124 DEMONSTRATIONS of 



By theor. 6. 2(XP-f-XG*-f XHM-XK 1 ) = 2(NF*+NG*+NH* 

 +NK 2 )-f 8NX*. But 2(XF 2 +XG 2 +XH 2 +XK 2 ) = 

 2(XP+XQH-XR 2 +XS J )+2(.X« 2 +X^ 2 -fX^+X^ 1 ). Therefore 

 2(NP+NG*-fNH*+NK 2 )+8NX* - 2 (XP 1 +XQy-XR 2 +XS 2 )+ 

 2(X^-fX<H-Xc*+X^). But fince, 



2(E/4-EG a +EH a +E/& a ) - 4 (EL 2 +EM 2 ), and from the point 

 X parallels to Cf, CG, CH, Ck, and to CL, CM, are drawn, 

 cutting the perpendiculars from E, to thefe lines, in a, b, c, d, and 

 in Y, Z, therefore, by Cor. Theor. 12. 

 2(E^+E£ 2 -|-E^+E^) = 4(EY 2 H-EZ Z ), and confequently 

 2{Xa*+X6*+Xc*+Xd*) = 4(XY 2 +XZ 2 ) = 8(NY 2 +NX 2 ), 

 (Prop. 1.). Therefore, 



2(NP-fNG*+NHM-NK 2 ) = 2(XP 2 -fXQ*+XR J -|-XS a )+ 

 8NY\ But by Theor. 6. 



2(EFM-EG*-f-EH 2 +EK*) = 2(NF*+NG a -fNH a +NK a )-h 

 8NE a . Therefore, 



2(EP+EG a -fEH a +EK a ) = 2(XPM-XQ a J f-XR a +XS a )-f 

 8(NY a +NE a ) ; or, 



2(EP-fEG a +EH a +EK 2 ) - 2 (XP 2 -f-XQ t 2 +XR 2 -fXS 2 )-f 

 4 (EY 2 +EZ 2 ), (Prop. 1.). 



It remains to demonflrate that X is a given point, and that 

 XY, XZ, are lines given in pofition. 



The point O may be found, by bifecting (Fig. X. No. 2.) 

 GH in if, joining gk, and dividing it in m, fo that^ra ~ ?gk, 

 and joining fm and dividing it in O, fo that mO = \mfi and 

 in the fame manner the point N may be found by joining gK, 

 and making ^/z ~ ks¥*-> and joining »F, and making «N = i»F; 

 let mn be joined, through O draw Op, and through N draw Nq, 

 both parallel to EF, and meeting mn in p, q; let EF meet mn in 

 r, join ON, and through O draw Oj- parallel to mn, meeting 

 N^ in s. 



Then becaufe gm z= %gk y and gn zr j^K, the line mn is 

 parallel and equal to jKL Becaufe alfo N» rz £F», Ng zz flrj 



and. 



