Problem (2010ARMOg10p6): Into triangle $ABC$ gives point $K$ lies on bisector of $ \angle BAC$. Line $CK$ intersect circumcircle $ \omega$ of triangle $ABC$ at $M \neq C$. Circle $ \Omega$ passes through $A$, touch $CM$ at $K$ and intersect segment $AB$ at $P \neq A$ and $\omega $ at $Q \neq A$. Prove, that $P$, $Q$, $M$ lies at one line.
AlphaGeometry formalization: a b c = triangle; k = angle_bisector b a c; m = on_line c k, on_circum a b c; o = on_tline k c m, on_bline a k; p = on_line a b, on_circle o a; q = on_circle o a, on_circum a b c ? coll p q m
HAGeo formalization: A B C = triangle; la = angle_bisector B A C; K = on_line la; CK = line C K; (ABC) = circle A B C; M = intersection CK (ABC); CM = line C M; l1 = perpendicular_line K CM; l2 = perpendicular_bisector A K; O = intersection l1 l2; (O) = circle_center_point O A; AB = line A B; P = intersection AB (O); Q = intersection (ABC) (O); Prove: collinear P Q M
Difficulty score: 2.4