In the previous post, we have proved the Perpendicular Bisector Theorem using reflection. In this post, we prove the same theorem using SAS Congruence.

**Theorem**

If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

**Given**

*P* is on the perpendicular bisector *l* of segment *AB*.

**What To Show **

**Proof**

Let be the intersection of and line .

. Definition of perpendicular bisector. (S)

Both angles measure (definition of perpendicular bisector). (A)

. A segment is congruent to itself (Reflexive property, Axiom E1). (S)

By SAS Congruence, triangle is congruent to triangle .

Now, and are corresponding sides of triangles and .

Since corresponding parts of congruent triangles are congruent, . This completes the proof.