Example $4 P$ is a fixed point inside the acute angle $\angle A$, and a line $l$ passing through $P$ intersects the two sides of $\angle A$ at points $B$ and $C$. Find:
(1) the maximum value of $\frac{1}{P B}+\frac{1}{P C}$ and the minimum value of $P B \cdot P C$;
(2) the line $l$ passing through $P$ such that $S_{\triangle A B C}$ is minimized.
(1) As shown in Figure 4, take
the constants $A P=a, \angle C A P=\alpha, \angle B A P=\beta$ and the variable $\angle A P C=\theta \in(\beta, \pi-\alpha)$ as the basic quantities.
Notice that
$$
\begin{array}{l}
\frac{1}{P B}+\frac{1}{P C}=\frac{\sin (\theta-\beta)}{a \sin \beta}+\frac{\sin (\theta+\alpha)}{a \sin \alpha} \\
=\frac{\sin (\alpha+\beta) \cdot \sin \theta}{a \sin \alpha \cdot \sin \beta},
\end{array}
$$

Therefore, when $\theta=90^{\circ}$, $\frac{1}{P B}+\frac{1}{P C}$ reaches its maximum value
$$
\begin{array}{l}
\frac{\sin (\alpha+\beta)}{a \sin \alpha \cdot \sin \beta} . \\
\text { Also, } P B \cdot P C=\frac{a \sin \beta}{\sin (\theta-\beta)} \cdot \frac{a \sin \alpha}{\sin (\theta+\alpha)} \\
=\frac{2 a^{2} \sin \alpha \cdot \sin \beta}{\cos (\alpha+\beta)-\cos (2 \theta+\alpha-\beta)} .
\end{array}
$$

It is easy to see that the denominator is always positive, thus, when $2 \theta+\alpha-\beta=180^{\circ}$, i.e., $\theta=90^{\circ}+\frac{\beta-\alpha}{2}$ (at this time, $l$ is perpendicular to the angle bisector of $\angle B A C$), $P B \cdot P C$ reaches its minimum value
$$
\frac{a^{2} \sin \alpha \cdot \sin \beta}{\cos ^{2} \frac{\alpha+\beta}{2}} .
$$
(2) Notice that
$$
\begin{array}{l}
2 S_{\triangle A B C}=A C \cdot A B \sin \angle B A C \\
=\frac{a \sin \theta}{\sin (\theta+\alpha)} \cdot \frac{a \sin \theta}{\sin (\theta-\beta)} \sin (\alpha+\beta) \\
=\frac{a^{2} \sin (\alpha+\beta)}{(\cos \alpha+\sin \alpha \cdot \cot \theta)(\cos \beta-\sin \beta \cdot \cot \theta)} \\
=\frac{a^{2} \sin (\alpha+\beta)}{-\sin \alpha \cdot \sin \beta \cdot \cot ^{2} \theta+\sin (\alpha-\beta) \cdot \cot \theta+\cos \alpha \cdot \cos \beta},
\end{array}
$$

Since $S_{\triangle A B C}>0, a^{2} \sin (\alpha+\beta)>0$, the denominator of the above expression is positive.

Also, since $-\sin \alpha \cdot \sin \beta<0$, by the extremum property of quadratic functions, when $\cot \theta=\frac{\sin (\alpha-\beta)}{2 \sin \alpha \cdot \sin \beta}$, the denominator reaches its maximum value. Thus, $S_{\triangle A B C}$ is minimized. At this time,
$$
\begin{array}{l}
\frac{P B}{P C}=\frac{\frac{a \sin \beta}{\sin (\theta-\beta)}}{\frac{a \sin \alpha}{\sin (\theta+\alpha)}} \\
=\frac{\sin \beta(\sin \alpha \cdot \cot \theta+\cos \alpha)}{\sin \alpha(\cos \beta-\sin \beta \cdot \cot \theta)} \\
=\frac{\sin (\alpha-\beta)+2 \sin \beta \cdot \cos \alpha}{2 \sin \alpha \cdot \cos \beta-\sin (\alpha-\beta)}=1 .
\end{array}
$$

Therefore, $P B=P C$.
Thus, the construction method for $l$ is: extend $A P$ by one more length to point $A^{\prime}$, draw $A^{\prime} B / / C A$ intersecting $A B$ at $B$, and draw the line $l$ through $B$ and $P$.