179 



the centre of the circle, and draw the diameter COE, and join AE, 

 BE, DE. 



Then because ACDE is a quadrilateral inscribed in the circle, by 

 Ptolemy's Theorem, (Prop. D. Book VI. Simson's Euclid.) 



AD X CE n AC X DE + CD X AE. 



or because DE = BE| . ad x CE = AC x BE + BC x AE . . . I. 

 and , . CD = BC j 



Again, because ABCE is a quatrilateral inscribed in the circle. 



AC X BE = AB X CE + BC x AE 



or AB X CE = AC x BE — BC x AE . . II. 



andhencebyaddition| ..(ad f AB) x CE =2 AC xBE; 

 1. and II. . . . j ^ ' 



orbecauseCE = 2CO.(AD + AB)x2CO=2ACxBE; 



or . CAD + AB) x CO = ACxBE; which is the 

 3d Lemma : for CO is the radius of the circle, AD + AB the sum 

 of the chords of the two arches ABD and AB ; AC the chord of half 

 their sum ABC, and BE the supplemental chord of half their differ- 

 ence BC, or CD ; whence the truth of the 3d Lemma is manifest. 



In the same manner, by subtraction, I. and II. 



(AD— AB) X CO = BC X AE ; which is the 

 4th Lemma : for CO is the radius of the circle, AD — AB the dif- 

 ference of the chords of the two arches ADB and AB ; AE the 

 supplemental chord of half their sum, and BC the chord of half 

 their difference : whence the truth of the 4th Lemma is also ma- 

 nifest. 



Analysis. — See Plate, Fig. 4. — Suppose now in the semicircle 

 AEK, the semicircumference to be divided into seventeen equal 

 arches, of which the points B, C, D, &c. are the 1st, 3d, 5th, &c. 

 points of division, and AB, AC, AD, &c. the chords drawn from A, 

 the extremity of the diameter, to those points respectively. 



B b2 



