Let $ABCD$ be a parallelogram, and let $M$ and $N$ be the midpoints of $BC$ and $CD$, respectively. Let $P$and $Q$ be the intersections of $BD$ with $AM$ and $AN$, respectively. Then find the ratio of the area of quadrilateral $MNQP$ to the area of parallelogram $ABCD$. How do I find the ratio $AQ/QN$, else everything I have figured out.
Let the intersection of the diagonals of the parallelogram be O.
Then, Area(ΔCOQ) = Area(ΔAOQ) = α.
Also, Area(ΔNQC) = Area(ΔDQN)= β.
Area(ΔAQD) = Area(ABCD)/4 - β.
Area(ΔAOD) = Area(ABCD)/4 = Area(ABCD)/4 - β + α.
We can then conclude AQ/QN = 2