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 180^{o}. Since they are congruent angles, each must measure half of 180^{o}, which is 90^{o}. 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.