10. As shown in Figure 4, given $\angle B A D=\angle D A C=9^{\circ}, A D$ $\perp A E$, and $A B+A C=B E$. Then $\angle B=$
$10.48^{\circ}$.
As shown in Figure 9, extend $B A$ to point $F$, such that
$$
A F=A C .
$$

From the given conditions, we know
$$
\begin{array}{l}
B F=B E, \\
\angle 3=90^{\circ}-\angle 1 \\
=90^{\circ}-\angle 2 \\
=\angle 4 .
\end{array}
$$

Thus, $\triangle E A F \cong \triangle E A C$, which implies $\angle A E F=\angle A E C$ (let this be $\alpha$).
Furthermore, $\angle F=2 \alpha, \angle B=180^{\circ}-4 \alpha$.
Therefore, $90^{\circ}-\alpha=\angle 5=\angle 1+\angle B$
$$
=9^{\circ}+180^{\circ}-4 \alpha \text {. }
$$

Solving this, we get $\alpha=33^{\circ}, \angle B=180^{\circ}-4 \times 33^{\circ}=48^{\circ}$.