2. Given that $E$ is any point on side $C D$ of square $A B C D$, and $E F \perp A C$ at point $F$. Extend $B F$ to intersect line $A E$ at point $G$. Then $\angle B G C=(\quad)$.
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
2. B.

As shown in Figure 2, connect $D F$.
Since quadrilateral $A B C D$ is a square,
$$
\begin{array}{l}
\Rightarrow \angle D C F=\angle B C F=45^{\circ}, D C=B C \\
\Rightarrow \triangle D F C \cong \triangle B F C \Rightarrow \angle F B C=\angle F D C . \\
\text { Also, } \angle A D E=90^{\circ}, \angle E F A=90^{\circ} \\
\Rightarrow A, D, E, F \text { are concyclic } \Rightarrow \angle F A E=\angle F D C \\
\Rightarrow \angle C A G=\angle F A E=\angle F D C \\
=\angle F B C=\angle G B C \\
\Rightarrow G, A, B, C \text { are concyclic } \\
\Rightarrow \angle B G C=\angle B A C=45^{\circ} .
\end{array}
$$