RU-ALLRUSSIAN 2023 Problem 114

Let $\omega$ be the circumcircle of triangle $ABC$ with $AB<AC$ . Let $I$ be its incenter and let $M$ be the midpoint of $BC$ . The foot of the perpendicular from $M$ to $AI$ is $H$ . The lines $MH, BI, AB$ form a triangle $T_b$ and the lines $MH, CI, AC$ form a triangle $T_c$ . The circumcircle of $T_b$ meets $\omega$ at $B'$ and the circumcircle of $T_c$ meets $\omega$ at $C'$ . Prove that $B', H, C'$ are collinear.