In $\triangle A B C$, $\angle B A C=100^{\circ}, \angle A B C=$ $50^{\circ}$, point $D$ is on $B C$, $\angle B A D=50^{\circ}$, point $E$ is on $A C$, $\angle A B E=20^{\circ}$. Find the degree measure of $\angle A D E$.
Solution: As shown in Figure 1, construct a regular $\triangle A D F$ inside $\triangle A B D$, and draw the angle bisector of $\angle A F D$ to intersect $B E$ at $G$. Connect $G A$, $G D$, and $B F$. It is easy to see that
$$
\begin{array}{l}
D F=D A=D B, \\
\angle B D F=20^{\circ} .
\end{array}
$$

Therefore, $\angle D B F=\angle D F B=80^{\circ}$.
It is easy to see that $\angle D B G=30^{\circ}=\angle D F G$, so points $F$, $B$, $D$, and $G$ are concyclic. Thus,
$\angle F D G=\angle F B G=80^{\circ}-30^{\circ}=50^{\circ}$.
Hence, $\angle G A D=\angle G D A=60^{\circ}-50^{\circ}=10^{\circ}$.
It is easy to calculate that $\angle G A E=60^{\circ}=\angle A E G$, which means $\triangle A E G$ is an equilateral triangle. Therefore, $G D=G A=G E$.
Thus, $\angle G D E=\frac{1}{2} \angle B G D=\frac{1}{2} \angle B F D=40^{\circ}$.
Therefore, $\angle A D E=40^{\circ}-10^{\circ}=30^{\circ}$.