In a right triangle with sides $a$ and $b$, and hypotenuse $c$, the altitude drawn on the hypotenuse is $x$. Then: 
$\textbf{(A)}\ ab = x^2 \qquad \textbf{(B)}\ \frac {1}{a} + \frac {1}{b} = \frac {1}{x} \qquad \textbf{(C)}\ a^2 + b^2 = 2x^2 \\ \textbf{(D)}\ \frac {1}{x^2} = \frac {1}{a^2} + \frac {1}{b^2} \qquad \textbf{(E)}\ \frac {1}{x} = \frac {b}{a}$
The area of this triangle can be expressed in two ways; the first being $\frac{ab}{2}$ (as this is a right triangle), and the second being $\frac{cx}{2}$. But by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $c=\sqrt{a^2+b^2}$. Thus a second way of finding the area of the triangle is $\frac{x\sqrt{a^2+b^2}}{2}$. By setting them equal to each other we get $x=\frac{ab}{\sqrt{a^2+b^2}}$, and we can observe that the correct answer is $\boxed{\textbf{(D) \ }}$.
~anduran