PROPOSITION XX. THEOREM. The Angle at the Center of a Circle is double to the LET ABC be a Circle, at the Center, whereof is For join AE and produce it to F. Then because EA is equal to EB, the Angle EAB fhall be equal to the Angle EBA*. Therefore the ⋆ 5. 1. Angles EAB, EBA, are double to the Angle EAB; but the Angle BEF is + equal to the Angles EAB, † 32. 1. EBA; therefore the Angle BEF is double to the Angle EAB. For the fame Reason, the Angle EFC is double to EA C. Therefore the whole, Angle BEC is double to the whole Angle BA C. Again, let there be another Angle BDC, and join DE, which produce to G. We demonftrate in the fame Manner, that the Angle GEC is double to the Angle GDC; whereof the Part GEB is double to the Part GDB. And therefore B E C is double to BD C. Confequently, an Angle at the Center of a Circle is double to the Angle at the Circumference, when the Same Arc is the Bafe of the Angles; which was to be demonftrated. PROPOSITION XXI. THEOREM. Angles that are in the fame Segment in a Circle, are equal to each other. LET ABCDE be a Circle, and let BAD, BED, be Angles in the fame Segment BAED. Ifay, thofe Angles are equal. For let F be the Center of the Circle ABCDE, and join BF, FD. Now because the Angle BFD is at the Center, and the Angle BAD at the Circumference, and they ftand upon the fame Arc BCD; the Angle B FĎ *20of this. will be* double to the Angle BAD. For the fame Reafon, the Angle BFD is alfo double to the Angle BED. Therefore the Angle BAD will be equal to the Angle BED. † 32.1. # 15.1. * 32.1. If the Angles BAD, BED, are in a Segment lefs than a Semicircle, let AE be drawn; and then all the Angles of the Triangle ABG are equal to all the Angles of the Triangle DEG. But the Angles ABE, ADE, are equal, from what has been before prov'd, and the Angles AGB, DGE, are alfo equal, for they are vertical Angles. Wherefore the remaining Angle BAG is equal to the remaining Angle GED. Therefore, Angles that are in the fame Segment in a Circle, are equal to each other; which was to be demonstrated. PROPOSITION XXII. THEOREM. The oppofite Angles of any Quadrilateral Figure defcribed LET ABDC be a Circle, wherein is described the Then because the three Angles of any Triangle are * equal to two Right Angles, the three Angles of the Triangle ABC, viz. the Angles CAB, ABC, BCA, are equal to two Right Angles. But the Angle ABC +21 of this. is equal to the Angle ADC; for they are both in the fame Segment ABDC. And the Angle ACB is equal to the Angle ADB, because they are in the fame Segment ACDB: Therefore the whole Angle BDC is equal to the Angles ABC, ACB; and if the common Angle BAC be added, then the Angles BAC, ABC, ACB, are equal to the Angles BAC, BDC; but the Angles BA C, ABC, A CB, are 2 * equal * equal to the two Right Angles. Therefore likewife, * 32. 1. the Angles BAC, BDC fhall be equal to two Right Angles. And after the fame Way we prove, that the Angles A BD, ACD, are alfo equal to two Right Angles. Therefore the oppofite Angles of any Quadrilateral Figure defcribed in a Circle, are equal to two Right Angles; which was to be demonftrated. PROPOSITION XXIII. THEOREM. Two fimilar and unequal Segments of two Circles, cannot be fet upon the fame Right Line, and on the fame Side thereof. FOR if this be poffible, let the two fimilar and un equal Segments ACB, ADB, of two Circles stand. upon the Right Line AB on the fame Side thereof. Draw A CD, and let CB, BD, be joined. Now becaufe the Segment A CB is fimilar to the Segment ADB, and fimilar Segments of Circles are fuch * Def. 11. which receive equal Angles; the Angle ACB will of this. be equal to the Angle ADB, the outward one to the inward one; which is † abfurd. Therefore fimilar † 16. 1. and unequal Segments of two Circles, cannot be fet upon the fame Right Line, and on the fame Side thereof; which was to be demonftrated. PROPOSITION XXIV, THEOREM. Similar Segments of Circles being upon equal Right Lines, LET AEB, CFD be equal Segments of Circles, For the Segment AEB being apply'd to the Segment CFD, fo that the Point A co-incides with C, and the Line AB with CD; then the Point B will co-incide with the Point D, fince A B and CD are equal. And fince the Right Line A B co-incides with G 3 CD, CD, the Segment AEB will coincide with the Segment CFD. For if at the fame Time that AB coincides with CD, the Segment A EB fhould not coincide with the Segment CFD, but be otherwise, as CGD; then a Circle would cut a Circle in more Points than two, viz. in the Points C, G, D; which 10 of this. is impoffible. Wherefore if the Right Line A B co-incides with CD, the Segment AEB will co-incide with and be equal to the Segment CFD. Therefore fimilar Segments of Circles being upon equal Right Lines, are equal to one another; which was to be demonftrated. * 10. 1. +II.1. 23.1. PROPOSITION XXV. PROBLEM, A Segment of a Circle being given, to describe the Circle whereof it is the Segment. LE ET ABC be a Segment of a Circle given. It is requir'd to describe a Circle whereof ABC is a Segment. Bifect AC in D, and let DB be drawn + from t the Point D at Right Angles to AC, and join AB. Now the Angle ABD is either greater, equal, or lefs than the Angle BAD. And firft let it be greater, and make the Angle BAE at the given Point A, with the Right Line BA, equal to the Angle ABD; produce DB to E, and join EC. * Then becaufe the Angle ABE is equal to the Angle BAE, the Right Line BE will be equal to EA, And because A D is equal to DC, and DE common, the two Sides AD, DE, are each equal to the two Sides CD, DE; and the Angle ADE is equal to the Angle CDE; for each is a Right one. Therefore the Bafe AE is equal to the Bafe EC. But AE has been prov'd to be equal to EB. Wherefore BE is alfo equal to EC. And accordingly the three Right Lines AE, EB, EC, are equal to each other. Therefore a Circle defcrib'd about the Center E, with either of the Distances AE, EB, EC, fhall pass thro the other Points, and be that requir'd to be defcrib'd. But it is manifeft that the Segment ABC is less than 2 a Se |