Ambiguous case in euclidian geometry

Question

Is it possible to locate $4$(or more than $4$) points on a plane such that every point is at an equal distance from every other point?

Answer 1

No. Assume that such points exists and call them $A$, $B$, $C$, $D$.

It is clear that $ \triangle ABC$ is an equilateral triangle, and point $D$ is at an equal distance from $A$, $B$ and $C$, which means that $D$ is the centroid since they are one a single plane.

However, if $\overline {AB}=a$, then this implies that $\overline {AD}=\frac{\sqrt{3}a}{3}$ since $D$ is the centroid. However, this is a contradiction, as $\overline{AB}=\overline {AD}$.