Školjka

%V0 Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB$ and $A$ is located between $P$ and $C$. Suppose that there exists and interior point $D$ of segment $BQ$ for which $PD = PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.