Let $A B C D$ be a parallelogram such that $A C=B C$. A point $P$ is chosen on the extension of the segment $A B$ beyond $B$. The circumcircle of the triangle $A C D$ meets the segment $P D$ again at $Q$, and the circumcircle of the triangle $A P Q$ meets the segment $P C$ again at $R$. Prove that the lines $C D, A Q$, and $B R$ are concurrent. Common remarks. The introductory steps presented here are used in all solutions below. Since $A C=B C=A D$, we have $\angle A B C=\angle B A C=\angle A C D=\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain $$\angle C R A=180^{\circ}-\angle A R P=180^{\circ}-\angle A Q P=\angle D Q A=\angle D C A=\angle C B A,$$ so the points $A, B, C$, and $R$ lie on some circle $\gamma$.