Suppose that $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively of $\triangle ABC$.The perpendiculars to the sides $AB$and $AC$ at $P$ and $Q$ respectively meet at $D$, an interior point of $\triangle ABC$. If $M$ is the midpoint of $BC$, prove that $PM = QM$ if and only if $\angle BDP = \angle CDQ$.
I'm hopeless at geometry. All I could do is figure out that $\triangle PDB \sim \triangle QDC$ and that $\square APDQ$ is cyclic. Any help would be really appreciated. Thanks!
Requested picture from Geogebra shows that $\alpha\neq\beta$.