Question 157
A triangle $ABC$ is given. The circle $w$ with the centre at the point $Q$ touches the side $BC$ and is internally tangent to the circumscribed circle of $\triangle ABC$ in the point $A$. Let $M$ be the midpoint of the side $BC$, and $N$ be the middle of the arc $BAC$ of the circumscribed circle of $\triangle ABC$. On the side $BC$ the point $S$ is selected such that $\angle BAM = \angle SAC$. Prove that the points $N$, $Q$ and $S$ are collinear.
(Mykhaylo Plotnykov)
(Mykhaylo Plotnykov)