Example 1 Given $n(n \geqslant 4)$ points on a plane such that no three points are collinear, and a line segment is drawn between any two points. Find the maximum number of line segments that do not intersect any other line segments (except at endpoints). ${ }^{[1]}$
(23rd Turkish Mathematical Olympiad)
【Analysis】When $n=4$, the maximum number of line segments that do not intersect with any other line segments (except at endpoints) is 6, as shown in Figure 1.
Figure 1
Figure 2
When $n=5$, the maximum number of line segments that do not intersect with any other line segments (except at endpoints) is 8, as shown in Figure 2.

For $n$ points forming a convex triangle, consider a regular $\triangle ABC$. Let the center of the triangle be $O$, and the arc $\overparen{BC}$ of the circumcircle of $\triangle BOC$ lies inside $\triangle ABC$. Take any $n-3$ points $D_{1}, D_{2}, \cdots, D_{n-3}$ on this arc. Then, the $n$ points $A, B, C, D_{1}, D_{2}, \cdots, D_{n-3}$ have the following line segments that satisfy the condition: $AB, BC, CA, AD_{1}, AD_{2}, \cdots, AD_{n-3}, BD_{1}, D_{1}D_{2}, D_{2}D_{3}, \cdots, D_{n-4}D_{n-3}, D_{n-3}C$, totaling $2n-2$ segments. By providing examples, we can determine the maximum number of line segments that satisfy the condition for $n$ points on a plane, and then prove the conclusion using mathematical induction.
Proof Let $S_{n}$ be the number of line segments that satisfy the condition.
We need to prove: $S_{n} \leqslant 2n-2$.
When $n=4$, the conclusion is obviously true.
Assume the conclusion holds for $n$, i.e., $S_{n} \leqslant 2n-2$.
For $n+1$, if the convex hull of the $n+1$ points is a convex $n+1$-gon, then
$$
S_{n+1}=n+1 \leqslant 2(n+1)-2 \text{, }
$$

the conclusion holds.
If the convex hull $\Gamma$ of the $n+1$ points is not a convex $n+1$-gon, then there exists a point $P$ inside the convex hull $\Gamma$. For the remaining $n$ points, we can perform a triangular subdivision of the convex hull $\Gamma$, where each small triangle's vertices are among the $n$ points. Assume point $P$ is inside one of these triangles, say $\triangle ABC$.
If we remove point $P$, by the induction hypothesis, $S_{n} \leqslant 2n-2$.
(1) If one of $AB, BC, CA$ does not satisfy the condition, say $CA$ does not satisfy the condition, then there exists a line segment $EF$ that intersects $CA$ internally. When point $P$ is added, line segment $EF$ must intersect either $PA$ or $PC$. Hence,
$$
S_{n+1} \leqslant S_{n}+2 \leqslant 2(n+1)-2 \text{. }
$$
(2) If $AB, BC, CA$ all satisfy the condition, since $n+1 \geqslant 5$, there is still a point $Q$ outside $\triangle ABC$. The line segment $PQ$ must intersect one of the three sides of $\triangle ABC$, say $PQ$ intersects $CA$. Hence,
$$
S_{n+1} \leqslant S_{n}+3-1 \leqslant 2(n+1)-2 \text{. }
$$

Therefore, the maximum value of $S_{n}$ is $2n-2$.