Example 1 As shown in Figure 2.10.1, in $\triangle A B C$, $A X$ is the angle bisector of $\angle A$. Try to find a point $P$ on $A X$ such that the absolute value of the difference between $\angle P B A$ and $\angle P C A$ is maximized.
When $A B=A C$, any point on $A X$ makes $\angle P B A = \angle P C A$, thus the difference between $\angle P B A$ and $\angle P C A$ is always 0.

Assume $A B > A C$. For easier observation, move the vertex $C$ of $\angle P C A$ to $A B$, and take $A C^{\prime} = A C$ on $A B$, then connect $P C^{\prime}$. Thus, $\angle P C A - \angle P B A = \angle P C^{\prime} A - \angle P B A = \angle B P C^{\prime}$. When $P$ moves on $A X$, the size of $\angle B P C^{\prime}$ changes. Since $B C^{\prime}$ is a fixed-length segment, $\angle B P C^{\prime}$ is maximized only when the circumcircle of $\triangle P B C^{\prime}$ is tangent to $A X$ at point $P$. Therefore, a point $P$ can be taken on $A X$ such that $A P^{2} = A C^{\prime} \cdot A B$. This point $P$ is not difficult to construct.