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

 untranslated text remains the same as the source text.
6. Solution: Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$. First, consider an example:

If $n$ is odd, connect the segments $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{n}; A_{2} A_{3}, A_{4} A_{5}, \cdots, A_{n-1} A_{n}$.

If $n$ is even, connect the segments $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{n}; A_{2} A_{3}, A_{4} A_{5}, \cdots, A_{n-2} A_{n-1}, A_{2} A_{n}$.

Clearly, the segments connected by the above method satisfy the conditions, so the minimum value of $m$ is less than or equal to $\left[\frac{3 n-2}{2}\right]$. (Note: $\left[\frac{3 n-2}{2}\right]$ is the number of segments connected in the above method, where $[x]$ denotes the greatest integer not exceeding $x$.)

Next, we prove that these $n$ points require at least $\left[\frac{3 n-2}{2}\right]$ segments to meet the requirements of the problem.

In fact, if each of $A_{1}, A_{2}, \cdots, A_{n}$ is connected to at least 3 segments, then $m \geqslant \frac{3 n}{2} > \left[\frac{3 n-2}{2}\right]$;

If one of them (let's say $A_{1}$) is connected to no more than 2 segments, there are two cases:
(1) $A_{1}$ is connected to only 1 segment, 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 $A_{1}$ is not connected to any segment, it also leads to a contradiction.
(2) $A_{1}$ is connected to exactly 2 segments, $A_{1} A_{2}$ and $A_{1} A_{3}$. In this case, the segment $A_{2} A_{3}$ must be connected (because for the two points $A_{1}$ and $A_{2}$, only $A_{3}$ can be connected to both $A_{1}$ and $A_{2}$).

Considering the points $A_{1}$ and $A_{i} (i \geqslant 4)$, $A_{i}$ must be connected to either $A_{2}$ or $A_{3}$. Also, from (1), $A_{i}$ must be connected to at least 2 segments, so the number of segments connected from $A_{4}, A_{5}, \cdots, A_{n}$ (a total of $n-3$ points) is at least $n-3 + \left[\frac{n-2}{2}\right]$. Therefore, the total number of segments connected from these $n$ points is at least $3 + (n-3) + \left[\frac{n-2}{2}\right] = \left[\frac{3 n-2}{2}\right]$.
Thus, the minimum value of $m$ is $\left[\frac{3 n-2}{2}\right]$.