Example 9 Given $n$ points in the plane with no three points collinear, $m$ line segments are drawn between these $n$ points as endpoints. It is known that for any two points $A$ and $B$, there is a point $C$ such that $C$ is connected to both $A$ and $B$ by line segments. Find the minimum value of $m$.

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Solution: First, we make an overall estimate of the number of line segments to be connected, to see how few line segments are needed to meet the requirements of the problem.

Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$. If any point among $A_{1}, A_{2}, \cdots, A_{n}$ has at least three line segments, then the number of line segments $m \geqslant \frac{3 n}{2}$.
If there is a point among $A_{1}, A_{2}, \cdots, A_{n}$ (let's assume it is $A_{1}$) that has at most two line segments, then there are the following two scenarios:
Scenario 1. $A_{1}$ only has one line segment, let's say $A_{1} A_{2}$, then there is no point that is connected to both $A_{1}$ and $A_{2}$, which is a contradiction. Similarly, if $A_{1}$ does not have any line segments, a contradiction can also be derived.

Scenario 2. $A_{1}$ has exactly two line segments $A_{1} A_{2}$ and $A_{1} A_{3}$, in this case, the line segment $A_{2} A_{3}$ must be connected (because for the point pair $A_{1}, A_{2}$, only $A_{3}$ can be connected to both $A_{1}$ and $A_{2}$).

Considering the point pair $A_{1}, A_{i}(i \geqslant 4)$, by the condition, $A_{i}$ must be connected to either $A_{2}$ or $A_{3}$. Also, from the proof of Scenario 1, we know that $A_{i}$ must have at least two line segments, so the total number of line segments starting from $A_{3}, A_{4}, \cdots, A_{n}$ is at least $2(n-3)$, and at least $n-3$ of these line segments are from either $A_{2}$ or $A_{3}$. If $n$ is odd, then the remaining $n-3$ line segments are counted at most twice, so in this case, the number of line segments $m \geqslant 3+(n-3)+\frac{n-3}{2}=\frac{3 n-3}{2}=\left[\frac{3 n-2}{2}\right]$; if $n$ is even, then at most $n-4$ of the remaining $n-3$ line segments are counted twice, so in this case, the number of line segments $m \geqslant 3+(n-1)+\frac{n-4}{2}+1=\frac{3 n-2}{2}=\left[\frac{3 n-2}{2}\right]$. Note that $\frac{3 n}{2}>\left[\frac{3 n-2}{2}\right]$, so in any case, we have $m \geqslant\left[\frac{3 n-2}{2}\right]$.

Secondly, if $n$ is odd, then we connect the line segments $A_{1} A_{2}, A_{1} A_{3} \cdots, A_{1} A_{n} ; A_{2} A_{3}, A_{4} A_{5}, \cdots, A_{n-1} A_{n}$. In this case, the number of line segments is $n-1+\frac{n-1}{2}=\frac{3 n-3}{2}=\left[\frac{3 n-2}{2}\right]$; if $n$ is even, then we connect the line segments $A_{1} A_{2}$, $A_{1} A_{3}, \cdots, A_{1} A_{n} ; A_{2} A_{3}, A_{4} A_{5}, \cdots, A_{n-2} A_{n-1}, A_{2} A_{n}$, in this case, the number of line segments is $n-1+\frac{n-2}{2}+1=\frac{3 n-2}{2}=\left[\frac{3 n-2}{2}\right]$, and it is clear that the line segments connected in the above method meet the conditions of the problem.
In summary, the minimum number of line segments $m$ is $\left[\frac{3 n-2}{2}\right]$.