is less than a Semicircle, because the Centre thereof is without the same. But if the Angle A B D be equal to the Angle BAD; then if A D be made equal to B D, or DC, the three Right Lines AD, BD, DC, are equal between themselves, and D will be the Centre of the Circle to be delcribed, and the Segment A B C is a Semicircle. But if the Angle A B D is less than the Angle BAD; let the Angle BAE be made, at the given Point A, with the Right Line BA, within the Sego ment ABC, equal to the Angle A B D. Then the Point E, in the Right Line D B, will, by arguing as beiore, appear to be the Centre, and A B C a Segment greater than a Semicircle. Therefore, a Circle is described, whereof a Segment is given ; which was to be done. PROPOSITION XXVI, THEOREM. Circumferences, whetker they be at iheir Cen- BGC, EHF, be equal Angles at their Centres ; and B AC, EDF, equal Angles at their Circumferences. I say, the Circumference BKC is equal to the Circumference EL F. For, let B C, E F be joined. Because A B C, DEF, are equal Circles, the Lines drawn from their 1 Def. 1. Centres will I be equal. Therefore the two Sides BG, GC, are equal to the iwo Sides E H, HF;' and the Angle G is equal to the Angle H. Wherefore the Bafe B C is * equal to the Base E F. Again, because the Angle at A is equal to that at D, the † Def. 11. Segment BAC will be † similar to the Segment EDF; and they are upon equal Right Lines B C, EF. But those similar Segments of Circles, that are 1 24 of tbis. upon equal Right Lines, are I equal to each other. # Def. 11. Therefore the Segment BAC will be + equal to the Segment EDF. But the whole Circumference ABKCA is equal to the whole Circumference DELFD DELFD. Therefore the remaining Circumference BKC shall be equal to the remaining Circumference EL F. Therefore, in equal Circles, equal Angles fland upon equal Circumferences, whether they be at their Centres, or at their Circumferences ; which was to be demonstrated. PROPOSITION XXVII. THEOREM. equal Circles, are equal to each other, whether LET the Angles BGC,EHF, at the Centres of the equal Circles A B C D E F, and the Angles BAC, ED F, at their Circumferences, stand upon the equal Circunferences B C, EF. I say, the Angle BGC is equal to the Angle EHF; and the Angle B A C to the Ange EDF. For if the Angle BGC be equal to the Angle EHF, it is manifest that the Angle BAC is also equal to the Angle EDF: But if the Angle GBC be not equal to the Angle E H F, let one of them be the greater, as BGC, and make * the Angle B GK, 23. 1. at the Point G, with the Line B G, equal to the Angle EHF. But equal Angles stand + upon equal + 26 of ebis. Circumferences, when they are at the Centres. Wherefore the Circumference BK is equal to the Circumference EF. But the Circumference E F is equal to the Circumference BC. Therefore BK is equal to B C, a less to a greater, which is absurd. Wherefore the Angle B GC is not unequal to the Angle E H F, and so it must be equal to it. But the Angle at A is one half of the Angle BGC; and the Angle at D is one half of the Angle EHF. Therefore the Angle at A is equal to the Angle at D. Wherefore, Angles, that stand upon equal Circumferences in equal Circles, are equal to each other, whether they be at their Centres, or Circumferences; which was to be demonstrated. PRO. PROPOSITION XXVIII. THEOREM. Paris of the Circumferences ; the greater Part lelser. LET ABC, DEF, be equal Circles, in which are the equal Right Lines BC, EF, which cut off the greater Circumferences B A C, EDF, and the leffer Circumferences BGC, EHF. I say, the greater Circumference B A C is equal to the greater Circumference EDF, and the lefier Circumference BGC to the lefler Circumference EHF.. For assume the Centres K and L of the Circles; and join BK, KC, EL, LF. Because the Circles are equal, the Lines drawn I Def. 1. from their Centres * are also equal. Therefore the two Sides B K, K C, are equal to the two Sides E L, LF; and the Base B C is equal to the Base E F. + 8.1. Therefore the Angle BKC is + is equal to the Angle 1 26 of ibis. EL F. But equal Angles I itand upon equal Cir- THEORE M.' equal Circumferences, are equal. and let the equal Circumferences BGC, EHF, be assumed in them, and B C, EF, joined. I say, the the Right Line B C is equal to the Right Line EF. For, find * K and L, the Centres of the Circles ; * 1 of ebis. and join B K, KC, EL, L F. Then, because the Circumference B GC is equal to the Circumference EHF, the Angle BKC fall be f equal to the Angle ELF. And because the † 27 of tbis. Circles ABGCA, DEHFD, are equal, the Lines drawn from their Centres shall be equal. I Def. 1. Therefore the two Sides B K, KC, are equal to the two Sides E L, LF; and they contain equal Angles : Wherefore the Base B C is f equal to the Base E F. † 4. I. And so, in equal Circles, the Right Lines which subtend equal Circumferences, are equal; which was to be demonfrated. PROPOSITION XXX. PROBLEM. Parts. L'ET the given Circumference be AD B: It is re quired to cut the same into two equal Parts. Join A B, which bisect * in C; and let the Right * 10. I. Line C D be drawn from the Point C at Right Angles to A B +; and join AD, DB. +11. I. Now, - because A C is equal to CB, and C D is common, the two Sides A C, CD are equal to the two Sides BC, CD; but the Angle A CD is equal to the Angle BCD; for each of ihem is a Right Angle: Therefore the Base A D is + equal to the trir Base BD. But equal Right Lines cut off equal | 28 of biso Circumferences. Wherefore the Circumference A D shall be equal to the Circumference B D. Therefore, a given Circumference is cut into two equal Parts ; which was to be done. PRO. PROPOSITION XXXI. THEOREM. Right Angle; but the Angle in a greater Seg- a lesser Segment is less than a Right Angle. LET. there be a Circle ABDC, whose Diameter is BC, and Centre E; and join BA, A C, AD, DC. I say, the Angle which is in the Semicircle BAC is a Right Angle; that which is in the Seginent ABC being greater than a Semicircle, viz. the Angle A B C is less than a Right Angle; and that which is in the Segment ADC being less than a Semicircle ; that is, the Angle ADC is greater than a Right Angle. Farjoin A E, and produce B A to F. Then, because B E is equal to E A, the Angle EAB fhall be * equal to the Angle E BA. And because A E is equal to EC, the Angle A CE will be * equal to the Angle CAE. Therefore the whole Angle B AC, is equal to the two Angles A B C, ACB; but the Angle FAC, being without the Tri+ 32. 1. angle ABC, is + equal to the two Angles A B C, ACB; therefore the Angle BAC is equal to the | Def. 10. 1. Angle FAC; and fo each of them is I a Right An gle. Wherefore, the Angle B AC, in a Semicircle, is a Right Angle. And because the two Angles A B C, B AC, of the Triangle ABC*, are less than two *17.1. Right Angles, and B A C is a Right Angle; then, And since A BCD is a quadrilateral Figure in a Circle, and the opposite Angles of any quadrilate.al + 22 of ibis. Figure described in a Circle are + equal to two Right Angles; the Angles A B C, ADC, are equal to two will 5. I, |