Example: $2 P$ is a point inside a regular $\triangle A B C$, $\angle A P B=113^{\circ}, \angle A P C=123^{\circ}$. Prove that a triangle can be formed with $A P, B P, C P$ as sides, and determine the degree measures of the interior angles of the formed triangle.
Proof: As shown in Figure 3, with point $C$ as the center, rotate $\triangle A P C$ counterclockwise by $60^{\circ}$. Since $A C = B C$, $\angle A C P = \angle B C P_{1} = 60^{\circ} - \angle P C B$, $C P = C P_{1}$, and $\angle P C P_{1} = 60^{\circ}$, we have:
$\triangle A P C \cong \triangle B P_{1} C$.
Therefore, $A P = B P_{1}$, $\angle B P_{1} C = \angle A P C = 123^{\circ}$.
Given $C P = C P_{1}$ and $\angle P C P_{1} = 60^{\circ}$, we know that $\triangle P C P_{1}$ is an equilateral triangle. Thus,
$$
P P_{1} = C P, \angle C P P_{1} = \angle C P_{1} P = 60^{\circ}.
$$

At this moment, $\triangle B P P_{1}$ is a triangle formed by the sides $B P$, $A P (= B P_{1})$, and $C P (= P P_{1})$. It is easy to see that
$\angle B P_{1} P = \angle B P_{1} C - \angle C P_{1} P$
$= \angle A P C - 60^{\circ} = 63^{\circ}$.
Also, $\angle B P C = 360^{\circ} - 113^{\circ} - 123^{\circ} = 124^{\circ}$,
so, $\angle B P P_{1} = \angle B P C - \angle C P P_{1} = 64^{\circ}$.
Therefore, $\angle P B P_{1} = 180^{\circ} - 63^{\circ} - 64^{\circ} = 53^{\circ}$.
Explanation: By rotating $\triangle A P C$ counterclockwise by $60^{\circ}$ around point $C$, a new configuration is formed, concentrating the segments $A P$, $B P$, and $C P$. This intuitively proves the existence of the triangle $\left(\triangle B P P_{1}\right)$ with these three segments as sides, and it is easy to calculate that the three interior angles of the triangle are $63^{\circ}$, $64^{\circ}$, and $53^{\circ}$.