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:
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{8}\qquad \textbf{(C)}\ \frac{1}{9}\qquad \textbf{(D)}\ \frac{1}{12}\qquad \textbf{(E)}\ \frac{1}{16}$
Let $[NME] = a$ and $[BMD] = b$.  The altitudes and bases of $\triangle ANM$ and $\triangle NME$ are the same, so $[NME] = [ANM] = a$.  Since the base of $\triangle BEM$ is twice as long as $\triangle NME$ and the altitudes of both triangles are the same, $[BEM] = 2a$.
Since the altitudes and bases of $\triangle BMD$ and $\triangle CMD$ are the same, $[BMD] = [CMD] = b$.  Additionally, since $\triangle AME$ and $\triangle EMB$ have the same area, $\triangle AMC$ and $\triangle BMC$ also have the same area, so $[AMC] = 2b$.
Also, the altitudes and bases of $\triangle ABD$ and $\triangle ADC$ are the same, so the area of the two triangles are the same.  Thus,
\[a+a+2a+b=2b+b\]
\[4a=2b\]
\[2a = b\]
Thus, the area of $\triangle ABC$ is $12a$, so the area of $\triangle NME$ is $\frac{1}{12}$ the area of $\triangle ABC$.  The answer is $\boxed{\textbf{(D)}}$.