AHSME 1961 Problem 21

Medians $AD$ and $CE$ of $\triangle ABC$ intersect in $M$. The midpoint of $AE$ is $N$. Let the area of $\triangle MNE$ be $k$ times the area of $\triangle ABC$. Then $k$ equals: $\textbf{(A)}\ \frac{1}{6} \textbf{(B)}\ \frac{1}{8} \textbf{(C)}\ \frac{1}{9} \textbf{(D)}\ \frac{1}{12} \textbf{(E)}\ \frac{1}{16}$