2. Let $A B C D$ be a rectangle with an area of 2, and let $P$ be a point on side $C D$. Let $Q$ be 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 $A Q \cdot B Q$.
(Ruo Zengru)
Thus, $\frac{1}{2} P A \cdot P B \sin \angle A P B=1$,
which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$.
Equality holds only when $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, and this circle should intersect with $C D$,
Therefore, when $P A \cdot P B$ takes its minimum value, we should have $B C \leqslant \frac{A B}{2}$, i.e., $A B \geqslant 2 B C$.
(2) Let the inradius of $\triangle A P B$ be $r$, then
$$
\begin{array}{l}
P A \cdot P B=(r+A Q)(r+B Q) \\
=r(r+A Q+B Q)+A Q \cdot B Q .
\end{array}
$$

And $P A \cdot P B=2 S_{\triangle A P B}, r(r+A Q+B Q)=S_{\triangle A P B}$,
Thus, $A Q \cdot B Q=S_{\triangle A P B}=1$.