Example 14. As shown in Figure 14; $P$ is a point inside an equilateral triangle $ABC$, and $PA=3, PB=4, PC=5$. Find the side length of this equilateral triangle.
Prove that outside $\triangle ABC$ there is a point $D$, such that $AD=AP, CD=CP$.
It is easy to know that $\triangle APB \cong \triangle ADC$.
$\therefore \angle 1=\angle 2$.
Therefore, $\angle PAD=60^{\circ}$.
$\therefore \triangle APD$ is an equilateral triangle.

Also, $\because PD=3, DC=4, PC=5$,
$\therefore \triangle PDC$ is a right triangle,
i.e., $\angle PDC=90^{\circ}$.
$\because \angle ADP=60^{\circ}$,
$\therefore \angle ADC=\angle ADP+\angle PDC=150^{\circ}$.
In $\triangle ADC$, by the cosine rule we get:
$$
AC=\sqrt{25+12 \sqrt{3}} \text{. }
$$

Thus, the side length of this equilateral triangle is
$$
\sqrt{25+12 \sqrt{3}} \text{. }
$$