For example, $\triangle ABC$ is an equilateral triangle with side length 8, $M$ is a point on side $AB$, $MP \perp AC$ at point $P$, $MQ \perp BC$ at point $Q$, and connect $PQ$.
(1) Find the minimum length of $PQ$;
(2) Find the maximum area of $\triangle CPQ$.
Solution: (1) Let the height of $\triangle ABC$ be $h$, then $h=4 \sqrt{3}$.
From $S_{\triangle C M}+S_{\triangle B C M}=S_{\triangle B B C}$, we get
$$
M P+M O=h=4 \sqrt{3} \text {. }
$$

As shown in Figure 11, draw perpendiculars from points $P$ and $Q$ to side $AB$, with the feet of the perpendiculars being $P_{1}$ and $Q_{1}$, respectively. Since
$$
\begin{array}{l}
\angle P M A=\angle Q M B \\
=30^{\circ},
\end{array}
$$

we have
$$
\begin{array}{l}
P_{1} M=P M \cos 30^{\circ}=\frac{\sqrt{3}}{2} P M, \\
Q_{1} M=Q M \cos 30^{\circ}=\frac{\sqrt{3}}{2} Q M, \\
P Q \geqslant P_{1} Q_{1}=P_{1} M+M Q_{1} \\
=\frac{\sqrt{3}}{2}(P M+Q M)=6 .
\end{array}
$$

When $M$ is the midpoint of $AB$, the equality holds. Therefore, the minimum value of $P Q$ is 6.
(2) Since $\angle P M A=\angle Q M B=30^{\circ}$, we have
$$
\begin{array}{l}
A P+B Q=\frac{1}{2} A M+\frac{1}{2} B M=\frac{1}{2} A B=4, \\
C P+C Q=16-(A P+B Q)=12 .
\end{array}
$$

Thus, $S_{\triangle C P Q}=\frac{1}{2} C P \cdot C Q \sin C=\frac{\sqrt{3}}{4} C P \cdot C Q$
$$
\leqslant \frac{\sqrt{3}}{4} \cdot \frac{(C P+C Q)^{2}}{4}=9 \sqrt{3} \text {. }
$$

When $M$ is the midpoint of $AB$, the equality holds. Therefore, the maximum area of $\triangle C P Q$ is $9 \sqrt{3}$.