Example 5. In $\triangle A B C$, if there exists a point $P$ such that $\angle P A C=\angle P B A=\angle P C B=\theta$. Find the range of values for $\theta$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}=4 S_{\triangle A B C} \operatorname{ctg} \theta, \\
\therefore \operatorname{ctg} \theta=\frac{a^{2}+b^{2}+c^{2}}{4 S_{\triangle A B C}} . \text { Let } S_{\triangle A B C}=S, \\
S=\sqrt{p(p-a)(p-b)(p-c)} \\
\leqslant \sqrt{p} \sqrt{\left[\frac{(p-a)+(p-b)+(p-c)}{3}\right]^{3}} \\
=\frac{4 p^{2}}{3 \sqrt{3}}=\frac{(a+b+c)^{2}}{4 \cdot 3 \sqrt{3}}, \\
\because \quad(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c \\
\quad+2 c a \leqslant 3\left(a^{2}+b^{2}+c^{2}\right) \\
\therefore \quad \frac{a^{2}+b^{2}+c^{2}}{4 S} \geqslant \sqrt{3}, \text { i.e., } \operatorname{ctg} \theta \geqslant \sqrt{3},
\end{array}
$$

Also, $\theta$ is an acute angle, $\therefore \theta \in\left[0, \frac{\pi}{6}\right]$.
Solving the problem with the line segment lengths set as follows:
As shown in the figure. Note that
$$
\begin{aligned}
\operatorname{ctg} A & =\frac{\cos A}{\sin A} \\
& =\frac{b^{2}+c^{2}-a^{2}}{4 s},
\end{aligned}
$$

Applying the above conclusion in $\triangle P A B, \triangle P A C, \triangle P B C$, we have
$$
\left\{\begin{array}{l}
y^{2}+c^{2}-x^{2}=4 S_{\triangle P A B} \operatorname{ctg} \theta, \\
x^{2}+b^{2}-z^{2}=4 S_{\triangle P A C} \operatorname{ctg} \theta, \\
z^{2}+a^{2}-y^{2}=4 S_{\triangle P B C} \operatorname{ctg} \theta .
\end{array}\right.
$$