3. As shown in Figure 1, in trapezoid $A B C D$, $A D / / B C, A D$ $=3, B C=9, A B=6, C D$ $=4$. If $E F / / B C$, and the perimeters of trapezoids $A E F D$ and $E B C F$ are equal, then the length of $E F$ is
(A) $\frac{45}{7}$
(B) $\frac{33}{5}$.
(C) $\frac{39}{5}$
(D) $\frac{15}{2}$
3. C.

From the given information, we have
$$
A D+A E+E F+F D=E F+E B+B C+C F \text {. }
$$

Then $A D+A E+F D=E B+B C+C F$
$$
=\frac{1}{2}(A D+A B+B C+C D)=11 \text {. }
$$

Since $E F \parallel B C$, we have
$E F \parallel A D, \frac{A E}{E B}=\frac{D F}{F C}$.
Let $\frac{A E}{E B}=\frac{D F}{F C}=k$. Then
$A E=\frac{k}{k+1} A B=\frac{6 k}{k+1}$,
$D F=\frac{k}{k+1} C D=\frac{4 k}{k+1}$.
Thus, $A D+A E+F D$
$$
=3+\frac{6 k}{k+1}+\frac{4 k}{k+1}=\frac{13 k+3}{k+1} \text {. }
$$

Therefore, $\frac{13 k+3}{k+1}=11 \Rightarrow k=4$.
Draw $A H \parallel C D$, intersecting $B C$ and $E F$ at points $H$ and
$G$ respectively. Then
$$
\begin{array}{l}
G F=H C=A D=3, \\
B H=B C-C H=9-3=6 .
\end{array}
$$

By $\frac{E G}{B H}=\frac{A E}{A B}=\frac{4}{5}$, we get
$$
E G=\frac{4}{5} B H=\frac{24}{5} \text {. }
$$

Thus, $E F=E G+G F=\frac{24}{5}+3=\frac{39}{5}$.