Question

Let $s$ denote the semi-perimeter of a triangle $ABC$ in which $BC=a,CA=b$ and $AB=c$. If a circle touches the sides $BC,CA$ and $AB$ at $D,E$ and $F$ respectively, prove that $BD=BF=s-b$ and $CD=CE=s-c$.

Answer 1

R.E.F image

Let BD = X as shown in the figure.

BD = BF = X (Tangents drawn from the same point B to the in circle)

Similarly, CD = CE = z and AF = AE = Y

Perimeter of the triangle is 2(x+y+z) = 2s

$\Rightarrow x+y+z=s$

$\Rightarrow x=s-(y+z)$

$\Rightarrow x=s-AC$

$\Rightarrow x=s-b$ (Proved)