Example 6 As shown in Figure 5, in Rt $\triangle C A B$, $\angle A=$ $90^{\circ}, \angle B 、 \angle C$ are bisected and intersect at $F$, and intersect the opposite sides at points $D 、 E$. Find $S_{\text {quadrilateral } B C D E}: S_{\triangle B F C}$.
Solution: Let $A B=c, A C=b, B C=a$. By the property of the internal angle bisector, we have
$$
\frac{A E}{E B}=\frac{b}{a} \text {. }
$$

Then $\frac{A E}{c}=\frac{b}{a+b}$, i.e., $A E=\frac{b c}{a+b}$.
Thus, $B E=c-A E=\frac{a c}{a+b}$.
Therefore, $S_{\triangle C B E}=\frac{1}{2} B E \cdot b=\frac{a b c}{2(a+b)}$.
And $\frac{C F}{F E}=\frac{a}{B E}=\frac{a+b}{c}$, so
$$
\frac{C F}{C E}=\frac{a+b}{a+b+c}, \frac{S_{\triangle C B F}}{S_{\triangle C B E}}=\frac{a+b}{a+b+c},
$$
$$
S_{\triangle B P C}=\frac{a b c}{2(a+b+c)} \text {. }
$$

Similarly, we can get $A D=\frac{b c}{a+c}$.
Then $S_{\text {quadrilateral } B C D E}=\frac{1}{2} b c-\frac{1}{2} \cdot \frac{b c}{a+c} \cdot \frac{b c}{a+b}$
$$
=\frac{a b c(a+b+c)}{2(a+b)(a+c)} \text {. }
$$

Noting that $a^{2}=b^{2}+c^{2}$, then
$$
\begin{array}{l}
\frac{S_{\text {quadrilateralBCDE }}}{S_{\triangle B F C}}=\frac{(a+b+c)^{2}}{(a+b)(a+c)} \\
=\frac{a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a}{a^{2}+a b+a c+b c} \\
=\frac{2 a^{2}+2 a b+2 b c+2 c a}{a^{2}+a b+a c+b c}=2 .
\end{array}
$$