Problem (2016RMMp1): Let $ABC$ be a triangle and let $D$ be a point on the segment $BC, D\neq B$ and $D\neq C$. The circle $ABD$ meets the segment $AC$ again at an interior point $E$. The circle $ACD$ meets the segment $AB$ again at an interior point $F$. Let $A'$ be the reflection of $A$ in the line $BC$. The lines $A'C$ and $DE$ meet at $P$, and the lines $A'B$ and $DF$ meet at $Q$. Prove that the lines $AD, BP$ and $CQ$ are concurrent (or all parallel).
AlphaGeometry formalization: a b c = triangle; d = on_line b c; o1 = circumcenter a b d; e = on_line a c, on_circle o1 a; o2 = circumcenter a c d; f = on_line a b, on_circle o2 a; a1 = reflect a b c; p = on_line a1 c, on_line d e; q = on_line a1 b, on_line d f; x = on_line a d, on_line b p ? coll x c q
HAGeo formalization: A B C = triangle; BC = line B C; D = on_line BC; O1 = circumcenter A B D; (O1) = circle_center_point O1 A; AC = line A C; E = intersection (O1) AC; O2 = circumcenter A C D; (O2) = circle_center_point O2 A; AB = line A B; F = intersection (O2) AB; A1 = reflect_point_wrt_line A BC; A1C = line A1 C; DE = line D E; P = intersection A1C DE; A1B = line A1 B; DF = line D F; Q = intersection A1B DF; AD = line A D; BP = line B P; X = intersection AD BP; Prove: collinear X C Q
Difficulty score: 3.0