Example 2-3 Let quadrilateral $A B C D$ be a rectangle with an area of 2, $P$ a point on side $C D$, and $Q$ the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant 2 B C$;
(2) Find the value of $A Q \cdot B Q$.
(2001, China Western Mathematical Olympiad)
Solution 1: (1) From the given information,
$S_{\triangle P A B}=\frac{1}{2} S_{\text {rectangle } A B C D}=1$.
Thus, $P A \cdot P B=\frac{2 S_{\triangle P A B}}{\sin \angle A P B} \geqslant 2$.
Equality holds if and only if $\angle A P B=90^{\circ}$. At this time, point $P$ lies on the circle with $A B$ as its diameter, meaning this circle intersects $C D$ at point $P$.

By the property that "in a right triangle, the median to the hypotenuse is not less than the altitude to the hypotenuse," when $P A \cdot P B$ reaches its minimum value, we have
$$
B C \leqslant \frac{A B}{2} \Rightarrow A B \geqslant 2 B C \text {. }
$$
(2) In the right triangle $\triangle P A B$, by the property that "the lengths of the tangents from a point outside a circle to the circle are equal," we get
$$
|P A-P B|=|A Q-B Q| .
$$

Squaring both sides, we obtain
$$
P A^{2}+P B^{2}-2 P A \cdot P B=A B^{2}-4 A Q \cdot B Q \text {. }
$$

Thus, $A Q \cdot B Q=\frac{1}{2} P A \cdot P B=1$.
Solution 2: As shown in Figure 3, let the incircle of $\triangle P A B$ touch $P A$ and $P B$ at points $M$ and $N$, respectively. Let
$$
\begin{array}{l}
A Q=A M=z, \\
B Q=B N=x, \\
P N=P M=y .
\end{array}
$$

Then $S_{\triangle P A B}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{x y z(x+y+z)} \\
=\frac{1}{2} S_{\text {rectangle } A B C D}=1 .
\end{array}
$$

Thus, $x y z(x+y+z)=1$.
$$
\begin{array}{l}
\text { Also, } P A \cdot P B=(y+z)(x+y) \\
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2,
\end{array}
$$

Therefore, when $x z=y(x+y+z)$, $P A \cdot P B$ reaches its minimum value of 2.
(1) Multiplying both sides of $x z=y(x+y+z)$ by 2 and adding $x^{2}+z^{2}$, we get
$$
(x+z)^{2}=(y+z)^{2}+(x+y)^{2},
$$

which means $P A^{2}+P B^{2}=A B^{2}$.
Thus, $\triangle P A B$ is a right triangle, and $B C$ is the altitude to the hypotenuse of the right triangle. Hence,
$$
2 B C=\frac{2 P A \cdot P B}{A B} \leqslant \frac{P A^{2}+P B^{2}}{A B}=A B .
$$
(2) Adding $x z$ to both sides of $x z=y(x+y+z)$, we get
$$
2 x z=(x+y)(y+z) \text {, }
$$

which means $2 A Q \cdot B Q=P A \cdot P B=2$.
Thus, $A Q \cdot B Q=1$.