Example 7. $\triangle A B C$ is an isosceles triangle, with vertex angle $A=$ $80^{\circ}, O$ is a point inside $\triangle A B C$, $\angle O B C=10^{\circ}$, $\angle O C B=20^{\circ}$. Find $\angle O A C$.
Solve: As shown in the figure, construct the symmetric point $N$ of $O$ with respect to $AC$, and connect $AN, BN, CN$.
$$
\begin{array}{l}
\because \angle OCA = 30^{\circ}, \angle NCA = 30^{\circ}, \\
\therefore \angle OCN = 60^{\circ}.
\end{array}
$$

Therefore, $\triangle OCN$ is an equilateral triangle.
Thus, $ON = OC$.
$$
\begin{array}{l}
\text{Also, } \angle BOC = 180^{\circ} \\
- 20^{\circ} - 10^{\circ} \\
= 150^{\circ}, \\
\therefore \angle BON = 360^{\circ} \\
- 150^{\circ} - 60^{\circ} \\
= 150^{\circ} = \angle BOC.
\end{array}
$$
$\therefore \triangle BON \cong \triangle BOC$. Hence, $BC = BN$.
$$
\begin{array}{l}
\therefore \angle BNC = \angle BCN = 20^{\circ} + 60^{\circ} = 80^{\circ}. \\
\because \angle BAC = 80^{\circ},
\end{array}
$$
$\therefore B, A, N, C$ are concyclic.
$$
\begin{array}{l}
\text{Thus, } \angle NBC = \angle NAC = 2 \times 10^{\circ} = 20^{\circ}. \\
\therefore \angle OAC = 20^{\circ}.
\end{array}
$$