AUT_ABooklet_2023

Since $ACQ$ is an isosceles triangle, the perpendicular bisector of $AQ$ is the angle bisector of $\angle QCA$. But the perpendicular bisector of $AQ$ also passes through the circumcenter $O$ of the triangle $APQ$.

Therefore, $O$ lies on the angle bisector of $\angle QCA$ which is the exterior angle bisector of $\angle ACB$ by definition of $Q$.

Analogously, the point $O$ lies also on the exterior angle bisector of $\angle CBA$. Therefore, the point $O$ is the intersection of the two exterior angle bisectors which makes it the excenter of the excircle of $ABC$ tangent to $BC$. This excenter lies on the angle bisector of $\angle BAC$ as desired.

(Theresia Eisenkölbl) $\square$