Given in $\triangle A B C$, $A B=A C, \angle A=20^{\circ}$, $D$ is on $A C$ and $E$ is on $A B$. If $\angle A B D=10^{\circ}, \angle B D E$ $=20^{\circ}$, find the degree measure of $\angle A C E$.
$$
\begin{array}{l}
\text { Solution: } \because \angle A=20^{\circ}, A B=A C, \\
\therefore \angle A B C=\angle A C B=80^{\circ} .
\end{array}
$$

As shown in the figure, take point $O$ as the circumcenter of $\triangle B D E$, and connect $O D, O E, O B, O C$. We have
$$
\begin{array}{c}
\angle B O E=2 \angle B D E=40^{\circ}, \\
\angle E O D=2 \angle E B D=20^{\circ}, \\
O B=O E=O D . \\
\therefore \angle O E D=\angle O D E= \\
80^{\circ}, \\
\angle B D O=\angle O D E- \\
\angle B D E=60^{\circ} . \\
\text { Therefore, } D O=D B . \\
\text { Also, } \angle B D C=30^{\circ}, \\
\therefore \angle B D C=\angle O D C .
\end{array}
$$

Thus, $B$ and $O$ are symmetric with respect to line $A C$.
$$
\therefore \angle D C O=\angle D C B=80^{\circ}=\angle D E O \text {. }
$$

Therefore, $D, E, C, O$ are concyclic.
$$
\text { Hence, } \angle A C E=\angle D O E=20^{\circ} \text {. }
$$