8. As shown in Figure $3, A C=$ $B C, A C \perp B C$ at point $C, A B=A D=B D$, $C D=C E=D E$. If $A B=\sqrt{2}$, then $B E=$
8. 1 .

In $\triangle A D C$ and $\triangle B D C$, given $A D=D B, D C=D C$, $A C=B C$, we can conclude that
$\triangle A D C \cong \triangle B D C, \angle A D C=\angle B D C$.
Since $\angle A D B=60^{\circ}$, it follows that,
$\angle A D C=\angle B D C=\angle E D B=30^{\circ}$.
Therefore, $D B \perp C E, B C=B E$.
Moreover, $\triangle A C B$ is an isosceles right triangle, and $A B=\sqrt{2}$, so, $B C=B E=1$.