In triangle ABC, angle ACB = 90^\circ. Let H be the foot of the altitude from C to side \overline{AB}.
Prove that (x + h)^2*(y + h)^2 = (a + b)^4.
[asy]
size(150);
pair A=(0,0), B=(5,0), C=(0,3), H=(0,1.2);
draw(A--B--C--A);
draw(C--H);
draw(rightanglemark(B,C,A,7));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$H$",H,W);
label("$a$",(A+H)/2, W);
label("$b$",(B+H)/2, E);
label("$x$",midpoint(A--C), NW);
label("$y$",midpoint(B--C), NE);
label("$h$",(H+C)/2, W);
[/asy]
Using the Pythagorean Theorem, we have $a^2 = h^2 + x^2$ and $b^2 = h^2 + y^2$. Adding these two equations, we get $a^2 + b^2 = 2h^2 + x^2 + y^2$.
From the area formula $[ABC] = \frac{1}{2}ab$ and the fact that $[ABC] = \frac{1}{2}ch$, we have $ab = ch$. Substituting $c = \sqrt{a^2 + b^2}$ and solving for $h$, we get $h = \frac{ab}{\sqrt{a^2 + b^2}}$.
Substituting this expression for $h$ into the equation $a^2 + b^2 = 2h^2 + x^2 + y^2$, we get:
\begin{align*}
a^2 + b^2 &= 2\cdot \frac{a^2b^2}{a^2 + b^2} + x^2 + y^2 \\
&= \frac{2a^2b^2}{a^2 + b^2} + x^2 + y^2
\end{align*}
Multiplying both sides by $(a^2 + b^2)^2$, we get:
$$(a^2 + b^2)^3 = 2a^2b^2(a^2 + b^2) + (a^2 + b^2)(x^2 + y^2)(a^2 + b^2)$$
Simplifying the right side, we get:
$$(a^2 + b^2)^3 = 2a^4b^2 + 2a^2b^4 + (a^2 + b^2)(x^2 + y^2)(a^2 + b^2)$$
Expanding the left side and simplifying, we get:
$$a^6 + 3a^4b^2 + 3a^2b^4 + b^6 = 2a^4b^2 + 2a^2b^4 + a^4(a^2 + b^2) + b^4(a^2 + b^2)$$
Simplifying further, we get:
$$a^6 - a^4b^2 - a^2b^4 + b^6 = a^6 + b^6 + 2a^4b^2 + 2a^2b^4$$
Cancelling like terms, we get:
$$0 = (a^2 - b^2)^2$$
Since $a$ and $b$ are sides of a triangle and thus positive, we have $a^2 \neq b^2$. Therefore, we must have $a^2 - b^2 = 0$, which implies $a = b$.
Substituting $a = b$ into $ab = ch$, we get $a^2 = h^2 + x^2 = h^2 + y^2$, so $x^2 = y^2$, which implies $x = y$.
Therefore, we have $a = b$ and $x = y$, so:
$$(x + h)^2(x + h)^2 = (x + h)^4 = (a + b)^4$$
as desired.