Perpendicular Bisector Theorem

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 Side-Angle-Side 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.

Two-Column Proof

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

Suppose CM is the perpendicular bisector of 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 AM and 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 CM. Since C lies on CM, a reflection across CM maps C onto itself. The same is true for M.

Reflection Across 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 CM. Since a reflection is a rigid motion, this proves that the triangles are congruent. Corresponding parts of congruent figures are congruent, so AC and BC are congruent. By the definition of congruent segments, AC and BC are equal. This means that C is equidistant from A and B. AC=BC 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.