Proof: To prove this, I must first prove that the three perpendicular bisectors of a triangle are concurrent. From Figure 1, consider triangle ABC, I know that the perpendicular bisector of AB, passing through midpoint M of AB, is the set of all points that have equal distances to A and B. Let’s prove this: Consider P, a point on the perpendicular bisector. Because, PM is congruent to PM, MA is congruent to MB and <PMA is congruent to <PMB, I know that triangle PMB is congruent to triangle PMA by SAS. And I can conclude that all of the points on the perpendicular bisector are equal distances from A and B. So, we know that the perpendicular bisectors of AB is the set of all points P such that PA is congruent to PB and this relation is true for all sides of the triangle. Consider D, which is the intersection of the perpendicular bisectors, then I know that DA is congruent to DB which is congruent to DC. So, D lies on the perpendicular bisector of BC and AC also, thus the three perpendicular bisectors of a triangle are concurrent.