Let $O$ be the circumcenter of triangle $APQ$. Then, $\angle BAO=\angle PAO-\angle PAB=(90^\circ-\angle AQP)-\angle PAB=90^\circ-\dfrac{\angle B+\angle C}{2}=\dfrac{\angle A}{2},$ as desired.

The perpendicular bisectors of $AP$ and $AQ$ are the bisectors of the exterior angles of $\widehat{B}$ and $\widehat{C}$, respectively, therefore the circumcenter of $APQ$ is the $A-excenter$ of triangle $ABC$.

Let N be the intersection of the angle bisector in A with the line perpendicular to AP through P. Simple angle chasing gives that APQN is cyclic. By Thales' theorem the center of this circle is the midpoint of AN, which is on the angle bisector.