In \(\triangle ABC\), \(\angle B >\angle C\). Let \(P\) and \(Q\) be two points on line \(CA\) such that \(\angle PBA=\angle QBA=\angle ACB\) and \(A\) is located between \(P\) and \(C\). Suppose that there exists an interior point \(D\) of segment \(BQ\) for which \(PD=PB\). Let the ray \(AD\) intersect the circumcircle of \(\triangle ABC\) at \(R\neq A\). Prove that \(QB=QR\).