Prove that the perpendicular bisectors of the sides of a triangle are concurrent. [br][br]Proof: Consider the triangle [math]\Delta[/math]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 [math]\Delta[/math]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):[br][br][math]\Delta[/math]AOY [math]\cong[/math] [math]\Delta[/math]COY [AY [math]\cong[/math] CY, OY [math]\cong[/math] OY, and [math]\angle[/math]E = 90[math]^\circ[/math]][br][math]\Delta[/math]AOZ [math]\cong[/math] [math]\Delta[/math]BOZ [AZ [math]\cong[/math] CZ, OZ [math]\cong[/math] OZ, and [math]\angle[/math]Z = 90[math]^\circ[/math]][br][br]Let X be the midpoint of BC. Note that CO [math]\cong[/math] AO and AO [math]\cong[/math] BO implies that BO [math]\cong[/math] CO by transitivity. Since BO [math]\cong[/math] CO, BX [math]\cong[/math] CX, OX[math]\cong[/math]OX, then [math]\angle[/math]X = 90[math]^\circ[/math] by proposition 8 and proposition 13. Thus, we know that [math]\Delta[/math]BOX [math]\cong[/math] [math]\Delta[/math]COX and OX is the perpendicular bisector of BC. Therefore, all perpendicular bisectors of [math]\Delta[/math]ABC intersect at point O.