Question 35
An acute-angled triangle $ABC$ ($AC \gt CB$) has a circumscribed circle with point $N$ the midpoint of the arc $ACB$. Let points $A_1$ and $B_1$ be the feet of the perpendiculars to the line $NC$ drawn from the points $A$ and $B$ respectively (the segment $NC$ lies inside $A_1B_1$). The altitude $A_1A_2$ of triangle $A_1AC$ and the altitude $B_1B_2$ of triangle $B_1BC$ intersect at point $K$. Prove that $\angle A_1KN = \angle B_1KM$, where $M$ is the midpoint of the segment $A_2B_2$.