II. (25 points) In $\triangle A B C$, $D$ is the midpoint of $B C$, $G$ is a moving point on $A D$ (excluding $A, D$), the extensions of $B G, C G$ intersect $A C, A B$ at $E, F$ respectively. Find the maximum value of $\frac{S_{\triangle B G F}+S_{\triangle C G E}}{S_{\triangle A B C}}$.
As shown in Figure 4, it is evident that
$$
\frac{B D}{C D}=\frac{S_{\triangle A E D}}{S_{\triangle A C D}}=\frac{S_{\triangle B C D}}{S_{\triangle C G D}}=\frac{S_{\triangle A C B}}{S_{\triangle A C C}},
$$
$$
\frac{C E}{E A}=\frac{S_{\triangle B C E}}{S_{\triangle B A E}}=\frac{S_{\triangle C C E}}{S_{\triangle A C E}}=\frac{S_{\triangle B C C}}{S_{\triangle A C B}},
$$
$$
\frac{A F}{F B}=\frac{S_{\triangle A C F}}{S_{\triangle B C F}}=\frac{S_{\triangle A G F}}{S_{\triangle B G F}}=\frac{S_{\triangle A C C}}{S_{\triangle B O C}} \text {. }
$$
(1) $\times$ (2) $\times$ (3) gives
$$
\frac{B D}{C D} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=1 \text {. }
$$

Substituting $B D=D C$ into equation (1) yields
$$
S_{\triangle A C B}=S_{\triangle A C C} \text {. }
$$

Let $\frac{C E}{E A}=x$, from equation (4) we get $\frac{A F}{F B}=\frac{1}{x}$.
For convenience, let $S_{\triangle B G F}=a, S_{\triangle O C E}=b$.
From equation (2), we have $S_{\triangle M C E}=\frac{b}{x}$; from equation (3), we have $S_{\triangle M G F}=\frac{a}{x}$.
Therefore, $S_{\triangle A G B}=S_{\triangle A G F}+S_{\triangle B G F}=\frac{a(x+1)}{x}$,
$$
S_{\triangle A C C}=S_{\triangle A C E}+S_{\triangle C C E}=\frac{b(x+1)}{x} \text {. }
$$

From equations (5), (6), and (7), we get $a=b$.
From equations (2) and (6), with $\frac{C E}{E A}=x$, we get $S_{\triangle B C C}=a(x+1)$.
Thus, $S_{\triangle A B C}=S_{\triangle A C B}+S_{\triangle A C C}+S_{\triangle B C C}$
$$
=\frac{a\left(x^{2}+3 x+2\right)}{x} \text {. }
$$

Therefore, $\frac{S_{\triangle B C F}+S_{\triangle O C E}}{S_{\triangle A B C}}=\frac{2 x}{x^{2}+3 x+2}=\frac{2}{x+\frac{2}{x}+3}$
$$
\begin{array}{c}
=\frac{2}{\left(\sqrt{x}-\sqrt{\frac{2}{x}}\right)^{2}+3+2 \sqrt{2}} \\
\leqslant \frac{2}{3+2 \sqrt{2}}=6-4 \sqrt{2} .
\end{array}
$$

When $\sqrt{x}=\sqrt{\frac{2}{x}}$, i.e., $x=\sqrt{2}$,
$$
\frac{S_{\triangle B G F}+S_{\triangle O C E}}{S_{\triangle A B C}}=6-4 \sqrt{2} \text {. }
$$

Thus, the maximum value of the original expression is $6-4 \sqrt{2}$.