$18 \cdot 37$ In the figure, $ABCD$ is a parallelogram, line $DP$ bisects $BC$ at $N$ and intersects the extension of $AB$ at $P$, $CQ$ bisects side $AD$ at $M$ and intersects the extension of $BA$ at $Q$. $DP$ and $CQ$ intersect at $O$. If the area of parallelogram $ABCD$ is $k$, then the area of $\triangle QPO$ is
(A) $k$.
(B) $\frac{6}{5} k$.
(C) $\frac{9}{8} k$.
(D) $\frac{5}{4} k$.
(E) $2 k$.
(25th American High School Mathematics Examination, 1974)
[Solution] Since $D M=A M, \angle Q M A=\angle D M C$ and $\angle C D M=$ $\angle Q A M$, then $\triangle Q A M \cong \triangle C D M$.
Similarly, $\triangle B P N \cong \triangle C D N$.
Also,

and
$S_{\triangle Q P O}=S_{\triangle A B C D}+S_{\triangle D O K}$,

Therefore
Hence, the answer is $(C)$.
$$
\begin{array}{c}
S_{\triangle D O C}=\frac{1}{4}\left(\frac{1}{2} S_{\square \triangle A B C D}\right)=\frac{k}{8}, \\
S_{\triangle Q P O}=k+\frac{k}{8}=\frac{9 k}{8} .
\end{array}
$$