2. As shown in Figure 1, the side length of the equilateral $\triangle ABC$ is $6$, and $D$, $E$ are points on sides $BC$, $AC$ respectively, such that $CD = AE$. Let $BE$ and $AD$ intersect at point $F$, and connect $CF$. Draw $EG \parallel CF$ intersecting $AD$ at point $G$. If $EF = 1$, then $AG$ $=(\quad)$.
(A) $\frac{1}{6}$
(B) $\frac{1}{2}$
(C) 1
(D) 2
2. C.

Notice,
$$
\begin{array}{l} 
\triangle A B E \cong \triangle A C D \Rightarrow \angle E F D=120^{\circ} \\
\Rightarrow E, F, D, C \text { are concyclic } \\
\Rightarrow \angle F D E=\angle F C E=\angle G E A \\
\Rightarrow \triangle A E G \backsim \triangle A D E \\
\Rightarrow A E^{2}=A G \cdot A D . \\
\text { Also } \angle F A E=\angle A B E \Rightarrow \triangle A F E \backsim \triangle B A E \\
\Rightarrow A E^{2}=E F \cdot E B .
\end{array}
$$

From equations (1) and (2), combined with $A D=B E$, we get $A G=E F=1$.