Problem (2018ARMOg10p2): Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic.
AlphaGeometry formalization: a b c = triangle; m = midpoint a b; n = midpoint a c; d = foot a b c; k = on_line m n, on_bline b c; q = on_line k d, on_circum a b c ? cyclic c n k q
HAGeo formalization: A B C = acute_triangle; M = midpoint A B; N = midpoint A C; BC = line B C; D = foot A BC; l1 = perpendicular_bisector B C; MN = line M N; K = intersection MN l1; DK = line D K; (ABC) = circle A B C; Q = intersection DK (ABC); Prove: concyclic C N K Q
Difficulty score: 2.5