177 



CE in B andC, the angle IBD is equal to ICE, (29 B. 1. Euc.) 

 and for a like reason, because FA meets the parallels, the angle 

 BDI is equal to CEI ; therefore in the two triangles BDI and CEI, 

 the two angles IBl), BDI, are equal to the two ICE, CEI, and the 

 side BI is equal to IC, therefore the two triangles are equal in all 

 respects, (26, B. I. Euc.) and the side BD to CE*.— Q. E. D. 



LEMMA II. 



If the semi-circumference of a circle be divided into any odd 

 number of equal arches, and chords be drawn connecting the opposite 

 points of division, the difference between the sums of those chords, 

 alternately taken, will be equal to the radius of the circle. See 

 Plate, Fig. 2. 



Let the semi-circumference AEK be divided into any odd number 

 of equal arches AB, BC, «&;c. and chords BI, CH, ^c. be drawn, 

 connecting the opposite points of division B, I; C,H,&c. ; then will the 

 difference between the sums of those chords, alternately taken, be 

 equal to the radius of the circle ; that is 



. . (BI + DG) — (CH + EF) = OK, the radius of the circle. 



For draw the radius OF to either extremity, as F, of the least 

 chord EF, cutting tlie several other chords in the points L. M. N. 



Then because the diameter AK, and chords BI, CH, <&c. inter- 

 cept the equal arches AB, KI; BC, IH ; &c. they are parallel to 

 each other, as is well known, and therefore, because the arch EF 

 == FG, the right line EF is = NG, by the foregoing Lemma, and 

 for the same reason DN is = MH, and CM = LI, &c. 



* The demonstration of this property might conveniently be made a part of that in the 2d 

 Lemma ; I have however, for the sake of distinctness, given it a separate demonstration. 



VOL. zni. B B 



