Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}$ if $\measuredangle C = 90^{\circ}$

Question

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?

Answer 1

The first part:

$$(a+b+c)(a+b-c)=(a+b)^2-c^2 = a^2+b^2 + 2ab -c^2=2ab$$ by Pythagora's theorem.

The second part: $r_c=\frac{A}{s-c}=\frac{ab}{a+b-c}$ and it works with absolutely the same calculation above.

Answer 2

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