II. (25 points) As shown in Figure 2, $EF$ intersects the diagonal $AC$ of $\square ABCD$ at point $G$,
intersects $AB$, $AD$ at points
$E$, $F$, and connects $CE$,
$CF$, $BG$. If $\frac{1}{S_{\triangle ACE}}+$
$\frac{1}{S_{\triangle ACF}}=\frac{\lambda}{S_{\triangle ABG}}$, find
the value of $\lambda$.
II. As shown in Figure 8, draw $EM \parallel BC$ intersecting $AC$ at $M$, then $\frac{AB}{AE}=\frac{AC}{AM}$.
From $\frac{AD}{AF}=\frac{BC}{AF}$
$=\frac{BC}{EM} \cdot \frac{EM}{AF}$
$=\frac{AC}{AM} \cdot \frac{GM}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AM}+\frac{AC}{AM} \cdot \frac{GM}{AG}$
$=\frac{AC}{AM}\left(1+\frac{GM}{AG}\right)=\frac{AC}{AM} \cdot \frac{AM}{AG}=\frac{AC}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AG}$
$\Rightarrow \frac{AB \cdot AG}{AC \cdot AE}+\frac{AD \cdot AG}{AC \cdot AF}=1$.
Since $\frac{AB \cdot AG}{AC \cdot AE}=\frac{S_{\triangle ABG}}{S_{\triangle ACE}}, \frac{AD \cdot AG}{AC \cdot AF}=\frac{S_{\triangle ADG}}{S_{\triangle ACF}}$, thus
$\frac{S_{\triangle ABG}}{S_{\triangle ACE}}+\frac{S_{\triangle ADG}}{S_{\triangle ACF}}=1$.
It is easy to prove that $\triangle ABC \cong \triangle CDA$, then
$S_{\triangle ABC}=S_{\triangle ADC}$.
Draw $BP \perp AC$, and $DQ \perp AC$, with feet of the perpendiculars at $P$ and $Q$ respectively.
From $S_{\triangle ABC}=\frac{1}{2} AC \cdot BP, S_{\triangle ADC}=\frac{1}{2} AC \cdot DQ$, we get
$BP=DQ$.
Also, $S_{\triangle ABG}=\frac{1}{2} AG \cdot BP, S_{\triangle ADG}=\frac{1}{2} AG \cdot DQ$,
thus $S_{\triangle ABG}=S_{\triangle ADG}$
$\Rightarrow \frac{S_{\triangle ABG}}{S_{\triangle ACE}}+\frac{S_{\triangle ABG}}{S_{\triangle ACF}}=1 \Rightarrow \lambda=1$.