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 $E = F$.


 * Perpendicular-Bisector-Equidistance.png

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

By Pythagoras's Theorem:
 * $AP^2 = AC^2 + PC^2 = BC^2 + PC^2 = BP^2$

and so:
 * $AP = BP$

That is, $P$ is equidistant from $A$ and $B$, and so:
 * $P \in E$

Hence:
 * $F \subseteq E$

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

By Pythagoras's Theorem:
 * $AC^2 = PA^2 - PC^2$
 * $BC^2 = PB^2 - PC^2$

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:
 * $E \subseteq F$

We have:
 * $E \subseteq F$

and:
 * $F \subseteq E$

and so by definition of set equality:
 * $E = F$

Hence the result.