Suppose $\overleftrightarrow{CM}$ is the perpendicular bisector of $\overline{AB}.$ Then $M$ is the midpoint of $\overline{AB}.$

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 $\overline{AM}$ and $\overline{BM}.$ Since all right angles are congruent, $\angle AMC\cong \angle BMC.$ Furthermore, by the Reflexive Property of Congruence, $\overline{CM}$ is congruent to itself.

By the Side-Angle-Side Congruence Theorem, the triangles are congruent. Therefore, since corresponding parts of congruent figures are congruent, their hypotenuses $\overline{AC}$ and $\overline{BC}$ are also congruent. By the definition of congruent segments, $\overline{AC}$ and $\overline{BC}$ 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.

Two-Column Proof

The proof can be summarized in the following two-column table.

Suppose $\overleftrightarrow{CM}$ is the perpendicular bisector of $\overline{AB}.$

Using the given points $A,$ $B,$ $C,$ and $M$ as vertices, two triangles can be formed. The resulting triangles, $△ACM$ and $△BCM,$ can be proven to be congruent by identifying a congruence transformation that maps one triangle onto the other.

Since $\overline{AM}$ and $\overline{BM}$ are congruent, the distance between $A$ and $M$ is equal to the distance between $B$ and $M.$ Therefore, $A$ is the image of $B$ after a reflection across $\overleftrightarrow{CM}.$ Since $C$ lies on $\overleftrightarrow{CM},$ a reflection across $\overleftrightarrow{CM}$ maps $C$ onto itself. The same is true for $M.$

Reflection Across $\overleftrightarrow{CM}$
Preimage	Image
$B$	$A$
$C$	$C$
$M$	$M$

The above table shows that the images of the vertices of $△BCM$ are the vertices of $△ACM.$ Therefore, $△ACM$ is the image of $△BCM$ after a reflection across $\overleftrightarrow{CM}.$ Since a reflection is a rigid motion, this proves that the triangles are congruent. Corresponding parts of congruent figures are congruent, so $\overline{AC}$ and $\overline{BC}$ are congruent. By the definition of congruent segments, $AC$ and $BC$ are equal. This means that $C$ is equidistant from $A$ and $B.$ The same reasoning can be applied to any point on a perpendicular bisector, showing that the point is equidistant from the endpoints of the segment.