178 



Hence therefore since NG = EF 



DN =MH 



LI = CM 



and ... BL = OK 



by adrlilion . . BI + DG = OK + CH + EF 



or. . (BI + DG>-(CH + EF)=OK..Q.E.D. 



And in the same manner whatever be the number of equal arches. 



Corollary. — Hence as the above chords are the chords subtend- 

 ing respectively 1, 3, 5, &c. of the equal arches, if therefore the 

 semi-circumference be divided into any odd number of equal arches, 

 the difference between the sums of the chords alternately taken, 

 originating in the same point, and subtending 1, 3, 5, &c. of those 

 arches, is equal to the radius of the circle : the greater sum being 

 always that in which the greatest chord is included. 



LEMMA HI. 



In a circle, the rectangle under the radius and the sum of the 

 chords of two arches, is equal to the rectangle under the chord of 

 half their sum, and the supplemental chord of half their difference : 

 and, 



LEMMA IV. 



The rectangle under the radius and the difference of the chords 

 of two arches, is equal to the rectangle under the supplemental 

 chord of half their sum, and chord of half their difference. See 

 Plate, Fig. 3. 



Let AB and ABD be two arches of which AB, and AD are their 

 chords ; bisect the arch BD in C, and join AC, BC, CD ; take O 



