AHSME 1974 Problem 25

In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$. From vertex $C$, line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q$. Lines $DP$ and $CQ$ meet at $O$. If the area of parallelogram $ABCD$ is $k$, then the area of the triangle $QPO$ is equal to

$\mathrm{(A)\ } k \mathrm{(B) \ }\frac{6k}{5} \mathrm{(C) \ } \frac{9k}{8} \mathrm{(D) \ } \frac{5k}{4} \mathrm{(E) \ }2k$