Equa! QUAL circles (ABD, EGH,), are those of which the diameters (AD, EH,) or the rays (CB, FG,) are equal. Fig. 1. If the circles be applied to one another, so that their centers coincide, when their rays are equal, the circles must likewise coincide. II. Similar segments of circles (ABC, DEF,), are those in which the angles (ABC, DEF,), are equal. Fig. 2. Circles are similar figures. If then the two segments (ABC, DEF,) be taken away by subftituting the equal angles (ABC, DEF,), those segments are Similar. TO PROPOSITION I. PROBEEMI. Sought The center F of this O. Resolution. 1. Draw the chord AB. Pof. 1. 2. Bisect it in the point D. P. 10. B. i. 3. At the point Din AB, erect the L DE & produce it to E. P.11.B. 1. 4. Bisect CE in F. P. 10. B. The point F will be the center fought of the given O ACBE. DEMONSTRATION. Some other point as H or G taken in the line, or without the Cafe I. D. 15.B.I. But FE being = to FC (Ref. 4.) & HC <FC (Ax. 8. B. 1.). 2. HC will be allo < FE, & a fortiori < HE. 3. Therefore HE is not = to HC. 4. Consequently, the point H taken in the line EC different from the point F, cannot be the center of the O ACBE. Café II. Preparation. Pol. 1. Becau GB ( Prep. & D. 15. B. 1.), the fide GD common to the two A, & the base AD= to the base DB (Ref. 2.). 1. The adjacent Vatb&c to which the equal lides GA, GB, are opposite, are = to one another, P. 8. B.I. 2. Therefore Va t b is a L. D. 10. B. 1, But V a being also a L (Ref. 3.). 3. It follows, that Va+b is = to Va, which is impossible. Ax. 8.B.1. 4. Therefore the point G taken without the line EC, cannot be the cen- rent from F (Case 1.) nor without the line EC in a point G (Case. 11.) 5. The center fought of the O ACBE, will be necessarily in F. Which was to be done, COROLLA RY. IF F in a circle ACBE, a chord EC bisects another chord AB at right angles; chord CE is a diameter, & consequently passes thro' the center of the circle, (D. 17. B. 1.). PROPOSITION II. THEOREM I. If any two points (A & B) be taken in the circumference of a circle BECAUSE (AEB); the straight line (AB) which joins them, shall fall within the circle. Hypothesis. Thesis. The two points A & B are taken The firaight line AB falls in the O AEB. within tbe O AEB. Preparation, 1. Find the center of O AEB. P. 1. B. 3. 2. Draw the straight lines CA, CD, CB. Pol. 1. DEMONSTRATION. ECAUSE in the A ACB, the side CA is = to the side CB, ( Prep. 2. & D. 15. B. 1.). 1. The V CAD, CBD, are = to one another. P. 5. B. 1. But V CDA being an exterior V of A CDB. 2. It is>than its interior CBD. P. 16. B. 1. And because the V CBD is = to the V CAD ( Arg. 1.). 3. This V CDA will be also > than V CAD. 4. Consequently, the side CA opposite to the greater V CDA, is > the fide CD opposite to the lesser Ÿ CAD. P. 19. B. 1. 5. From whence it follows, that the extremity D of this side CD falls within the O AEB. in the line AB. Which was to be demonstrated. N PROPOSITION III. THEOREM II. F a diameter (CD) bise&s a chord (AB) in (F); it shall cut it at right angles, & reciprocally if a diameter (CD) cuts a chord (AB) at right angles, it shall bisect it. I. Thesis. The diameter CD is I upen which bifects AB in F. the chord AB. Preparation. Pol. 1. DEMONSTRATION. N the A AEF, BEF, the side EA is = to the side EB ( Prep. & D. 15. B. 1.), the side EF is common to the two A, & the base AF is = to the base BF (Hyp.). 1. Consequently, the adjacent V m & n, to which the equal fides EA, EB, are opposite, are = to one another. P. 8. B. s. 2. Wherefore, the straight line CD, which stands upon AB making the adjacent y m&n=to one another, is I upon AB. D. 10.B... Which was to be demonstrated. IN . II. Thesis. AF is = to FB. ihe chord AB ; or which makes m= V n. DEMONSTRATION. He sides EA, EB, of the A AEB being = to one another (Prep. & D. 15. B. 1.). 1. The V EAF, EBF, will be also = to one another, P. 5. B. 1 Since then in the AEF, BEF, the VEAF, EBF, are = ( Arg. 1.), as also the m&n (Hyp.), & the side EF cominon to the two A. 2. The base AF will be = to the base FB. P. 26.B.I. Which was to be demonstrated, T T |