In triangle $ABC$ ($AB\ne AC$), where all angles are greater than $45^\circ$, the altitude $AD$ is drawn. Let $\omega_1$ and $\omega_2$ be-- circles with diameters $AC$ and $AB$, respectively. The angle bisector of $\angle ADB$ secondarily intersects $\omega_1$ at point $P$, and the angle bisector of $\angle ADC$ secondarily intersects $\omega_2$ at point $Q$. The line $AP$ intersects $\omega_2$ at the point $R$. Prove that the circumcenter of triangle $PQR$ lies on line $BC$.
