30. [15] Let $A B C$ be a triangle with $A B=8, A C=12$, and $B C=5$. Let $M$ be the second intersection of the internal angle bisector of $\angle B A C$ with the circumcircle of $A B C$. Let $\omega$ be the circle centered at $M$ tangent to $A B$ and $A C$. The tangents to $\omega$ from $B$ and $C$, other than $A B$ and $A C$ respectively, intersect at a point $D$. Compute $A D$.
Answer:
16
Solution: Redefine $D$ as the reflection of $A$ across the perpendicular bisector $l$ of $B C$. We prove that $D B$ and $D C$ are both tangent to $\omega$, and hence the two definitions of $D$ align. Indeed, this follows by symmetry; we have that $\angle C B M=\angle C A M=\angle B A M=\angle B C M$, so $B M=C M$ and so $\omega$ is centered on and hence symmetric across $l$. Hence reflecting $B A C$ across $l$, we get that $D B, D C$ are also tangent to $\omega$, as desired.
Hence we have by Ptolemy that $5 A D=12^{2}-8^{2}$, so thus $A D=16$.