The Pythagorean Theorem by Reductio ad Absurdum

Other proofs by contradiction can be found in cut-the-knot.org (see proof #122) and in Loomis' collection (see proofs # 16/2 and # 32).

Let $\triangle{ABC}$ be a right-triangle with $\angle{ACB}=90^\circ$. Let $D$, $E$ and $F$ be the contact points of the incircle with $BC$, $AC$ and $AB$, respectively. Also, let $AE=AF=x$; $BD=BF=y$; $CD=CE=r$, where $r$ is the inradius of $\triangle{ABC}$.

Assume to the contrary that $a^2+b^2>c^2$. Then, 

$$(r+y)^2+(x+r)^2>(x+y)^2.$$

Expanding, collecting like terms and simplifying we get

$$ry+rx+r^2>xy.$$

Notice that $ry+rx+r^2=r(y+x+r)=rs=\Delta$, where $\Delta$ denotes the area of $\triangle{ABC}$ and $s$ its semiperimeter. Moreover, it is well-known that $xy=\Delta$. So $ry+rx+r^2>xy$ is equivalent to write $\Delta>\Delta$, which is a contradiction. A similar situation arise if you assume $a^2+b^2<c^2$.

There is a simple direct proof starting from $\Delta=rs$ and $\Delta=xy$, can you find it? My sincerest thanks to Andrius Navas and José Hernández for pointing me out this.