AHSME 1950 Problem 47

A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then: $\textbf{(A)}\ x=\dfrac{1}{2}h \textbf{(B)}\ x=\dfrac{bh}{b+h} \textbf{(C)}\ x=\dfrac{bh}{2h+b} \textbf{(D)}\ x=\sqrt{\dfrac{hb}{2}} \textbf{(E)}\ x=\dfrac{1}{2}b$