$\begin{array}{l}\text { 10. As shown in Figure } 3, D C \\ / / A B, \angle B A E= \\ \angle B C D, A E \perp D E, \\ \angle D=130^{\circ} \text {. Then } \angle B \\ =\end{array}$
$10.40^{\circ}$.
Extend $A E$ and $C D$ to intersect at point $F$. Given $\angle B A E=\angle B C D$ and $D C / / A B$, it is easy to see that quadrilateral $A B C F$ is a parallelogram.
$$
\text { Also, } \angle F=\angle C D E-\angle D E F=130^{\circ}-90^{\circ}=40^{\circ} \text {, }
$$

Therefore, $\angle B=\angle F=40^{\circ}$.