Example 8 On the plane: there is an $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$. The area is $S$, and there is another point $M$. $M$ rotates around $A$ by an angle $\alpha$ to get points $M_{i}(i=1,2, \cdots, n)$. Find the area of the $n$-sided polygon $M_{1} M_{2} \cdots M_{n}$.
(30th IMO Preliminary Problem)
Prove: As shown in Figure 8, in the isosceles $\triangle A_{i} M M_{i}$, since
$$
\angle A_{i} M M_{i}=\frac{180^{\circ}-\alpha}{2} \text {, }
$$
$\frac{M M_{i}}{M A_{i}}=2 \sin \frac{\alpha}{2}$, it can be seen that point $A_{i}$ undergoes a homothetic rotation transformation $S\left(M,-\frac{180^{\circ}-\alpha}{2}, 2 \sin \frac{\alpha}{2}\right)$ to become $M_{i}$. Therefore, the convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, after the homothetic rotation transformation $S\left(M,-\frac{180^{\circ}-\alpha}{2}, 2 \sin \frac{\alpha}{2}\right)$, becomes the convex $n$-sided polygon $M_{1} M_{2} \cdots M_{n}$, with a homothety ratio of $2 \sin \frac{\alpha}{2}$. Thus, the area of the $n$-sided polygon $M_{1} M_{2} \cdots M_{n}$
$$
=\left(2 \sin \frac{\alpha}{2}\right)^{2} S=4 S \sin ^{2} \frac{\alpha}{2} \text {. }
$$