Prove that the three perpendicular bisectors are concurrent.

To prove that the perpendicular bisectors are concurrent, we have to show that they all intersect at the same point.

Now lets define perpendicular bisector: a line segment that perpendicular to another line segment that passes through its midpoint.

To prove this idea we will also need the following theorem: The perpendicualr bisector of a line segment is the locus of all points that are equidistant from its endpoints.

Let ABC be a triangle.

Let points D, E, and F be the midpoints of the sides of the triangle.

Let DG, EH and FI be the perpendicular bisectors of AC, CB, and BA respectively.

Now, suppose that DG and EH do not meet at some point P. Then DG and EH are parallel. Therefore, AC and BC must be parallel or they must be the same segment. However, ABC is a triangle, since this is impossible, and therefore there is a contradiction. So DG and EH must meet at some point P.

Because P lies on the perpendicular bisectors of AC and BC, by the beginning statement, we know that n(AP) = n(BP) and n(AP) = n(CP). Therefore by the property of transitivity, n(BP) = n(CP).

Because n(BP) is equal to n(CP), P lies on the perpendicular bisector of the segment BC, by the beginning statement. Therefore P lies on the line EH. Therefore P lies on EH, DG, and FI, so the lines EH, DG and FI are concurrent, which means that the perpendicular bisectors of the triangle ABC are concurrent.

By Carolyn Amos