176 THE ROYAL SOCIETY OF CANADA 
Let E, F and G be the feet of perpendiculars let fall from D on 
BC, CA and AB respectively and let a= ED, B= FD and y=GD, then 
will 
aa : B : cy : 2A : : (CDB) : (ADC) : (BDA) : (ABC) 
sth :u:viÀ+utky 
ses aae +bBe+Cyes 
2A 
Ifx:y:3:1::X:u:v:1+u+tv then will 
€= Xe, + Ver +263 
wherein x+y+z=1. 
II. Lemma. Let ABC be a triangle, d a straight line in the plane 
of ABC, P, Q and R the feet of perpendiculars let fall on d from A, 
B and C respectively, and p, g and 7 the numerics of (PA), (QB) and 
(RC), then will 
a(p—q) (b—r)+b(g—b) (g—r)+c(r—p) (r—g) = 40? 
er 
(OR ZBC) = 
(OR'Z CA) =6.=A+B+6, 
(ORZAB)=6,=8 ee. 
2 . 0—0=7—C 
03—0= 7 — A 
and 03—0=7+B 
and a sin 4 —=QB—-RC=q—7r 

b sin @=RC—PA=r—p 
c sin 4 =PA—-QB=p—qQ _ 
‘. ap sin 6,+bq sin 6:.+cr sin 6;=0. 
Also 
a cos #,=QR=QP+PR 
b cos =RP=RQ+QP 
c cos = PQ=PR+RQ 
.". ap cos 0,+bg cos 62+cer cos 63=2A. 
". dp?+bqg +cr? —2bcqgr cos À —2acpr cos B 
— 2abpq cos C=4A? 
On substituting b?-+c?—a* for 2bc cos A, etc., and reducing, this 
becomes 
epg) (Pon) SD) (a=7)- erp) 4g) ane 
Gora ii 
PA): OB TRG 2A oben 2) 
then will 
2A’= J {aX(l—m) (l—n) +0(m—]) (m—n)+c(r—1) (r—m)}. 
