Problem (2013CTSTp1): The quadrilateral $ABCD$ is inscribed in circle $\omega$. $F$ is the intersection point of $AC$ and $BD$. $BA$ and $CD$ meet at $E$. Let the projection of $F$ on $AB$ and $CD$ be $G$ and $H$, respectively. Let $M$ and $N$ be the midpoints of $BC$ and $EF$, respectively. If the circumcircle of $\triangle MNG$ only meets segment $BF$ at $P$, and the circumcircle of $\triangle MNH$ only meets segment $CF$ at $Q$, prove that $PQ$ is parallel to $BC$.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; f = on_line a c, on_line b d; e = on_line b a, on_line c d; g = foot f a b; h = foot f c d; m = midpoint b c; n = midpoint e f; o1 = circumcenter m n g; p = on_line b f, on_circle o1 m; o2 = circumcenter m n h; q = on_line c f, on_circle o2 m ? para p q b c
HAGeo formalization: A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; F = intersection AC BD; AB = line A B; CD = line C D; E = intersection AB CD; G = foot F AB; H = foot F CD; M = midpoint B C; N = midpoint E F; BF = line B F; O1 = circumcenter G M N; (O1) = circle_center_point O1 M; P = intersection BF (O1); CF = line C F; O2 = circumcenter H M N; (O2) = circle_center_point O2 M; Q = intersection CF (O2); PQ = line P Q; BC = line B C; Prove: parallel PQ BC
Difficulty score: 2.7