Example 3 In $\triangle A B C$, it is known that $A B=A C$, $\angle B A C=100^{\circ}, P$ is a point on the angle bisector of $\angle A C B$, and $\angle P B C=10^{\circ}$. Find $\angle A P B$.
【Analysis 1】First, identify which angles are known.
As shown in Figure 3, given $A B=A C, \angle B A C=100^{\circ}$, we have
$$
\angle A B C=\angle A C B=\frac{180^{\circ}-100^{\circ}}{2}=40^{\circ} \text {. }
$$

Thus, $\angle 1=\angle 2=20^{\circ}, \angle 3=10^{\circ}, \angle 4=30^{\circ}$.
Furthermore, since $\angle 1=\angle 2$, $C P$ is the axis of symmetry. Therefore, construct an isosceles $\triangle B C D$ with the extension of $C P$ as the angle bisector (axis of symmetry), as shown in Figure 3. Extend $C A$ to point $D$ such that $C D=C B$, and connect $P D$ and $B D$.
From $\triangle P B C \cong \triangle P D C$
$\Rightarrow B P=D P, \angle 6=\angle 3=10^{\circ}$.
From $\angle D B C=\angle B D C=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ}$
$\Rightarrow \angle D B P=70^{\circ}-10^{\circ}=60^{\circ}$.
Since $P B=P D$, $\triangle P B D$ is an equilateral triangle.
Thus, $\angle D B P=60^{\circ}, \angle 4=30^{\circ}$
$\Rightarrow \angle 4=\angle 5 \Rightarrow \triangle A B D \cong \triangle A B P$
$\Rightarrow \angle A P B=\angle A D B=70^{\circ}$.
【Analysis 2】From $\angle 1=\angle 2$, consider constructing a triangle with $P$ as the incenter.

As shown in Figure 4, take point $D$ on side $A C$ such that $\angle D B P=\angle 3$, then $P$ is the incenter of $\triangle D B C$.
Extend $C P$ to intersect $A B$ at point $E$, and connect $D E$.
Since $\angle D B C=20^{\circ}$, $B D$ is the angle bisector of $\angle A B C$.

Also, $C E$ is the angle bisector of $\angle A C B$, and by the symmetry of the isosceles triangle, $D E \parallel B C$.
In $\triangle D B C$, since
$\angle B D C=180^{\circ}-\left(20^{\circ}+20^{\circ}\right)-\left(10^{\circ}+10^{\circ}\right)=120^{\circ}$,
we have $\angle P D C=60^{\circ}$.
In $\triangle A E C$, since
$$
\angle A E C=180^{\circ}-\left(100^{\circ}+20^{\circ}\right)=60^{\circ}=\angle P D C \text {, }
$$

$A, E, P, D$ are concyclic.
Thus, $\angle 5=\angle 6=\frac{1}{2}\left(180^{\circ}-100^{\circ}\right)=40^{\circ}$.
Since $\angle 4=\angle 2+\angle 3=20^{\circ}+10^{\circ}=30^{\circ}$, we have $\angle A P B=\angle 4+\angle 5=70^{\circ}$.