AHSME 1961 Problem 16

An altitude $h$ of a triangle is increased by a length $m$. How much must be taken from the corresponding base $b$ so that the area of the new triangle is one-half that of the original triangle? $\textbf{(A)}\ \frac{bm}{h+m} \textbf{(B)}\ \frac{bh}{2h+2m} \textbf{(C)}\ \frac{b(2m+h)}{m+h} \textbf{(D)}\ \frac{b(m+h)}{2m+h} \textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}$