F G is not perpendicular to DE. And in the fame Manner we prove, that no other Right Line but F C is perpendicular to DE. Wherefore FC is perpendicular to DE. Therefore, if any Right Line touches a Circle, and from the Center to the Point of Contact a Right Line be drawn; that Line will be perpendicular to the Tangent; which was to be demonftrated. PROPOSITION XIX. THEORE M. If any Right Line touches a Circle, and from the LE ET any Right Line DE touch the Circle ABC in C, and let CA be drawn from the Point C at Right Angles to DE., I fay, the Circle's Center is in A C. For if it be not, let F be the Center, if poffible, and join CF. Therefore F is not the Then because the Right Line DE touches the Circle ABC, and F C is drawn from the Center to the Point of Contact; FC will be perpendicular to * 18 of this. DE*. And fo the Angle FCE is a Right one. From the But ACE is also a Right Anglet: Therefore the Hyp Angle FCE is equal to the Angle ACE, a less to a greater; which is abfurd. Center of the Circle ABC. prove, that the Center of the Circle can be in no other Line, unless in A C. Wherefore, if any Right Line touches a Circle, and from the Point of Contact a Right Line be drawn at Right Angles to the Tangent, the Center of the Circle fhall be in the faid Line; which was to be demonstrated. PRO PROPOSITION XX. THEOREM. The Angle at the Center of a Circle is double to the Angle at the Circumference, when the fame Arc is the Bafe of the Angles. ET ABC be a Circle, at the Center whereof is the Angle BEC, and at the Circumference the Angle BAC, both of which stand upon the fame Arc BC. I fay, the Angle BEC is double to the Angle BAC. 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 EA B; 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 FEC is double to EAC. Therefore the whole Angle BEC is double to the whole Angle BAC. 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 GE B is double to the Part GDB. And therefore BEC is double to BDC. Confequently, an Angle at the Center of a Circle is double to the Angle at the Circumference, when the fame 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. ET ABCDE be a Circle, and let BAD, BED, be Angles in the fame Segment BAED. I say, 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 BFD *20 of this. will be double to the Angle BAD. For the fame Reason, the Angle BFD is alfo double to the Angle BED. Therefore the Angle BAD will be equal to the Angle BED. + 32. I. 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 DE G. But the Angles ABE, ADE, are equal, from what has been before proved, 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 demonftrated. PROPOSITION XXII. THEOREM. The oppofite Angles of any Quadrilateral Figure defcribed in a Circle, are equal to two Right Angles. ET ABDC be a Circle, wherein is defcribed the quadrilateral Figure ABCD. I fay, two oppofite Angles thereof are equal to two Right Angles. For join AD, BC. * 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, A B C, 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 AD B, 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 B A C be added, then the Angles BAC, ABC, ACB, are equal to the Angles BAC, BDC; but the Angles BAC, ABC, ACB, are 2 * equal |