Example 9 On the square $A B C D$: there are 10 points, 8 of which are inside $\triangle A B C$, and 2 points are on the side $\mathrm{I}$ of the square (not at the vertices). [L These 10 points, together with points $A, B, C, D$, are not collinear. Find how many small triangles these 10 points, along with the 4 vertices of the square, can divide the square into.
Solution: Since adding a point on the plane is equivalent to adding a full angle, the total angle of the figure formed by 14 vertices is $360^{\circ} \times 14=5040^{\circ}$. We have $\}^{\circ}, B, C, D$ as the vertices of the square, so $\angle A+\angle B+\angle C+\angle D=$ $90^{\circ} \times 4=360^{\circ}$. Two of the points are on the sides of the square, so the sum of the interior angles of the figure formed by these two points is $180^{\circ} \times 2$. Thus,
$$
\begin{aligned}
& 5040^{\circ}-360^{\circ} \times 4+360^{\circ}-360^{\circ} \times 2 \\
+ & 180^{\circ} \times 2 \\
= & 3600^{\circ} .
\end{aligned}
$$

Therefore, there are $20\left(=\frac{3600^{\circ}}{180^{\circ}}\right)$ small