$35 \cdot 37$ Using the 3 vertices of a triangle and 7 points inside it (a total of 10 points) as vertices, the number of small triangles into which the original triangle can be divided is
(A) 11.
(B) 15.
(C) 19.
(D) cannot be determined.
(China Shanghai Junior High School Mathematics Competition, 1985;
China Jiangsu Junior High School Mathematics Competition, 1988)
[Solution 1] Suppose the original triangle can be divided into $m$ small triangles. Consider the sum of all their interior angles:

On one hand, the sum of the interior angles of $m$ triangles is $m \cdot 180^{\circ}$; on the other hand, the lines connecting the points inside the triangle form a full angle at each point, so the 7 points contribute $360^{\circ} \cdot 7$, plus the original triangle's interior angle sum of $180^{\circ}$, totaling
$$
360^{\circ} \cdot 7 + 180^{\circ} \text{.}
$$

Thus, $m \cdot 180^{\circ} = 360^{\circ} \cdot 7 + 180^{\circ}$, solving for $m$ gives $m = 15$.
Therefore, the answer is (B).

[Solution 2] First, consider the general case: a triangle with $n$ points inside.

If the number of small triangles formed by $n-1$ points is denoted as $a_{n-1}$, we examine the scenario when adding the $n$-th point $p_{n}$.
(1) If $p_{n}$ is inside a small triangle (as shown in Figure (1)), connecting $p_{n}$ to the three vertices of the small triangle divides it into three, thus adding 2 small triangles;
(2) If $p_{n}$ is on the common edge of two small triangles (as shown in Figure (2)), connecting $p_{n}$ to the vertices of the two small triangles divides them into four, also adding 2 small triangles.
Thus, we have $a_{n} = a_{n-1} + 2$, and $a_{0} = 1$.
$$
\begin{aligned}
\therefore a_{n} & = a_{n-1} + 2 = a_{n-2} + 4 = \cdots \\
& = a_{0} + 2n = 2n + 1.
\end{aligned}
$$

When $n = 7$, $a_{7} = 15$.
Therefore, the answer is $(B)$.