Question

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

Answer 1

Let $BD=x$

From fig $BD=BF=x$

Similarly $CD=CE=z$ and $AF=AE=y$

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

$x+y+z=s$

$x=s-(y+z)$

$x=s-AC$

$x=s-b$