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center normal and center normal plane as loci

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\textbf{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.\\

{\em Proof.}\, Let $A$ and $B$ be arbitrary given distinct points.
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 $1^\circ.$\; 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 \PMlinkname{side}{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 $AB$.


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$2^\circ.$\; 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 $B$.\\


\textbf{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.\\

{\em Proof.}\, Change ``center normal'' in the preceding proof to ``center normal plane''.
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