17.156 In $\triangle ABC$, the medians $AD$ and $CE$ intersect at $M$, and the midpoint of $AE$ is $N$. If the area of $\triangle MNE$ is $k$ times the area of $\triangle ABC$, then $k$ equals
(A) $\frac{1}{6}$.
(B) $\frac{1}{8}$.
(C) $\frac{1}{9}$.
(D) $\frac{1}{12}$.
(E) $\frac{1}{16}$.
(12th American High School Mathematics Examination, 1961)
[Solution] $S_{\triangle M N E}=\frac{1}{2} S_{\triangle A M E}=\frac{1}{2}\left(\frac{1}{3} S_{\triangle A E C}\right)$ $=\frac{1}{2}\left[\frac{1}{3}\left(\frac{1}{2} S_{\triangle A B C}\right)\right]$ $=\frac{1}{12} S_{\triangle A B C}$.

Therefore, the answer is $(D)$.