As shown in Figure 3, in $\triangle ABC$, let $D_{1}, D_{2}, \cdots, D_{m}$ be the internal division points on side $BC$ in sequence, and $E_{1}, E_{2}, \cdots, E_{n}$ be the internal division points on side $AC$ in sequence. Connect $A D_{1}, A D_{2}, \cdots, A D_{m}, B E_{1}, B E_{2}, \cdots, B E_{n}$. Find the number of triangles in the figure.
(1) As shown in Figure 4, the number of triangles in the figure is $\mathrm{C}_{m+2}^{2}$.
(2) As shown in Figure 5, after drawing a line segment $B E_{i}$, the number of triangles added above $B E_{i}$ (including the $A$ part) is $\mathrm{C}_{m+2}^{2}$, and the number of triangles added below $B E_{i}$ (including the $C$ part) is $m+1$. Therefore, the total number of triangles in Figure 5 is $\mathrm{C}_{m+2}^{2}+\mathrm{C}_{m+2}^{2}+m+1$.
(3) As shown in Figure 6, when $B E_{1}, B E_{2}, \cdots, B E_{n}$ are all drawn, observe $\triangle B E_{i} E_{j}(i, j=1,2, \cdots, n, i \neq j)$, the number of triangles added is $m+1$. Therefore, the total number of triangles in Figure 6 is
$$
\begin{array}{l}
\mathrm{C}_{m+2}^{2}+n\left[\mathrm{C}_{m+2}^{2}+(m+1)\right]+\mathrm{C}_{n}^{2}(m+1) \\
=(n+1) \mathrm{C}_{m+2}^{2}+n(m+1)+(m+1) \mathrm{C}_{n}^{2} \\
=(n+1) \frac{(m+2)(m+1)}{2}+n(m+1)+ \\
(m+1) \frac{n(n-1)}{2} \\
= \frac{1}{2}(m+1)(n+1)(m+n+2) .
\end{array}
$$