186 THE ROYAL SOCIETY OF CANADA 
XIV. Given the coplanar triangles ABC and EFG and the points 
A”=BC ..FG, B’=CA.GE and C’=AB.EF, to determine the 
ratio (A”B’C”) : (EFG). 
Join A and E, B and F, C and G and let A’=AE. BC, B’= 
BF. CA and C’=CG. AB, then employing the notation of XIII, 
A € € VIE 
ind B= Powe es 
ptutn 
fa ae Net cea 1e 
A+o+ 2 
and ind G = Nappe sure, 
ATE T 
If ind C’=xea+(1—x)e, then will 
{xe+(1 —x)e} (pe tuerie) (Ne + ae + ve) =0 
x A1—x O0 | =0, 
Pp pb Vy 
No: V2 
Pye Avi— pre 
(+ o)r1 — (ut p)ve 
Andre Hdi QT pn) e1 = (um avi) €2 
(A+ 0) 1 — (u-F p) P2 
In like manner it may be shown that 
Ar (u—o)e—(m—7T)e 
BO Peo. 
and cia ey Rue 
vy —T—A+p 
AB Cl = 0 p-o —mw+Tr|ABC 
—)-+ p 0 Vi —T 610263 
NY — pv — M2 or) 0 
= (ve Fa V1) p K V] ABC 
À a va | 418283 
Nu T 
510031" B’C" = (ve—r1) (pt uti) (A+o-+7) (A+u+r7)EFG 
in which 6; =u—0—v2+7T 
d2=7,—T—A+pP 
and d3= (A+ o)r1 — (ut+p)v 
