2. In $\triangle A B C$, $D$ and $E$ are points on sides $A B$ and $B C$ respectively, and $\frac{B E}{E C}+\frac{B D}{D A}=2 \sqrt{2}, A E$ intersects $C D$ at $F$, and the extension of $B F$ intersects $A C$ at $G$. Then $\frac{B F}{F G}=(\quad)$.
(A) $\sqrt{2}$
(B) $2 \sqrt{2}$
(C) $4 \sqrt{2}$
(D) 8
2.B.

As shown in Figure 1, notice that
$$
\begin{array}{l}
\frac{B F}{F G}=\frac{S_{\triangle A B F}+S_{\triangle B F C}}{S_{\triangle A F C}} \\
=\frac{S_{\triangle B F F}}{S_{\triangle A F C}}+\frac{S_{\triangle B F C}}{S_{\triangle A F C}} . \\
\text { Also, } \frac{B E}{E C}=\frac{S_{\triangle A B F}}{S_{\triangle A F C}}, \\
\frac{B D}{A D}=\frac{S_{\triangle B F C}}{S_{\triangle A F C}},
\end{array}
$$

Therefore, $\frac{B F}{F G}=\frac{B E}{E C}+\frac{B D}{A D}=2 \sqrt{2}$.