124 DEMONSTRATIONS of 



By theor. 6. 2(XF^+XG^+XH^+XK') - 2(NF^+NG^+NH' 

 +NK0+8NX^ But 2{XF^+XG^+XH^+XK^) = 

 2(XP=+XQH-XR>+XS0+2(Xfl^+X3^+X^^+X^^). Therefore 

 2(NF^+NG^+NH^+NK0+8NX^ - 2(XP^+XQ^+XR^+XS^)+ 

 2(X^'+X^^+Xc'+Xrf''). But fince, 



2(E/^+EG'+EH'+E/('0 = 4(EL^+EM0, and from the point 

 X parallels to C/, 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(Ea^+Ei^+E6-^+Ea'^) = 4(EY^+EZ'), and confequently 

 2(Xfl>+X^^+X^'+X^^) = 4(XY'+XZ0 = 8(NY = 4-NX = ), 

 (Prop. I.). Therefore, 



2(NF^+NG^-+NH^+NK^) - 2(XP^+XQ^+XR^+XS0+ 

 8NY\ But by Theor. 6. 



2(EF^+EG^+EH'+EK^) = 2(NF^+NGHNH^4-NK^)+ 

 8NE^ Therefore, 



2(EF^+EG^+EH'+EK^) zz 2(XP^+XQ^+XR^+XS0+ 

 8(NY^+NE^) ; or, 

 2(EF^+EG^+EH^+EK0 = 2(XP^+XQ^+XR'+XS')+ 



4(EY = +EZ = ), (Prop, i-)- 



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

 XY, XZ, are lines given in pofition. 



The point O may be found, by bifedting (Pig. X. No. 2.) 

 GH in g;, joining gk, and dividing it in m, fo that^/w = '^gk, 

 and joining /ot and dividing it in O, fo that mO — \mf; and 

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

 and making^;; := t^K., and joining «F, and making hN = i«Fj 

 let mn be joined, through O draw O/, and through N draw Ny, 

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

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

 Ny in /. 



Then becaufe gm — '^gk^ and gn — |^K, the line mn is 

 parallel and equal to ^K^. Becaufe alfo N« =: iF«, Ny = \^r; 



and. 



