Example 3 In a regular $\triangle A B C$ with side length 8, take a moving point $P$ on side $B C$. Find the maximum and minimum values of the segment $M N$ intercepted by the perpendicular bisector of $A P$ on $\triangle A B C$.
As shown in Figure 3, take the side length 8 and the variable $\angle B A P = \alpha \in \left[0, \frac{\pi}{3}\right]$ as the basic quantities. Then
$$
\begin{array}{l}
A D = \frac{1}{2} A P \\
= \frac{1}{2} \cdot \frac{8 \sin 60^{\circ}}{\sin \left(60^{\circ} + \alpha\right)} \\
= \frac{2 \sqrt{3}}{\sin \left(60^{\circ} + \alpha\right)}, \\
M N = A D \left[\tan \alpha + \tan \left(60^{\circ} - \alpha\right)\right] \\
= \frac{2 \sqrt{3}}{\sin \left(60^{\circ} + \alpha\right)} \cdot \frac{\sin 60^{\circ}}{\cos \alpha \cdot \cos \left(60^{\circ} - \alpha\right)} \\
= \frac{3}{\sin \left(60^{\circ} + \alpha\right) \cdot \cos \alpha \cdot \cos \left(60^{\circ} - \alpha\right)} \\
= \frac{6}{\sin \left(60^{\circ} + \alpha\right) \left[\cos 60^{\circ} + \cos \left(60^{\circ} - 2 \alpha\right)\right]}.
\end{array}
$$

The two factors in the denominator of equation (1) increase and decrease in the same manner. Therefore, when $\alpha = 0^{\circ}, 60^{\circ}$ (i.e., $A P$ coincides with $A B$ or $A C$), the line segment $M N$ has a maximum value of $4 \sqrt{3}$; when $\alpha = 30^{\circ}$ (i.e., $P$ is the midpoint of $B C$), the line segment $M N$ has a minimum value of 4.