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 

$\mathrm{(A)\ } k \qquad \mathrm{(B) \ }\frac{6k}{5} \qquad \mathrm{(C) \  } \frac{9k}{8} \qquad \mathrm{(D) \  } \frac{5k}{4} \qquad \mathrm{(E) \  }2k$
Note that \[[QPO]=[QAM]+[PBN]+[AMONB]=[AMONB]+[MDC]+[NCD]\]
\[=[AMONB]+[MDC]+[NOC]+[DOC]=[ABCD]+[DOC]=k+[DOC].\]
Also, note that $DCNM$ is a parallelogram, and so $[DOC]=\frac{1}{4}[DCNM]=\frac{1}{8}[ABCD]=\frac{k}{8}$.
Therefore, $[QPO]=k+\frac{k}{8}=\frac{9k}{8}, \boxed{\text{C}}$.