Problem (2020POGCHAMPp1): Let $ABC$ be a triangle and $P$ be a point on its circumcircle. Lines $AP$ and $BC$ intersect at $D$ and point $T$ is the reflection of $D$ over the midpoint of $\overline{BC}$. Let $AT$ meet the circumcircle of $\triangle PDT$ again at $G$. The circumcircle of $\triangle AGP$ meets $AB$ and $AC$ again at $E$ and $F$ respectively. Lines $EF$ and $GP$ intersect at $Q$. Prove that $AQ \parallel BC$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; p = on_circle o a; d = on_line a p, on_line b c; m = midpoint b c; t = mirror d m; o1 = circumcenter p d t; g = on_line a t, on_circle o1 p; o2 = circumcenter a g p; e = on_line a b, on_circle o2 a; f = on_line a c, on_circle o2 a; q = on_line e f, on_line g p ? para a q b c
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; P = on_circle (O); AP = line A P; BC = line B C; D = intersection AP BC; M = midpoint B C; T = reflect_point_wrt_point D M; AT = line A T; O1 = circumcenter D P T; (O1) = circle_center_point O1 P; G = intersection AT (O1); O2 = circumcenter A G P; (O2) = circle_center_point O2 A; AB = line A B; E = intersection (O2) AB; AC = line A C; F = intersection (O2) AC; EF = line E F; GP = line G P; Q = intersection EF GP; AQ = line A Q; Prove: parallel AQ BC
Difficulty score: 2.9