2. On the hypotenuse $AB$ of the right triangle $\triangle ABC$, take $BP = BC$, $AQ = AC$. Construct $PM \perp BC$ at point $M$, $QN \perp AC$ at point $N$, and let $PM$ intersect $QN$ at point $G$. Try to compare the area of rectangle $GMCN$ with the area of $\triangle PQG$.
From Fig. 4, we know
$$
\begin{array}{l}
C M \cdot C N=\left(a-\frac{a^{2}}{c}\right)\left(b-\frac{b^{2}}{c}\right) \\
=\frac{a b}{c^{2}}(c-a)(c-b) .
\end{array}
$$

And $S_{\triangle P Q G}=\frac{1}{2} Q G \cdot P G$
$$
\begin{array}{l}
=\frac{1}{2}(a+b-c) \frac{a}{c}(a+b-c) \frac{b}{c} \\
=\frac{a b}{c^{2}} \cdot \frac{1}{2}(a+b-c)^{2}
\end{array}
$$

It is easy to prove $(c-a)(c-b)=\frac{1}{2}(a+b-c)^{2}$.
Therefore, $S_{\text {quadrilateral } C M C N}=S_{\triangle P Q G \text {. }}$.