Problem (2015ARMOg10p7): In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$ Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$. The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$
AlphaGeometry formalization: a b c = triangle; m = midpoint b c; h = foot a b c; q = on_line a b, on_tline m a c; p = on_line a c, on_tline m a b; o1 = circumcenter p m q; x = on_line b c, on_circle o1 p ? cong b h c x
HAGeo formalization: A B C = acute_triangle; M = midpoint B C; BC = line B C; H = foot A BC; AC = line A C; l1 = perpendicular_line M AC; AB = line A B; Q = intersection AB l1; l2 = perpendicular_line M AB; P = intersection AC l2; O1 = circumcenter M P Q; (O1) = circle_center_point O1 P; X = intersection BC (O1); Prove: cong B H C X
Difficulty score: 2.7