Proof

Perpendicular Bisector Theorem

Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.

This can be proven using congruent triangles.
Suppose $\overline{C M}$ is the perpendicular bisector of $\overline{A B},$ and that $M$ is the midpoint of $\overline{A B} .$

Two triangles can be created by connecting points $A$ and $C,$ and $B$ and $C .$

These triangles both have a right angle and one of the legs measures half of $A B .$ They also share one leg, $\overline{C M} .$

According to the SAS Congruence Theorem, the triangles are congruent. Thus, their hypotenuses are also congruent.

Therefore, any point on a perpendicular bisector is equidistant from the endpoints of the segment. This can be summarized in a two-column proof.

Statement	Reason
$\overline{A M} ≅ \overline{M B} \overline{A B} ⊥ C M$	Given
$△ A M C ≅ △ B M C$	SAS congruence theorem
$\overline{A C} ≅ \overline{B C}$	$△ A M C ≅ △ B M C$
$A C = B C$	Definition of congruent segments

Note that $A M C$ and $M B C$ are not triangles if $C$ is the point of intersection, $M .$ However, since $M$ is the midpoint of $\overline{A B}$ it is, by definition, equidistant from $A$ and $B .$

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Proof

Perpendicular Bisector Theorem

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