a line segment has at most one midpoint

(this proof is not correct yet)

Theorem 1.

In an ordered geometry a line segment has at most one midpoint.

Proof.

Let $[p,q]$ be a closed line segment and suppose $m$ and $m^{\prime }$ are midpoints. If $m:p:q$ then so $m$ is not a midpoint. Similarly we cannot have $p:q:m$, so we have $p:m:q$. And also, $p:m^{\prime }:q$. Suppose $m\ne m^{\prime }$. Without loss of generality we can assume $p:m:m^{\prime }$ and $m:m^{\prime }:q$. But then $[p,m^{\prime }]>[p,m]\cong [m,q]>[m^{\prime },q]$ so that $[p,m^{\prime }]\cong ̸[m^{\prime },q]$, a contradiction. Hence $m=m^{\prime }$. ∎

Title	a line segment has at most one midpoint
Canonical name	ALineSegmentHasAtMostOneMidpoint
Date of creation	2013-03-22 17:17:37
Last modified on	2013-03-22 17:17:37
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	9
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 51G05