Note 6. Proof of Lemma 2

Suppose satisfies the condition: . Then it is easy to see that

 

.

 

But this implies that

 

.

 

Thus

 

.

 

So it follows that

 

,

 

or that (for the midpoint of the segment)

 

.

 

But this puts Z on the orthogonal bisector of the segment. Each step of the argument is reversible, and so the orthogonal bisector is equal to the set of points with equal interval from . ¨