The Perpendicular Bisector Theorem

We have discussed this before, and now we will give a precise proof:

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

There are two things to prove here, a statement and its converse, so we will split the proof into two parts, using generic diagrams for each.

Statement:

If  then  is on the perpendicular bisector of .

Proof:

Let  be the midpoint of  and draw :

Then  by SSS.  Therefore  since these are corresponding parts of the congruent triangles.  But these angles are a linear pair, so their measures add to 180o.  Since they are congruent angles, each must measure half of 180o, which is 90o.  Therefore .  So  is on the perpendicular bisector of  (by definition of "perpendicular bisector").

Converse:

If  is the perpendicular bisector of  then .

Proof:

Because  is the perpendicular bisector of , , and  since both are right angles.   is a common side to  and , so these triangles are congruent by SAS.  Therefore  since they are corresponding parts of the congruent triangles.