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# Perpendicular Bisector Theorem

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, is the perpendicular bisector of Therefore, is equidistant from and

### Proof

Geometric Approach

Suppose is the perpendicular bisector of Then is the midpoint of

Consider a triangle with vertices and and another triangle with vertices and and

Both and have a right angle and congruent legs and Since all right angles are congruent, Furthermore, by the Reflexive Property of Congruence, 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 and are also congruent. By the definition of congruent segments, and have the same length. This means that is equidistant from and

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.

 Statements Reasons and are right angles Definition of a perpendicular bisector. All right angles are congruent. Reflexive Property of Congruence. SAS Congruence Theorem. Corresponding parts of congruent figures are congruent. Definition of congruent segments.

### Proof

Perpendicular Bisector Theorem

Suppose is the perpendicular bisector of and that is the midpoint of

Two triangles can be created by connecting points and and and

These triangles both have a right angle and one of the legs measures half of They also share one leg,

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 Given SAS congruence theorem Definition of congruent segments
Note that and are not triangles if is the point of intersection, However, since is the midpoint of it is, by definition, equidistant from and