Example 3.2.6 (Training Selection Competition 2001) Given an equilateral $\triangle ABC$, and $D$ is any point on side $BC$. The circumcenter and incenter of $\triangle ABD$ are $O_{1}$ and $I_{1}$, respectively, and the circumcenter and incenter of $\triangle ADC$ are $O_{2}$ and $I_{2}$, respectively. The line $O_{1} I_{1}$ intersects $O_{2} I_{2}$ at point $P$. Try to find: When point $D$ moves on $BC$, the locus of point $P$.

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Solution 1 As shown in the figure, by adding auxiliary lines, from $\angle A O_{2} D=2 \angle C=120^{\circ}$, $\angle A I_{2} D=90^{\circ}+1 / 2 \cdot \angle C=120^{\circ}$, $\angle B=60^{\circ}$. We know that $O_{2} 、 I_{2}$ are both on $\odot O_{1}$.
Similarly, $O_{1} 、 I_{1}$ are both on $\odot O_{2}$.
Obviously, $\odot O_{1}$ and $\odot O_{2}$ are equal circles, $\angle O_{1} D O_{2}=60^{\circ}$.
$$
\begin{array}{l}
\because \angle A I_{2} O_{2}=\angle A D O_{2}=30^{\circ}=\angle I_{1} A I_{2}, \\
\therefore A I_{1} / / O_{2} P, \angle O_{1} P O_{2}=\angle O_{1} I_{1} A=30^{\circ}=1 / 2 \cdot \angle O_{1} D O_{2},
\end{array}
$$

Thus, $D$ is the circumcenter of $\triangle O_{1} P O_{2}$. [ $D$ is inside $\triangle O_{1} P O_{2}$, and on the perpendicular bisector of $O_{1} O_{2}$, $\angle O_{1} P O_{2}=$ $1 / 2 \cdot \angle O_{1} D O_{2}$, then $D$ is the circumcenter of $\triangle O_{1} P O_{2}$. ]
In $\triangle O_{2} D I_{2}$, by $\angle D I_{2} O_{2}=150^{\circ} \Rightarrow \angle O_{2} D I_{2}+\angle D O_{2} I_{2}=30^{\circ}$,
$$
\begin{array}{l}
\because \angle O_{2} D C=\angle I_{2} D C+\angle O_{2} D I_{2}=\angle A D I_{2}+\angle O_{2} D I_{2} \\
=\angle A D O_{2}+\angle O_{2} D I_{2}+\angle O_{2} D I_{2} \\
=\angle A B O_{2}+2 \angle O_{2} D I_{2}=30^{\circ}+2 \angle O_{2} D I_{2} . \\
D O_{2}=D P=D O_{1}, \angle D O_{2} P=\angle D P O_{2} \text {. } \\
\end{array}
$$

From (1)(2) we get $\angle P D C=180^{\circ}-\angle O_{2} D C-2 \angle D O_{2} P=150^{\circ}-2 \angle O_{2} D I_{2}-2 \angle D O_{2} P=90^{\circ}$. That is, $P D \perp B C$, and $A D=\sqrt{3} D O_{1}=\sqrt{3} D P$.
Now, establish a Cartesian coordinate system with the line of $B C$ as the $x$-axis and the midpoint of $B C$ as the origin, and assume the side length of the equilateral $\triangle A B C$ is 2, and the coordinates of point $P$ are $(x, y)$.
Then, in the right $\triangle A O D$, we have $A D^{2}-O D^{2}=A O^{2}$, that is, $(\sqrt{3} D P)^{2}-O D^{2}=A O^{2}$,
$\therefore(\sqrt{3} y)^{2}-x^{2}=(\sqrt{3})^{2}$, the trajectory equation is:
$y^{2}-\frac{x^{2}}{3}=1,-1<x<1, y<0$.
Solution 2 Establish a Cartesian coordinate system with the midpoint of $B C$ as the origin and the line of $B C$ as the $x$-axis.
Let $B C=2, C(1,0), B(-1,0)$. Suppose the coordinates of point $D$ are $(a, 0)$, and assume $-1<a<1$.
$B E \perp A C, C F \perp A B$, then $O_{2} 、 I_{1}$ are on $B E$ or its extension, $O_{1} 、 I_{2}$ are on $C F$ or its extension.
Since $O_{2}$ is on the perpendicular bisector of $D C$, $\angle C B O_{2}=30^{\circ}$, from this we get the coordinates of $O_{2}$ as: $\left(\frac{a+1}{2}, \frac{3+a}{2} \cdot \frac{\sqrt{3}}{3}\right)$, similarly, we can derive the coordinates of $O$ as $\left(\frac{a-1}{2}, \frac{3-a}{2} \cdot \frac{\sqrt{3}}{3}\right)$;

Draw $I_{2} G \perp B C$ intersecting $B C$ at $G$. Then by $\angle G C I_{2}=30^{\circ}$. The x-coordinate of point $I_{2}$ is $1-O G$, from this we can calculate and express, getting the coordinates of $I_{2}$ as:
$$
\left(\frac{1}{2}\left(a-1+\sqrt{a^{2}+3}\right), \frac{\sqrt{3}}{6}\left(3-a-\sqrt{a^{2}+3}\right)\right),
$$

Similarly, the coordinates of $I_{1}$ are: $\left(\frac{1}{2}\left(a+1-\sqrt{a^{2}+3}\right), \frac{\sqrt{3}}{6}\left(3+a-\sqrt{a^{2}+3}\right)\right)$,
Thus, we can derive the equation of $O_{2} I_{2}$, and the equation of $O_{1} I_{1}$. Let the coordinates of the intersection point $P$ be $(x, y)$,
From these two equations, we solve to get $x=a$, substituting this we get $y=-\sqrt{1+\frac{a^{2}}{3}}$. Eliminating $a$ we get the trajectory equation of point $P$ as:
$$
y^{2}-\frac{x^{2}}{3}=1,-1<x<1, y<0 \text {. }
$$