3. As shown in Figure 2, the side length of square $ABCD$ is $1, E$ is a point outside side $CD$, satisfying $CE // BD$, $BE=BD$. Then $CE=$
3. $\frac{\sqrt{6}-\sqrt{2}}{2}$.

As shown in Figure 4, draw $E F \perp$
$B C$ at point $F$. Then
$$
E F=C F \text {. }
$$

Let $C F=x$. Then
$$
\begin{aligned}
2 & =B D^{2}=B E^{2} \\
& =B F^{2}+E F^{2} \\
& =(1+x)^{2}+x^{2} \\
\Rightarrow & 2 x^{2}+2 x-1=0
\end{aligned}
$$
$\Rightarrow x=\frac{\sqrt{3}-1}{2}$ (negative root discarded).
$$
\Rightarrow C E=\sqrt{2} x=\frac{\sqrt{6}-\sqrt{2}}{2} .
$$