You could find some solutions in Russian here https://youtu.be/JV_JkzqBLPE

Let line $CS$ and $DS$ intersect the line through $M$ at $X$ and $Y$ respectively. Also, let $I$ be the intersection of radical axes of three circles. Then as the circles are tangent at $S$, we have the two circles are homothetic. Letting $CS$ and $DS$ intersect $(ASB)$ at $J$ and $K$ respectively, we have $JK \parallel DC.$ Let $SK \cap AB = P.$ Then, $JP.PS = AP.AB$, so it suffices to show $JP.PS = PX.PC$. Now, notice that as $XY \parallel AS$, we have homothety centered at $P$ that sends $X$ to $S$ and $M$ to $I$. So, $$\frac{MP}{PI} = \frac{PX}{PS}.$$ Similarly also as $JM \parallel ID$, we have $$\frac{MP}{PI} = \frac{JP}{PC}.$$ So, $$\frac{PX}{PS} = \frac{JP}{PC} \implies JP.PS = PX.PC$$ as desired.

Call the perpendicular line through M to MN "l". assume $SC\cap l=X$ and $DS\cap l=Y.$ With angle chasing it can be easily shown that DCYX is cyclic. Notice that XSY is a right triangle and SM is the altitude to its hypotenuse. Thus MX.MY=MS^2=BM.MA. So BYAX is cyclic too. Consider circumcircle of quadrilaterals (CDXY),(XYAB) and (ABCD). By radical axis theorem we'd have XY,CD and AB are concurrent. So M lies on CD; Which is a contradiction .So these three circles are actually one circle and AXBYCD is cyclic. Therefore X lies on (ABCD). QED.