# center normal and center normal plane as loci

Theorem 1. In the Euclidean plane^{}, the center normal of a line segment^{} is the locus of the points which are equidistant from the both end points^{} of the segment.

Proof. Let $A$ and $B$ be arbitrary given distinct points.

${1}^{\circ}.$ Let $P$ be a point equidistant from $A$ and $B$. If $P\in AB$, then $P$ is trivially on the center normal of $AB$. Thus suppose that $P\notin AB$. In the triangle^{} $PAB$, let the angle bisector^{} of $\mathrm{\angle}P$ intersect the side (http://planetmath.org/Triangle) $AB$ in the point $D$. Then we have

$$\mathrm{\Delta}PDA\cong \mathrm{\Delta}PDB\mathit{\hspace{1em}}\text{(SAS)},$$ |

whence

$$\mathrm{\angle}PDA=\mathrm{\angle}PDB={\mathrm{\hspace{0.33em}90}}^{\circ},DA=DB.$$ |

Consequently, the point $P$ is always on the center normal of $AB$.

${2}^{\circ}.$ Let $Q$ be any point on the center normal and $D$ the midpoint^{} of the line segment $AB$. We can assume that $Q\ne D$. Then we have

$$\mathrm{\Delta}QDA\cong \mathrm{\Delta}QDB\mathit{\hspace{1em}}\text{(SAS)},$$ |

implying that

$$QA=QB.$$ |

Thus $Q$ is equidistant from $A$ and $B$.

Theorem 2. In the Euclidean space, the center normal plane of a line segment is the locus of the points which are equidistant from the both end points of the segment.

Proof. Change “center normal” in the preceding proof to “center normal plane”.

Title | center normal and center normal plane as loci |

Canonical name | CenterNormalAndCenterNormalPlaneAsLoci |

Date of creation | 2013-03-22 18:48:51 |

Last modified on | 2013-03-22 18:48:51 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 12 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M15 |

Classification | msc 51N05 |

Classification | msc 51N20 |

Synonym | center normal as locus |

Synonym | center normal plane as locus |

Related topic | SAS |

Related topic | CircumCircle^{} |

Related topic | AngleBisectorAsLocus |