MATMLM \TICS. PLAN'E GEOMETRY. [BOOK in PROP, xx. XXIY. 



ri:< (POSITION XX. THEOREM. 



The angle (B K C) a< ffe centre (E) of 

 a circle (A B C) u <JuM o/ tfo am/fe 

 (BAG) at the circumference wjxm 

 Owam are(BC). 



Draw A E, and prolong it to F. 

 And fint let the centre E bo within 

 I] A C. Because K A - E B, the 

 . i i. angle E A B - E B A,* . . 

 EAB-fEBAU double of E A B j 

 + MI. but BEF=E AB+EBA.t.'. BEF is 

 double <>f K A B. For a like reason F EC double of 

 EAC, . . l; V. C it double of B AC. 



Next let the centre E be without 

 BAG. It may be demonstrated, 

 as in the first case, that F EC is 

 double of F AC, and that FEE, 

 a part of the first, is double of 

 F A B, a part of the other, . . the 

 remaining angle BEG it double of 

 the remaining angle BAG;.', the 

 3 --- -% angle at the centre, <tc. Q. E. D. 



PROPOSITION XXI. THEOREM. 



J he angle* (BA.D, to ED) in the tame tegment (B A ED) 



of a circle are equal. 



. i in. Take F the centre of a circle :* draw F B, F 1), 

 as also the diameter A C ; and join 

 E, C. Because B F C, at the 

 centre, is upon the same arc B C, 

 as B A C at the circumference, . . 

 + 10 in. BFCis double of BAC.f 

 Again, because B F C, at the cen- 

 tre, is upon the same arc B C, as 

 B E C at the circumference, . . 

 B F C is double of B E C. But it 

 was shown that B F C is also double 

 of B A C, /. B A C - B E G. In 

 like manner it may be proved that C A D = C E D, .". the 

 whale ant/It B A D = tlte whole angle BED; .'. the angles, 

 A-c. Q. E. D. 



PROPOSITION XXII. THEOREM. 

 The oppotite angles of any quadrilateral firrtirt (A BCD) 

 intenbtd in a, circle (ABC D), are together equal to two 

 right ai.gle*. 



Draw the diagonals A C, B D. Then 

 because the three angles of every 

 triangle are together = two right 

 sa I. angles,*.'. CAB -I- ABC 

 -f- B C A - two right angles : but 

 + 21 in. CAB = CDB,fandBCA 

 = BDA, .-.CAB + BCA=CDA. 

 To each of these add ABC, .'. ABC 



. . ABC+CDA= two right angle*: and as the four 

 8J I. angles of a quadrilateral = four right angles,* 

 Cor. i. . ' . the other two oppotite angle* are also = two 

 right angles ; . . the oppotite angles, 4c. Q. E. D. 



MOT. -It appeara from the foregoing- property, that if one tide (aa 

 A II) of a quadrilateral In a circle be prolonicrd, the exterior angle 

 will be equal to the Interior and opposite angle (D). 



PROPOSITION XXIIL THEOREM. 



Upon the nime itrniylit line (AB), and on the same tide of 

 it, thrre cannot be two similar teyments of circlet not 

 coinciding with one another. 



For if it be supposed possible, let A C B , A D B be two 

 imilar segments, not coinciding with one another. Then, 

 because the circles cut in the two points A, B, they can- 

 10 in. not cut in any other jwint,* . . one segment 

 must be within the other. Let 

 A C B be within A D B : draw 

 BCD, as also C A, DA. Then, 

 because the segment* are simi- 

 Hyp. lar,f and that similar 

 segments of circles contain equal 

 10 in angles.* .' . the angle 

 * AC B A D B, the exterior to 



+ IB I the interior and opposite, which is im possible, f 

 . . thtre cannot be two similar segments of circlet upon tht 

 tame tide of the tame straight line, which do not coincide. 

 Q.E.D. 



PROPOSITION XXIV. THEOREM. 



Similar tegmrnts of circlet (A E B, C F D) -vpon equal 

 choi ds (A B, C D) are equal to one another. 



For if the segment 

 A K II bo applied to 

 CFD, so that A may 

 ~j) be on C, and A B 

 on CD, B shall 

 coincide with D, because A B = C D, . . A B coinciding 

 with C D, the segment A E B must coincide with the tey- 



23 III. ment C F D.* and it . . equal to it ; . . similar 

 tegments, &c. Q. E. D. 



NOT*. It obrioiuly followa from thi, that ilmilar legmenta of 

 circle* upon equal chorda hare equal area. 



PROPOSITION XXV. PROBLEM. 



An arc (A B K) f a circle being given, to describe the 

 circle of which it it an arc. 



Take any point B in the arc, from which draw any 

 two chords B C, B D. Bisect 

 101. them in E and F,* and 

 t u I. draw E G, F H at right 

 angles to B C, B D. These per- 

 pendiculars will cut each other in 

 a point P, which will \m the cen- 

 tre of the circle. For tha centre 

 of the circle to which the arc B C 

 belongs, is in the line E G, bisect- 

 I in. POT. ing the chord C B at right angles :* it is also 

 in the line F H, bisecting the chord B D at right angles : 

 the centre being thus in both the lines E G, F H, must be 

 at P, where they intersect, . . with P as centre, and P M 

 as radius, if a circle be described, it will be that of which 

 the given are is a part. Which was to be done. 



PROPOSITION XXVI. THEOREM. 



In equal circlet (A B C, D E F) equal angles ttand u/>on 

 equal arcs, whether they be at the centre! or at the cir- 

 cumferencet. 

 First, let the angles B G C, E H F at the centres be 



equal. Take any 



point, A, in the 



arc B A C, and 



any point, D, in 



the arc E D F: 



draw A B, AC, 



D E, D F, B C, 



E F. Then, be- 

 cause the circles 



are equal, the 



radii G B, G C are equal to the radii HE, H F, each to 



Hyp. each; and the angle G = H,* .'. BC = 

 + 41. EF:t andbecause BAC= EOF,' the seg- 



20 in. inent B A C is similar to the segment E D F ; t 

 + Def. 10 III. and they have equal chords, . . the seg- 



"i in. ment B A C segment E D F ; * but the 

 whole circle A B K C - the whole circle D K L F, .'. the 

 remaining segment BKC-ELF, .-.</* arc B K C = 

 arc E L F. 



Next let the angles B A C, E D F at the circumfcrcncei 



i l III be equal. Find the centres G, H ; f and 



draw G B, G C, B C ; HE, H F, E F : then if B A be 



less than a right angle, the angle B G C, at the centre, 



will stand on the same arc B K C as BAG, and will . . 



Join, be double of B A C.* In like manner E H F 

 will be double of E D F, .-.the angles G, H at tho 

 centres are equal, and.'., as before proved, the arrt 

 B K C, ELF must be equal. But if B A C bo not leas 



+ 91. than a right angle, bisect it,t as also the equal 

 angle E D F, by the straight lines A K, D L, then the 

 angles B A K, E D L being equal, and each less than a 

 right angle, the arc B K. is, by tho first case, = the arc 



