1955 AHSME Problem 48

Given \triangle ABC with medians AE, BF, CD; FH parallel and equal to AE; BH and HE are drawn; FE extended meets BH in G. Which one of the following statements is not necessarily correct?

\begin{array}{l} \textbf{(A)}\ AEHF\text{ is a parallelogram}\qquad\textbf{(B)}\ HE=HG\\ \textbf{(C)}\ BH=DC\qquad\textbf{(D)}\ FG=\frac{3}{4}AB\qquad\textbf{(E)}\ FG\text{ is a median of triangle }BFH \end{array}

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