IF a ftraight line touches a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the center of the circle fhall be in that line. Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE; the center of the circle is in CA. For if not, let F be the center, if poffible, and join CF. Because DE touches the circle ABC, and FC Sce N. B A that no other point which is not in CA, is the center; that is, the center is in CA. Therefore if a straight line, &c. Q. E. D. THE angle at the center of a circle is double of the angle at the circumference, upon the fame base, that is, upon the fame part of the circumference. Let Let ABC be a circle, and BEC an angle at the center, and BAC Book III. an angle at the circumference, which have the fame circumference BC for their base; the angle BEC is double of the angle BAC. a First, Let E the center of the circle be within the angle BAC, and join AE, and produce it to F. Because EA is equal to EB, the angle EAB is equal to the angle EBA; therefore the angles EAB, EBA are double of the angle EAB; but the angle BEF is equal b to the angles EAB, EBA; therefore alfo the angle BEF is double of the angle EAB. double of the angle EAC. of the whole angle BAC. B F A. E a. 5. I. for the fame reason, the angle FEC is therefore the whole angle BEC is double remaining angle BDC. Therefore the angle at the center,&c. Q. E. D. PRO P. XXI. THEO R. b. 32. I. THE angles in the fame segment of a circle are equal see N. to one another. Let ABCD be a circle, and BAD, BED angles in the fame fegment BAED; the angles BAD, BED are equal to one another. Take F the center of the circle ABCD. and, firft, let the fegment BAED be greater than a femicircle, B and join BF, FD. and because the angle BFD is at the center, and the angle BAD at the circumference, and A E D that a Book III. that they have the fame part of the circumference, viz. BCD for their base, therefore the angle BFD is double of the angle BAD. a. 20. 3. for the fame reason, the angle BFD is double of the angle BED. therefore the angle BAD is equal to the angle BED. But if the segment BAED be not greater than a femicircle, let BAD, BED be angles in it; these also are equal to one another. draw AF to the center, and produce it to C, and join CE. therefore the fegment B A E D F C Wherefore the angles in the fame segment, &c. Q. E. D. THE 'HE oppofite angles of any quadrilateral figure described in a circle, are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD; any two of its oppofite angles are together equal to two right angles. Join AC, BD; and because the three angles of every triangle are 3. 32. 1. equal to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles. but the angle b. 21. 3. CAB is equal to the angle CDB, be A D B cause they are in the fame fegment fhewn fhewn to be equal to two right angles. Therefore the oppofite Book III. angles, &c. Q. E. D. PROP. XXIII. THEOR. UPON the fame ftraight line, and upon the fame fide see N. of it, there cannot be two fimilar fegments of circles, not coinciding with one another. If it be poffible, let the two fimilar fegments of circles, viz. ACB, ADB be upon the same side of the same straight line AB, not coinciding with one another. then because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point a. one of the fegments must therefore fall within the other; let ACB fall within ADB, and draw the ftraight line BCD, and join CA, DA. and because the seg- A ment ACB is fimilar to the segment ADB, B 2.10.3. and that fimilar fegments of circles contain b equal angles; the b.11.Def.3. angle ACB is equal to the angle ADB, the exterior to the interior, which is impoffible . Therefore there cannot be two fimilar feg- c. 16. 1. ments of a circle upon the fame fide of the fame line, which do not coincide. Q. E. D. PROP. XXIV. THEOR. SIMILAR fegments of circles upon equal straight lines, see N. are equal to one another. Let A EB, CFD be fimilar fegments of circles upon the equal ftraight lines AB, CD; the fegment AEB is equal to the fegment CFD. line AB upon CD, the point B fhall coincide with the point D, be caufe Book III. caufe AB is equal to CD. therefore the straight line AB coinciding with CD, the fegment AEB must coincide with the fegment 2. 23.3. CFD, and therefore is equal to it. Wherefore fimilar segments, &c! Q. E. D. See N. a. 1o. I. bi 11. 1. c. 6. 1. A PROP. XXV. PROB. Segment of a circle being given, to describe the circle of which it is the fegment. Let ABC be the given fegment of a circle; it is required to defcribe the circle of which it is the segment. C Bifecta AC in D, and from the point D draw DB at right angles to AC, and join AB. First, let the angles ABD, BAD be equal to one another; then the ftraight line BD is equal to DA, and therefore to DC. and because the three ftraight lines DA, DB, d. 9. 3. DC are all equal, D is the center of the circle . from the center D, at the distance of any of the three DA, DB, DC defcribe a circle; this shall pafs thro' the other points; and the circle of which ABC is a fegment is defcribed. and because the center D is in AC, C. 23. I. the segment ABC is a femicircle. but if the angles ABD, BAD are not equal to one another, at the point A in the ftraight line AB make the angle BAE equal to the angle ABD, and produce BD to E, and join EC. and because the angle ABE is equal to the angle BAE, the ftraight line BE is equal to EA. and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two fides AD, DE are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, for each of them is a right f. 4. 1. angle; therefore the bafe AE is equal f to the bafe EC. but AE was shewn to be equal to EB, wherefore alfo BE is equal to EC;' and the three straight lines AE, EB, EC are therefore equal to one another; |