We have $r=\frac{\frac{ab}2}{\frac{a+b+c}2}=\frac{ab}{a+b+c}$, so it suffices to show that $2ab=(a+b+c)(a+b-c)=(a+b)^2-c^2\Leftrightarrow a^2+b^2=c^2$, which is just the Pythagorean theorem.

Alternatively, we also know that if $I$ is the incenter, and $D$ and $E$ are the touchpoint of the incircle with $BC$ and $AC$ respectively, then $ID=IE$ and $\angle C=\angle DIE=90$ which makes $IDCE$ a square, and $ID=CD$. It is fairly well known (can be proven with straight algebra) that $CD=\frac{a+b-c}{2}$ and we are done