# center normal and center normal plane as loci

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  $P\not\in AB$.  In the triangle  $PAB$, let the angle bisector  of $\angle P$ intersect the side (http://planetmath.org/Triangle) $AB$ in the point $D$.  Then we have

 $\Delta PDA\;\cong\;\Delta PDB\quad\mbox{(SAS)},$

whence

 $\angle PDA\;=\;\angle PDB\;=\;90^{\circ},\quad DA\;=\;DB.$

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  $Q\neq D$.  Then we have

 $\Delta QDA\;\cong\;\Delta QDB\quad\mbox{(SAS)},$

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