For $\triangle ABC$ with right angle at $A$ and inradius $r$, show that $2r=b+c-a$

Question

I am trying to solve this problem:

The angle $A$ of $\triangle ABC$ is a right angle, and the sides of $BC$, $CA$ and $AB$ are of lengths $a$, $b$, and $c$, respectively. Each side of the triangle is tangent to a circle of radius r. Show that $2r=b+c-a$.

I have no idea on how to solve this problem, and could you guys help me with to solve it?

Thank you so much for you guys’ replies.

Answer 1

Use $r=(S-a)tan(\frac{A}{2})$ $$ $$ Where $\angle A =90°$ hence $$r=s-a$$ $$2r=b+c-a$$