Consider $\overline{AB}$ and a point $C$ equidistant from $A$ and $B.$

To prove that $C$ lies on the perpendicular bisector of $\overline{AB},$ it will be shown that the line perpendicular to $\overline{AB}$ through $C$ bisects $\overline{AB}.$ If $M$ is the point of intersection between the line and the segment, it must be proven that $AM=MB.$ This line forms two right triangles that share a common leg $\overline{CM}.$ Because all right angles are congruent, $\angle AMC$ is congruent to $\angle BMC.$ Also, by the Reflexive Property of Congruence, $\overline{CM}$ is congruent to itself. Since $AC$ is equal to $BC,$ $\overline{AC}$ is congruent to $\overline{BC}.$ By the Hypotenuse Leg Theorem, $△AMC$ and $△BMC$ are congruent triangles. Because corresponding parts of congruent figures are congruent, $\overline{AM}$ is congruent to $\overline{BM}.$

Additionally, it was already known that $\overleftrightarrow{CM}$ and $\overline{AB}$ are perpendicular.

By the definition of a perpendicular bisector, $\overleftrightarrow{CM}$ is the perpendicular bisector of $\overline{AB}.$ Therefore, $C$ lies on the perpendicular bisector of $\overline{AB}.$