3. In the right trapezoid $A B C D$, it is known that $\angle A=$ $\angle D=90^{\circ}, D E \perp A C$ at point $E, \angle A C D=$ $\angle E B C=30^{\circ}, A D=\sqrt{3}$. Then $B C=$ $\qquad$
3. 3 .

As shown in Figure 3, from the given conditions we know
$$
\begin{array}{l}
\angle A D E=\angle C A B \\
=30^{\circ} \\
\Rightarrow \triangle C B E \backsim \triangle C A B \\
\Rightarrow \frac{C B}{C E}=\frac{C A}{C B} \\
\Rightarrow C B^{2}=C E \cdot C A . \\
\text { Also, } C D^{2}=C E \cdot C A, \text { thus } \\
C B=C D=\sqrt{3} A D=3 .
\end{array}
$$