I have found the following problem: There is an acute $\triangle ABC$. Denote its circumcircle as $\omega$. The angle bisector of $\angle BAC$ intersects $BC$ and $\omega$ in points respectively $A_1$ and $M$, angle bisector of $\angle ABC$ intersects $AC$ and $\omega$ in respectively $B_1$ and $N$. Let $t$ be the tangent to $\omega$ in point $C$. Prove that $t$, $MN$ and $A_1B_1$ intersect in one point.
Here is the problem drawn with GeoGebra:
What I have found out: It is obvious that $M$ is the middle point of the arc $BC$ and $N$ is the middle of $AC$. I know that I can solve the problem if I choose two of the given lines to intersect and prove that the third goes through their intersection point, but I have tried all pairs, but I can't find a way to prove it. Does anyone have ideas?