Olympiad Geometry · Hard problem

Triangle $ABC$ has $AB = 13$, $BC = 14$, and $CA = 15$. Let $D$ be the reflection of $A$ over the perpendicular bisector of $\overline{BC}$ (so $DB = AC$ and $DC = AB$, with $D$ on the same side of $\overline{BC}$ as $A$). Find $AD$.