Denote by $\Gamma$ the circumscribed circle of $\triangle ABC$. Let $W$ be the middle of the smaller arc $BC$ of the circle $\Gamma$, and $O$ be the centre of $\Gamma$ (Fig. 46). Obviously, points $N$, $M$, $W$, $O$ lie on the median perpendicular to the segment $BC$. By Lemma of Archimedes points of contact of $w$ to $BC$, to $\Gamma$ and the point $W$ are collinear. As $w$ is tangent to the circle $\Gamma$ in the point $A$, then the circle $w$ touches $BC$ in the point $L$, which is the intersection of $BC$ and $AW$. Note that

Fig. 46
$L$ is the base of the bisector of the angle $\angle BAC$. Let us draw the rays $AM$ and $AS$ until they intersect with the circle $\Gamma$ in the points $X$ and $P$ respectively. By the statement of the problem, $\angle BAX = \angle PAC$, hence $P$ and $X$ are symmetrical with respect to $WN$. Also $\angle XAW = \angle WAP$. $WN$ is the diameter of $\Gamma$, therefore $\angle NML = \angle NAL = 90^\circ$, and thus the quadrangle $ALMN$ is inscribed. Then $\angle WNP = \angle WAP = \angle XAW = \angle MNL$, and hence $N$, $L$, $P$ are collinear. Note that $LM$, $NA$, $WP$ are the heights of $\triangle NLW$, their intersection point is the orthocentre $K$. Let $T$ be the intersection point of the ray $AN$ and $w$. There exists a homothetic transformation transforming $\Gamma$ into $w$. By this homothetic transformation point $N$ transforms into $T$, point $W$ into $L$, i.e. $TL$ is the diameter of $w$, because $WN$ is the diameter of $\Gamma$. Hence, $Q$ is the midpoint of the segment $TL$. Let us use Menelaus' theorem to $\triangle TLK$ and points $N$, $Q$, $S$. Then to prove $N$, $Q$, $S$ are collinear, it is sufficient to show that
$$\frac{TN}{NK} \cdot \frac{KS}{SL} \cdot \frac{LQ}{QT} = 1, \text{ i.e. (as } TQ = QL) \frac{TN}{NK} = \frac{SL}{KS}.$$
Let us prove this equality. It is clear that the lines $AL$ and $AK$ are the internal and the extremal bisectors of the angle $\angle MAS$, hence $\frac{ML}{LS} = \frac{MK}{SK}$. Then $\frac{ML}{MK} = \frac{SL}{SK}$. Obviously, $TL \parallel NW$, hence by Thales' theorem $\frac{TN}{NK} = \frac{LM}{MK}$, therefore $\frac{TN}{NK} = \frac{SL}{SK}$, Q.F.D.