Problem (2017CzechAPSlovakp4): Let ${ABC}$ be a triangle. Line l is parallel to ${BC}$ and it respectively intersects side ${AB}$ at point ${D}$, side ${AC}$ at point ${E}$, and the circumcircle of the triangle ${ABC}$ at points ${F}$ and ${G}$, where points ${F,D,E,G}$ lie in this order on l. The circumcircles of triangles ${FEB}$ and ${DGC}$ intersect at points ${P}$ and ${Q}$. Prove that points ${A, P,Q}$ are collinear.
AlphaGeometry formalization: a b c = triangle; d = on_line a b; e = on_line a c, on_pline d b c; o = circumcenter a b c; f = on_line d e, on_circle o a; g = on_line d e, on_circle o a; o1 = circumcenter f e b; o2 = circumcenter d g c; p = on_circle o1 f, on_circle o2 d; q = on_circle o1 f, on_circle o2 d ? coll a p q
HAGeo formalization: A B C = triangle; AB = line A B; D = on_line AB; AC = line A C; BC = line B C; l = parallel_line D BC; E = intersection l AC; O = circumcenter A B C; (O) = circle_center_point O A; F G = intersection l (O); O1 = circumcenter B E F; (O1) = circle_center_point O1 F; O2 = circumcenter C D G; (O2) = circle_center_point O2 D; P = intersection (O1) (O2); Q = intersection (O1) (O2); Prove: collinear A P Q
Difficulty score: 2.0