Example 1 In an equilateral $\triangle ABC$, there is a point $P$ inside, and its distances to the three vertices $A, B, C$ are $1, \sqrt{2}, \sqrt{3}$, respectively. Find the degree of $\angle APB$.

Analysis: The three segments $PA, PB, PC$ form a "Y" shape, and their lengths are exactly the three sides of a right-angled triangle (known as Pythagorean numbers). How can we concentrate $PA, PB, PC$ to form a right-angled triangle?

Method: Construct an equilateral $\triangle PBD$ with $PB$ as a side, such that point $D$ is on the opposite side of $A$ with respect to $BC$, and connect $CD$ (as shown in Figure 1). It is easy to prove that $\triangle PCD$ is a right-angled triangle, thus allowing us to find the degree of $\angle APB$.
Solution: As shown in Figure 1, construct an equilateral $\triangle P B D$ with $P B$ as a side, such that point $D$ is on the opposite side of $A$ with respect to $B C$, and connect $C D$.
Since $\angle A B C=\angle P B D=60^{\circ}$, we have
$$
\begin{array}{l}
\angle A B C-\angle P B C=\angle P B D-\angle P B C \\
\Rightarrow \angle A B P=\angle C B D .
\end{array}
$$

Also, $A B=B C, P B=B D$, thus
$\triangle A B P \cong \triangle C B D$.
Therefore, $D C=P A=1, \angle A P B=\angle B D C$.
In $\triangle P C D$,
$$
\begin{array}{l}
C D^{2}+P D^{2}=P A^{2}+P B^{2} \\
=1^{2}+(\sqrt{2})^{2}=(\sqrt{3})^{2}=P C^{2} .
\end{array}
$$

Thus, $\triangle P C D$ is a right triangle, i.e., $\angle P D C=90^{\circ}$.
Hence, $\angle A P B=\angle B D C=\angle B D P+\angle P D C$
$$
=60^{\circ}+90^{\circ}=150^{\circ} \text {. }
$$

Note: For any "Y"-shaped three segments whose lengths form a Pythagorean triple, after constructing an equilateral triangle, they can form a right triangle.