hello, we have $ {\frac{ab}{2}=\frac{a^2+ab+ac+ab+b^2+bc-ac-bc-c^2}{4}=\frac{(a+b+c)(a+b-c)}{4}=p(p-c)}$ $ {\frac{ab}{2}=\frac{-a^2+ab+ac+ab-b^2-bc-ac+bc+c^2}{4}=\frac{(-a+b+c)(a-b+c)}{2}=(p-a)(p-b)}$ so we have $ S=p(p-c)=(p-a)(p-b)$. Sonnhard.

Let $ r$ be the radius of the incircle of the right triangle. Then: $ S = rp$. Hence: $ r = \frac {S}{p} = \frac {ab}{a + b + c} = \frac {ab(a + b - c)}{(a + b + c)(a + b - c)} = \frac {ab(a + b - c)}{(a + b)^2 - c^2}$ $ = \frac {ab(a + b - c)}{a^2 + b^2 - c^2 + 2ab} = \frac {a + b - c}{2} = p - c$. So $ p(p - c) = rp = S$. Since $ S^2 = p(p - a)(p - b)(p - c)$, so $ (p - a)(p - c) = S = p(p - c)$.

If ABC is the triangle, with C=90 degs, I - incenter, M and N the contact point of the incircle with BC, CA, then CNIM is a square, hence r=CN=p-c; further, follow shobber's. Best regards, sunken rock