How many diagonals does a regular $n$-sided polygon have? I guess the answer is: $C(n,n-3)/2$ since for $n$-sided polygon,there are $n$ vertices, and for each vertex, it cannot form diagonal with the adjacent points and itself. So, each vertex can form $n-3$ points. And half of the vertices have repeated diagonals and we need to divide by $2$.
It holds true for quadrilateral and pentagon,but why it does not fit triangles?

Your reasoning, which is correct, implies that the number of diagonals is
$$\frac{n(n-3)}2$$
and this fits for triangles, that have no diagonals.
