*Given any five points on a flat surface that are in general position, i.e. no two of them coinciding and no three of them on a straight line, prove that four of these points will always form a convex quadrilateral.*

This is the latest proof of mine:

1. Let us start with the observation that a line can intersect three edges of a triangle at most (see below). The number of intersected edges can be zero, two or three, but the number of intersection points can be zero or two (point of entry and point of exit).

a) The number of intersection points is six, i.e. the maximum, and

b) Any vertex of any of the triangles is collinear with two other vertices.

For those who think the explanation in it. 3 above is not clear enough I gladly provide the following elaboration based on an example that is easy to understand.

In the picture below there are 3 quite obvious intersection points. Suppose that (contrary to what we see) the useful edge of the intersecting blue triangle (or its extension) intersected the useless edge of the intersected green triangle. Then the number of intersection points would be 5, since the useful edge (or its extension) not only intersects the useless edge but also intersects another edge of the green triangle. Therefore, the third edge of the green triangle is not intersected by the useful edge (or its extension) and does not intersect it either by itself or by its extension, since if it did we would have 7 intersection points (where the theoretical maximum is 6).