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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.

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  • $\begingroup$ And what is P, huh? $\endgroup$ – Stefan4024 Mar 3 '15 at 12:13
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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.

Therefore α=β.

We can then conclude AQ/QN = 2

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  • $\begingroup$ Can you tell how you figured out the last step? "We can then conclude AQ/QN = 2" $\endgroup$ – Stav Alfi Apr 17 '16 at 19:22

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