Without loss of generality, let no three points be collinear. The other cases are just limit cases of the following cases. Let the points be $A, B, C, D, E$, with the distance between $D, E$ being the maximum possible between any two points in the configuration. The triangle $ABC$ has a greatest angle, let's say it's $A$. We make two cases: 1. $ABC$ is not isosceles. wlog $AB<AC$. Then shift the point $E$ to a location such that $AEBC$ is an isosceles trapezoid with $AE || BC$. Now note that $AE <BC < DE$ because $B, C$ are necessarily acute. So we can put $D$ on the perpendicular bisector of $BC$, which will be the line of symmetry of the configuration. 2. $ABC$ is isosceles at $A$. In that case, shift $E$ to the location such that $EBAC$ is an isosceles trapezoid, with $AB || EC$. Then note that the distance from $E$ to the perpendicular bisector of $AB$ is $0.5 EC < 0.5(AE + AC) = 0.5(BC + AC) \le 0.5(DE + DE) = DE$, so we can place $D$ on the perpendicular bisector of $AB$ and we will be done.