Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}$ if $\measuredangle C = 90^{\circ}$ 
In$\triangle ABC$, $\angle C$ is a right angle. Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}.$ Legs are named in traditional way. 

My Work
As $r = \frac{A}{s}$. So, here $r = \frac{ab}{a+b+c}$. Is there any way to prove this is equal to $\frac{a+b-c}{2}$?. Or else how to solve this is also the second part? 

$$r=\frac{2S}{a+b+c}=\frac{ab}{a+b+c}=\frac{ab(a+b-c)}{a^2+2ab+b^2-c^2}=\frac{a+b-c}{2}$$
$$r_c=\frac{2S}{a+b-c}=\frac{ab}{a+b-c}=\frac{ab(a+b+c)}{a^2+2ab+b^2-c^2}=\frac{a+b+c}{2}$$
