# Proving the Perpendicular Bisector Theorem Using SAS Congruence

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 $\overline{PA} \cong \overline{PB}$

Proof

Let $M$ be the intersection of $\overline{AB}$ and line $l$. $\overline{AB} \cong \overline{MB}$. Definition of perpendicular bisector. (S) $\angle AMP \cong \angle BMP$ Both angles measure $90^\circ$ (definition of perpendicular bisector). (A) $\overline{PM} \cong \overline{PM}$. A segment is congruent to itself (Reflexive property, Axiom E1). (S)

By SAS Congruence, triangle $PMA$ is congruent to triangle $PMB$.

Now, $\overline{PA}$ and $\overline{PB}$ are corresponding sides of triangles $AMP$ and $BMP$.

Since corresponding parts of congruent triangles are congruent, $\overline {PA} \cong \overline{PB}$. This completes the proof.