2. As shown in Figure 2, in rhombus $A B C D$, $\angle A B C = 120^{\circ}, A B = \sqrt{3}$, $E$ is a point on the extension of $B C$, $A E$ intersects $C D$ at point $F$, and $B F$ is extended to intersect $D E$ at point $G$. Then the range of the length of segment $B G$ is $\qquad$
2. $\sqrt{3}<BG \leqslant 2$.

As shown in Figure 6, extend $BG$ and $AD$ to intersect at point $M$, and connect $BD$. Thus, $\triangle BCD$ is an equilateral triangle, and
$$
\frac{BC}{DM}=\frac{CF}{DF}=\frac{CE}{AD}.
$$

Since $AD=BC=CD$, we have $\frac{BD}{DM}=\frac{CE}{CD}$.
Since $\angle BDM=\angle DCE=120^{\circ}$, it follows that
$\triangle BDM \backsim \triangle ECD \Rightarrow \angle M=\angle CDE$.
Also, $\angle M=\angle EBG$, so $\angle CDE=\angle EBG$.
Therefore, points $B$, $C$, $G$, and $D$ are concyclic.
The diameter $d$ of the circumcircle of $\triangle BCD$ is $2 \times \frac{2}{3} \times \frac{\sqrt{3}}{2} \times \sqrt{3}=2$, so, $\sqrt{3}<BG \leqslant 2$.