Proof: Perpendicular Bisector Concurrence
Proof: Perpendicular Bisector Concurrence
Prove that the perpendicular bisectors of the sides of a triangle are concurrent.
Proof: Consider the triangle ABC. Let O be the point at which the perpendicular bisector of AB and perpendicular bisector of AC intersect. Note that by constructing line segments from O to each vertex of ABC, congruent triangles are formed using the logic of SAS (in the picture below, you can see that all aspects of SAS are satisfied and result in congruent triangles):
AOY COY [AY CY, OY OY, and E = 90]
AOZ BOZ [AZ CZ, OZ OZ, and Z = 90]
Let X be the midpoint of BC. Note that CO AO and AO BO implies that BO CO by transitivity. Since BO CO, BX CX, OXOX, then X = 90 by proposition 8 and proposition 13. Thus, we know that BOX COX and OX is the perpendicular bisector of BC. Therefore, all perpendicular bisectors of ABC intersect at point O.