HMMT-Nov 2017 guts, Problem 12

Trapezoid $A B C D$, with bases $A B$ and $C D$, has side lengths $A B=28, B C=13, C D=14$, and $D A=15$. Let diagonals $A C$ and $B D$ intersect at $P$, and let $E$ and $F$ be the midpoints of $A P$ and $B P$, respectively. Find the area of quadrilateral $C D E F$.

Solution: Note that $E F$ is a midline of triangle $A P B$, so $E F$ is parallel to $A B$ and $E F=\frac{1}{2} A B=14=C D$. We also have that $E F$ is parallel to $C D$, and so $C D E F$ is a parallelogram. From this, we have $E P=P C$ as well, so $\frac{C E}{C A}=\frac{2}{3}$. It follows that the height from $C$ to $E F$ is $\frac{2}{3}$ of the height from $C$ to $A B$. We can calculate that the height from $C$ to $A B$ is 12 , so the height from $C$ to $E F$ is 8 . Therefore $C D E F$ is a parallelogram with base 14 and height 8 , and its area is $14 \cdot 8=112$. $\fbox{112}$.

Answer: 112