*Given any five points on a flat surface, with no three of them in a straight line, prove that four of these points will always form a convex quadrilateral.*

This is my proof of Esther Klein’s theorem based on the 2+2+1 partition of 5. Let us think about the five points as of two lines and an extra point. Line

**l1**is defined by

**P1**and

**P2**, line

**l2**is defined by

**P3**and

**P4**and

**P5**is the extra point. As any of

**l1**and

**l2**divides the plane into two halves, there are more points (not on the line) in the one half of the plane than in the other half (according to the Pigeonhole principle). Let us call the former half of the plain

*more populated*and the latter

*less populated*. There are only 3 cases to consider:

__Case 1.__Neither line intersects the segment between the points on the other line. One can build a convex quadrilateral by connecting the points on

**l1**and

**l2**.

__Case 2.__Each line intersects the segment between the points of the other line. One can build a convex quadrilateral by connecting the points on

**l1**and

**l2**.

__Case 3.__Only one of the lines intersects the segment on the other line. Let us assume without loss of generality that

**l1**intersects the

**P3P4**segment of

**l2**. Regardless of the exact position of

**P5**we are sure that it is in a more populated half of the plane (with respect to

**l1**). One can build a convex quadrilateral by connecting

**P5**with the other point in the more populated half of the plane (

**P4**below) and the two points on

**l1**.