Picture proof that the area of a right triangle is $xy$

Question

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 $y$.

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$^1$ 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.

Answer 1

Let A be the area of the original triangle cut up in the obvious way as on the picture. Then the area of the rectangle is $A_r = 2 A = ab = xy + A$. Thus $xy = A$.

Answer 2

My solution uses a little bit of algebra.

If you superpose the triangles as shown in the figure, the area of the blue square is counted twice, but the green area is not counted.

The green area is (using Pythagoras) $(x-z)(y-z) = 2z^2$. One $z^2$ is the area of the triangles that we haven't count yet, and the other $z^2$ comes from the double counting.

Answer 3

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, $W = (x-z)(y-z) = 2A -S -A = 2z^2$.