Plane trigonometry, by S.L. Loney.

RADIUS OF THE INCIRCLE.

231

Hence, by addition, we have
-r. a + r. b + ~r. c= sum of the areas of the triangles
IBC, ICA, and JAB
= area of the A ABC.
a+b +c
z.e. r = S
so that r.s =S.
- S
S,e r=-.
203. Since the angles IBD and IDB are respectively
equal to the angles IBF and IFB, the two triangles IDB
and IFB are equal in all respects.
Hence BD = BF, so that 2BD = BD + BF.
So also AE = AF, so that 2AE = AE +AF,
and CE = CD, so that 2CE = CE + CD.
Hence, by addition, we have
2BD + 2AE+2CE=(BD +CD)+ (CE+ AE) + (AF+ FB),
i.e. 2BD + 2A C = BC+CA + AB..'. 2BD + 2b = a + b + c =2s.
Hence BD = s - b = BF;
so CE= s - c = CD,
and AF=s-a=AE.
ID B
Now tBD = tan IBD = tan B
BD 2B
B B
r= ID=BDtan -=(s-b)tan-.
2i 2 -