Prove: With $A$ as the inversion center and $r$ (where $r$ is any real number) as the inversion radius, the inversion of Figure 4 results in Figure 5.

Among them, $\odot T_{1}$ and $\odot T_{2}$ become two parallel lines $T_{1}^{\prime}$ and $T_{2}^{\prime}$, $\odot T$ becomes a circle $\odot T^{\prime}$ tangent to the two parallel lines, the tangent line $MN$ inverts to a line parallel to the line $T_{\mathrm{t}}^{\prime}$, and point $D^{\prime}$ is outside $\odot T^{\prime}$.

To prove that $A$ is the incenter of $\triangle BCD$, it is sufficient to show that the distances from point $A$ to $BC$, $CD$, and $BD$ are equal, and that $A$ is inside $\triangle BCD$, i.e., the circumradii of $\triangle AB^{\prime}C^{\prime}$, $\triangle AB^{\prime}D^{\prime}$, and $\triangle AC^{\prime}D^{\prime}$ are equal.
By the Law of Sines, it is sufficient to prove that
$\angle AB^{\prime}C^{\prime} = \angle AD^{\prime}C^{\prime}$, $\angle AC^{\prime}B^{\prime} = \angle AD^{\prime}B^{\prime}$.
Let $M^{\prime}N^{\prime}$ intersect $B^{\prime}C^{\prime}$ at point $P$. By the power of a point theorem, we have
$\left(\frac{AM + AN}{2}\right)^{2} - AD^{2} = AM \cdot AN$,

i.e., $\square$
$\frac{1}{4}\left(\frac{AM^{\prime} + AN^{\prime}}{AM^{\prime} \cdot AN^{\prime}}\right)^{2} - \left(\frac{1}{AD^{\prime}}\right)^{2} = \frac{1}{AM^{\prime} \cdot AN^{\prime}}$.
Thus, $\left(\frac{AM^{\prime} - AN^{\prime}}{2AM^{\prime} \cdot AN^{\prime}}\right)^{2} = \left(\frac{1}{AD^{\prime}}\right)^{2}$.
Then $AD^{\prime} = \frac{2AM^{\prime} \cdot AN^{\prime}}{AM^{\prime} - AN^{\prime}} = \frac{AM^{\prime} \cdot AN^{\prime}}{AP}$.
Therefore, $AD^{\prime} \cdot AP = AM^{\prime} \cdot AN^{\prime}$.
Hence, $AP \cdot D^{\prime}P = AM^{\prime} \cdot AN^{\prime} + AP^{2}$
$= M^{\prime}P \cdot N^{\prime}P = B^{\prime}P \cdot C^{\prime}P$.
Thus, $\frac{AP}{PB^{\prime}} = \frac{PC^{\prime}}{PD^{\prime}}$, i.e.,
$\tan \angle AB^{\prime}C^{\prime} = \tan \angle AD^{\prime}C^{\prime}$.
Since $\angle AB^{\prime}C^{\prime}$ and $\angle AD^{\prime}C^{\prime} \in \left(0, \frac{\pi}{2}\right)$, then
$\angle AB^{\prime}C^{\prime} = \angle AD^{\prime}C^{\prime}$.
Similarly, $\angle AC^{\prime}B^{\prime} = \angle AD^{\prime}B^{\prime}$.