**Concurrency of Perpendicular Bisectors**

**By Mary
Negley**

I want to prove that the perpendicular bisectors of
any triangle are concurrent.

Letbe any triangle, let *D* and* E* be the midpoints of
*AB* and *BC*, respectively and let *l* and *m* be
the perpendicular bisectors of *AB*
and *BC*, respectively. Let *P* be the point of intersection of *l* and* m*. The triangle is pictured below.

Now, make segments connecting *P* to each of the vertices of the triangle.

Notice that four new right triangles are formed (, , , and ). First look at
and . Since *D* is the midpoint of *AB*, *AD* = *BD* and since *l*
is the perpendicular bisector of *AB*,
and : so . By reflexivity
. So by SAS, . By definition
of congruent triangles, . Similarly, and. Since and , by transference, .

Now look at . Since , . Let *F* be the midpoint of *AC* and make a segment between *P* and *F*.

By
construction, . Since , , and , we have SAS, so . By definition of congruent triangles, . Since can be extended
to be a line, . So if , we find that:

Since , , so the segment connecting *P* to *F*
must be part of the perpendicular bisector of *AC*. I have
shown that this perpendicular bisector goes through *P* also.
Therefore, the perpendicular bisectors of are concurrent.