Proof As shown in Figure 2, let $B C=a, C A=b, A B=c$, $p$ be the semiperimeter of $\triangle A B C$, and $\odot O_{1}, \odot O_{2}$ be the incircles of $\triangle A B D, \triangle A D C$ with radii $r_{1}, r_{2}$, respectively.

Extend $B O_{1}, C O_{2}$ to intersect at point $O$, and draw $O G \perp B C$, $O_{1} E \perp B C, O_{2} F \perp B C$.
Clearly, $O_{1} E=r_{1}, O_{2} F=r_{2}$.
Let $O G=r, O$ be the incenter of $\triangle A B C$.
By $r_{1}=r_{2} \Rightarrow O_{1} O_{2} \parallel B C$.
Then $O_{1} O_{2}=E F=D E+D F$
$=\frac{1}{2}(A D+B D-c)+\frac{1}{2}(A D+D C-b)$
$=\frac{1}{2}(2 A D+a-b-c)$.
Hence $\triangle O O_{1} O_{2} \sim \triangle O B C$
$\Rightarrow \frac{O_{1} O_{2}}{B C}=\frac{r-r_{1}}{r}$.
Substituting and rearranging, we get
$r_{1}=\frac{r(p-A D)}{a}$.
Also, $S_{\triangle A B D}+S_{\triangle A D C}=S_{\triangle A B C}$
$\Rightarrow r_{1}(A D+B D+c)+r_{2}(A D+D C+b)$
$=r(a+b+c)$
$\Rightarrow r_{1}(2 A D+a+b+c)=r(a+b+c)$
$\Rightarrow 2(p+A D) r_{1}=2 p r$
$\Rightarrow r_{1}=\frac{r p}{p+A D}$.
Comparing equations (1) and (2), we get
$a p=(p+A D)(p-A D)=p^{2}-A D^{2}$
$\Rightarrow A D^{2}=p(p-a)$.
It is easy to prove that $B, O, M, C, O, N$ are collinear, respectively.
Draw $M M^{\prime} \perp B C, N N^{\prime} \perp B C$.
Let $M M^{\prime}=m, N N^{\prime}=n$.
By the tangent segment theorem,
$B G=p-b, C G=p-c, B M^{\prime}=p=C N^{\prime}$.
By $\triangle B M M^{\prime} \sim \triangle B O G$
$\Rightarrow \frac{m}{r}=\frac{B M^{\prime}}{B G}=\frac{p}{p-b}$
$\Rightarrow m=\frac{p r}{p-b}=\frac{S_{\triangle A B C}}{p-b}$;
By $\triangle C N N^{\prime} \sim \triangle C O G$
$\Rightarrow \frac{n}{r}=\frac{C N^{\prime}}{C G}=\frac{p}{p-c}$
$\Rightarrow n=\frac{p r}{p-c}=\frac{S_{\triangle A B C}}{p-c}$.
Thus, $m n=\frac{S_{\triangle A B C}^{2}}{(p-b)(p-c)}$
$=\frac{p(p-a)(p-b)(p-c)}{(p-b)(p-c)}$
$$=p(p-a) \text {. }$$

Comparing equations (3) and (4), we get $A D^{2}=m n$, i.e., $A D$ is the geometric mean of $m$ and $n$.
(Huang Quanfu, Jiangzhen Middle School, Huaining County, Anhui Province, 246142)