[192 
192] 
GENERAL TBILINEAR EQUATION OF THE SECOND DEGREE. 
147 
19—2 
iO = o, 
Writing a, b, c (instead of l, m, n) for the sides of the fundamental triangle and 
A, B, G for the angles, the equation in question is 
r 2 = be cos A tj 2 + ca, cos B rf + ab cos C £ 2 . 
Now writing a, /3, 7 for the inclinations of the line r to the sides of the triangle, we 
have 
A = /3 - 7, 
B = 7 - a, 
G =7r + OL — /3. 
Moreover taking for a moment X, /a, v to denote the perpendiculars from the angles 
on the opposite sides, we have 
X = c sin B = b sin G, 
and 
n — a sin G = c sin A, 
v = b sin A = a sin B, 
f = 
r sm a 
r sin ¡3 
. _ r sm 7 
v 
the values of £' 3 , ?? 2 , £ 2 consequently are 
r 2 sin 2 a. r 2 sin 2 ¡3 r 2 sin 2 7 
be sin B sin G ’ ca sin G sin A ’ ab sin A sin B ' 
and the equation to be proved becomes 
_ cos A sin 2 a. cos B sin 2 /3 cos G sin 2 7 
sin B sin C sin G sin A sin A sin B ’ 
or, what is the same thing, 
sin A sin B sin G — sin A cos A sin 2 a + sin B cos B sin 2 ¡3 + sin G cos C sin 2 7, 
or again 
4 sin A sin BsinC = sin 2 A (1 — cos 2a) + sin 2 B (1 — cos 2/3) + sin 2 C (1 — cos 27), 
or putting for A, B, G their values in terms of a, ¡3, 7 this is 
gation 
— 4 sin (/3 — 7) sin (7 — a) sin (a — /3) = sin (2/3 — 27) (1 — cos 2a ) 
+ sin (27 - 2a ) (1 - cos 2/3) 
+ sin (2a — 2/3) (1 — cos 27),