1975 AHSME Problem 28

In \triangle ABC shown in the adjoining figure, M is the midpoint of side BC, AB=12 and AC=16. Points E and F are taken on AC and AB, respectively, and lines EF and AM intersect at G. If AE=2AF then \frac{EG}{GF} equals

\begin{array}{l} \textbf{(A)}\ \frac{3}{2} \qquad \textbf{(B)}\ \frac{4}{3} \qquad \textbf{(C)}\ \frac{5}{4} \qquad \textbf{(D)}\ \frac{6}{5}\\ \qquad \textbf{(E)}\ \text{not enough information to solve the problem} \end{array}

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