Q:

Find the area of a quadrilateral in a parallelogram.

Assume quadrilateral $ABCD$ is a parallelogram and its area is $S$. And satisfy the following conditions: $AE=BE$, $BF=FC$, $AQ // PC$. Find the area of quadrilateral $APCQ$

A:

You have tried a special case and you can get some hint from it. To calculate the area of $APCQ$, we can divide it into two parts: $\triangle{APQ}$ and $\triangle{CPQ}$. Let the intersection point of line $(ED)$ and line$(BC)$ be $T$. By the Menelaus' theorem we have $\frac{AE}{EB} \cdot \frac{BT}{TF} \cdot \frac{FP}{PA}=1$, since $E,F$ are midpoints, we may get that $\frac{FP}{PA}=\frac{3}{2}$. Let the intersection point of the line $(CP)$ and line$(AB)$ be $S$. Apply the Menelaus' theorem again we can get $\frac{AS}{SB} \cdot \frac{BC}{CF} \cdot \frac{FP}{PA}=1$, which implies that $\frac{AS}{SB}=\frac{1}{3}$. Since E is the midpoint of segment $AB$, $S$ is the midpoint of $AE$. As $AQ \parallel CS$, $P$ is the midpoint of $EQ$. The rest is just simple calculation and you can try by yourself. The answer should be the same as what you get in the special case.

