Example 8. In isosceles $\triangle A B C$, the vertex angle $B=20^{\circ}$. Points $D, E$ are taken on sides $B C, A B$ respectively, such that $\angle D A C=$ $60^{\circ}, \angle A C E=50^{\circ}$. Find $\angle A D E$.
Given the figure, let $F$ be the point symmetric to $D$ with respect to the angle bisector of $\angle B$, and $O$ be the intersection of $A D$ and $C F$. Connect $E O$.

To prove $\triangle E F D \cong \triangle E O D$, we only need to prove $F D=O D$ and $E F=E O$.

In $\triangle O F D$, all interior angles are $60^{\circ}$, so $F D=O D$.
$$
\begin{array}{l}
\because \angle A C E=50^{\circ}, \\
\angle A E C=180^{\circ}-80^{\circ}-50^{\circ}=50^{\circ}, \\
\therefore E A=A C=A O .
\end{array}
$$

Thus, $\angle E O A=\angle O E A$
$$
=\frac{1}{2}\left(180^{\circ}-\angle E A O\right)=80^{\circ} \text {. }
$$
$$
\begin{array}{l} 
\text { Also, } \angle B F D=80^{\circ}=\angle E O A, \\
\angle D F O=60^{\circ}=\angle F O D, \\
\therefore \angle E F O=180^{\circ}-\angle B F D-\angle D F O \\
=180^{\circ}-\angle E O A-\angle F O D=\angle E O F .
\end{array}
$$

Therefore, $E F=E O$.
This proves that $\triangle E F D \cong \triangle E O D$.
$$
\therefore \angle A D E=\frac{1}{2} \angle O D F=30^{\circ} \text {. }
$$