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This is the famous result of Esther Klein which introduced Erds and Szekeres to Ramsay theory.... Clearly, there is nothing to prove if the convex hull of the five points is a quadrilateral or a pentagon. Since the points are not collinear, we only have to deal with the case that the convex hull is a triangle, say $ABC$, with two of the points, say $D,E$ in its interior. Now, the line $DE$ meet two side of the triangle $ABC$ (recall that no three points are collinear), say $AB$ and $AC$. Thus $B,C,D,E$ are in some order the vertices of a convex quadrilateral, and we are done. Pierre.

Isn't this rather combinatorics? I just can't believe I'd solve a Geometry longlisted question so easy If point E is outside ABCD, the problem is trivially solved. If it's inside, draw the diagonal AC. Since no 3 are collinear, E is on one side of AB, say WLOG on the side towards D. Then it follows immediately that ABCE is convex Peter

Ah darn, Pierre is once again faster. And this time with a <1 page solution this must be really easy!

Let the $5$ points be $A$, $B$, $C$, $D$, and $E$. We first consider $A$, $B$, $C$, and $D$. If each of these is in the exterior of the triangle determined by the other $3$ points, we are done by $ABCD$. If one of these is in the interior, however, WLOG let it be $A$. We now proceed with casework based on the position of $E$. If $E$ is in the exterior of triangle $BCD$, the lines $AB$, $AC$, and $AD$ split the plane into $3$ regions. If $E$ is in the region determined by $AB$ and $AC$, then $ABEC$ is convex, and analogous reasoning applies to the other $2$ regions. If $E$ is in the interior of triangle $BCD$, extend $BA$ bast $A$ to intersect $CD$ at $E$, and define $F$ and $G$ similarly. Then, if $E$ is in the interior of triangle $ACG$ or $ADF$, $AECD$ or $AEDC$ is convex respectively. Analogous reasoning applies to the other $2$ pairs of triangles. $\blacksquare$

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