2. In an equilateral $\triangle A B C$, it is known that $D$ and $E$ are points on sides $A B$ and $A C$ respectively, and satisfy $A D=C E, B E$ intersects $C D$ at point $F$. Then $\angle B F C=$ $\qquad$ .
2. $120^{\circ}$.

As shown in Figure 3, note that,
$$
\begin{array}{l}
\triangle A D C \cong \triangle C E B \\
\Rightarrow \angle A C D=\angle C B E . \\
\text { Then } \angle E F C \\
=\angle F B C+\angle F C B \\
=\angle E C F+\angle F C B \\
=60^{\circ} .
\end{array}
$$

Therefore, $\angle B F C=120^{\circ}$.