Example 2 When a point $P$ inside a non-obtuse $\triangle ABC$ makes the value of $P A+P B+P C$ the smallest, find the degree measures of $\angle A P B$, $\angle B P C$, and $\angle C P A$.

Analysis: Let $Q$ be any point inside $\triangle ABC$, and $Q A$, $Q B$, $Q C$ form a "Y" shape. According to the shortest distance between two points being a straight line, to minimize the sum of the segments $Q A+Q B+Q C$, it is best to connect $Q A$, $Q B$, and $Q C$ end-to-end first.
Solution: Without loss of generality, let $\angle B A C \leqslant 60^{\circ}$.
As shown in Figure 2, construct an equilateral
$\triangle A Q D$ with $A Q$ as a side, such that point
$D$ is on the opposite side of $B$ with respect to $A C$,
and construct an equilateral $\triangle A C E$ (where $E$ is a fixed point) outside $\triangle A B C$ with $A C$ as a side, and connect $D E$.
It is easy to prove that $\triangle A Q C \cong \triangle A D E \Rightarrow Q C=D E$.
Therefore, $Q A+Q B+Q C=Q D+Q B+D E$
$$
=B Q+Q D+D E \geqslant B E \text {. }
$$

Thus, when $B Q+Q D+D E$ is minimized to $B E$, points $Q$ and $D$ should lie on segment $B E$. At this time,
$$
\angle A Q B=\angle A D E=120^{\circ} \text {. }
$$

Therefore, point $Q$ is the desired point $P$.
Hence, $\angle A P B=\angle B P C=\angle C P A=120^{\circ}$, meaning that point $P$ forms an angle of $120^{\circ}$ with each side of $\triangle A B C$.