Example 5 As shown in Figure 9, in the right triangle $\triangle ABC$, $\angle C = 90^{\circ}, CD \perp AB$, the angle bisector of $\angle B$ intersects $CD$ and $CA$ at points $E$ and $F$, respectively. $G$ is the midpoint of $EF$, and $CG$ is connected. Let the perimeters of $\triangle CFG$, $\triangle BED$, and $\triangle BFC$ be $c_{1}$, $c_{2}$, and $c_{3}$, respectively. Find the maximum value of $\frac{c_{1}+c_{2}}{c_{3}}$.
(2001, Shanghai Junior High School Mathematics Competition)
Solve: From $\angle C E F=\frac{1}{2} \angle B+\angle B C D$
$$
=\frac{1}{2} \angle B+\angle A=\angle C F E \text {, }
$$

we get $C E=C F$.
Also, since $G$ is the midpoint of $E F$, then $C G \perp E F$.
It is also easy to see that
$$
\begin{array}{l}
\text { Rt } \triangle C F G \backsim \mathrm{Rt} \triangle C E G \backsim \\
\text { Rt } \triangle B E D \backsim \mathrm{Rt} \triangle B F C . \\
\text { Let } \triangle F=x, B F=k x(k>0) . \text { Then } \\
G F=\frac{C F^{2}}{B F}=\frac{x}{k}, E F=2 G F=\frac{2 x}{k}, \\
B E=B F-E F=\frac{k^{2}-2}{k} x . \\
\text { Therefore, } \frac{c_{1}+c_{2}}{c_{3}}=\frac{C F+B E}{B F} \\
=-2\left(\frac{1}{k}\right)^{2}+\frac{1}{k}+1 \\
=-2\left(\frac{1}{k}-\frac{1}{4}\right)^{2}+\frac{9}{8} \\
\leqslant \frac{9}{8} .
\end{array}
$$

When $k=4$, i.e., $B F=4 C F$, $\frac{c_{1}+c_{2}}{c_{3}}$ reaches its maximum value of $\frac{9}{8}$.