$1 \cdot 87$ A square piece of paper has 1965 holes pierced inside it. No three of these holes and the vertices of the square are collinear. Draw several non-intersecting line segments, whose endpoints are these holes or the vertices of the square, to divide the square into some triangles, and ensure that there are no more holes inside or on the sides of these triangles. How many line segments were drawn in total? How many triangles were obtained?
[Solution] We denote the set of 1965 small holes and the 4 vertices of the square as $M$. Clearly, the points in $M$ are the common vertices of some triangles. We calculate the sum of the interior angles of all triangles from two perspectives.

Assume that a total of $n$ triangles are formed, so their sum of interior angles is $n \cdot 180^{\circ}$. On the other hand, the vertices of these interior angles are all points in $M$. In other words, their interior angles are provided by the points in $M$. Each vertex of the square provides a $90^{\circ}$ angle, and each hole point provides a $360^{\circ}$ angle. Therefore, the sum of the interior angles of the $n$ triangles should be
$$
4 \times 90^{\circ} + 1965 \times 360^{\circ} = 1966 \times 360^{\circ}.
$$

Thus, we have $n \cdot 180^{\circ} = 1966 \times 360^{\circ}$. Solving for $n$ gives $n = 3932$, meaning a total of 3932 triangles are formed.
These 3932 triangles have a total of $3932 \times 3$ sides, of which 4 sides are the 4 sides of the original square, and do not need to be drawn separately. The other sides are all line segments drawn, but each line segment is a common side of two triangles, so the total number of line segments drawn is
$$
(3932 \times 3 - 4) \div 2 = 5896.
$$