Example 3 Given in $\triangle A B C$, $A B=A C$, $\angle B A C=100^{\circ}, P$ is a point on the angle bisector of $\angle C$, $\angle P B C=10^{\circ}$. Find the degree measure of $\angle A P B$.
Solution: As shown in Figure 6, take a point $D$ on the extension of $CA$ such that $CD = CB$, and connect $DP$ and $DB$. Let the intersection of $DB$ and $CP$ be $E$. It is easy to see that $B$ and $D$ are symmetric with respect to $PC$. Therefore, $\triangle PBD$ is an equilateral triangle, and $\angle PDA = \angle PBC = 10^{\circ}$.

In $\triangle CBD$, we know that $\angle DBC = 70^{\circ}$. Thus, $\angle ABD = 30^{\circ} = \angle ABP$, so $AB$ is the perpendicular bisector of $PD$.
$$
\therefore AP = AD.
$$

Therefore, $\angle APD = \angle ADP = 10^{\circ}$.
$$
\therefore \angle APB = 70^{\circ}.
$$

Here, we focus on the exterior angle $\angle EPB = 30^{\circ}$ of $\triangle PBC$, and construct an equilateral triangle, which creates conditions for solving the problem.