Example 5. Using the three vertices of a triangle and nine points inside it (a total of 12 points) as vertices, the number of small triangles that the original triangle can be divided into is ( ).
( A ) 15. ( B ) 19 .
( C ) 22. ( D ) Cannot be determined.
( 1988, Jiangsu Junior High School Competition)
Solve: First, consider the case of placing $n$ points inside a triangle. Suppose the $n-1$ points inside $\triangle ABC$ can divide the original triangle into $a_{n-1}$ small triangles. Next, we examine the situation after adding the $n$-th internal point $P_{a}$. (1) If the point is inside a small triangle, the three vertices of the original small triangle, along with point $P_{a}$, will divide this small triangle into three, i.e., two new small triangles are added; (2) If point $P_{\text {a }}$ is on the common edge of two small triangles (as shown in Figure (2)), 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 into which $n$ points inside $\triangle ABC$ divide the original triangle is $a_{n}=a_{n-1}+2$. And it is easy to know that $a_{0}=1$.
$$
\begin{aligned}
\therefore a_{n} & =a_{n-1}+2=a_{n-2}+4=a_{n-3}+6 \\
& =\cdots=a_{0}+2 n \\
& =2 n+1 .
\end{aligned}
$$

In the original problem, $n=9, 2 n+1=19$. Therefore, the answer should be ( B ).