Problem (2018USAMOp5): In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; e = on_line a c, on_line b d; f = on_line a b, on_line c d; g = on_line b c, on_line d a; o1 = circumcenter a b e; p = on_line c b, on_circle o1 a; o2 = circumcenter a d e; q = on_line c d, on_circle o2 a; m = on_line f p, on_line g q ? perp m a a c
HAGeo formalization: A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; E = intersection AC BD; AB = line A B; CD = line C D; F = intersection AB CD; BC = line B C; AD = line A D; G = intersection BC AD; O1 = circumcenter A B E; (ABE) = circle_center_point O1 A; P = intersection BC (ABE); O2 = circumcenter A D E; (ADE) = circle_center_point O2 A; Q = intersection CD (ADE); FP = line F P; GQ = line G Q; M = intersection FP GQ; AM = line A M; Prove: perpendicular AM AC
Difficulty score: 3.7