Four points lie in a plane. They are partitioned into two pairs so that the sum of the lengths of the segments joining the points of each pair has the minimal possible value.

Prove that these segments have no common points.

Let segments AB and CD have common point E. Then AC<AE+EC and BD<BE+ED.

AE+EC+BE+ED=AB+CD

To prove that the segments formed by partitioning the four points into two pairs have no common points, we can use contradiction.

Let's assume that the segments have a common point. Without loss of generality, let's label the four points as A, B, C, and D. Suppose segment AB and segment CD have a common point P.

If P lies on the same side of segment AB as point A, then P must lie inside triangle ABC. Otherwise, if P lies on the same side of segment AB as point B, then P must lie inside triangle BCD.

First, let's consider the case where P lies inside triangle ABC. In this case, the sum of the lengths of the segments AP, BP, CP, and DP is greater than the sum of the lengths of the segments AB and CD. This is because segment AP + segment BP > segment AB, and segment CP + segment DP > segment CD. Therefore, this contradicts the assumption that the sum of the lengths of the segments joining the points of each pair has the minimal possible value.

Next, let's consider the case where P lies inside triangle BCD. Similarly, in this case, the sum of the lengths of the segments AP, BP, CP, and DP is greater than the sum of the lengths of the segments AB and CD. Again, this contradicts the assumption that the sum of the lengths of the segments joining the points of each pair has the minimal possible value.

Since both cases lead to a contradiction, we can conclude that the segments formed by partitioning the four points into two pairs have no common points.