1961 AHSME 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}\qquad \textbf{(B)}\ \frac{bh}{2h+2m}\qquad \textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad \textbf{(D)}\ \frac{b(m+h)}{2m+h}\qquad \textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.