How do you derive that the inradius in a right triangle is $r=\frac{a+b-c}2$?

Let the tangent points be $A'$, $B'$ and $C'$ labelled in the usual way. Then, since tangents from a point to a circle have equal length, $CB'=r=CA'$.

Therefore, for the same reason, $$AB'=AC'=b-r$$ and $$BA'=BC'=a-r$$ And since $AC'+BC'=c$, we get $a-r+b-r=c$ and hence the result.

$$\large \Delta=\color{green}\blacktriangle+\color{red}\blacktriangle+\color{blue}\blacktriangle$$ $$\frac{ab}2=\frac{ar}2+\frac{br}2+\frac{cr}2$$ $$ab=(a+b+c)r$$ $$ab(a+b-c)=(a+b+c)(a+b-c)r$$ $$ab(a+b-c)=[(a+b)^2-c^2]r$$ $$ab(a+b-c)=(\underbrace{a^2+b^2-c^2}_{0}+2ab)r$$ $$ab(a+b-c)=2ab\cdot r$$ $$r=\frac{a+b-c}2$$

Here's an approach that I don't think I've seen before (but don't doubt exists in the literature):

By the squares method, $$r^2+r(a-r)+r(b-r)=r\,\dfrac{a+b+c}2,$$ $$2(a+b-r)=a+b+c,$$ $$\color{green}{\mathbf{r=\dfrac{a+b-c}2.}}$$

Then $$R+r=\dfrac c2+r=\dfrac{a+b}2\le\sqrt{ab\mathstrut}\,=\sqrt{2S}\,.$$

$$\begin{align*}r&=\frac{\Delta}s\\&=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \\&=\sqrt{\frac{({a+b-c})({a-b+c})({-a+b+c})}{8s}} \\&=\sqrt{\frac{({a+b-c})[c^2-({a-b})^2]}{8s}} \\&=\sqrt{\frac{({a+b-c})(2ab)}{8s}} \\&=\sqrt{\frac{({a+b-c})}{2}.\frac{\Delta}{s}} \\&=\sqrt{\frac{({a+b-c})}{2}.r}\end{align*}$$ $$\implies r=\frac{a+b-c}2,$$