I got $\boxed{60^\circ}$ I had let CD be the perpendicular bisector and the point of concurrency be E. I let CE be 2x and ED be x. $\triangle{AED} \cong \triangle{AEH}$ by AAS similarity. As a result, $\triangle{EHC}$ has lengths of $x$ and $2x$ with a $90^\circ$ angle corresponding to the $2x$. This means $\triangle{EHC}$ a 30-60-90 triangle with $\angle{HEC}$ being $60^\circ$. Since $\triangle{AED} \cong \triangle{AEH}$, $\angle{AED} = \angle{AEH}$. Notice that $\angle{AED} + \angle{AEH} + 60 = 180^\circ$. This means both of those angles are $60^\circ$ since they are equivalent. $\triangle{AED}$ and $\triangle{AEH}$ are then 30-60-90 triangles. $\angle{BAC} = \angle{DAE} + \angle{CAE} = 30 + 30 = \boxed{60^\circ}$. *With some more experimenting, you can find out that $\triangle{ABC}$ is an equilateral triangle.*