In Rt $\triangle A B C$, it is known that $\angle A=$ $20^{\circ}, \angle B=90^{\circ}, A D$ is the bisector of $\angle B A C$, point $E$ is on side $A B$, and lines $C E$ and $D E$ are connected. If $\angle D C E=30^{\circ}$, find the degree measure of $\angle A D E$.
Solve As shown in Figure 2, construct $\angle A E F=20^{\circ}, E F$ intersects $A D$ at point $F$. Take point $G$ on side $A C$ such that $A G=A E$, and connect $F G$, $E G$, and $C F$.
Since $A D$ bisects $\angle E A G$, by its symmetry we know
$$
F E=F G,
$$

and
$$
\begin{array}{l}
\angle G F D=\angle E F D \\
=\angle A E F+\angle E A F \\
=20^{\circ}+10^{\circ}=30^{\circ} .
\end{array}
$$

Therefore, $\triangle F E G$ is an equilateral triangle.
Thus, $G F=G E$.
It is easy to see that $\angle A G E=80^{\circ}, \angle G C E=40^{\circ}$.
Then $\angle G E C=\angle A G E-\angle G C E$
$$
=40^{\circ}=\angle G C E \text {. }
$$

Therefore, $G C=G E=G F$.
Thus, $\angle G C F=\frac{1}{2} \angle A G F=\frac{1}{2} \times 20^{\circ}=10^{\circ}$.
Hence $\angle F C E=\angle G C E-\angle G C F=30^{\circ}$.
Since $\angle E F D=\angle E C D=30^{\circ}$, points $E$, $F$, $C$, and $D$ are concyclic.
Therefore, $\angle A D E=\angle F C E=30^{\circ}$.
(Wan Xiren, Hunan Library Training Building Wanxi School, 410011)