Example 3 A square piece of paper contains 1000 points, and no three points among these points and the vertices of the square are collinear. Then, some line segments are drawn between these points and the vertices of the square to divide the square entirely into small triangles (using the drawn line segments and the sides of the square as sides, and the drawn line segments have no common points except at their endpoints). How many line segments are drawn in total? How many triangles are obtained in total?
Let's assume a total of $l$ line segments are connected, resulting in $k$ triangles.
On one hand, the sum of the interior angles of the $k$ triangles is $k \cdot 180^{\circ}$.
On the other hand, in the $k$ triangles, the sum of all interior angles with 1000 interior points as vertices is $1000 \times 360^{\circ}$, and the sum of all interior angles with the vertices of the square as vertices is $4 \times 90^{\circ}$. Therefore,
$$
k \cdot 180^{\circ}=1000 \times 360^{\circ}+4 \times 90^{\circ} \text {, }
$$

So $k=2002$.
Next, each triangle has 3 sides, and $k$ triangles have a total of $3 k$ sides. On the other hand, each connected line segment is a common side of 2 triangles, and each side of the square is a side of one triangle, so the $k$ triangles have a total of $2 l+4$ sides, thus $3 k=2 l+4$.
So $l=\frac{1}{2}(3 k-4)=\frac{1}{2}(3 \times 2002-4)=3001$.