3. As shown in Figure 1, given that $E$ is the midpoint of side $AD$ of rectangle $ABCD$, $BE \perp AC$ at point $F$, and $AF=2$. Then $DF=$ ( ).
(A) $2 \sqrt{3}$
(B) 3
(C) $3 \sqrt{3}$
(D) $2 \sqrt{2}$
3. A.

Since $\triangle A E F \sim \triangle C B F$, we have $C F=2 A F=4$.
In Rt $\triangle A B C$ and Rt $\triangle A B E$, by the projection theorem, we get
$$
\begin{array}{l}
B F^{2}=A F \cdot C F=8, B F=2 \sqrt{2}, \\
E F=\frac{1}{2} B F=\sqrt{2} ; \\
A E^{2}=E F \cdot E B, A E=\sqrt{6} . \\
\text { Therefore, } \frac{E D}{E F}=\frac{A E}{E F}=\frac{E B}{A E}=\frac{E B}{E D} . \\
\text { Also, } \angle B E D=\angle D E F \Rightarrow \triangle B E D \backsim \triangle D E F \\
\Rightarrow \frac{E D}{E F}=\frac{B D}{D F} \Rightarrow D F=\frac{B D \cdot E F}{E D}=\frac{6 \times \sqrt{2}}{\sqrt{6}}=2 \sqrt{3} .
\end{array}
$$