Problem (2012ARMOg9p3): Consider the parallelogram $ABCD$ with obtuse angle $A$. Let $H$ be the feet of perpendicular from $A$ to the side $BC$. The median from $C$ in triangle $ABC$ meets the circumcircle of triangle $ABC$ at the point $K$. Prove that points $K,H,C,D$ lie on the same circle.
AlphaGeometry formalization: a b c = triangle; d = parallelogram a b c d; h = foot a b c; m = midpoint a b; o = circumcenter a b c; k = on_line c m, on_circle o a ? cyclic k h c d
HAGeo formalization: A B C D = parallelogram; BC = line B C; H = foot A BC; M = midpoint A B; CM = line C M; O = circumcenter A B C; (ABC) = circle_center_point O A; K = intersection CM (ABC); Prove: concyclic K H C D
Difficulty score: 3.2