Example 6 In $\triangle ABC$, $\angle CAB = \angle CBA = 50^{\circ}$, $O$ is a point inside the triangle, $\angle OAB = 10^{\circ}$, $\angle OBC = 20^{\circ}$. Find the degree measure of $\angle OCA$.
Solution: As shown in Figure 6, draw a perpendicular from point $C$ to $A B$, intersecting the extension of $B O$ at $E$. Connect $A E$.

Given $\angle C A B=\angle C B A=50^{\circ}$, we know that point $A$ is symmetric to point $B$ with respect to $C E$. Also, given $\angle O B C=20^{\circ}$, $\angle E C B=40^{\circ}$, we have $\angle C E A=\angle C E B=120^{\circ}$. Therefore,
$$
\angle O E A=120^{\circ}=\angle C E A.
$$

Given $\angle E A B=\angle E B A=30^{\circ}$, $\angle O A B=10^{\circ}$, we know that $A E$ bisects $\angle C A O$. Thus, point $C$ is symmetric to point $O$ with respect to $A E$.
$$
\begin{array}{l}
\text { Then } \angle O C A=\angle C O A \\
=\frac{1}{2}\left(180^{\circ}-\angle O A C\right)=70^{\circ}.
\end{array}
$$

From the accurate diagram, we can conjecture that $A O = A C$, or that the axis of symmetry between point $O$ and point $C$ passes through point $A$. Since the diagram does not provide the axis of symmetry, we can "fold" the line $B O$ to the position of $A E$ through the perpendicular bisector of $A B$, thus solving the problem of the angle bisector of $\angle C A O$. This approach is ingenious.

In summary, when dealing with angle bisectors in a diagram, we should consider using symmetric points. When the diagram lacks an angle bisector, we should also try to adjust the diagram to make the angle bisector appear in a timely manner, facilitating the determination of the symmetric relationship.