408 Demonstration^ of Stewards 



figure. Through V, draw the diameter PQ,. With the centre 

 O, and distance OV, describe a circle. Draw OA, OF, &.c. to 

 all the points of contact, from the centre. From V, draw V^, 

 Vs, kc. perpendicular to O/V, OF, &;c. respectively. From the 

 points a,f, &,c., where the lines OA, OF, Sic. cut the circumfer- 

 ence of the smaller circle, draw ap,fq, &c. perpendicular to the 

 diameter PQ. 



Since 01 is perpendicular to KG, it is parallel to VH ; and 

 because Yh is perpendicular to OA, it is parallel to KG. There- 

 fore, VHAA is a parallelogram ; and YH—PsJi. And because 

 the perpendiculars Yh, and ap, are drawn from the opposite ends 

 of the same arch Ya, to the radii, OA and OP, passing through 

 the extremities of the same arch ; therefore P/? = A/t = VH. In 

 like manner, P9'=VS ; and so of all the other perpendiculars. 



Now it is manifest, that the radii OA, OF, &,c. divide the cir- 

 cumference of the smaller circle, at the points o,/, he. into as 

 many equal parts, as they do that of the greater ; and moreo- 

 ver, that the lines Yp, Yq, &c. are respectively equal to perpen- 

 diculars drawn from V, to the sides of a regular figure circum- 

 scribed about the smaller circle, by drawing tangents to it at 

 the points a, f, he. ; as was shewn in the demonstration of Lem- 

 ma 4th. 



Now Pp=PY+Yp=zPO-YO+Yp=R-d+Vp. Hence, 



Vp^ ={R-^-^Yp)^ ={U-d)^ +3{R-(}Y 'Yp+3{R-d)-Yp* 

 + V>3. ^In like manner, Pq' =(R.-dy + 3{'R-d)^'Yq-{' 

 3{R—d) 'Yq^+Yq^ : and so of all the rest. 



}ience,Pp'+Pq'+-^-c. = {Ii-d)^+{J{-dy-\-L^c.-\-3['R-dy 

 • V/'TV^-Wf^. + 3(R - £?) -Y^+Y^^h^. +Yp"^+Yq^ + 4^c. 



But {R-dy + {R-dy+8^c.=n{n~dy; and (Lemma 3d,) 

 Yp-{-yq-\-^c.=nd, (where d denotes the radius of the smaller 

 circle;) and (Cor. Prop. IV.) Yp^+Yq^' + ^c. =-nd", (m, in 



the general expression, being in this case 2 ;) and (Prop. III.) 

 Y2}''+Yq^-\-SfC.=^—nd' 



n 



Therefove,Vp^+Pq^+hc.=n{R-dy+3nd{R-dy+-nd^ 



2 



{ll-d) + ^nd'' =:n-lV -3il^d+3Rd^ -d^ + 3R^d-QRd^ + 



3d^+~Rd^ -^d' +-d^ =n R^ -^--dm. 



2 2 2 2 



Hence, 2-Pp3-|-P2^ + .^c.=2-VH3+VS3+4"C. =2nll3 + 



