Let $A B C D$ be a parallelogram. Let $E$ be the midpoint of $A B$ and $F$ be the midpoint of $C D$. Points $P$ and $Q$ are on segments $E F$ and $C F$, respectively, such that $A, P$, and $Q$ are collinear. Given that $E P=5, P F=3$, and $Q F=12$, find $C Q$.
Solution: Triangles $P F Q$ and $P E A$ are similar, so $A E=F Q \cdot \frac{P E}{P F}=12 \cdot \frac{5}{3}=20$. Now, $C Q=C F-Q F=$ $20-12=8$. $\fbox{8}$.