# a line segment has at most one midpoint

(this proof is not correct yet)

###### Theorem 1.

###### 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
$[p,{m}^{\prime}]\cong \u0338[{m}^{\prime},q]$, 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 |