Problem (2019SilkRoadp1): The altitudes of the acute-angled non-isosceles triangle $ ABC $ intersect at the point $ H $. On the segment $ C_1H $, where $ CC_1 $ is the altitude of the triangle, the point $ K $ is marked. Points $ L $ and $ M $ are the feet of perpendiculars from point $ K $ on straight lines $ AC $ and $ BC $, respectively. The lines $ AM $ and $ BL $ intersect at $ N $. Prove that $ \angle ANK = \angle HNL $.
AlphaGeometry formalization: a b c = triangle; h = orthocenter a b c; c1 = foot c a b; k = on_line c1 h; l = foot k a c; m = foot k b c; n = on_line a m, on_line b l ? eqangle a n n k h n n l
HAGeo formalization: A B C = acute_triangle; H = orthocenter A B C; AB = line A B; C1 = foot C AB; C1H = line C1 H; K = on_line C1H; AC = line A C; L = foot K AC; BC = line B C; M = foot K BC; AM = line A M; BL = line B L; N = intersection AM BL; Prove: equal_angle A N K H N L
Difficulty score: 2.5