Perpendicular Bisector is Locus of Points Equidistant from Endpoints
Theorem
Let $AB$ be a straight line segment.
The locus of points which are equidistant from $A$ and $B$ is the perpendicular bisector of $AB$.
Proof
Let $E$ denote the set of points equidistant from $A$ and $B$.
Let $F$ denote the set of points on the perpendicular bisector of $AB$.
We are to show that $E = F$.
First we show that $F \subseteq E$.
Let $P \in F$.
Hence by definition $P$ is on the perpendicular bisector of $AB$.
Let $C$ be the intersection of $AB$ and the perpendicular bisector.
We have:
- $\angle PCA = \angle PCB = 90 \degrees$
- $AC = CB$
- $PC$ is common
- $AP^2 = AC^2 + PC^2 = BC^2 + PC^2 = BP^2$
and so:
- $AP = BP$
That is, $P$ is equidistant from $A$ and $B$, and so:
- $P \in E$
Hence:
- $F \subseteq E$
$\Box$
Let $P \in E$ be arbitrary.
That is, let $P$ be equidistant from $A$ and $B$.
Drop a perpendicular $PC$ to $AB$.
We inspect the triangles $\triangle PCA$ and $\triangle PCA$
We have:
- $PA = PB$
- $\angle PCA = \angle PCB = 90 \degrees$
- $PC$ is common
- $AC^2 = PA^2 - PC^2$
- $BC^2 = PB^2 - PC^2$
But as $PA = PB$ it follows that:
- $AC = BC$
Thus $PC$ is the perpendicular bisector of $AB$.
That is:
- $P \in F$
Hence by definition of subset:
- $E \subseteq F$
$\Box$
We have:
- $E \subseteq F$
and:
- $F \subseteq E$
and so by definition of set equality:
- $E = F$
Hence the result.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The circumcentre
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): equidistant