6. As shown in Figure 3, in rhombus $A B C D$, $\angle A B C=$ $120^{\circ}, A B=\sqrt{3}, E$ is a point on the extension of side $B C$, $A E$ intersects $C D$ at point $F$, connect $B F$ and extend it, intersecting $D E$ at point $G$. Then the maximum length of line segment $B G$ is ( ).
(A) $\sqrt{3}$
(B) 2
(C) $2 \sqrt{3}$
(D) 4
6. B.

Draw $E H / / C D$, intersecting the extension of $B G$ at point $H$, and connect $C H$.
Then $\angle H E C=180^{\circ}-\angle A B C=60^{\circ}$,
$\frac{C E}{B C}=\frac{F E}{A F}=\frac{H E}{A B}$.
By $B C=A B \Rightarrow C E=H E$
$\Rightarrow \triangle H C E$ is an equilateral triangle
$\Rightarrow C H=C E$.
It is easy to prove that $\triangle B C H \cong \triangle D C E$.
Therefore, $\angle G B C=\angle G D C$, which means point $G$ lies on the circumcircle of $\triangle B C D$.

From the given conditions, it is easy to find that the diameter of this circle is 2, so the maximum length of segment $B G$ is 2.