Example 1. Using the three vertices of a triangle and several points inside it as vertices, the number of small triangles into which the original triangle can be divided is ( .
(A)15; (B)19; (C) 22; (D) cannot be determined・ (88 Jiangsu Province Junior High School Mathematics Competition Question)
Solution 1: To find the number of small triangles after the division, we first calculate the sum of all the interior angles of these small triangles. Since each of the $n$ interior points of the original triangle is a common vertex of several of the divided small triangles, the sum of the interior angles of these small triangles with that point as the vertex is $360^{\circ}$; and because the sum of the interior angles of each point in the original triangle is $180^{\circ}$.

Therefore, the sum of the interior angles of all the small triangles after division is $n \cdot 360^{\circ} + 180^{\circ}$. The total number of small triangles formed is $\frac{n \cdot 360^{\circ} + 180^{\circ}}{180^{\circ}} = 2n + 1$.

The thinking method of Solution 1 reflects the principle of wholeness, which involves observing the sum of all the interior angles of these small triangles as a whole, thereby solving the problem.

Solution 2: Suppose $(n-1)$ points inside the triangle divide the original triangle into $a_{n-1}$ small triangles. Let's examine the situation after adding the new $n$-th internal point $P_{n}$.
(1) If point $P_{n}$ is inside one of the small triangles (as shown in Figure 1-1), then the three vertices of the original small triangle, along with point $P_{n}$, will divide this small triangle into three, i.e., two new small triangles are added.
(2) If point $P_{n}$ is on the common side of two small triangles (as shown in Figure 1-2), then the vertices of these two small triangles, along with point $P_{n}$, will divide these two small triangles into four, i.e., two new small triangles are also added.

Therefore, the number of small triangles formed by $n$ points inside the triangle is $a_{n} = a_{n-1} + 2$, and it is easy to see that $a_{0} = 1$.
Thus, $a_{n} = 2n + 1$.