3. Given that $E$ is the midpoint of side $BC$ of square $ABCD$, and perpendiculars are drawn from points $B$ and $D$ to $AE$, with the feet of the perpendiculars being $F$ and $G$ respectively. Then $\angle FBG=(\quad)$.
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
3. B.

As shown in Figure 3, it is easy to see that,
$$
\begin{array}{l}
\text { Rt } \triangle A B F \cong \text { Rt } \triangle D A G \\
\Rightarrow A G=B F . \\
\text { From Rt } \triangle A B F \sim \\
\text { Rt } \triangle A E B \\
\Rightarrow \frac{B F}{A F}=\frac{E B}{A B}=\frac{1}{2} \\
\Rightarrow B F=\frac{1}{2} A F \Rightarrow B F=F G \\
\Rightarrow \angle F B G=45^{\circ} .
\end{array}
$$