Incircle Right Triangle Proof

You don’t need to find the radius of the incircle.

It’s enough to note that $a^2+b^2=(a-r+b-r)^2$

$a^2+b^2=a^2+b^2+4r^2+2ab-4ar-4br$

$0=4r^2-4ar-4br+2ab$

$2ab= 4r^2-4ar-4br+4ab$

$\frac{ab}{2}= r^2-ar-br+ab$

$\frac{ab}{2}= (a-r)(b-r)$

Since you have rendered $r=(a+b-c)/2$, simply substitute into the product $(a-r)(b-r)$:

$(a-r)(b-r)=\left(a-\dfrac{a+b-c}2\right)\left(b-\dfrac{a+b-c}2\right)$

$=\left(\dfrac{a-b+c}2\right)\left(\dfrac{-a+b+c}2\right)$

$=\left(\dfrac{c+(a-b)}2\right)\left(\dfrac{c-(a-b)}2\right)$

You then have $[c+(a-b)][c-(a-b)]=c^2-(a-b)^2=(c^2-a^2-b^2)+2ab$, where $c^2-a^2-b^2=0$ (why?) and thus your product reduces to $2ab/4=ab/2$. This matches the familiar "half the base times altitude" formula for the area, since each leg of a right triangle is the altitude to the other leg.

For a general triangle $\triangle ABC$: if the incircle divides side $\overline{AB}$ into pieces measuring $l$ and $m$, then the area of the triangle is

$S=lm\cot(\frac12\angle C).$

With a right angle at $C$ the argument of the cotangent is $45°$ so that factor reduces to $1$.