What is the area of △ QPO?

Question

Parallelogram $ABCD$ has area of 1.

Centers of the $\overline{AD}$ and $\overline{BC}$ sides are labeled M, N.

The $\overline{CM}$ and $\overline{DN}$ lines intersect at $O$ while their intersections with the $\overline{AB}$ line are marked $Q$, $P$.

What is the area of $\triangle$ $QPO$?

EDIT:

Here's my solution with the help of @Dubs and @Vasya:

Answer 1

Using proportional segments theorem: $BP:AP=BN:AD=1:2$. Thus, $BP=AB$ similarly, $QA=AB$. So $QP=3AB$ and the height is $\frac{3}{4}$ of the height to $AB$. Thus, $A_{\triangle{QPO}}=3 \cdot \frac{3}{4} \cdot \frac{1}{2}=\frac{9}{8}$