their surfaces. Of their volumes. 72. The edge of a cube is 3. What is the edge of a cube twice as large ? Four times as large ? Half as large ? 73. An edge of a tetrahedron is 6. What is the edge of a similar tetrahedron three times as large? Eight times as large? Nine times as large ? One third as large ? 74. An edge of a regular icosahedron is 3 in. What is the edge of a similar solid five times as large ? Ten times as large ? Fifty times as large? A thousand times as large? 75. The edges of a trunk are 2 ft., 3 ft., 5 ft. Another trunk is twice as long (the other edges 2x3 ft.). How do their volumes compare? A third trunk has each dimension double those of the first. How does its volume compare with the first ? How do their surfaces compare ? 336 SOLID GEOMETRY 76. If the altitude of a certain regular pyramid is doubled, but the base remains unchanged, how is the volume affected? If each edge of the base is doubled, but the altitude unchanged, how is the volume affected ? If the altitude and each edge of the base are all doubled, how is the volume affected? 77. If the slant height (only) of a regular pyramid is doubled, how is the lateral area affected ? If each edge of the base is doubled, how is the lateral area affected? If both are doubled, what is the effect? 78. A pyramid is cut by a plane parallel to the base and bisecting the altitude. What part of the entire pyramid is the less pyramid cut away by this plane ? 79. The volume of a certain pyramid, one of whose edges is 7, is 686. Find the volume of a similar pyramid whose homologous edge is 8. 80. A certain polyhedron whose shortest edge is 2 in. weighs 40 Ib. What is the weight of a similar polyhedron whose shortest edge is 5 in. ? 81. An edge of a polyhedron is 5 in. and the homologous edge of a similar polyhedron is 7 in. The entire surface of the first is 250 sq. in. and its volume is 375 cu. in. Find the entire surface and volume of the second. 82. Find the edge of a cube whose volume equals that of a rectangular parallelepiped whose edges are 3 x 4 x 18. 83. A pyramid and an equivalent prism have equivalent bases. How do their altitudes compare ? 84. A pyramid and a prism have the same altitude and equivalent bases. Compare their volumes. 85. Solve: V=%B.hiorB. Fork. 86. Solve : L = \ P . s for P. For s. 87. Solve : L = \ (P +p) . s for s. 88. A prism whose altitude is 8 and whose base is an equilateral tri- angle whose side is 9 in. is transformed into a regular pyramid whose base is 10 in. square. Find its altitude. 89. An altitude of a pyramid is 10 m. How far from the vertex will a plane parallel to the base divide the pyramid into two equivalent parts ? 90. The altitude of a pyramid is 12, and two planes are passed parallel to the base and dividing the pyramid into three equivalent parts. At what distances from the vertex are they ? BOOK VIII CYLINDERS, CONES CYLINDERS 649. A cylindrical surface is a surface generated by a moving straight line which continually intersects a given curved line in a plane, and which is always parallel to a given straight line not in the plane of the curve. The generating line is the generatrix. The directing curve is the directrix. An element of a cylindrical surface is the generating line in any position. / ULn CYLINDRICAL RIGHT OBLIQUE CYLINDER SURFACE CIRCULAR CYLINDER OF REVOLUTION CYLINDER 650. A cylinder is a solid bounded by a cylindrical surface and two parallel planes. The bases of a cylinder are the parallel plane sections. The lateral area of a cylinder is the area of the cylindrical surface included between the planes of the bases. The total area of a cylinder is the sum of the lateral area and the areas of the bases. The altitude of a cylinder is the perpendicular distance between the planes of the bases. 337 338 SOLID GEOMETRY 651. A right cylinder is a cylinder whose elements are perpendicular to the planes of the bases. A circular cylinder is a cylinder whose base is a circle. An oblique cylinder is a cylinder whose elements are not perpendicular to the planes of the bases. A right circular cylinder is a right cylinder whose base is a circle. A cylinder of revolution is a cylinder generated by the revolution of a rectangle about one of its sides as an axis. Similar cylinders of revolution are cylinders generated by similar rectangles revolving on homologous sides. 652. A right section of a cylinder is a section made by a plane perpendicular to all the elements. A plane is tangent to a cylinder if it contains one element of the cylindrical surface and only one, however far it may be extended. A prism is inscribed in a cylinder if its lateral edges are elements of the cylinder arid the bases of the prism are in- scribed in the bases of the cylinder. A prism is circumscribed about a cylinder if its lateral faces are tangent to the cylinder and the bases of the prism are circumscribed about the bases of the cylinder. PRELIMINARY THEOREMS 653. THEOREM. Any two elements of a cylinder are equal and parallel. (See 524 and 511.) 654. THEOREM. A line drawn through any point in a cylindrical surface, parallel to an element, is itself an element. (See 92.) 655. THEOREM. A right circular cylinder is a cylinder of revolu- tion. Ex. If a plane be defined as a surface generated by a moving straight line, what would the directrix be? BOOK VIII 339 THEOREMS AND DEMONSTRATIONS 656. THEOREM. Every section of a cylinder made by a plane con taining an element is a parallelogram. Given : Cylinder AE ; plane CE containing element CD. To Prove : CE is a O. Proof : At E draw EF II to CD in plane CE. Also, EFis an element of the cylinder (654). .*. EF is the intersection of the plane and the cylindrical surface (?) (482). Also, CFis II to DE (?) (500). /. CDEFis aO (?) (126). 657. THEOREM. The bases of a cylinder are equal. Given: (?). To Prove : (?) Proof : Suppose R, 8, and T any three points in the perim- eter of base AC. Draw ele- ments RR', ss', TT'. Also, draw RS, 8T, RT, R'S', S'T', R'T'. Now, Q.E.D. 8S'is = and II to TT .'. RS' is a O, RT' is a O, ST' is a O (?) (135). .'.BS=R'S'; 8T= S'T'',RT=R'T'(1W)', A RST= A fl's'r' (?). .*. base u4C may be placed upon base BD so that ft, /s, and T coincide with R 1 , -s', and T f , respectively. But S is any point on the perimeter ; hence, every point on perimeter of AC will coincide with a corresponding point on perimeter of BD. Therefore, base AC = base BD (?) (28). Q.E.D. 340 SOLID GEOMETRY 658. COR. Parallel plane sections of a cylinder (cutting all the elements) are equal. 659. COR. Any section of a circular cylinder made by a plane parallel to the base is a circle. 660. THEOREM. Every section of a right cylinder made by a plane containing an element is a rectangle (?). 661. THEOREM. If a regular prism be inscribed in, or circum- scribed about, a right circular cylinder and the number of sides of the base be indefinitely in- creased, the lateral area of the cylinder is the limit of the lat- eral area of the prism. Given : A regular prism inscribed in and a regular prism circumscribed about a right circular cylinder ; the lateral area of the cyl- inder = L, and of the prisms, L t and i c , respectively. To Prove: That as the number of sides of the bases of the prisms is indefinitely increased, L is the limit of both i z and L c . Proof : If the number of sides of the bases of the prisms is indefinitely increased, their perimeters will approach the circumference of the base of the cylinder as a limit (?). Hence, it is obvious that the lateral area of the cylinder is the limit of the lateral area of either prism. Q.E.D. Ex. 1. If the cylindrical surface of a cylinder be cut along an element, and this surface be placed in coincidence with a plane, what plane geometrical figure will it become ? Ex. 2. What two lines determine the size of aright circular cylinder? BOOK VIII 341 662. THEOREM. If a prism having a regular polygon for a base be inscribed in, or circumscribed about, any circular cylinder and the number of the sides of the base of the prism be indefinitely in- creased, the volume of the cylinder is the limit of the volume of the prism. Proof : If the number of sides of the base of either prism be indefi- nitely increased, the area of the base of the prism will approach the area of the base of the cylinder (440, II). .'.it is obvious that the volume of the cylinder is the limit of the volume of either prism. Q.E.D. 663. THEOREM. The lateral area of a right circular cylinder is equal to the product of the circumference of the base by an element. Given : A right circular cylinder, the circumference of whose base = (7, and whose element = E. To Prove : Lateral area L = C E. Proof : Inscribe in the cylinder a regular prism, the perimeter of whose base is P, whose lateral edge is E, and whose lateral area is L'. Then L' = P E (?). If the num- ber of sides of the base of the prism is indefinitely increased, L 1 will approach L as a limit (?). P will approach C as a limit (?). P E will approach C E as a limit. ,'.L=C-E (?) (242). Q.E.D. 342 SOLID GEOMETRY NOTE. The lateral area of an oblique circular cylinder is equal to the product of the perimeter of a right section of the cylinder by an element. The right section of an oblique circular cylinder is not a circle. The right section of an inscribed prism, having a regular polygon for a base, is not a regular polygon. Since elementary geometry does not deal with curves other than the circle, any proof or application of this theorem is omitted. 664. V 664.\ THEOREM. The volume of a circular cylinder is equal to the prdcttict of its base by its altitude. Given: A circular cylinder whose base = .B, altitude = A, and volume = F. To Prove: V=B - h. Proof : Inscribe a prism having a regular polygon for its base, whose base = B r and volume = V 1 . Then, F'= ' A(?) (603). If the number of sides of the base of the prism is indefinitely increased, V 1 approaches F as a limit (?), B r approaches B as a limit (?), and B' h approaches B - h as a limit. /. V = B h (?) (242). Q.E.D. FORMULAS Let B area of base. h = altitude. T = total area. E = element. L = lateral area. F = volume. C = circumference of base. R = radius of base. 665. Lateral area of right circular cylinder, L = C E (?). 666. Total area of right circular cylinder, T=i + 2 B (?). /. T=27rM + 27TE 2 (?). .-. T= 21TJB(A+.B). 667. Volume of circular cylinder, VB h (?). BOOK VIII 343 668. THEOREM. Of two similar cylinders of revolution : ~i. The lateral areas are to each other as the squares of altitudes or as the squares of the radii of their bases. II. The total areas are to each other as the squares of altitudes or as the squares of the radii of their bases. III. The volumes are to each other as the cubes of their alti- tudes or as the cubes of the radii of their bases. Given : Two similar cylin- ders of revolution whose lat- eral areas L and l\ whose total areas = T and t\ whose volumes = V and v ; their their whose altitudes = H and A, and whose radii are E and r. To Prove : I. L: l = H 2 : A 2 =,R 2 : r 2 . II. T: t = H*: A 2 =E 2 - r 2 . III. Vi v = H*: A3 = jR 3. ,* Proof: The generating rectangles of the cylinders similar (?) (651). ' are Hence, H + R : h+ r= H : h = R : r (?) (301). L 2 EH R H H H II I 27rrh -.- = -.- = ?_. (Explain.) ^^ } III - = 7rK2ff R(g+ff) _ R H+R = H H r(A-f-r) ~ r h + r h h 7T2 7?2 r 2 h r 3 Q.E.D. 344 SOLID GEOMETRY ORIGINAL EXERCISES (NUMERICAL) TT = 3f 1 bu. = 2150.42 cu. in. 1 gal. = 231 cu. in. In a cylinder of revolution, 1. If R = 5 in., h = 14 in., find L; T\ V. 2. If 12 = 7m., A = 10m., find Z; T 7 ; F. 3. If 12 = 4f ft., A = 18 ft., find Z; T\ V. 4. If '12 = 6 in., L = 792 sq. in., find h ; 7 7 ; F. 5. If R = 4, !T = 352, find A; Z; I 7 . 6. If 12 = 2, F= 22, find A; L; T. 7. If A = 5.6, Z = 352, find 12; 7"; F. 8. If A = 9, !F = 440, find 12; L; F. 9. If h = 9, F= 66, find 12; Z; T 7 . 10. If Z = 440, T 7 = 1672, find 12 ; h ; F. 11. If L = 198, F = 594, find 12; A; T 7 . 12. How many square inches of tin will be required to make a cylin- drical pail 10 in. in diameter and a foot in height, without any lid? How many gallons will it contain? 13. The diameter of a well is 5| ft. and the water is 14 ft. deep. How many gallons of water in the well? 14. In a cylinder of revolution generated by a rectangle 30 x 14 in. revolving about its shorter side as an axis, find L ; T ; F. 15. In a cylinder of revolution generated by the rectangle of No. 14, revolving about its longer side as an axis, find L ; T ; F. 16. A Cylindrical vessel 9 in. high, closed at one end, required 361f sq. in. of tin in its construction. Find its radius. 17. A cylindrical pail 12 in. high holds exactly two gallons. Find 12. 18. How many cu. ft. of metal in a hollow cylindrical tube 42 ft. long, whose outer and inner diameters are 10 in. and 6 in., respectively? 19. A tunnel whose cross section is a semicircle 18 ft. high is one mi. long. How many cu. yd. of material were removed in the excavation ? 20. An irregular stone is placed in a cylindrical vessel a in. in diameter and partly full of water. The water rises b in. Find volume of stone. 21. A rod of copper 18 ft. long and 2 in. square at the end is melted and formed into a wire in. in diameter. Find the length of the wire. 22. If a cylindrical bushel measure has for its altitude the diameter of the base, find the altitude. BOOK VIII 345 CONES l A conical surface is a surface generated by a moving straight line that continually intersects a given curve in a plane, and passes through a fixed point not in this plane. The generating line is. the generatrix. The directing curve is the directrix. The fixed point is the vertex of the conical surface. An element of a conical surface is the generating line in any position. ob*-l m I m* 3^- CONICAL RIGHT OBLIQUE CONE OF FRUSTUM SURFACE CIRCULAR CONE REVOLUTION OF A CONE CONE 670. A cone is a solid bounded by a conical surface and a plane cutting all the elements. The base of a cone is its plane surface. The lateral area of a cone is the area of the conical surface. The total area of a cone is the sum of the lateral area and the area of the base. The altitude of a cone is the perpendicular distance from the vertex to the plane of the base. 671. A circular cone is a cone whose base is a circle. The axis of a circular cone is the line drawn from the vertex to the center of the base. A right circular cone is a circular cone whose axis is per- pendicular to the plane of the base. 346 SOLID GEOMETRY An oblique circular cone is one whose axis is oblique to the plane of the base. A cone of revolution is a cone generated by the revolution of a right triangle about one of the legs as an axis. Similar cones of revolution are cones generated by the revolution of similar right triangles revolving about homol- ogous sides. The slant height of a cone of revolution is any one of its elements. 672. A frustum of a cone is the portion of a cone between the base and a plane parallel to the base. The altitude of a frustum of a cone is the perpendicular distance between the planes of its bases. The slant height of a frustum of a cone is the portion of an element included between the bases. The lateral area of a frustum is the area of its curved surface. The total area of a frustum is the sum of the lateral area and the area of the bases. 673. A plane is tangent to a cone if it contains one ele- ment of the conical surface and only one, however far it may be extended. A pyramid is inscribed in a cone if its base is inscribed in the base of the cone, and its vertex is the vertex of the cone. A pyramid is circumscribed about a cone if its base is cir- cumscribed about the base of the cone, and its vertex is the vertex of the cone. The frustum of a pyramid is inscribed in, or circumscribed about, the frustum of a cone if the bases of the pyramid are inscribed in, or circumscribed about, the bases of the cone. The midsection of a frustum of a cone is the section made by a plane parallel to the bases, and midway between them. BOOK VIII 347 PRELIMINARY THEOREMS 674. THEOREM. The elements of a right circular cone are all equal. (See 520, II.) 675. THEOREM. A right circular cone is a cone ot revolution. The altitude, any element, and the radius of the base of a right circular cone form a right triangle, and in the same cone all such triangles are equal (53). 676. THEOREM. The altitude of a cone of revolution is the axis of the cone. 677. THEOREM. A straight line drawn from the vertex of a cone to any point in the perimeter of the base is an element. (See 39.) 678. THEOREM. The lateral edges of a pyramid inscribed in a cone are elements of the cone. (See 677.) 679. THEOREM. The lateral faces of a pyramid circumscribed about a cone are tangent to the conical surface. (Explain.) 680. THEOREM. The slant height of a regular pyramid circum- scribed about a right circular cone is the same as the slant height of the cone. 681. THEOREM. The slant height of the frustum of a regular pyra- mid circumscribed about the frustum of a right circular cone is the same as the slant height of the frustum of the cone. (See 680.) 682. THEOREM. The radius of the mid-section of a frustum of a right circular cone is equal to half the sum of the radii of the bases. Proof : The radius of the mid-section is the median of a trapezoid whose bases are the radii of the bases of the frustum. That is, m = (B + r). (See 149.) Ex. 1. What two lines determine the size of aright circular cone? What two lines determine the total area? Ex. 2. Find the slant height of a right circular cone whose altitude is 8 and whose radius is 6. V 348 SOLID GEOMETRY THEOREMS AND DEMONSTRATIONS 683. THEOREM. Any section of a cone made by a plane passing through the vertex is a triangle. Given: Cone O-AB; plane OCD. To Prove : Section OCD is a A. Proof: Draw straight lines OC, OD, in plane OCD. They are ele- ments (?) (677). /. OC and OD compose the inter- section of the plane and the conical surface (?) (482). Also CD is a straight line (?). .'. OCD is a A (?) (23). Q.E.D. 684. THEOREM. Any section of a circular cone made by a plane parallel to the base is a circle. Given: Cone o AB; circle C its base; section A'B' 11 to base. To Prove : A'B' also a O. Proof: Draw the axis OC and pass planes OCD, OCE intersecting the base in CD, CE respectively, and the section in C'D', C f E r . In A OCD and OCE, D'C' is II to DC; C'E' is II to CE (?) (500). /. A OC'D r is similar to A OCD; A OC'E' is similar to A OCE (316). oc f OC C'D' CD ^- = 11^- (323, 3). /.^- = OC CE v CD CE But CD= CE (?) (200). .-. C'D' = C'E' (Ax. 3). That is, all points on the perimeter of A'B' are equally distant from C ! . .'. perimeter of section A'B' is a circumference (?) (192). Hence, A'B' is a O (?) (193). Q.E.D. BOOK VIII 349 685. THEOREM. If a regular pyramid be inscribed in, or circum scribed about, a right circular cone and the number of sides of the base be indefinitely increased, the lateral area of the cone is the limit of the lateral area of the pyramid. (See Fig. A.) Demonstration is similar to that of 661. FIG. A FIG. B 686. THEOREM. If a pyramid having a regular polygon for a base be inscribed in, or circumscribed about, any circular cone and the number of sides of its base be indefinitely increased, the volume of the cone is the limit of the volume of the pyramid. (See Fig. B.) Demonstration is similar to that of 662. 687. THEOREM. If a frustum of a regular pyramid be inscribed in, or circumscribed about, the frustum of a right circular cone and the number of sides of the bases be indefinitely increased, the lateral area of the frustum of the cone is the limit of the lateral area of the frustum of the pyramid. 688. THEOREM. If the frustum of a pyramid having regular polygons for its bases be inscribed in, or circumscribed about, a frustum of any circular cone and the number of sides of the bases of the frustum be indefinitely increased, the volume of the frustum of the cone is the limit of the volume of the frustum of the pyramid. 350 SOLID GEOMETRY 689. THEOREM, The lateral area of a right circular cone is equal to half the product of the circumference of the base by the slant height. Given : Right circular cone O-AD, the circumference of whose base = C and whose slant height = s. To Prove : Lateral area = J C s. Proof : Circumscribe a regu- lar pyramid and denote the lateral area by L f and the perimeter of the base by P. Slant height OA =s (?) (680). Then ' = J P (?) (617). Now indefinitely increase the number of sides of the base of the pyramid and, L r will approach L as a limit (?) (685). P will approach C as a limit (?) (440, I). P-s will approach J C*s as a limit (?). Hence, L = J c s (?) (242). Q.E.D. 690. THEOREM. The lateral area of the frustum of a right circu- lar cone is equal to half the sum of the circumferences of the bases multiplied by the slant height. Given: Frustum of right circular cone, whose lateral area is L\ whose slant height is s\ and the circumferences of whose bases are C and c. To Prove: L = J (c + c) s. Proof : Circumscribe a frustum of a regular pyramid and denote its lateral area by L', the perimeters of its bases by P and p. BOOK VIII 351 The slant height of frustum of pyramid = s (?) (681). Now, i' = J(P +;>)(?) (618). Indefinitely increase the number of the sides of the bases of the frustum of the pyramid, and L f will approacli L as a limit (?), also P will approach c as a limit (?), and p will approach c as a limit (?), and J (P -\-p) -s will approach J (c + ]. Hence, F = J- A[> + fl + ViTT] (?) (242). Q.E.D. Ex. 1. Could you prove the theorem of 690 by inscribing a frustum of a pyramid ? Could you prove the theorem of 692 by circumscribing a frustum of a pyramid ? Give reasons for your answer. Ex. 2. Could you prove the theorem of 689 by inscribing a pyramid? Could you prove the theorem of 691 by inscribing a pyramid V Give reasons. Ex. 3. The frustum of a circular cone is 15 in. high and the bases are circles whose radii are 3 in. and 5 in. Find the volume. Ex. 4. The frustum of a right circular cone has a slant height of 9 ft. and the radii are 5 ft. and 7 ft. Find the lateral area and the total area. What is the length of the altitude of this frustum? BOOK VIII 353 FORMULAS Let B = area of base. m = radius of mid-section of frustum. b = area of less base. R = radius of base. C circumference of base. r radius of less base. c = circumf. of less base. s = slant height. h = altitude. T = total area. L = lateral area. V = volume. 693. Lateral area of right circular cone, 694. Total area of right circular cone, T = TTKS + TrR 2 (?). + JB). 695. Volume of circular cone, V= J B h = ^ TrR 2 h (?). 696. Lateral area of frustum of right circular cone, /. L =ir(R + r)s. Also, L = 7r(2 m)s = 2 irms (682). 697. Total area of frustum of right circular cone, T= TT(B + r)9 .-. T = ir[(JB 698. Volume of frustum of circular cone, + Trr 2 + Ex. 1. State in words each of the final formulas of the paragraphs of this page. Ex. 2. Find the radius of a circle having the same area as the lateral area of a cone of revolution whose radius is 4 and slant height, 9. 354 SOLID GEOMETRY 1 699. THEOREM. Of two similar cones of revolution : I. The lateral areas are to each other as the squares of their alti- tudes, or as the squares of their radii, or as the squares of their slant heights. II. The total areas are to each other as the squares of their alti- tudes, or as the squares of their radii, or as the squares of their slant heights. III. The volumes are to each other as the cubes of their altitudes, or as the cubes of their radii, or as the cubes of their slant heights. Given: Two similar cones of revolution, whose respec- tive lateral areas are L and Z, total areas are T and , volumes are V and v, alti- tudes are H and A, radii are R and r, slant heights are S and s. To Prove y _L/ a K o TT L _ III. Proof: The generating Hence, L _H 2 _&_ s 2 l = h 2 r 2 s 2 ' T H 2 R 2 _ s 2 7" h 2 r 2 ' V H B B 3 s 3 V h s r 3 s 3 o Tp cirnilfiT (?\ H R 8 (9} . dl \5 oliUlitll ^' } " h r s U === (?) (301 and Ax. 1). r+s r s h L^TTRS^^R S^H H == n_ = R_ = S^ (Explain.) I TTTS r s h h h 2 r 2 s 2 II. - r+s h~ h 2 r 2 s 2 (Explain.) 7rr 2 h r 2 h h 2 h \\ BOOK VIII 355 ORIGINAL EXERCISES In a cone of revolution, OA (?) (520, I). .*. X lies without the sphere (717). Hence, every point of plane MN, except A, is without the sphere; that is, plane MN is tangent to the sphere (?) (703). Q.E.D. 720. THEOREM. A plane tangent to a sphere is perpendicular to the radius drawn to the point of contact. Given : Plane MN tangent to sphere O at A ; radius OA. To Prove : OA is _L to plane MN. Proof: Every point in MN, except A, is without the sphere (?) (703). Take any point X in MN and draw OX. Now, OX is > OA (?) (717). That is, OA is the shortest line from O to MN. .'. OA is J_ to MN (?) (520, I). Q.E.D. 364 SOLID GEOMETRY 721. THEOREM. A line tangent to a sphere lies in the plane tan- gent to the sphere at the same point. Proof: The line is _L to the radius drawn to the point of contact (?) (216). The plane also is -L to the radius (?). /. the line is in the plane (502). 722. THEOREM. At a point on the surface of a sphere there can be only one tangent plane. (Explain.) N 723. THEOREM. All points in the circumference of a circle of a sphere are equally distant from either pole ; that is, the polar distances of all points in the circumference of a circle are equal. Given : P and z,, the poles of O C on sphere O, and great (D PAL, PEL. To Prove : Arc PA = arc PB ; arc AL = arc BL. Proof : Draw the axis PL meeting plane of O C at C. Draw AC, AP, BC, BP. PC is _L to plane DAB (?) (Def . of axis, 702). .*. chord PA = chord PB (?). Hence, arc PA = arc PB (?). Likewise, arc AL = arc BL. Q.E.D. 724. COR. The polar distance of a great circle is a quadrant. 725. THEOREM. If a point on the surface of a sphere is at the distance of a quadrant from two other points on the surface, not the ends of a diameter, it is the pole of the great circle containing these two points. Given : P, a point, and R and 2V, two other points, all on the surface of sphere O ; arcs PR and P7V, quadrants ; great circle ARNB. BOOK IX To Prove : P is the pole of O AENB. Proof : Draw the radii OP, OB, O-ZV. A PON and FOR are rt. A (245). /. PO is _L to plane AB (?) (501). .-. PO is the axis of O AENB (702). .-. P is the pole of OAENB (702). Q.E.D, 365 72$. THEOREM. A spherical angle is measured by the arc of a great circle having the vertex of the angle as a pole and intercepted by the sides of the angle. Given: Spherical /. A VB; arc AB of great O whose pole is F, on sphere O. To Prove : /. A VB is measured by arc AB. Proof : Draw radii OA, OB, OF, and at F draw VC tangent to O VA, and FD tangent to O VB. VB is a quadrant (724) . OF is _L to FD (?), and OF is _L to OB (?) (245). Likewise, OF is J_ to VC and to OA (?). .-. FD is II to OB, and VC is II to OA (?) (93). .\^ CVD = Z.AOB (?) (515). And Z. CVD is the spherical Z. AVB (706). But /.AOB is measured by arc AB (245). .'./. CVD is measured by arc AB (Ax. 6). That is, /. AVB is measured by arc AB (Ax. 6). Q.E.D. 727. COR. All arcs of great circles containing the pole of a great cir- cle are perpendicular to the circumference of the great circle. (See 540.) 728. COR. A spherical angle is equal to the plane angle of the dihe- dral angle formed by the planes of the sides of the angle. (See 530.) 729. COR. If two great circles are perpendicular to each other, each of their circumferences contains the pole of the other. (See 543; 702.) 366 SOLID GEOMETRY 730. THEOREM. A sphere may be inscribed in any tetrahedron. Given : Tetrahedron A-BCD. To Prove: (?). Proof : Pass plane OAB bisecting dih. Z AB, and plane OBC bisecting dih. Z BC, and plane OCD bisecting dih. Z CD, the three planes meeting at point O. Point O, in plane OAB, is equally dis- tant from faces ABC and ABD (?) (551). Point O, in plane OBC, is equally dis- tant from faces ABC and BCD (?). Point O, in plane OCD, is equally distant from faces BCD and ACD (?). /. O is equally distant from all four faces (Ax. 1). Hence, using O as a center and the perpendicular distance OB as a radius, a sphere can be inscribed (?) (704). Q.E.D. 731. COR. The six planes bisecting the six dihedral angles of any tetrahedron meet in a point. 732. THEOREM. A sphere may be circumscribed about any tetra- hedron. Given: (?). To Prove: (?). Proof : Take E and F, the centers of circles circumscribed about the faces ACD and BCD, respectively. Erect EG, and FH, J_ to these faces. Find M, the midpoint of edge CD. EG is the locus of points equally distant from points A, D, C (?) (526). FH is the locus of points equally distant from points B, C, D (?). That is, all points in EG and FH are equally distant from C and D (Ax. 1). BOOK IX 367 But all points equally distant from C and D are in a plane J. to CD at M (?) (525). .'.EG and FH are in one plane and are not parallel. (NotJ_ to the same plane). That is, EG and FH must intersect at O. Hence, O is equally distant from A, JS, C, and D (?) (Ax. 1). That is, using O as a center and OA, or (XB, or OO, or OD, as a radius, a sphere may be circumscribed about the tetra- hedron A-BCD (?) (704). Q.E.D. 733. COR. Through any four points not in the same plane a sphere may be described. 734. COR. The six planes perpendicular to the edges of any tetra hedron at their midpoints meet in a point. (Explain.) 735. THEOREM. The intersection of two spherical surfaces is a circumference. Given : Two intersecting circum- ferences O and O r ; common chord CD ; line of centers XF, intersecting CD at M. To Prove : The spherical surfaces generated by the revolution of these (D, intersect in a cir- cumference. Proof: If these be revolved upon XT as an axis, they will generate spheres (?) (714). CM=MD (?) (232). Point (7, common to both CD, will generate the intersection of the spherical surfaces (?) (482). CM is always _L to XY (?) (232). .'. the curved line generated by C is in one plane (?) (502). .'. the intersection is a circumference (?) (192). Q.E.D. 368 SOLID GEOMETRY CONSTRUCTIONS 736. PROBLEM. To find the radius of a material sphere. P P' Given: A material sphere. Required: To find its radius. Construction : First, place one point of the compasses at P, and using any opening of the compasses, as AP, with the other point draw a circumference on the surface of the sphere. Upon this circumference take three points, A and B and C, and by means of the compasses measure the straight lines AB, AC, BC. Second, construct a A A'B'C', whose sides are AB, AC, BC. Circumscribe a circle about this A, and draw the radius A f D f . Third, construct a right A P'A"D", whose hypotenuse is the known line PA and whose leg is the known radius A r D f . At A" erect A n R ! _L to P'A" meeting P'D", produced, at E' . Bisect P'n f at O r . Statement : o r P r = the required radius. Q.E.F, Proof : Points P and o are equally distant from the points of the circumference ABC (Const, and 707). .*. P is in the line _L to plane ABC at the center D (?) (526). If this J_ could be drawn within the solid sphere, it would be a diameter (?) ; arid the Z PDA would be a rt. Z (?) (489). AA'B'C r (?) (58). .'.DA = J) f A' (227,201). BOOK IX 369 Also, A PAD = A P'A"D" (?) (73). .'.Zp = Z p' (?) Z P^t# = a rt. Z (?). .-. A PAR = A P'A"R' (?) (74). .\PR = P f Rf (?). Hence, OP = o'p' (Ax. 3). That is, O'P' = the radius. Q.E.D. 737, PROBLEM. To find the chord of a quadrant of a material sphere. Given: (?). Required: (?). Construction : Find the radius of the sphere (by 736). Using this radius OP and any center O, describe a semicircu in- ference PMR. Erect radius OM J_ to diameter P#, and draw PM. Statement: Arc PM is a quadrant of great circle of the given sphere and chord PM is the required chord. Q.E.F. Proof : Arc PM is a quadrant (?). ; 738. PROBLEM. To describe the circumference of a great circle through two given points on the surface of a sphere. Given : The points A and B on the surface of the sphere O. Required: (?). Construction : Find the chord of a quadrant of the given sphere (by 737). Place one point of the compasses at -4, and using the chord just found as an opening, describe an arc on the surface of the sphere. Similarly, place one point of the compasses at .B, and using the same opening, describe an arc, meeting the former arc at P. Now, place one point of the compasses at P and describe the circumference BAG, using the chord as before. Statement: (?). Proof: (Use 725.) 370 SOLID GEOMETRY 739. PROBLEM. To draw an arc of a great circle through a given point on the surface of a sphere and perpendicular to a given great circle. Given : Point A on sphere O, and great circle BC, whose pole is P. Required : To draw through A an arc of a great circle J_ to the great circle BC. Construction : Place one point of the compasses at A, and with an opening equal to the chord of a quadrant of the given sphere describe an arc of a great circle intersecting the given great circle at D. Now, place one point of the compasses at D and similarly draw arc of great circle PAE. Draw PD, the arc of great circle (by 738). Statement: Arc PAE is J_ to circumference BEDC. Q.E.F. Proof: ED, EP, and PD are quadrants (?) (724). .-. E is the pole of arc PD (?) (725). Hence, Z. PED is measured by quadrant PD (?) (726). /. ^ PED is a right angle (247). That is, arc PAE is _L to circumference BEDC. Q.E.D. SPHERICAL TRIANGLES 740. A spherical triangle is a portion of the surface of a sphere bounded by three arcs of great circles. The bounding arcs are the sides of the triangle. The intersections of the sides are the vertices of the triangle. The angles formed by the sides are the angles of the triangle. Spherical triangles are equilateral, equiangular, isosceles, scalene, acute, right, obtuse, under the same conditions as in plane triangles. BOOK ix 371 741. A birectangular spherical triangle is a spherical triangle, two of whose angles are right angles. A trirectangular spherical triangle is a spherical triangle all of whose angles are right angles. The unit usually employed in measuring the sides of a spherical triangle is the degree. It is obvious that the circumferences of three great circles divide the surface of a sphere into eight spherical triangles. 742. Two spherical triangles are mutually equilateral if their sides are equal each to each; and they are mutually equiangular if their angles are equal each to each. MUTUALLY EQUILATBRAL SPHERICAL TRIANGLES MUTUALLY EQUIANGULAR SPHERICAL TRIANGLES POLAR TRIANGLES 743. If three great circles are described, having as their poles the vertices of a spherical triangle, one of the eight triangles thus formed is the polar triangle of the first. The polar triangle is the one whose vertices are nearest the vertices of the original triangle. 872 SOLID GEOMETRY 744. If the diameters of a sphere are drawn to the vertices of a spherical triangle, the original triangle, and the triangle whose vertices are the opposite ends of these diame- ters, are symmetrical spherical triangles. 745. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. Two spherical polygons are equal if they can be made to coincide. The diagonal of a spherical polygon is the arc of a great circle connecting two vertices not in the same side. Only convex spherical polygons are considered in this book. PRELIMINARY THEOREMS 746. THEOREM. The planes of the sides of a spherical triangle form a trihedral angle : I. Whose vertex is the center of the sphere (?) (702). II. Each of whose face angles is measured by the intercepted side of the triangle (?)(245). III. Each of whose dihedral angles is equal to the corresponding angle of the triangle (?) (728). 747. THEOREM. A pair of symmetrical spherical triangles are mutu- ally equilateral and mutually equiangular. (See 51, 206, 537, 728.) 748. THEOREM. The homologous parts of a pair of symmetrical spherical triangles are arranged in reverse order. Proof : If the eye is at the center of the sphere, the order of the vertices A, B, c is the same in direction as the motion of the hands of a clock. But the order of A f , B r , C f is in the opposite direction. (See 556, Note.) Hence, the parts are arranged in reverse order. Q.E.D. 749. THEOREM. The homologous parts of two symmetrical spheri- cal triangles are equal. (747.) BOOK IX 373 750. THEOREM. Two symmetrical isos- celes spherical triangles can be superposed and are equal. Proof : The method of superposition, as in the case of plane triangles. k751. .ngle is THEOREMS AND DEMONSTRATIONS THEOREM. One side of a spherical tri- angle is less than the sum of the other two. Given: (?). To Prove : AB 180; II. Z^t + Z + Zc< 540. Proof: I. Construct A A'B'C', the polar triangle of A ABC. Z.A + a' = 180, Z B + b f = 180, Z c + a' = 180 (?) (758). Adding, Z^i + Zz? + ZC'+ a' + &'+ 180 (Ax. 9). ' = 540 (Ax. 2). '> 0(?)(74Q). Z^ + ZC < 540 (Ax. 9). Q.E.D. 761. COR. The sum of the angles of a spherical triangle is greater than two right angles and less than six right angles. 762. COR. A spherical triangle may have one, two, or three obtuse angles. V763.) THEOREM. Two symmetrical spherical triangles are equiva- lent.^ Given: Two symmetrical spheri- cal A ABC and A'B'C'. To Prove: A ABC o A A r B f C r . Proof: Suppose p is the pole of the circle containing A, 5, C. Draw the diameters AA r , BB f , CC 1 , pp', and the arcs of great circles, PA, PB, PC, P'A', P'B', P'C'. Z POA = Z P'OA' (?). /. arc PA = arc P'A' (?) (206). BOOK IX 377 Likewise, arc PB = arc P'B' and arc PC = arc P ! C f . But PA = PB = PC(?) (723). .'.p'A f = P f B'=p'c' (Ax. 1). Hence, A APB = AA f P r B f } A APC = A A'P'C' | (?) (750). Adding, A ABC ^ A A r B r C f (Ax. 2). Q.E.D. NOTE. If the pole P should be without the triangle ABC, one of the pairs of equal triangles would be without the original triangles and would be subtracted from the sum of the others to obtain triangles ABC and A'B'C'. 764. THEOREM. Provided two spherical triangles on the same sphere (or on equal spheres) have their parts arranged in the same order, they are equal: I. If two sides and the included angle of one are equal respec- tively to two sides and the included angle of the other. II. If a side and the two angles adjoining it of one are equal respectively to a side and the two angles adjoining it in the other. III. If three sides of the one are equal respectively to three sides of the other. Given: (?). A R To Prove: (?). Proof: I and II. Superposition as in plane triangles. III. Draw radii of the sphere to all the vertices of the triangles. The face A of the trih. Z O = the face A of the trih. Z N, respectively. (Explain.) Hence, trih. Z o = trih. Z jy (?) (561). .*. dih. Z o^ = dih. Z NR-, dih. Z o# = dih. Z NS; etc. .'. the A are mutually equiangular (?) (746, III). Hence, the A can be made to coincide. .'. AABC= A BST (?) (28). Q.E.D. 378 SOLID GEOMETRY 765. THEOREM. Provided two spherical triangles on the same sphere (or on equal spheres) have their parts arranged in reverse order, they are symmetrical: I. If two sides and the included angle of one are equal respectively to two sides and the included angle of the other. II. If a side and the two angles adjoining it of one are equal respec- tively to a side and the two angles adjoining it of the other. III. If three sides of one are equal respectively to three sides of the other. Proof: In each of these cases construct a third spherical tri- angle, R r s r T r , symmetrical to the A EST. Then A R'S'T* will have its parts equal to the parts of A ABC and arranged in the same order. (Explain.) .'. A E'S'T' = A ABC (?) (764). Hence, A EST is symmetrical to A ABC (Ax. 6). Q.E.D. 766. COR. Two mutually equilateral spherical triangles are mutu- ally equiangular and are equal or symmetrical. When are they equal? When are they symmetrical? 767. THEOREM. Two mutually equiangular spherical triangles on the same sphere (or on equal spheres) are mutually equilateral, and are equal or symmetrical. Given : A A and A', mutually /'*""\ X'""x equiangular. To Prove : A A and A' mutu- [ ally equilateral, and equal or \ symmetrical. Proof : Construct A E and E', the polar A of A and A r . The sides of E are supplements of the A of A 1 x 9 v and the sides of E' are supplements of the A of A 1 \ BOOK IX 379 But the A of A are = respectively to the A of A 1 (Hyp.). .'. A E and E' are mutually equilateral (?) (49). Hence, A E and tf are mutually equiangular (?) (766). Again, A A and A' are the polar A of E and ^ (?) (757). .*. the sides of A are supplements of the A of E \ s^ and the sides of A 1 are supplements of the A of E' } Hence, A A and A' are mutually equilateral (?). Also they are equal (when?); or symmetrical (when?). Q.E.D. 768. THEOREM. The angles opposite the equal sides of an isosceles spherical triangle are equal. Given : (?). To Prove : Z B = Z c. Proof : Suppose X the midpoint of BC. Draw AX, the arc of a great circle. Now the two spherical A ABX and ACX are mutually equilateral. (Explain.) .'. they are mutually equiangular and symmetrical (?) (766). = Z c(?) (749). 769. COR. The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the midpoint of the base bisects the vertex- angle and is perpendicular to the base. (See 749.) 770. THEOREM. If two angles of a spherical triangle are equal, the sides opposite are equal. Given: (?). To Prove: (?). Proof: Construct A A'B'C', the polar triangle of A ABC. Theii,.4/Ir is the supplement ofZ(7 j .^ : C' and A' a' is the supplement of Z B /. A'B' = A'C 1 (?) (49). Therefore, Z B' = Z. C f (?) (768). Again, A ABC is the polar triangle of A A'B'C' (?) (757). And AB is the supplement of Z c', and AC of Z B' (?). .'.AB = AC (?). Q.E.D 380 SOLID GEOMETRY 771. THEOREM. If two angles of a spherical triangle are unequal, the sides opposite are unequal and the greater side is opposite the greater angle. Given : A ABC ; Z ABC > Z C. A To Prove: AC > AB. Proof : Suppose BR, the arc of a great circle, drawn, making Z GBR = Z C and ~. meeting AC at R. Now AR + BR > AB (?) (751). But BR = CR (?) (770). .'. AR + CR > AB (Ax. 6). That is, AC > AB. Q.E.D. 772. THEOREM. If two sides of a spherical triangle are unequal, the angles opposite are unequal and the greater angle is opposite the greater side. To Prove : Z ABC > Z C. Proof : Z ABC is either < Z C or = Z C or > Z C. Continue by method of exclusion (88). 773. THEOREM. If the circumferences of two circles on a sphere contain a point on the arc of a great circle that joins their poles, they have no other point in common. Given : Point P on the arc AB of a great circle of a sphere, and P common to two circumferences whose poles are A and B. To Prove: P is the only point common to these circumferences. Proof: Suppose X is another common point. Draw arcs of great CD, AX and BX. Then AX + BX > APB (?) (751). But AX = AP (?) (723). Subtracting, BX > BP (?) (Ax. 7). That is, X is without the O B and cannot be in both the circumferences. Q.E.D. BOOK IX 381 774. COR. If two circumferences on a sphere touch each other at one point, and only one, the arc of a great circle joining their poles con- tains their common point. 775. THEOREM. The shortest line that can be drawn on the sur- face of a sphere, between two points on the surface, is the less arc of the great circle containing the two points. Given: Points A and B, and AB the arc of a great O joining them; line ADEB, any other line on the surface of the sphere, between A and B. To Prove : Arc AB is the short- est line on the surface, that can be drawn connecting A and B. Proof : Take on arc AB any point C, and describe two circum- ferences through c, having A and B as their poles, and inter- secting ADEB at Z> and E. Point C is the only point common to these two (?) (773). No matter what kind of line AD is, a line of equal length can be drawn from A to C, on the surface ; and a line can be drawn from B to C equal in length to BE. [Imagine ^4 D revolved on the surface of the sphere, using A as a pivot, and D will move along the circumference to coincide with point C. Similarly with BE.'] There is now a line from A to JB, through C, < ADEB. That is, whatever the nature of ADEB, there is a, shorter line from A to -B, which contains C, any point of arc AB. Thus the shortest line contains all the points of AB and therefore is the line An. Q.E.D. NOTE. This theorem justifies the definition of the " distance " be- tween two points, etc., in 705. 5 P \ 382 SOLID GEOMETRY 776. THEOREM. Any point in the arc of a great circle that bisects a spherical angle is equally distant from the sides of the angle. Given : Spherical Z EAG\ arc AT bisect- ing it ; any point P, of arc AT\ PD and P#, JjL arcs of great (D J_ to AB and AC\ respec- ..-* / >N tively. To Prove : Arc PD = arc P^. \ Proof : The arcs PD and PE, if pro- c . longed, will pass through R and s, re- spectively, the poles of AB and AC (?) (729). Draw arcs AR and AS. Now, Z DAR = Z.EAS. [Each is a right Z ; (727).] Z D^P = Z EAP (Hyp.). Subtracting, Z RAP = Z SAP (Ax. 2). Also RA = 8A (?) (724$ and AP = ^P. .-.A R^P is symmetrical to A SAP (?) (765, I). .\RP = SP (?). But ED = SE. (Each is a quadrant.) .*. PD = PE (Ax. 2). Q.E.D. 777. THEOREM. Any point on the surface of a sphere and equally distant from the sides of a spherical angle is in the arc of a great circle that bisects the angle. (The proof is similar to the proof of 776.) ORIGINAL EXERCISES 1. Vertical spherical angles are equal. 2. If two spherical triangles, on the same or equal spheres, are mutu- ally equilateral, their polar triangles are mutually equiangular. 3. The polar triangle of an isosceles spherical triangle is isosceles. 4. The polar triangle of a birectangular spherical triangle is birec- tangular. 5. If two dihedral angles of a trihedral angle are equal, the opposite face angles also are equal. Proof : Construct a sphere having the vertex as center, etc. 6. If two face angles of a trihedral a-ngle are equal, the opposite di- hedral angles also are equal. BOOK IX 383 7. A trirectangular spherical triangle is its own polar triangle. 8. Two symmetrical spherical polygons are equivalent. 9. Any side of a spherical polygon is less than the sum of the other sides. [Draw diagonals from a vertex.] 10. If the three face angles of a trihedral angle are equal, the three dihedral angles also are equal. 11. State and prove the converse of No. 10. 12. A straight line cannot meet a spherical surface in more than two points. 13. If two dihedral angles of a trihedral angle are unequal, the oppo- site face angles are unequal, and the greater face angle is opposite the greater dihedral angle. 14. State and prove the converse of No. 13. 15. Two lines tangent to a sphere at a point determine a plane tan- gent to a sphere at the same point. 16. All the tangent lines drawn to a sphere from an external point are equal. 17. The volume of any tetrahedron is equal to one third the product of its total surface by the radius of the inscribed sphere. 18. Every point in the circumference of a great circle that is perpen- dicular to an arc at its midpoint is equally distant from the ends of the arc. 19. The points of contact of all lines tangent to a sphere from an ex- ternal point lie in the circumference of a circle. 20. The arcs of great circles perpendicular to the sides of a spherical triangle at their midpoints meet in a point equally distant from the vertices. 21. If the opposite sides of a spherical quadrilateral are equal, the opposite angles are equal. 22. If the opposite sides of a spherical quadrilateral are equal, the diagonals bisect each other. 23. If the diagonals of a spherical quadrilateral bisect each other, the opposite sides are equal. 24. The exterior angle of a spherical triangle is less than the sum of the opposite interior angles. 25. The sum of the angles of a spherical quadrilateral is more than four right angles and less than eight right angles. 384 SOLID GEOMETRY 26. If two spheres are tangent to each other, the straight line joining their centers passes through the point of contact. 27. The sum of the angles of a spherical polygon is more than 2 n 4 right angles and less than 2 n right angles. 28. The arcs of great circles bisecting the angles of a spherical tri- angle meet in a point. 29. A circle may be inscribed in any spherical triangle. 30. If a tangent line and a secant be drawn to a sphere from an ex- ternal point, the tangent is a mean proportional between the whole secant and the external segment. 31. The product of any secant that can be drawn to a sphere from an external point, by its external segment, is constant for all secants drawn through the same point. 32. If two spherical surfaces intersect and a plane be passed contain- ing their intersection, tangents from any point in this plane to the two spherical surfaces are equal. 33. Find the distance from the center of a sphere whose radius is 15 to the plane of a small circle whose radius is 8. 34. The polar distance of a small circle is 60 and the radius of the sphere is 12 in. Find the radius of the circle. 35. The total surface of a tetrahedron is 90 sq. m., and the radius of the inscribed sphere is 4 m. Find the volume of the tetrahedron. , 36. Find the radius of the sphere inscribed in a tetrahedron whose volume is 250 and total surface is 150. 37. Find the total surface of a tetrahedron whose volume is 320, if the radius of the inscribed sphere is 8. 38. Find the radius of the sphere inscribed in a regular tetrahedron whose edges are each 10 in. 39. Find the radius of the sphere circumscribed about a regular tetrahedron whose edges are each 18 in. 40. Find the radii of the spheres inscribed in and circumscribed about a cube whose edges are each 10 in. 41. The sides of a spherical triangle are 60, 80, 110. Find the angles of its polar triangle. 42. The angles of a spherical triangle are 74, 119, 87. Find the sides of its polar triangle. 43. The chord of the polar distance of the circle of a sphere is 12, and the radius of the sphere is 9. Find the radius of the circle. BOOK IX 385 44. The polar distance of a circle is 60 and the diameter of the circle is 8. Find the diameter of the sphere. [Denote by,#, each side of an equilateral triangle whose altitude is 4.] 45. The radii of two spherical surfaces are 11 in. and 13 in., and their centers are 20 in. apart. Find the radius of the circle of their intersec- tion. Find also the distances from the centers of the spheres to the center of this circle. 46. The radii of two spherical surfaces are 20 m. and 37 in., and the distance between their centers is 19 m. What is the length of the diame- ter of their intersection ? 47 To bisect an arc of a great circle. 48. To draw an arc of a great circle perpendicular to a given arc of a great circle through a given point in the arc. 49. To bisect a spherical angle. 50. To bisect an arc of a small circle. 51. To circumscribe a circle about a given spherical triangle. 52. To construct a spherical angle equal to a given spherical angle at a given point on the same sphere. 53. To construct a spherical triangle having the three sides given. 54. To construct a spherical triangle having the three angles given. 55. To construct a plane tangent to a sphere at a given point on the surface. 56. To construct a spherical surface having the radius given and con- taining three given points. 57. To construct a spherical surface that shall have a given radius, touch a given plane, and contain two given points. 58. To construct a spherical surface that shall have a given radius, shall be tangent to a given sphere, and contain two given points. 59. To construct a spherical surface that shall contain four given points. 60. To construct a plane that shall contain a given line and be tan- gent to a given sphere. 61. To construct a plane tangent to a given sphere and parallel to a given plane. 62. What is the locus of points on the surface of a sphere : (a) Equally distant from two given points on the surface? (6) Equally distant from two given points not on the surface? (c) Equally distant from two given intersecting arcs of great circles ? ROBBINS' SOLID GEOM. 25 386 SOLID GEOMETRY AREAS AND VOLUMES 778. A lune is a portion of the surface of a sphere bounded by two semicircumferences of great circles. The points of intersection of the sides of a lune are the vertices of the lune. The angles made at the vertices by the sides are the angles of the lune. SPHERICAL SPHERICAL SPHERICAL SECTOKS PYRAMID SEGMENT SPHERICAL CONE 779. A zone is a portion of the surface of a sphere bounded by the circumferences of two circles whose planes are parallel. The bases of a zone are the circumferences bounding it. The altitude of a zone is the perpendicular distance between the planes of its bases. If one of the planes is tangent to the sphere the zone is a zone of one base. 780. A spherical degree is the one seven-hundred-and twentieth part of the surface of a sphere. If the surface of a sphere is divided into 720 equal parts, each part is a spheri- cal degree. The size of a spherical degree depends on the size of the sphere. It may be easily conceived to be half a lune whose angle is a degree, that is, a birectangular spherical triangle whose third angle is 1. How many spherical degrees in a trirectangular spherical triangle ? BOOK IX 387 781. The spherical excess of a spherical triangle is the sum of its angles less 180. That is, E = A + B + C 180. 782. A spherical pyramid is a portion of a sphere bounded by a spherical polygon and the planes of its sides. The vertex of a spherical pyramid is the center of the sphere. The base of a spherical pyramid is the spherical polygon. 783. A spherical sector is the solid generated by the revo- lution of the sector of a circle about any diameter of the circle as an axis. The base of the spherical sector is the zone generated by the arc of the circular sector. A spherical cone is a spherical sector whose base is a zone of one base. 784. A spherical segment is a portion of a sphere included between two parallel planes that intersect the sphere. The bases of a spherical segment are the circles made by the parallel planes. The altitude of a spherical segment is the perpendicular distance between the bases. A spherical segment of one base is a segment, one of whose bounding planes is tangent to the sphere. A hemisphere is a spherical segment of one base, and that base is a great circle. A spherical wedge is a portion of a sphere bounded by a lune and the planes of its sides. Ex. 1. What is the spherical excess of a spherical triangle whose angles are 60, 70, and 100? Ex. 2. Distinguish between a zone and a spherical segment. Ex. 3. Find the area of a spherical degree on a sphere whose surface is 3600 sq. in. Ex. 4. Find the area of a spherical triangle containing 80 spherical degrees, on a sphere whose surface is 450 sq. ft. 388 SOLID GEOMETRY PRELIMINARY THEOREMS 785. THEOREM. Either angle of a lune is measured by the arc of a great circle described with the vertex of the lune as a pole, and included between the sides of the lune. (See 726.) 786. COR. The angles of a lune are equal. 787. THEOREM. Every great circle of a sphere divides the sphere into two equal hemispheres, and the surface into two equal zones. 788. THEOREM. The spherical excess of a spherical n-gon is equal to the sum of its angles less (n - 2) 180. Proof : By drawing diagonals from any vertex, the polygon is divided into n 2 A. For one&,E=S- 180 (?) (781). For another A, E f = 8' - 180 (?). Etc. for (n - 2) A. /. by adding, the excess of all the A = the sura of all their A-(n-V) 180 (Ax. 2). That is, the excess of a spherical polygon = the sura of its / /s_(n-2) 180. Q.B.D. 789. THEOREM. If a regular polygon having an even number of sides be inscribed in, or circumscribed about, a circle, and the figure be made to revolve about one of the longest diagonals of the polygon, the surface generated by the polygon will be composed of the surface of cones, cylinders, and frustums, and the surface generated by the circle will be a spherical surface. 790. THEOREM. If a regular polygon having an even number of sides be inscribed in, or circumscribed about, a circle, and the figure be made to revolve about one of the longest diagonals of the polygon, the surface generated by the perimeter of the polygon will approach the surface of the sphere generated by the circle, as a limit, if the number of sides of the polygon is indefinitely increased, 791. THEOREM. If a polyhedron be circumscribed about a sphere and the number of its faces be indefinitely increased, the surface of the polyhedron will approach the surface of the sphere as a limit, and the volume of the polyhedron will approach the volume c-f the sphera as a limit. BOOK IX 389 THEOREMS AND DEMONSTRATIONS 792. THEOREM. The area of the surface generated by a straight line revolving about an axis in its plane is equal to the product of the projection of the line upon the axis by the circumference of a circle whose radius is the line perpendicular to the revolving line at its mid- point, and terminating in the axis. Given : Line AB revolving about axis XX ! ; CD projection of AB on XX ] \ M P a = JL erected at midpoint of AB and terminating in XX 1 '; MO = radius of midsection. To Prove : Surface generated by AB = CD - 2 ?ra. Proof : I. The surface generated by AB is the surface of the frustum of a right circu- lar cone whose bases are generated by AC and BD, and the midsection, by MO. Area of surface = 2 TT MO AB (696). Now, AABH and MOP are similar (?) (321). .'.MO: AH = MP:AB (?). Hence, MO - AB = An - MP = CD a (?). .*. area of surface 2 irCD - a CD - 2 TTO, (Ax. 6). II. If AB is II to XX 1 , the surface is cylindrical and equals CD'2-n-a (?) (665). III. If AB meets XX 1 at (7, the entire surface is conical and equals irBD - AB (?) (693). Now BD='2MO (?) (142) ; and MO AB = CD a (?). .'. 7T'BD'AB = 7r-2 MO'AB = 7r-2-CD-a = CD 2 Tra (Ax. 6). That is, the area of the surface = CD 2 ira (Ax. 6). Q.E.D. Ex. 1. Find the spherical excess of a polygon whose angles are 80, 110, 140, 130, 160. Ex. 2. The spherical excess of a spherical polygon is the difference between the sum of its angles and the sum of the angles of a plane poly- gon having the same number of sides. 390 SOLID GEOMETRY 793. THEOREM. The surface of a sphere is equivalent great circles, that is, to 4 irR 2 . Given : Semicircle ACF\ diameter AF-, S = surface of sphere generated by revolv- ing the semicircle about AF as an axis ; H = radius of this sphere. To Prove : S = 4 TrR 2 . Proof : Inscribe in this semicircle half of a regular polygon having an even number of sides. Draw the apothems and denote them by a. Draw the projections of the sides of the polygon on the diameter. Now, if the figure revolve on AF as an axis, the surface AB = AP ZTTO, the surface BC = PS - 2?ra the surface CD = ST etc. etc. to four (?) (T92). Adding, the entire surface = (AP + PS + -ST + etc.) -27ra (Ax. 2). = AF-Zira (Ax. 6). Now if the number of sides of the polygon is indefinitely increased, the entire surface generated by the polygon will approach S (?) (790), and a will approach R as a limit (437). Also AF 2 Tra will approach AF' 2 TTR. .'.S = AF-27TR (?) (242). But AF= %R (?). .'.S= 4irB 2 (Ax. 6). Q.E.D. 794. THEOREM. The area of a spherical degree equals ^ (780). 795. THEOREM. The areas of the surfaces of two spheres are to each other as the squares of their radii and as the squares of their diameters. (See 793.) Ex. 1. What is the area of the surface of a sphere whose radius is 10 in. ? What is the area of a spherical degree on this sphere ? BOOK IX 391 THEOREM. The area of a zone is equal to the product of its altitude by the circumference of a great circle. Proof: The area generated by chord BC (Fig. of 793) = PS-27ra (?) (792). If the number of sides of the inscribed polygon is indefi- nitely increased, the length of chord BC will approach arc BC and the surface generated by chord BC will approach the area of a zone. Also, PS 2 TTO, will approach PS 2 TTR (?). Hence, area of zone BC = PS-^irE (?) (242). If the altitude of the zone = A, area of zone = 2 irU h. Q.E.D. 797. THEOREM. The area of a zone of one base is equal to of a^circle whose radius is the chord of the generating arc. Given: Arc AB of semicircle ABC; diameter AC; chord AB. To Prove : Area of zone generated by arc AB = TT A&*. Proof: Area of zone AB=AD-27rR (796). That is, area of zone AB = IT - AD -2R. Draw chord BC. A ABC is a rt. A (?). ,'.AD'AC=AB 2 (i?) (342). That is, AD 2 E = AB 2 (Ax. 6). Hence, area of zone AB = TT AB 2 (Ax. 6). That is, area of zone of one base = ir(chord) 2 . the area Ex. 1. On a sphere whose radius is 6 in., find the area of a zone 2| in. in height. Ex. 2. What does the formula for the area of a zone become when the altitude becomes the diameter ? Ex. 3. What does the formula for the area of a zone of one base be- come when the generating arc becomes a semicircumference? 392 SOLID GEOMETRY 798. THEOREM. The area of a lune is to the area of the surface of its sphere as the angle of the lune is to 360? Given: Lune ABCDA on sphere O; L = area of lune ; 8 = area of sphere ; great O EB whose pole is A. To Prove : L : S = Z A : 360. Proof: I. If arc BD and the circum- ference of O EB are commensurable. There exists a common unit of meas- ure. Suppose this unit contained 5 times in BD ; 32 times in the circumference, /.arc BD : circumference = 5 : 32 (?). That is, arc BD : 360 = 5 : 32. Arc BD measures Z A (726). .'./.A :360 = 5 :32(Ax. 6). Pass great circles through the several points of division of circumference EB and vertex A, dividing the surface of the sphere into 32 equal lunes. Then, L : S = 5 : 32 (Ax. 3). Hence, L : 8 = Z A : 360 (Ax. 1). Q.E.D. II. If the arc and circumference are incommensurable. The proof is similar to that found in 244, 302, 368, 539, etc. 799. THEOREM. The number of spherical degrees in the area of a lune is double the number of degrees in its angle. Proof : Let L denote the area of the lune, expressed in spherical degrees. Then, L : S = Z A : 360 (?) (798). That is, L : 720 =Z A : 360 (Ax. 6). Hence, L = 2ZA. Q.E.D. jj2 800. FORMULA. The area of a lune = x z A. Proof : L : s = Z A : 360 (?) (798), and S = 4 TTR* (?). /. L : 4 7T72 2 = Z .4 : 360 (?). . . L = ^ - x Z A. Q.E.D. NOTE. The unit of measure in this formula is the square unit. BOOK IX 801. THEOREM. Two lunes on the same or equal spheres are to each other as their angles. Proof : L : s = Z A : 360, and L' : S = Z A' : 360 (?) (798). Dividing, L : L' = Z A : Z A' (Ax. 3). Q.E.D. 802. THEOREM. Two lunes on unequal spheres, having equal angles, are to each other as the squares of the radii of the spheres. Proof : L : S = Z A : 360, and L 1 : s' = Z A : 360 (?). Hence, L : S = L' : S' (Ax. 1). .-. L : L'= S : S' = R* : E'* (292 and 795). 803. THEOREM. The number of spherical degrees in a spherical triangle is equal to the spherical excess of the triangle. Given: Spherical A ABC on sphere O ; spherical excess of the A = E. To Prove : Number of spher- ical degrees in A ABCE. Proof : Continue the sides of the A ABC to form the lunes ABA'CA, BAB'CB, CAC'BC; draw diameters AA\ BB 1 , CC . A ABC' =0= A A'B'C (?) (763). Lune CAC'BC = A ABC + AAC'B =c^ A ABC + A A'B'C (Ax. 6). Now, A ABC + AA'B'Coluue CAC'BC, Pand A ABC + A A'BC=\ui\e ABA'CA, | .(^ x - 4). and A ABC + A AB'c = lune BAB'CB j Adding, 2 A ABC -{-A ABC -{-A A'B'C + AA'BC + A AB'C =o=lune A + lime B 4- lune C (Ax. 2). But, A ABC + AA'B'C + AA'BC + A ^!B'c=360 spherical degrees (?) (713). Lune A + lune B + luneC = 2Z^i + 2ZB + 2Zr7 (799). .'. 2 A ABC +360 = 2Z^ + 2ZJ5 + 2ZC (Ax. 6). Hence, A ABC=Z A + Z + Z C- 180 = # (Ax. 2, Ax. 3, 781). Q.E.D. 394 SOLID GEOMETRY 804. FORMULA. The area of a spherical triangle = ?2Q x E. Proof: Area of one spherical degree = JT (794). There are E spherical degrees in a spherical A (?) (803). /.the area of a spherical triangle = ^ Q x E. Q.E.D. NOTES. The unit of measure in 803 is a spherical degree. The unit of measure in 804 is a square unit (sq. in., sq. ft., etc.). The formula of 804 reduces to ^f 180 The area of a spherical triangle is determined by its angles. 805. THEOREM. The number of spherical degrees in a spherical polygon is equal to its spherical excess. Given : A spherical rc-gon. To Prove: The number of spherical de- grees in this w-gon = the excess of the poly- gon. Proof: From any vertex draw diagonals, dividing the polygon into (n 2) A ; the sums of the A of these A are denoted by s, s v 2 , . . . etc. Now, the number of sph. degrees in one A = s 180 (803); the number of sph. degrees in another A = s^ 180 (?). Etc., for (rc-2) A. Adding, the number of sph. degrees in the w-gon = the sum of its A-(n- 2) 180 (Ax. 2). Excess of the w-gon= sum of its z-(rc-2) 180 (788). .'. the number of spherical degrees in a spherical polygon = the excess of the polygon (Ax. 1). Q.E.D. Ex. 1. Find the area of a spherical triangle whose angles are 80, 125, and 95, on a sphere whose radius is 6.3 in. Ex. 2. Find the area of a spherical polygon whose angles are 135, 105, 85, 155, 120, on a sphere whose radius is 15 ft. BOOK IX 806. THEOREM. The volume of a sphere = 395 Given : Sphere O ; radius = R ; surface = 8 ; volume = V. 4 7TE 3 To Prove : V = o Proof: Suppose a polyhedron cir- cumscribed about the sphere, its surface denoted by 8 1 and its vol- ume by V 1 . Suppose planes be passed through the edges of the polyhedron and the center of the sphere, thus dividing the polyhe- dron into pyramids whose vertices are all at the center, and whose common altitude is R. The volume of one such pyramid = j R - its base (?) (625). .*. volume of all the pyramids = J R - sum of all their bases (Ax. 2) ; that is, v' = J R S 1 . Indefinitely increase the number of faces of the polyhe- dron, thus indefinitely decreasing each face, and v r will approach Fas a limit) , ? , fTQI^ and s r will approach S as a limit j Hence, ^ R - S r will approach i R S as a limit (?). Therefore, F= J R - S (?) (242). But s = 4 TTR 2 (?). :.V 8 6). Q.E.D. 807. THEOREM. The volumes of two spheres are to each other as the cubes of their radii or as the cubes of their diameters. Proof : ;,.-.. (Explain.) V '6 -6 X" ^JJ') D' 3 QED 808. THEOREM. The volume of a spherical pyramid is equal to one third the product of the polygon that is its base, by the radius of the sphere. Proof : Similar to the proof of 80C. 396 SOLID GEOMETRY 809. THEOREM. The volume of a spherical wedge is to the volume of the sphere as the angle of its base is to 360. Proof: Similar to the proof of 798. 810. THEOREM. The volume of a spherical sector is equal to one third the product of the zone that is its base by the radius of the sphere. Proof : Similar to the proof of 806. 811. FORMULAS. Vol. of a spherical sector = J R- zone (810). But the zone ^TrR-h (?). Therefore, 1. The volume af spherical sector = i irK 2 h (Ax. 6). 3 2. The volume of a spherical cone = f irJB 2 h (811, 1). 8 3. The volume of a spherical wedge = (from 809). 812. PROBLEM. To find the volume of a spherical segment. Ipherical segment of one base. Given : Spherical segment generated by the figure ACX\ semicircle XAY\ AC = r j radius of sphere = R ; altitude = CX = h. Required : To find the volume of the spherical segment. Computation: Draw chords AX, AY, and radius AO. The right A ACO will generate a cone of revolution (?) (671). The volume of spherical segment A CX = volume of spher- ical cone OA X minus volume of cone A CO. Volume of spherical cone OAX = | TrK 2 h (?) (811, 2) ; volume of cone ACO = J vrr 2 -CO (?) (695). Now r 2 = CX-CY=h (2 s-h) (?) (340,11); and CO = R - h. '.'. vol. ACO = J- Trh (2 R - h) (R - h) (Ax. 6). Hence, volume of spherical segment ACX = f 7rR 2 h - (f 77 R 2 h - TrRh 2 + J 7r 3 ) (Ax. 6). That is, volume of spherical segment of one base = j ivh 2 (3 R - h}. ^ BOOK IX 397 2. Spherical segment not including the center. Given : Spherical segment generated by figure ACDB:, semicircle XAEY\ AG r\ BD = r' ; radius of sphere = R ; altitude = Required: To find the volume of the spherical segment. Computation : The A AGO and BDO gen- erate cones of revolution (?) (671). The volume of spherical segment ACDB = volume of spherical sector ABO plus the volume of cone A CO minus the volume of cone BDO. Each of these volumes can be determined from formulas already established. 3. Spherical segment including the center. Given : Spherical segment generated by figure BDSR ; etc. Required : (?). Computation : The same as that of case 2, except that both cones, BDO and RSO, dreUdded. II ORIGINAL EXERCISES 1. Prove that the area of the surface of a sphere is equal to the square of the diameter multiplied by TT, that is, S = 7rZ> 2 . 2. Prove that the volume of a sphere is equal to one sixth the cube of the diameter multiplied by TT, that is, V = \ TrZ) 3 . 3. The surface of a sphere is equal to the cylindrical surface of the circumscribed cylinder. 4. The total surface of a hemisphere is three fourths the surface of the sphere. 5. The volume of a sphere is two thirds the volume of the circum- scribed cylinder. 398 SOLID GEOMETRY 6. Upon the same circle as a base are constructed a hemisphere, a cylinder of revolution, and a cone of revolution, all having the same altitude. Prove that their total areas are 3w/2 2 , lirR*, TrR\l + V2), respectively, and their volumes are f TrR 8 , TrR 8 , ^ TrR 3 , respectively. 7. Two zones on the same sphere, or on equal spheres, are to each other as their altitudes. 8. The area of the surface of a sphere is equal to the area of the circle whose radius is the diameter of the sphere. 9. Show that the formula for the volume of a spherical segment of one base reduces to the correct formula for the volume of a hemisphere, when the base of the segment is a great circle ; and to the correct formula for the volume of a sphere when the planes are both tangent. 10. In an equilateral triangle is inscribed a circle, and the figure is revolved about an altitude of the triangle as an axis. Prove, (a) That the surface generated by the circumfer- ence is two thirds the lateral surface generated by the triangle. (6) That the volume generated by the circle is four ninths the volume generated by the triangle. 11. Derive a formula for the surface of a sphere, containing only V andir. 12. Derive a formyja for the volume of a sphere, containing only S andir. 13. In a circle whose radius is R, there are inscribed a square and an equilateral triangle having their bases parallel ; the whole figure is then re- volved about the diameter perpendicular to the base of the triangle. Find, in terms of R, (a) The total areas of the three surfaces generated; (6) The volumes of the three solids generated. 14. If a cylinder of revolution having its altitude equal to the diam- eter of its base, and a cone of revolution having its slant height equal to the diameter of its base are both inscribed in a sphere, (a) The total area of the cylinder is a mean proportional between the area of the surface of the sphere and the total area of the cone; (6) The volume of the cylinder is a mean proportional between the volume of the sphere and the volume of the cone. BOOK IX 399 15. About a circle whose radius is R, there are circumscribed a square and an equilateral triangle having their bases in the same straight line. The whole figure is then re- volved about an altitude of the triangle. Find, in terms of R, (a) The total areas of the three surfaces gen- erated. (b) The volumes of the three surfaces generated. 16. If a cylinder of revolution having its altitude equal to the diam- eter of its base, and a cone of revolution having its slant height equal to the diameter of its base, be circumscribed about a sphere, (a) The total area of the cylinder is a mean proportional between the area of the surface of the sphere and the total area of the cone ; (ft) The volume of the cylinder is a mean proportional between the volume of the sphere and the volume of the cone. 17. The line joining the centers of two intersecting spherical sur- faces is perpendicular to the plane of the intersection at the center of the intersection. 18. A cube and a sphere have equal surfaces; show that the sphere has the greater volume. 19. Prove that the parallel of latitude through a point having 30 north latitude bisects the surface of the northern hemisphere. 20. Prove that in order that the eye may observe one sixth of the surface of a sphere, it must be at a distance from the center of the sphere equal to f of the radius. Proof: Zone TT = $ surface of sphere (Hyp.). :.AB = diam.^i /?. Hence, EC = %R. In rt. A ETC, TC 2 = EC-BC (?) ; .*. R* = EC . f R, or EC = f R (Explain.) Q.E.D. 21. How many miles above the surface of the earth (diameter of earth = 7960 mi.) must a person be in order that he may see one sixth of the earth's surface ? 22. If the area of a zone of one base is a mean proportional between the area of the remaining zone of the sphere and the area of the entire sphere, the altitude of the zone is R( Vo 1). 23. The area of a lune is to the area of a trirectangular spherical triangle as the angle of the lune is to 45. 24. A cone, a sphere, and a cylinder have the same diameters and altitudes. Prove that their volumes are in arithmetical progression. 400 SOLID GEOMETRY 25. The surface of a sphere bears the same ratio to the total surface of the circumscribed cylinder of revolution, as the volume of the sphere bears to the volume of the cylinder. 26. Ttee smallest circle upon a sphere, whose plane passes through a given point within the sphere, is the circle whose plane is perpendicular to the diameter through the given point. 27. What part of the surface of the earth could one see if he were at the distance of a diameter above the surface ? 28. Prove that if any number of lines in space be drawn through a point, and from any other point perpendiculars to these lines be drawn, the feet of all of these perpendiculars lie on the surface of a sphere. 29. The volume of a sphere is to the volume of the circumscribed cube as TT : 6. The volume of a sphere is to the volume of the inscribed cube as TT : | V3. 30. There are five spheres that touch the four planes of the faces of a tetrahedron. 31. If two angles of a spherical triangle are supplementary, the sides of the polar triangle, opposite these angles, are supplementary. 32. A square, whose side is a, is revolved about a diagonal, and also about an axis bisecting two opposite sides. Which of these figures con- tains the greater volume ? Which has the greater surface ? 33. Find the area of the surface, and the volume of a sphere whose radius is 6. 34. Find the area of a zone whose altitude is 4 on a sphere whose radius is 14. 35. Find the area of a lune whose angle is 30 on a sphere whose radius is 8 in. 36. Find the area of a spherical triangle whose angles are 110, 41, 92, on a sphere whose radius is 10. 37. Find the volume of a sphere whose radius is 5. 38. Find the volume of a spherical pyramid whose base is 35 sq. in., on SL sphere whose radius is 12 in. 39. Find the area of a spherical polygon whose angles are 87, 108, 121, 128, on a sphere whose radius is 25. 40. What is the radius of a sphere whose surface is 1386 sq. yd. ? 43,,. What is the radius of a sphere whose volume is - cu. in. ? BOOK IX 401 42. What is the area of the surface of a sphere whose volume is 288 TT cu. ft. ? 43. What is the volume of a sphere, the area of whose surface is 2464 sq. in. ? 44. Find the area of a zone whose altitude is 3J, if the radius of the sphere is 7|. 45. Find the volume of a spherical sector the altitude of whose base is 5 in. if the radius of the sphere is 6 in. 46. Find the diameter, the circumference of a great circle, and the volume of a sphere the area of whose surface is 25 TT sq. ft. 47. By how many cubic inches is a 9-in. cube greater than a 9-in. sphere ? 48. The radius of a sphere is 15, and the angles of the base of a spherical pyramid are 160, 127, 96, 145, and 117. Find the volume of the pyramid. 49. A cylindrical vessel 10 in. in diameter contains a liquid. A metal ball is immersed in the liquid and the surface rises in. What is the diameter of the ball? 50. If a sphere 3 ft. in diameter weighs 99 Ibs., what will a sphere of the same material 4 ft. in diameter weigh ? 51. The radii of the bases of a frustum of a cone of revolution are 5 in. and 6 in., and the altitude of the frustum is 19 in. What is the diameter of an equivalent sphere ? 52. What is the radius of a sphere whose surface is equivalent to the total surface of a right circular cylinder having an altitude equal to 21, and radius of the base equal to 6 V 53. Find the volume generated by the revolution of an equilateral triangle inscribed in a circle whose radius is 8, about an altitude of the triangle as an axis. (See Fig. of Ex. 55.) 54. In the figure of No. 55, find the volume of the segment generated by the figure A ED revolving about CD as an axis. 55. Find the area of the surface, and the volume of the sphere generated by a circle that is circum- scribed about an equilateral triangle whose side is 10. 56. Circumscribing a sphere whose radius is 18, is a cylinder of revolution. Compare their total areas. Their volumes. BOBBINS' SOLID GEOM, 26 402 SOLID GEOMETRY 57. Circumscribing a cylinder of revolution whose altitude and diameter are each 6 in., is a sphere. Find the volume and area of the surface of the sphere. 58. Circumscribing a cylinder whose altitude is 4 and diameter is 3, is a sphere. Find the radius and volume of the sphere. 59. Each edge of a cube is 8 in. What is the area of the surface, and the volume of the circumscribed sphere ? 60. Find the volume of one of the segments cut from a 10 in. sphere by the plane of one of the faces of the inscribed cube. 61. The volume of a certain sphere is 179f cu. ft. Find the radius of a sphere 8 times as large. Find the radius of a sphere 3 times as large. 62. The radius of a certain sphere is 5 in. What is the radius of a sphere twice as great ? Half as great? Two thirds as great ? 63. A hollow sphere has an outer diameter of 20 in., and an inner diameter of 16 in. Find the volume of the metal in the shell. 64. Find the diameter of that sphere whose volume is, numerically, equal to the area of its surface. 65. A projectile consists of a right circular cylinder having a hemi- sphere at each end. If the cylinder is 9 in. long and 7 in. in diameter, what is the volume of one projectile ? 66. Inscribed in a regular tetrahedron whose edge is 4, and circum- scribed about it are two spheres. Find their radii. 67. Find the radii of the spheres inscribed in and circumscribed about a regular hexahedron whose edge is 8 m. 68. Find the radii of the spheres inscribed in and circumscribed about a regular octahedron whose edge is 12 in. 69. How many spherical bullets | in. in diameter can be made from a cube of lead 5 inches on each edge ? 70. The area of a spherical triangle whose angles are 158, 77, 95, is 288|. Find the radius of the sphere. 71. The area of a spherical triangle whose excess is 75, is 135 TT. Find the radius of the sphere. 72. If the radius of a sphere is 2.5, and the sides of a triangle on it are 104, 115, 101, find the area of the polar triangle. BOOK IX 403 73. In a trihedral angle the plane angles of the dihedral angles are 75, 85, 110. Find the number of degrees of surface of a sphere, whose center is the vertex of the trihedral angle, inclosed by the faces of this trihedral angle. 74. What is the area of a spherical hexagon, each of whose angles is 145, on a sphere whose radius is 15? 75. How many miles above the earth would a person have to be in order that he may see a third of its surface ? One eighth of its surface? 76. Find the altitude of the zone whose area is equal to the area of a great circle of a sphere. 77. If the radius of a sphere is doubled, how is the amount of surface affected? The volume? The weight? 78. At a distance (= d) from the center of a sphere whose radius is r, is an illuminating point. What is the altitude of the zone illuminated? 79. On a sphere having a radius of 5 in. is an equiangular spherical triangle whose area is 5 TT sq. in. Find the angles of the triangle. 80. Find the area of the surface of a sphere whose volume is a cu. yd. 81. Find the volume of a sphere whose surface is a sq. yd. 82. If a circumference is described on the surface of a sphere, by a pair of compasses whose points are 2f in. apart, what is the area of the zone bounded by this circumference ? 83. On a sphere the area of whose surface is 288 sq. ft. is a birectan- gular spherical triangle whose vertex angle is 100. Find the area of this triangle. 84. Five inches from the center of a sphere whose diameter is two feet, a plane is passed. Find the areas of the two zones formed. Find the chords of their generating arcs. 85. The diameter of the moon is about 2000 mi. ; that of the earth, about 8000 mi. How do their surfaces compare? Their volumes? 86. The radii of two concentric spheres are 12 and 13 in. A plane is tangent to the inner sphere. Find area of section of outer sphere. 87. If a solid sphere 4 ft. in diameter weighs 500 Ibs., what is the weight of a spherical shell whose external diameter is 10 ft, made of the same material and a foot thick? 88. The sun's diameter is about 109 times the diameter of the earth. How do the areas of their surfaces compare? Their volumes? 89. How many quarter-inch spherical bullets can be made from a sphere of lead a foot in diameter ? 404 SOLID GEOMETRY 90. The angles of a spherical triangle are 80, 90, 100. Find the angle of an equivalent lune. 91. Find the angles of an equiangular spherical triangle equivalent to the sum of three equiangular spherical triangles (upon the same sphere) whose angles are each 75. 92. What is the radius of a sphere equivalent to the sum of two spheres whose radii are 3 in. and 4 in., respectively? 93. What is the radius of a sphere equivalent to the difference of two spheres whose radii are 5 in. and 4 in., respectively? 94. The area of an equiangular spherical triangle is TT, and the radius of the sphere is 4. Find the angles of the triangle. 95. The volumes of two spheres are to each other as 64 : 343. What is the ratio of their surfaces? 96. Find the volumes of the segments of a sphere whose radius is 12, formed by a plane whose distance from the center is 9. 97. If the radius of a sphere is 20, find : (a) The area of its surface. (6) The area of a zone whose altitude is 2. (c) The edge of a cube inscribed in the sphere. (d) The area of a lune whose angle is 80. (e ) The area of a spherical triangle whose angles are 75, 53, 72. (/) The area of a spherical polygon whose angles are 68, 119, 128, 147, 150. (gr) The area of a birectangular spherical triangle whose vertex- angle is 54. (A) The area of a zone of one base whose altitude is 5. ( i ) The radius of a sphere whose surface is four times as large. (j) The volume of the sphere. (fr) The volume of a wedge whose angle is 36. (/) The volume of a spherical pyramid whose base is the triangle of exercise (e). (w) The volume of the spherical sector whose base is the zone of exercise (6). (n) The volume of the spherical cone whose base is the zone of exercise (A). (o) The volume of a spherical segment of one base, whose altitude is 6. (/>) The radius of a sphere whose volume is four times as large. INDEX OF DEFINITIONS (The numbers refer to pages.) Abbreviations, 16. Acute angle, 13. Acute triangle, 15. Adjacent angles, 12. Alternate-exterior angles, 38. Alternate-interior angles, 38. Alternation, 141. Altitude, of cone, 345. of cylinder, 337. of frustum, 308, 346. of parallelogram, 46. of prism, 292. of pyramid, 307. of trapezoicl, 46. of triangle, 33. of zone, 386. Analysis, 127. Angle, 12. acute, 13. between intersecting curves, 361. birectangular trihedral, 286. bisector of, 14, 33. central, 80. central, of regular polygon, 220. complement of, 13. convex polyhedral, 284. degree of, 13. dihedral, 273. exterior, of polygon, 54. exterior, of triangle, 42. inscribed, in circle, 80. inscribed, in segment, 102. isosceles trihedral, 286. measure of, 102. oblique, 13. obtuse, 13. plane, 12. plane angle of dihedral, 273. polyhedral, 284. rectangular trihedral, 286. right, 12. sides of, 12. spherical, 361. Angle, supplement of, 13. trihedral, 286. trirectangular trihedral, 286. vertex of, 12. Angles, adjacent, 12. alternate-exterior, 38. alternate-interior, 38. complementary, 13. corresponding, 38. dihedral, 273. equal, 15. face, of polyhedral, 284. homologous, of polygons, 55. of lune, 386. of polygons, 54. of spherical triangle, 370. of triangle, 14. polyhedral, 284. right dihedral, 273. supplementary, 13. symmetrical polyhedral, 285. vertical, 12. Antecedents, 140. Apothem, 220. Arc, 80. degree of, 102. intercepted, 81. subtended, 81. Arcs, similar, 228. Area, 187. Auxiliary lines, 25. Axiom, 16. parallel, 36. Axioms, 16. Axis, of circle of sphere, 361. of circular cone, 345. of symmetry, 58. Base, of cone, 345. of figure, 45. of pyramid, 307. of spherical pyramid, 387. of spherical sector, 387. 405 406 INDEX OF DEFINITIONS Base, of triangle, 14. Bases, of cylinder, 337. of parallelogram, 45. of prism, 292. of spherical segment, 387. of trapezoid, 45. of zone, 386. Birectangular spherical triangle, 371. Birectangular trihedral angle, 286. Bisector of angle, 14, 33. Boundary, 12. Center, figure symmetrical with re- spect to, 58. of circle, 79. of circumference, 79. of regular polygon, 220. of sphere, 360. of symmetry, 58. Central angle/ 80. of regular polygon, 220. Chord, 79. Circle, 79. angle inscribed in, 80. center of, 79. circumscribed about polygon, 91. diameter of, 79. inscribed in polygon, 92 radius of, 79. sector of, 80. segment of, 80. tangent to, 79. Circles, concentric, 80. equal, 80. escribed, 124. of spheres, 361. tangent externally, 80. tangent internally, 80. Circular cone, 345. axis of, 345. right, 345. Circular cylinder, 338. right, 338. Circumference, 79. center of, 79. Circumscribed circle, 91. Circumscribed frustum of pyramid, 346. Circumscribed polygon, 92. Circumscribed prism, 312. Circumscribed sphere, 361. Commensurable quantities, 97. Common tangent, 92. Complementary angles, 13. Complement of angle, 13. Composition, 142. Composition and division, 143. Concave polygon, 54. Concentric circles, 80. Conclusion, 23. Concurrent lines, 57. Cone, 345. altitude of, 345. base of, 345. circular, 345. circular, axis of, 345. frustum of, 346. lateral area of, 345. lateral area of frustum of, 346. oblique circular, 346. of revolution, 346. of revolution, slant height of, 346. plane tangent to, 346. right circular, 345. spherical, 387. total area of, 345. total area of frustum of, 346. Cones, 345. similar, of revolution, 346. Conical surface, 345. Consequents, 140. Constant, 97. Construction, 115. Continued proportion, 140. Converse of theorem, 25. Convex polyhedral angle, 284. Convex polyhedron, 291. Corollary, 17. Corresponding angles, 38. Cube, 293, 324. Curved line, 79. Cylinder, 337. altitude of, 337. bases of, 337. circular, 338. lateral area of, 337. oblique, 338. of revolution, 338. right, 338. right circular, 338. right section of, 338. total area of, 337. Cylinders, similar, of revolution, 338. Cylindrical surface, 337. Decagon, 55. Degree, of angle, 13. of arc, 102. spherical, 386. Demonstration, 17. elements of, 23. INDEX OF DEFINITIONS 407 Determination of straight line, 11. of a circle, 116. Determined, plane, 252. Diagonal, 45. of polyhedron, 291. of spherical polygon, 372. Diameter of circle, 79. of sphere, 360. Dihedral angle, 273. edge of, 273. faces of, 273. plane angle of, 273. Dihedral angles, 273. right, 273. Dimensions of parallelepiped, 293. Direct proportion, 160. Directrix, 337, 345. Discussion, 115. Distance between points of surface of sphere, 361. between two points, 25. from point, to a line, 32. from point to a plane, 268. Division, 142. Dodecagon, 55. Dodecahedron, 291. regular, 324. Edge of dihedral angle, 273. Edges of polyhedral angle, 284. of polyhedron, 291. Element of conical surface, 345. of cylindrical surface, 337. Equal angles, 15. Equal circles, 83. Equal dihedral angles, 273. Equal figures, 15. Equal polygons, 55. Equal polyhedral angles, 285. Equal solids, 293. Equal spheres, 361. Equal spherical polygons, 372. Equiangular polygon, 54. Equiangular spherical triangles, 371. Equiangular triangle, 15. Equilateral polygon, 54. Equilateral spherical triangles, 371. Equilateral triangle, 15. Equivalent figures, 187. Equivalent solids, 293. Escribed circles, 124. Exterior angle, of polygon, 54. of triangle, 42. Extreme and mean ratio, 182. Extremes, 142. Face angle of polyhedral angle, 284. Faces, of dihedral angle, 273. of polyhedral angle, 284. of polyhedron, 291. Figure, base of, 45. rectilinear, 11. symmetrical, 58. Figures, equal, 15. equivalent, 187. isoperimetric, 245. Foot of line, 251. of perpendicular, 13. Fourth proportional, 140. Frustum of cone, 346. altitude of, 346. lateral area of, 346. midsection of, 346. slant height of, 346. total area of, 346. Frustum of pyramid, 308. altitude of, 308. circumscribed, 346. inscribed, 346. slant height of, 307. Generatrix, 337, 345. Geometrical solid, 294. Geometry, 11. plane, 11. - solid, 251. Great circle of sphere, 361. axis of, 361. Harmonic division, 149. Hemisphere, 387. Heptagon, 55. Hexagon, 55. Hexahedron, 291. regular, 324. Homologous angles, in polygons, 55. Homologous parts, 15. Homologous sides, in polygons, 55. in triangles, 155. Hypotenuse, 15. Hypothesis, 23. [cosahedron, 291. regular, 324. Inclination, 268. [ncommensurable quantities, 97. [ndirect method, 36. Inscribed angle, in circle, 80. in segment, 102. Inscribed circle, 92. Inscribed frustum of pyramid, 346. 408 INDEX OF DEFINITIONS Inscribed polygon, 91. Inscribed prism, 312. Inscribed prism in cylinder, 338. Inscribed sphere in polyhedron, 361. Intercept, to, 51, 81. Intercepted arc, 81. Intersect, to, 51. Intersection, 251. Inverse proportion, 160. Inversion, 142. Isoperimetric figures, 245. Isosceles trapezoid, 45. Isosceles triangle, 15. Isosceles trihedral angle, 286. Lateral area, of cone, 345. of cylinder, 337. of frustum of cone, 346. of prism, 292. of pyramid, 307. Lateral edges, of prism, 292. of pyramid, 307. Lateral faces, of prism, 292. of pyramid, 307. Legs, of isosceles trapezoid, 45. of isosceles triangle, 15. of right triangle, 15. Limit of variable, 97, 99. Limits, theorem of, 99. Line, 11. curved, 79. divided harmonically, 149. divided into extreme and mean ratio, 182. foot of, 251. inclination of, 268. projection of, 251. straight, 11. straight, oblique to plane, 251. straight, parallel to plane, 251. straight, perpendicular to plane, 251. tangent to circle, 79. tangent to circle of sphere, 361. tangent to sphere, 361. Lines, auxiliary, 25. concurrent, 57. divided proportionally, 147. oblique, 13. parallel, 36. perpendicular, 13. Locus, 60. Lune, 386. angles of, 386. vertices of, 386. Maximum, 245. Mean proportional, 140. Means, 140. Measure, of angle, 102. of quantity, 97. unit of, 97. Median, of trapezoid, 46. of triangle, 33. Method, indirect, 36. of exclusion, 36. Midsection of frustum of cone, 343. Minimum, 245. Motion, 12. Mutually equiangular polygons, 54. Mutually equilateral polygons, 54. Names of regular polyhedrons, 324. N-gon, 55. Normal, 251. Oblique angle, 13. Oblique circular cone, 346. Oblique cylinder, 338. Oblique lines, 13. Oblique parallelepiped, 293. Oblique prism, 292. Obtuse angle, 13. Obtuse triangle, 15. Octagon, 55. Octahedron, 291. regular, 324. Parallel axiom, 36. Parallelepiped, 293. dimensions of, 293. oblique, 293. rectangular, 293. right, 293. Parallel lines, 36. Parallel planes, 251. Parallelogram, 45. altitude of, 46. bases of, 45. Pentagon, 55 Pentedecagon, 55. Perimeter, 92. Perpendicular, 13. foot of, 13. Perpendicular planes, 273. Pi (*), 227. Plane, 11, 251. Plane angle, 13. Plane angle of dihedral angle, 273. Plane, determined, 252. distance from point to, 268. INDEX OF DEFINITIONS 409 Plane Geometry, 11. Plane, line oblique to, 251. line parallel to, 251. line perpendicular to, 251. tangent to cone, 346. tangent to cylinder, 338. tangent to sphere, 361. Plane section, 284. Planes, parallel, 251. perpendicular, 273. Point, 11. equally distant from two lines, 32. of contact, 79, 361. of tangency, 79, 361. projection of, 163, 252. Points, symmetrical, 58. Polar distance, 361. Polar triangle, 371. Poles of circle of sphere, 361. Polygon, 54. angles of, 54. center of regular, 220. central angle of regular, 220. circumscribed, 92. concave, 54. convex, 54. equiangular, 54. equilateral, 54. exterior angle of, 54. inscribed, 91. re-entrant, 54. regular, 217. vertices of, 54. Polygons, equal, 55. mutually equiangular, 54. mutually equilateral, 54. similar, 150. Polyhedral angle, 284. convex, 284. edges of, 284. face angles of, 284. faces of, 284. plane section of, 284. vertex of, 284. Polyhedral angles, 284. equal, 285. symmetrical, 285. vertical, 285. Polyhedron, 291. convex, 291. diagonal of, 291. edges of, 291. faces of, 291. regular, 323. sphere inscribed in, 361. Polyhedron, vertices of, 291. Polyhedrons, similar, 323. Postulate, 17. Prism, 292. altitude of, 292. bases of, 292. circumscribed, about cylinder, 338. circumscribed, about pyramid, 312 inscribed, in cylinder, 338. inscribed, in pyramid, 312. lateral area of, 292. lateral edges of, 292. lateral faces of, 292. oblique, 292. regular, 292. right, 292. right section of, 293. total area of, 292. triangular, 292. truncated, 292. Problem, 115. Projection, of line, 163, 252. of point, 163, 252. Proof, 17. Proportion, 140. antecedents of, 140. consequents of, 140. continued, 140. direct, 160. extremes of, 140. inverse, 160. means of, 140. reciprocal, 160. Proportional, fourth, 140. mean, 140. third, 140. Proposition, 115. Pyramid, 307. altitude of, 307. altitude of frustum of, 308. base of, 307. circumscribed, about cone, 346. circumscribed, frustum of, 346. inscribed, frustum of, 346. inscribed, in cone, 346. frustum of, 308. lateral area of, 307. lateral edges of, 307. lateral faces of, 307. *. regular, 307. slant height of frustum, 308. slant height of regular, 307. spherical, 387. triangular, 307. total area of, 307. 410 INDEX OF DEFINITIONS Pyramid, truncated, 308. vertex of, 307. Pyramids, 307. Q.E.D., 23. Q.E.F., 116. Quadrant, 80, 361. Quadrilateral, 45. angles of, 45. sides of, 45. vertices of, 45. Quantities, commensurable, 97. incommensurable, 97. Radius, of circle, 79. of regular polygon, 220. of sphere, 360. Ratio, 97, 140. extreme and mean, 182. series of equal, 140. Reciprocal proportion, 160. Rectangle, 45. Rectangular parallelepiped, 293. Rectangular trihedral angle, 286. Rectilinear figure, 11. Reductio ad absurdum, 36. Re-entrant polygon, 54. Regular polygon, 217. Regular polyhedrons, 323, 324. Regular prism, 292. Regular pyramid, 307. slant height of, 307. Revolution, cone of, 346. cylinder of, 338. similar cones of, 346. similar cylinders of, 338. Rhomboid, 45. Rhombus, 45. Right angle, 12. Right circular cone, 345. Right circular cylinder, 338. Right cylinder, 338. Right dihedral angle, 273. Right parallelepiped, 293. Right prism, 292. Right section, of cylinder, 338. of prism, 293. Right triangle, 15. Scalene triangle, 15. Secant, 79. Sector, of circle, 80. spherical, 387. Sectors, similar, 228. Segment, base of spherical, 387. Segment, of circle, 80. spherical, 387. Segments, of line, 148. similar, 228. Semicircle, 80. Semicircumference, 80. Series of equal ratios, 140. Sides, homologous, in polygons, 55. homologous, in triangles, 155. of angle, 12. of polygon, 54. of quadrilateral, 45. of spherical triangle, 370. of triangle, 14. Similar arcs, 228. Similar cones of revolution, 346. Similar cylinders of revolution, 338. Similar polygons, 150. Similar polyhedrons, 323. Similar sectors, 228. Similar segments, 228. Similar triangles, 150. Slant height, of cone, 346. of frustum of cone, 346. of frustum of pyramid, 308. of regular pyramid, 307. Small circle of sphere, 361. Solid, 11, 251. geometrical, 294. Solid geometry, 251. volume of, 293. Solids, equal, 293. equivalent, 293. Sphere, 360. axis of circle of, 361. center of, 360. circumscribed about polyhedron, 361. diameter of, 360. great circle of, 361. inscribed in polyhedron, 361. line tangent to, 361. plane tangent to, 361. radius of, 360. small circle of, 361. Spheres, equal, 361. tangent, 361. Spherical angle, 361. Spherical cone, 387. Spherical degree, 386. Spherical excess, 387. Spherical polygon, 372. diagonal of, 372. Spherical polygons, equal, 372. Spherical pyramid, 387. base of, 387. INDEX OF DEFINITIONS 411 Spherical pyramid, vertex of, 387. Spherical sector, 387. base of, 387. Spherical segment, 387. altitude of, 387. base of, 387. of one base, 387. Spherical surface, 360. Spherical triangle, 370. angles of, 370. birectangular, 371. sides of, 370. symmetrical, 372. trirectangular, 371. unit of measure of, 371. vertices of, 370. Spherical triangles, 370. mutually equiangular, 371. mutually equilateral, 371. Square, 45. Statement, 115. Straight angle, 13. Straight line, 11. divided, extreme and mean ratio, 182. divided harmonically, 149. oblique to plane, 251. parallel to plane, 251. perpendicular to plane, 251. tangent to circle, 79. tangent to sphere, 361. Straight lines, divided proportionally, 147. Subtend, to, 80. Subtended arc, 81. Superposition, 15. Supplement of angle, 13. Supplementary angles, 13. Surface, 11. conical, 345. cylindrical, 337. spherical, 360. Surfaces, 251. Symbols, 16. Symmetrical polyhedral angles, 285. Symmetrical spherical triangles, 372. Symmetry, 58. axis of, 58. center of, 58. Tangent, 79. circles, 80. spheres, 361. Tetrahedron, 291. Theorem, 17. converse of, 25. Theorem, elements of, 23. of limits, 99. Third proportional, 140. Total area, of cone, 345. of cylinder, 337. of frustum of cone, 346. of prism, 292. of pyramid, 307. Transversal, 38. Trapezium, 45. Trapezoid, 45. altitude of, 46. bases of, 45. isosceles, 45. legs of, 45. median of, 46. Triangle, 14. acute, 15. altitude of, 33. angles of, 14. base of, 14. equiangular, 15. equilateral, 15. exterior angle of, 42. isosceles, 15. median of, 33 obtuse, 15. right, 15. scalene, 15. sides of, 14. spherical, 370. vertex of, 14. vertices of, 14. Triangles, similar, 150. Triangular prism, 292. Triangular pyramid, 307. Trihedral angle, 286. birectangular, 286. isosceles, 286. trirectangular, 286. Trirectangular spherical triangle, 371, Trirectangular trihedral angle, 286. Truncated prism, 292. Truncated pyramid, 308. Unit, of measure, 97. of spherical triangle, 371. of surface, 187. of volume, 293. Variable, 97. limit of, 97, 99. Vertex angle, 14. Vertex, of angle, 12. of polyhedral angle, 284. 412 INDEX OF DEFINITIONS Vertex, of pyramid, 307. of spherical pyramid, 387. of triangle, 14. Vertical angles, 12. Vertical dihedral angles, 273. Vertical polyhedral angles, 285. Vertices, of June, 386. of polygon, 54. of polyhedron, 291. of quadrilateral, 45. of spherical triangle, 370. Vertices, of triangle, 14. Volume, of solid, 293. unit of, 293. Wedge, spherical, 387. Zone, 386. altitude of, 386. bases of, 386. of one base, 386. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. ''50BC 9NovDEAD 140ct'6JSL REC D LD rfttfjyg* 34-gpu LD 21-100m-ll, 1 49(B7146sl6)476 V C 22378 JUKI