7. [5] In $\triangle A B C, D$ is the midpoint of $B C, E$ is the foot of the perpendicular from $A$ to $B C$, and $F$ is the foot of the perpendicular from $D$ to $A C$. Given that $B E=5, E C=9$, and the area of triangle $A B C$ is 84 , compute $|E F|$.
Answer: $\frac{6 \sqrt{37}}{5}, \frac{21}{205} \sqrt{7585}$
Solution: There are two possibilities for the triangle $A B C$ based on whether $E$ is between $B$ and $C$ or not. We first consider the former case.
We find from the area and the Pythagorean theorem that $A E=12, A B=13$, and $A C=15$. We can then use Stewart's theorem to obtain $A D=2 \sqrt{37}$.
Since the area of $\triangle A D C$ is half that of $A B C$, we have $\frac{1}{2} A C \cdot D F=42$, so $D F=14 / 5$. Also, $D C=14 / 2=7$ so $E D=9-7=2$.
Notice that $A E D F$ is a cyclic quadrilateral. By Ptolemy's theorem, we have $E F \cdot 2 \sqrt{37}=(28 / 5) \cdot 12+$ $2 \cdot(54 / 5)$. Thus $E F=\frac{6 \sqrt{37}}{5}$ as desired.
The latter case is similar.