Yet Another Proof of the Pythagorean Theorem - Simpler version

In cut-the-knot.org you can find more than a hundred proofs of the Pythagorean theorem. Here I give another proof which I hope to be new.

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Below I will be using standard notations: $a$, $b$, $c$, for the side length, $s$ for the semiperimeter, $\Delta$ for the area and $r$ for the inradius.

$$s=c+r$$

$$s-\frac{c}{2}=\frac{c}{2}+r$$

$$\frac{a}{2}+\frac{b}{2}=\frac{c}{2}+r$$

$$a+b=c+2r$$

Squaring both sides, 

$$a^2+2ab+b^2=c^2+4cr+4r^2$$

It is well-known $\Delta=rs$, which implies $2ab=4r(c+r)$. So the equation $a^2+2ab+b^2=c^2+4cr+4r^2$ can be rewritten like this $a^2+b^2=c^2$, which is the Pythagorean theorem.

Below is the comment by John Molokach, whom I consider an expert on this topic (he has proved the theorem in more than ten different ways).