Problem (2014G3): Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; m = on_circle o a, angle_bisector a b c; p = on_dia b m, on_bline a b; q = on_dia b m, on_bline b c; r = on_line p q, on_bline b m ? para b r a c
HAGeo formalization: A B C = acute_triangle; O = circumcenter A B C; M = midarc_no A C B; (BM) = circle_diameter B M; AB = line A B; l1 = perpendicular_bisector A B; P = intersection l1 (BM); BC = line B C; l2 = perpendicular_bisector B C; Q = intersection l2 (BM); l3 = perpendicular_bisector B M; PQ = line P Q; R = intersection PQ l3; BR = line B R; AC = line A C; Prove: parallel BR AC
Difficulty score: 3.1