In the right triangle shown the sum of the distances $BM$ and $MA$ is equal to the sum of the distances $BC$ and $CA$. 
If $MB = x, CB = h$, and $CA = d$, then $x$ equals: 

$\textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h$
The question states that \[h+d = x+\sqrt{(x+h)^2+d^2}\]
We move $x$ to the left: \[h+d-x = \sqrt{(x+h)^2+d^2}\]
We square both sides: \[h^2 + d^2 + x^2 - 2xh - 2xd + 2hd = x^2 + 2xh + h^2 + d^2\]
Cancelling and moving terms, we get: \[4xh + 2xd = 2hd\]
Factoring $x$: \[x(4h+2d) = 2hd\]
Isolating for $x$: \[x=\frac{2hd}{4h+2d}=\frac{hd}{2h+d}\]
Therefore, the answer is $\fbox{A}$