Originally Posted by
Archie Meade Soroban was referring to the use of non-collinear points (no 3 points on a straight line).
In your image, you have 5 points.
If no 3 were collinear, you'd have 5C3 = 10 triangles.
However, in your diagram, 3 points are collinear along both diagonals,
so you lose 2 of those 10.
Unfortunately, where the diagonals cross has to be counted as a 5th point,
since you are using it in the formation of half of those 8 triangles.
(The formula counts triangles connected by points,
not including the formation of new points where lines cross).