Example 2 In $\triangle A B C$, $A B=A C, \angle B A C=$ $80^{\circ}, O$ is a point inside the triangle, $\angle O B C=10^{\circ}, \angle O C B=$ $30^{\circ}$. Find the degree measure of $\angle B A O$.
Solution: As shown in Figure 3, draw $A H \perp B C$ at $H$. Also, $A H$ is the angle bisector of $\angle B A C$, so $\angle B A H=\angle C A H=40^{\circ}$.

Extend $\mathrm{CO}$ to intersect $A H$ at $P$, then $\angle B O P=40^{\circ}=\angle B A P$. Connect $B P$, by symmetry we have $\angle P B C=\angle P C B=30^{\circ}$. Therefore, $\angle P B O=30^{\circ}-10^{\circ}=20^{\circ}$.
Thus, $\angle A B P=40^{\circ}-20^{\circ}=20^{\circ}$.
In $\triangle A B P$ and $\triangle O B P$,
$$
\begin{array}{l}
\angle B A P=\angle B O P=40^{\circ}, B P=B P, \\
\angle A B P=\angle O B P=20^{\circ} .
\end{array}
$$

Therefore, $\triangle A B P \cong \triangle O B P$. Hence $A B=O B$.
Since $\angle A B O=40^{\circ}$, we have
$$
\angle B A O=\frac{180^{\circ}-\angle A B O}{2}=70^{\circ} \text {. }
$$