Question

Prove that in the given figure r =$\frac{a+b-c}{2}$

Answer 1

Let the circle touches the sides BC, CA, AB of the right triangle ABC at D,E and F respectively,
where BC =a, CA =b and AB =c
Since length of trangents drawn from an external point are equal.
$\Rightarrow AE=AF$ and BD =BF.
Also CE=CD =r.
and b-r =AF, a-r =BF
Therefore,
AB =AF +BF
c=b-r+a-r AB =c =AF +BF =b-r+a-r
$\Rightarrow r=\frac{a+b-c}{2}$