(50 points) Let $ABCD$ be a rectangle with an area of 2, and let $P$ be a point on side $CD$. Let $Q$ be the point where the incircle of $\triangle PAB$ touches side $AB$. The product $PA \cdot PB$ varies with the changes in rectangle $ABCD$ and point $P$. When $PA \cdot PB$ is minimized,
(1) Prove: $AB \geqslant 2BC$;
(2) Find the value of $AQ \cdot BQ$.
(1) $S_{\triangle A P B}=\frac{1}{2} S_{\text {trapezoid } A B C D}=1 \Rightarrow \frac{1}{2} P A \cdot P B \sin \angle A P B=1$, which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$, equality holds if and only if $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, which should intersect with $C D$. Therefore, when $P A \cdot P B$ reaches its minimum value, it should have $B C \leqslant \frac{A B}{2}$;

(2) Let the inradius of $\triangle A P B$ be $r$, then $P A \cdot P B=(r+A Q)(r+B Q)=r(r+A Q+B Q)+A Q \cdot B Q$, and $P A \cdot P B=2 S_{\triangle A P B}, r(r+A Q+B Q)=S_{\triangle A P B}$, thus $A Q \cdot B Q=S_{\triangle A B P}=1$.