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Any point on a perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
Based on the characteristics of the diagram, $CM$ is the perpendicular bisector of $AB.$ Therefore, $C$ is equidistant from $A$ and $B.$
$AC=BC$
Suppose $CM$ is the perpendicular bisector of $AB.$ Then $M$ is the midpoint of $AB.$
Consider a triangle with vertices $A,$ $M,$ and $C,$ and another triangle with vertices and $B,$ $M,$ and $C.$
Both $△ACM$ and $△BCM$ have a right angle and congruent legs $AM$ and $BM.$ Since all right angles are congruent, $∠AMC≅∠BMC.$ Furthermore, by the Reflexive Property of Congruence, $CM$ is congruent to itself.
By the SideAngleSide Congruence Theorem, the triangles are congruent. Therefore, since corresponding parts of congruent figures are congruent, their hypotenuses $AC$ and $BC$ are also congruent. By the definition of congruent segments, $AC$ and $BC$ have the same length. This means that $C$ is equidistant from $A$ and $B.$
Using this reasoning it can be proven that any point on a perpendicular bisector is equidistant from the endpoints of the segment.
The proof can be summarized in the following twocolumn table.
Statements  Reasons 
$AM≅MB$ $∠AMC$ and $∠BMC$ are right angles 
Definition of a perpendicular bisector. 
$∠AMC≅∠BMC$  All right angles are congruent. 
$CM≅CM$  Reflexive Property of Congruence. 
$△ACM≅△BCM$  SAS Congruence Theorem. 
$AC≅BC$  Corresponding parts of congruent figures are congruent. 
$AC=BC$  Definition of congruent segments. 
Suppose $CM$ is the perpendicular bisector of $AB,$ and that $M$ is the midpoint of $AB.$
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 $AB.$ They also share one leg, $CM.$
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 twocolumn proof.
Statement  Reason 
$ AM≅MBAB⊥CM $

Given 
$△AMC≅△BMC$  SAS congruence theorem 
$AC≅BC$  $△AMC≅△BMC$ 
$AC=BC$  Definition of congruent segments 