3. As shown in Figure 1, $C$ is any point on line segment $A B$, $\triangle A C D$ and $\triangle B C E$ are equilateral triangles, $A F$ and $B G$ are tangents to the circle passing through points $D$, $C$, and $E$, with $F$ and $G$ being the points of tangency. Then ( ).
(A) $A F>B G$
(B) $A F=B G$
(C) $A F<B G$
(D) Uncertain
3. B.

Connect $A E$ and $B D$. Let $A E$ and $B D$ intersect at $P$, and $A E$, $B D$ intersect the circle at points $M$, $N$ respectively. It can be proven that $\triangle D C B \cong \triangle A C E$, then $A E = D B, \angle D P A = 60^{\circ}$.

Connect $M D$ and $N E$. It can be proven that $\triangle P D M$ and $\triangle P E N$ are equilateral triangles, then $M E = N D, A M = B N$.
$$
\begin{array}{l}
\text { Also, since } A F^{2} = A M \cdot A E, B G^{2} = B N \cdot B D, \\
\therefore A F = B G .
\end{array}
$$