Given a triangle $ ABC $, $ AD $ is its angle bisector. Let $ E, F $ be the centers of the circles inscribed in the triangles $ ADC $ and $ ADB $, respectively. Denote by $ \omega $, the circle circumscribed around the triangle $ DEF $, and by $ Q $, the intersection point of $ BE $ and $ CF $, and $ H, J, K, M $ , respectively the second intersection point of the lines $ CE, CF, BE, BF $ with circle $ \omega $. Let $\omega_1, \omega_2 $ the circles be circumscribed around the triangles $ HQJ $ and $ KQM $ Prove that the intersection point of the circles $\omega_1, \omega_2 $ different from $ Q $ lies on the line $ AD $. (Kivva Bogdan)