*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, and

c) No edges of the two triangles coincide.