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 **

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

**Proof**

Let $latex M$ be the intersection of $latex \overline{AB}$ and line $latex l$.

$latex \overline{AB} \cong \overline{MB}$. Definition of perpendicular bisector. (S)

$latex \angle AMP \cong \angle BMP$ Both angles measure $latex 90^\circ$ (definition of perpendicular bisector). (A)

$latex \overline{PM} \cong \overline{PM}$. A segment is congruent to itself (Reflexive property, Axiom E1). (S)

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

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

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