Problem (2013G4): 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 an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; p = on_line a c, on_aline p b a b c a; q = on_line a c, on_aline q b a a c b; d = on_line b q, on_circle p b; r = on_line a d, on_circle o a ? cong q b q r
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; AC = line A C; l1 = angle_equal1 B A B C A; P = intersection AC l1; l2 = angle_equal1 B A A C B; Q = intersection AC l2; BQ = line B Q; (P) = circle_center_point P B; D = intersection BQ (P); AD = line A D; R = intersection AD (O); Prove: cong Q B Q R
Difficulty score: 2.8