Calculate the number of secant lines and intersections Given a circle $k$, and a finite number of points $n$ on the circle, where every 2 points are connected with a secant line such that no point in circle exists where 3 secant lines intersect. Calculate the number of secant lines and the number of intersections. My attempt to solve this : The number of secant lines would simply be the number of lines defined by n points in a plane, $3+(n-3)(\frac{n}{2}+1)$. As for the second question, the lines on the circle form a convex polygon and the number of intersections would be equal to the number of this polygon diagonals i.e. $\frac{n*(n-3)}{2}$. My question is: Is my approach correct and if it is, how to prove that the number of intersections is the number of polygon diagonals ( I proved this empirically ).

If the question would read intersections inside the circle. Then the number of lines would be ${n \choose 2}$. For every line you need 2 points and the order doens't matter.

If you have an intersection you have $2$ lines and $4$ points. Note for every group of $4$ point you have $1$ intersection. So we get that there are $ {n\choose 4}$ intersections.