Plane trigonometry, by S.L. Loney.

238

TRIGONOMETRY.

Again, from the triangle ALM, we have
LM AL ABcosA
sin A sin AM1L cos PML
c cos A c cos A
cos PAL sin C.. LM= n sinA cosA
sin C
= a cos A (Art. 163).
So MK= b cos B, and KL = c cos B.
The sides of the pedal triangle are therefore a cos A,
b cosB, and c cos C; also its angles are the supplements
of twice the angles of the triangle.
211. Let I be the centre of the incircle and I1, I2 and
I3 the centres of the escribed circles
-which are opposite to A, B and C 1
respectively. As in Arts. 202 and A: —.... A
205 IC bisects the angle A CB, and \......
IC bisects the angle BCM.. I
z.. ICI, = z IC B+ 1,CB B I-
= - AGA CB + z MCB
= [z.ACB+ ZMCB] M
= ~.180~= a right angle.:..
Similarly Z ICI2 is a right
angle.
Hence 1GCI2 is a straight line to which IC is perpendicular.
So I2AI is a straight line to which IA is perpendicular, and I3BIT is a straight line to which IB is perpendicular.