 Concurrency of Perpendicular Bisectors

By Mary Negley

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

Let be 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.