BOOK m. PROP, xxvii. xxxni.] MATHEMATICS. PLANE GEOMETRY. 



5G9 



E L. In like manner K C = L F, . . the arcs B K C, 

 L F are equal, . . in equal circles, &c. Q. E. D. 



PROPOSITION XX VII. THEOREM. 



In equal circles (A B C, D E F) the angles (B A C, E D F), 

 which stand upon equal arcs (B G C, E H F), are equal, 

 whether they be at the centres or at the circumferences. 

 For if not, let one of them, as B A C, be the greater ; 



and make the angle 



BAK = EDF; 



23 I. then be- 

 cause in e jiial 

 circles equal angles 

 stand on equal 



+ 26 III. arcs,t .'. 



the arcs B G K, ^ \-^a & y* 



E H F are equal : 



but B G C = G 



Hyp. E H F,* , '. B G K = B G C, a part to 

 the whole, which is impossible, . . the angles BAG, 

 E D F cannot be the one greater than the other ; 

 that is, the angles are equal ; . . in equal circles, <tc. 

 Q. E. D. 



NOTK. The demonstrations of the last two propositions differ con- 

 siderably from those of Euclid, which, upon examination, will be 

 found to be incomplete, since they exclude the case in which the 

 angles B A C, D F, at the circumferences, are each not less than a 

 light angle. 



It will be observed that the demonstration of the second of these pro- 

 positions equally applies whether the angles B A C, E D F be at the 

 circumferences, as in the diagrams, or at the centres. These points 

 are left unmarked by letters, in order that, in going over the rea- 

 aoning, the letters A, D may be mentally transferred to the centres, 

 and thus both cases be included in the the same demonstration. 



PROPOSITION XXVIII. THEOREM. 

 In equal circles ( A B C, D E F) equal chords (B C, E F) cut off 

 equal ares (B A C, E D F, and B G C, E H F) Me greater 

 equal to the greater, and the less to the less- 



l in Tase K, L, the centres of the circles,* and 

 draw KB, KG, 



L E, L F. Then, 

 because the circles 

 are equal, K B, 

 K C are = L E, 

 L F, each to each, 

 and B C = E F.f 

 + Hyp. ' . the 

 angle B K C = 



si. ELF:* 



but equal angles at the centres stand on equal arcs,t ' 

 + 20 in. the arc B G C = E H F : but the whole cir- 



Hyp. cumference A B C = the whole E D F,* . . 

 the remaining part B A C = the remaining part E D F ; 

 .'.in eq\ial circles, &c. Q. E. D. 



PROPOSITION XXIX. THEOREM. 



In equal circles (A B C, D E F) equal arcs (B G C, E H F) are 



subtended by equal chords (B C, E F). 



[See preceding diagrams. ] 



If the arcs are semi-circumferences, then B C. E F are 

 diameters ; and as the circles are equal, these diameters 

 must be equal. But if the arcs be not semi-circum- 

 ferences, take K, L, the centres of the circles,* and 



i in. draw KB, K C, L E, L F. Then, because 

 the arc B G C = the arc E H F, the an^le B K C = 



+ 27 III. ELF;t and because the circles are equal, 

 K B, K C are equal to L E, L F, each to each, and the 

 41. angles K, L are also equal, . . B C = E F ; * 

 .'. in equal circles, <tc. Q. E. D 



NOTI. It is plain that what Is proved in the preceding propositions, 

 as to equal circlet, necessarily holds in reference to the lame 

 circle. 



PROPOSITION XXX. PROBLEM. 



To bisect a giren arc (A D B) ; that is, to divide it into two 



equal part*. 



10 i. Draw A B and bisect it in C : * from C draw 

 C D at right angles to A B : t the 



till. arc A D B shall be bi- 

 sected in D. Draw A D, B D. 

 Tlii-n, because A C = C B, and 

 C D common to the triangles A C D, 

 BCD, the two sides AC, C D are 



VOL. I. 



C B 



the two B C, C D, 



each to each ; and the angle A C D = B C D, each being 

 41. a right angle, .. AD = B D. * But equal 



chords cut off equal arcs, the greater are equal to the 

 + 28 III. greater, and the less to the less,f .'. arc AD 



= arc B D, each arc being less than a semicircle, because 



DC, or D C prolonged, passes through the centre ; . . 



the yiven arc is bisected in D. Which was to be done. 



PROPOSITION XXXI. THEOREM. 



The angle (B A C) in a semicircle (B A D C) is a right angle : 

 the angle in a segment (A B G C) greater than a semicircle, is 

 less than a right angle: and the anrjle (D) in a segment 

 (A D C) less than a semicircle, is greater than a right angle. 



From the centre, E, draw E A, and produce B A to F. 

 Then, because E B = E A, the 



si. angle EAB = EBA;* ? 



also because E A = E C, the angle 

 E A C = E C A, . . the whole angle 

 BAG = ABC + A CB : but FAC, 

 the exterior angle of the triangle 



+ 321. ABC,is = ABC-r-ACB,t 

 .-.BAG = FAC, .-. each is a right 



Def. 8 1. angle,* .'.the angle B A C 

 in a semicircle is a right angle. 

 Again, because the two angles B, C, 

 of the triangle ABC are together 



+ 171. less than two right angles, f 

 and that BAG has been proved to 

 be a right angle, . . ABC must be less than a right 

 angle, . . the angle B in a segment A B G C, greater Hum 

 a semicircle, ii less than a right angle. Lastly, because 

 A B C D is a quadrilateral in a circle, the opposite angles 

 2SIII. B, D are together = two right angles:* but B 

 has been proved to be less than a right angle, . . D, i a 

 segment ADC less than a semicircle, is greater than a right 

 angle ; .-.the angle, &c. Q. E. D. 



PROPOSITION XXXII. THEOREM. 



If a straight line (E F) touch a circle, and from the point of 

 contact (B) a straight line (B D) bt drawn cutting the circle, 

 the angles (DBF, D B E) which this lint mates with the 

 touching line shall be equal to the angles (A, C) in the alternate 

 segments of the circle, each to each. 



ill. From B draw B A at right angles to E F :* 



take any point C in the arc DCB, and draw AD, DC, CB. 



Then, because E F touches the circle in B, and that 

 B A is drawn at right angles 

 to E F, the centre of the circle, 



+ 19 in. is in B A,f .'. 

 A 1) B, being an angle in a 

 semicircle, is a right angle,* 



31 HI. .'. B A D + A B D 

 + 32 I. = a right angle ;t 



but A B F likewise is a right 

 Const. angle,* .'. ABF; 

 BAD + AB^D. Take 

 away the common angle A B D, 

 . . D B F = A, the angle in the 

 alternate segment. 



Again : because A C is a quadrilateral in a circle A -f- 

 + 22 in. C = two right angles :t but D B F + D B E 



13 1. _ two right angles,* .-.DBF + DBE = 

 A -f- C ; and it has been proved that D B F = A, .-. 

 DBE= C, the angle in the alternate segment; .-. if a 

 straight line, &c. Q. E. D. 



PROPOSITION XXXIII. PROBLEM 



Upon a given straight line 

 (AB) to describe a segment 

 of a circle capable of con- 

 taining an angle equal to a 

 given angle (C). 

 . 10 i. First let C be a right angle : bisect A B in F^ 

 and with F as centre, and radius F B, describe the semv 

 circle A H B : then the angle H, formed by H A, H B, 

 drawn from any point H in the arc, being in a semicircle, 

 + 31 III. w equal to the right angle C.f 



4 D 



