Example 8. Given two circles intersecting at $M$ and $N$, how can a line segment $A B$ be constructed, passing through point $M$ and with endpoints $A, B$ on the two circles respectively, such that $A M \cdot M B$ is maximized? (75 USA Mathematical Competition)

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Solution: Clearly, $M N$ is a fixed length while $\angle A, \angle B$ are both remote angles. Therefore,
$\alpha=\angle A N B=\pi-\angle A-\angle B$ is a fixed angle.
Let $\angle A N M=x$. Then $\angle B N M=\alpha-x$.
By the Law of Sines, we have
$$
\begin{array}{l}
A M=\frac{M N \sin x}{\sin A}, M B=\frac{M N \sin (\alpha-x)}{\sin B} . \\
\begin{aligned}
\therefore \quad A M & \cdot M B=\frac{M N^{2} \sin x \sin (\alpha-x)}{\sin A \sin B} \\
& =\frac{[\cos (2 x-\alpha)-\cos \alpha] M N^{2}}{2 \sin A \sin B} .
\end{aligned}
\end{array}
$$
The rest is obvious. Thus, $2 x-\alpha=0$, indicating that when $\angle M N A=\frac{\alpha}{2}=\pi-\angle A-\angle B$, the arc $A B$ is the one sought.