SILK-ROAD 2016 Problem 2

Around the acute-angled triangle $ABC$ ( $AC>CB$ ) a circle is circumscribed, and the point $N$ is midpoint of the arc $ACB$ of this circle. Let the points $A_1$ and $B_1$ be the feet of perpendiculars on the straight line $NC$ , drawn from points $A$ and $B$ respectively (segment $NC$ lies inside the segment $A_1B_1$ ). Altitude $A_1A_2$ of triangle $A_1AC$ and altitude $B_1B_2$ of triangle $B_1BC$ intersect at a point $K$ . Prove that $\angle A_1KN=\angle B_1KM$ , where $M$ is midpoint of the segment $A_2B_2$ .