China Girls' Mathematical Olympiad

Solution It is well known that $S,C,M$ are collinear. Indeed, consider the dilation centered at $S$ that sends ${\Gamma }_1$ to ${\Gamma }_2$. Then the line $AB$ is sent to the line $l$ parallel to $AB$ and tangent to ${\Gamma }_2$, i.e. the line tangent to ${\Gamma }_2$ at $M$ (the midpoint of $\widehat{AB}$). Thus, this dilation sends $C$ (the points of tangency of the line $AB$ and ${\Gamma }_1$) to $M$ (the points of tangency of the line $l$ and ${\Gamma }_2$), from which it follows that $S,C,M$ are collinear.

By the power-of-point theorem, we have $AC\times CB=SC\times CM$. It suffices to show that $$SC\times CM=2r\times MN\text{or}\frac{SC}{2r}=\frac{MN}{CM}.\stackrel{◯}{1}$$ Set $\angle MCN=\alpha$. Then $\angle SCA=\alpha$. By the extended sine law, we have $\frac{SC}{2r}=\sin \alpha$. In the right triangle $MNC$, we also have $\sin \alpha =\frac{MN}{CM}$. Combining the last two equations, we obtain ①.

(We can also derive ① by observing that the triangles $MNC$ and $CDS$ are similar.)