II. (25 points) As shown in Figure 2, point $D$ is on the side $BC$ of $\triangle ABC$, and does not coincide with $B$ or $C$. A line through point $D$ parallel to $AC$ intersects $AB$ at $E$, and a line through $D$ parallel to $AB$ intersects $AC$ at $F$. It is also known that $BC=5$.
(1) Let the area of $\triangle ABC$ be $S$. If the area of quadrilateral $AEDF$ is $\frac{2}{5} S$, find the length of $BD$.
(2) If $AC=\sqrt{2} AB$, and $DF$ passes through the centroid $G$ of $\triangle ABC$, find the distance between points $E$ and $F$.
(ii) $\because D E / / A C, D F / / A B$,
$\therefore \triangle B D E \backsim \triangle B C A \backsim \triangle D C F$.
Let $S_{- \text {B }}=S_{1}$,
$$
S_{\text {IUCF }}=S_{2} .
$$
$$
\begin{aligned}
\therefore S_{1}+S_{2} & =S-\frac{2}{5} S \\
& =\frac{3}{5} S .
\end{aligned}
$$
$$
\frac{\sqrt{S_{1}}}{\sqrt{S}}=\frac{B D}{B C}, \frac{\sqrt{S_{2}}}{\sqrt{S}}=\frac{C D}{B C} \text {. }
$$

Thus, $\frac{\sqrt{S_{1}}}{\sqrt{S}}+\frac{\sqrt{S_{2}}}{\sqrt{S}}=\frac{B D+D C}{B C}=\frac{B C}{B C}=1$,
which means $\sqrt{S_{1}}+\sqrt{S_{2}}=\sqrt{S}$.
Squaring both sides gives $S=S_{1}+S_{2}+2 \sqrt{S_{1} S_{2}}$.
Hence, $2 \sqrt{S_{1} \overline{S_{2}}}=S_{\text {pentagon continuation DF }}=\frac{2}{5} S$.
Thus, $S_{1} S_{2}=\frac{1}{25} S^{2}$.
Solving (1) and (2) gives $S_{1}=\frac{3 \pm \sqrt{5}}{10} S$, i.e., $\frac{S_{1}}{S}=\frac{3 \pm \sqrt{5}}{10}$.
Since $\frac{S_{1}}{S}=\left(\frac{B D}{B C}\right)^{2}$, we have $\frac{3 \pm \sqrt{5}}{10}=\left(\frac{B D}{5}\right)^{2}$,
$B D^{2}=\frac{15 \pm 5 \sqrt{5}}{2}=\left(\frac{5 \pm \sqrt{5}}{2}\right)^{2}$, thus $B D=\frac{5 \pm \sqrt{5}}{2}$.
(ii) Since $G$ is the centroid of $\triangle A B C$, $D F$ passes through point $G$, and $D F$
$A B$. We get $\frac{C D}{C B}=\frac{2}{3}$. Thus, $D F=\frac{2}{3} A B$.
Since $D E / / A C \cdot \frac{C D}{C B}=\frac{2}{3}$, we have $D E=\frac{1}{3} A C$.
$$
\begin{array}{l}
\because A C=\sqrt{2} A B, \\
\therefore \frac{A C}{A B}=\sqrt{2}, \frac{D F}{D E}=\frac{2 A B}{\sqrt{2} A B}=\sqrt{2} .
\end{array}
$$

Thus, $\frac{D F}{D E}=\frac{A C}{A B}$, which means $\frac{D F}{A C}=\frac{D E}{A B}$.
Also, $\angle E D F=\angle A$, so $\triangle D E F \backsim \triangle A B C$.
Thus, $\frac{E F}{B C}=\frac{D E}{A B}$. Therefore, $E F=\frac{5 \sqrt{2}}{3}$.