Example 24. $\triangle \mathrm{ABC} 4, \mathrm{AB}=\mathrm{AC}, \mathrm{O}$ is a point inside the shape,
$$
\begin{array}{c}
\angle \mathrm{A}=80^{\circ}, \angle \mathrm{OBC}=10^{\circ}, \angle \mathrm{OCB}=20^{\circ} . \\
\text { Find } \angle \mathrm{CAO}=\text { ? }
\end{array}
$$
Solve $\angle \mathrm{ACO}=30^{\circ}$,
$$
O \xrightarrow{\mathrm{S}(\mathrm{AC})} \mathrm{P} \text {, }
$$

then $\triangle \mathrm{CPO}$ is an equilateral triangle.
$$
\begin{array}{l}
\angle \mathrm{OAC}=\angle \mathrm{PAC} \text {. } \\
\angle \mathrm{BOC}=150^{\circ}, \angle \mathrm{BOP}=150^{\circ} .
\end{array}
$$
$$
\begin{array}{l}
\text { Draw } \mathrm{BP}, \triangle \mathrm{BOC}-\mathrm{S}(\mathrm{BO}) \\
\therefore \quad \angle \mathrm{PBO}=\angle \mathrm{CBO}=10^{\circ} \\
\Rightarrow \angle \mathrm{PBC}=20^{\circ} . \\
\angle \mathrm{BPO} \approx \angle \mathrm{BOP} . \\
\therefore \quad \angle \mathrm{BPC}=\angle \mathrm{BPO}+\angle \mathrm{OPC}=80^{\circ} \\
=\angle \mathrm{BAC} .
\end{array}
$$
$\therefore \quad B, A, P, C$ are concyclic.
$$
\therefore \quad \angle \mathrm{PAC}=\angle \mathrm{PBC}=20^{\circ} \text {. }
$$

Then $\angle \mathrm{CAO}=\angle \mathrm{PAC}=20^{\circ}$.