Let $I_{A}$ be the center of the excircle opposite to vertex $A$ of $\triangle ABC$. Let the composite transformation $h$: first perform an inversion with $A$ as the inversion center and $AB \cdot AC$ as the inversion power, then perform a reflection with $AI$ as the axis of symmetry.
Then $B \xrightarrow{h} C, C \xrightarrow{h} B, \odot O \xrightarrow{h}$ to line $BC, I \xrightarrow{h} I_{A}$.
As shown in Figure 4, let the circumcircles of $\triangle IAI_{1}$ and $\triangle IAI_{2}$ be $\odot O_{1}$ and $\odot O_{2}$, respectively.

Since $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $I$, we know that $O_{1}O_{2}$ is the perpendicular bisector of $AI$. Therefore,
$$\begin{array}{l} \angle AO_{1}O_{2} + \angle AO_{2}O_{1} = \angle AI_{1}I + \angle AI_{2}I \\ = \angle ABI_{1} + \angle BAI_{1} + \angle ACI_{2} + \angle CAI_{2} \\ = \frac{1}{2} \angle B + \frac{1}{2} \angle C + \frac{1}{2} \angle BAD + \frac{1}{2} \angle CAD \\ = \frac{1}{2}(\angle BAC + \angle B + \angle C) = 90^{\circ}. \end{array}$$

Thus, $\angle O_{1}AO_{2} = 90^{\circ}$, meaning $\odot O_{1}$ and $\odot O_{2}$ are orthogonal.
Let $M \xrightarrow{h} M^{\prime}, N \xrightarrow{h} N^{\prime}$.
Since points $M$ and $N$ lie on $\odot O$, points $M^{\prime}$ and $N^{\prime}$ lie on line $BC$.
Therefore, $\odot O_{1} \xrightarrow{h} I_{A}M^{\prime}, \odot O_{2} \xrightarrow{h} I_{A}N^{\prime}$.
Since $\odot O_{1}$ and $\odot O_{2}$ are orthogonal, by the property of inversion, we know that $I_{A}M^{\prime} \perp I_{A}N^{\prime}$.
Let the projection of point $I_{A}$ on $BC$ be $X$. Then $X$ is a fixed point.
In $\triangle I_{A}M^{\prime}N^{\prime}$, by the projection theorem,
$XM^{\prime} \cdot XN^{\prime} = XI_{A}^{2}$ (a constant).
Since $MN \xrightarrow{h}$ to the circumcircle $\Gamma$ of $\triangle AM^{\prime}N^{\prime}$, thus, $AX$ is the common radical axis of the moving circle $\Gamma$.
Let $AX$ intersect circle $\Gamma$ at point $Y$.
By $XA \cdot XY = XM^{\prime} \cdot XN^{\prime} = XI_{A}^{2}$, we know that $Y$ is a fixed point.
Let $Y_{0} \xrightarrow{h} Y$. Then $Y_{0}$ is also a fixed point, and $Y_{0}$ lies on line $MN$, meaning line $MN$ passes through a fixed point $\left(Y_{0}\right)$.

When the problem only involves multiple circles, it is advisable to use the radical center of the multiple circles, combined with the properties and theorems of circles, to solve by first performing an inversion and then a central symmetry.