HMMT-Nov 2019 team, Problem 8

In $\triangle A B C$, the external angle bisector of $\angle B A C$ intersects line $B C$ at $D . E$ is a point on ray $\overrightarrow{A C}$ such that $\angle B D E=2 \angle A D B$. If $A B=10, A C=12$, and $C E=33$, compute $\frac{D B}{D E}$.

Solution: Let $F$ be a point on ray $\overrightarrow{C A}$ such that $\angle A D F=\angle A D B . \triangle A D F$ and $\triangle A D B$ are congruent, so $A F=10$ and $D F=D B$. So, $C F=C A+A F=22$. Since $\angle F D C=2 \angle A D B=\angle E D C$, by the angle bisector theorem we compute $\frac{D F}{D E}=\frac{C F}{C E}=\frac{22}{33}=\frac{2}{3}$. $\fbox{\frac{2}{3}}$.

Answer: \frac{2}{3}