<s>[INST] Picture proof that the area of a right triangle is $xy$ I stumbled on the following result by accident:

Let $A, B, C$ be the vertices of a right triangle, with opposite side lengths $a, b, c$ respectively, where $\angle C = 90^\circ$ and $a^2 + b^2 = c^2$.
Draw the incircle, and let $x, y, z$ be the length of the tangent from the vertices $A$, $B$, and $C$ respectively to the incircle. (So $a = y + z$, $b = x + z$, and $c = x + y$.)
Then the area of the triangle is $\boldsymbol{xy}$.

I can prove this algebraically,$^1$ but is there a picture proof of this fact?
What I have in mind is that we cut up the triangle $ABC$ into finitely many pieces, and rearrange them into a rectangle with sides $x$ and $y$.

$^1$ Using $x = \frac{b+c-a}{2}$ and $y = \frac{a+c-b}{2}$, we get $xy = \frac14\left(c - (a-b)\right)\left(c + (a-b)\right) = \frac14\left(c^2 - a^2 - b^2 + 2ab\right)$. From $c^2 = a^2 + b^2$ this reduces to $\frac14 (2ab) = \frac12 ab$, which is the area of the triangle.
 [/INST] Here is an answer to @6005's comment "Maybe there is a way to also show $(x-z)(y-z)=z^2$ pictorally".
Referring to the pic below, it's immediate from the construction that, comparing it with the cut up area, A, of the triangle, the shaded area, $S$ equals $A-2z^2$. Then, $W = (x-z)(y-z) = 2A -S -A = 2z^2$.

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