Problem (2020ELMOp4): Let acute scalene triangle $ABC$ have orthocenter $H$ and altitude $AD$ with $D$ on side $BC$. Let $M$ be the midpoint of side $BC$, and let $D'$ be the reflection of $D$ over $M$. Let $P$ be a point on line $D'H$ such that lines $AP$ and $BC$ are parallel, and let the circumcircles of $\triangle AHP$ and $\triangle BHC$ meet again at $G \neq H$. Prove that $\angle MHG = 90^\circ$.
AlphaGeometry formalization: a b c = triangle; h = orthocenter a b c; d = foot a b c; m = midpoint b c; d1 = mirror d m; p = on_line d1 h, on_pline a b c; o1 = circumcenter a h p; o2 = circumcenter b h c; g = on_circle o1 a, on_circle o2 b ? perp m h h g
HAGeo formalization: A B C = acute_triangle; H = orthocenter A B C; BC = line B C; D = foot A BC; M = midpoint B C; D1 = reflect_point_wrt_point D M; l1 = parallel_line A BC; D1H = line D1 H; P = intersection D1H l1; O1 = circumcenter A H P; (O1) = circle_center_point O1 A; O2 = circumcenter B C H; (O2) = circle_center_point O2 B; G = intersection (O1) (O2); HM = line H M; GH = line G H; Prove: perpendicular HM GH
Difficulty score: 2.8