9. On the two legs $AB, AC$ of isosceles $\triangle ABC$ with base angles equal to $80^{\circ}$, points $D, E$ are taken such that $\angle BDC=$ $50^{\circ}, \angle BEC=40^{\circ}$. Then $\angle ADE=$ $\qquad$
$9.50^{\circ}$.
Given $\angle B A C=20^{\circ}, \angle B C D=50^{\circ}$, hence
$$
\begin{array}{l}
B C=B D, \\
\angle C B E=60^{\circ}, \angle A B E=20^{\circ} .
\end{array}
$$

Take a point $F$ on $C E$ such that $\angle C B F=20^{\circ}$, then
$$
\begin{array}{l}
\angle E B F=40^{\circ}, \\
B F=F E, \\
\angle D B F=60^{\circ}, \angle B F C=80^{\circ}, \\
B C=B F .
\end{array}
$$

From (1) and (3), we get $B D=B F$, so $\triangle B D F$ is an equilateral triangle. Thus,
$$
B F=D F .
$$

From (2) and (4), we get $D F=F E$, so $\triangle D F E$ is an isosceles triangle.
$$
\text{Also, } \angle B F D=60^{\circ} \text{. Thus, } \angle D F E=40^{\circ} \text{. }
$$

Therefore, $\angle F E D=70^{\circ}, \angle A D E=50^{\circ}$.