Triangle perpendicular bisectors meet

Let the perpendicular bisector of AB and AC of ∇ABC be DO and EO where O is their point of intersection.

Join the vertices to O

Consider the ∇ADO and ∇BDO

     AD = BD                (Construct)

     DO = DO                (Common)

   ^ADO = ^BDO              (Construct)

   ∇ADO = ∇BDO              (2 sides and incl. angle eq.)

∴    AO = BO

Similarly it can be shown by considering ∇AEO and ∇CEO that

     AO = CO

∴    AO = BO = CO 

Let F be the mid-point of BC

We now need to show that the line FO is perpendicular to BC

Consider ∇BFO and ∇CFO

    BO = CO                  (Proven above)

    BF = CF                  (Construct)

  ^OBF = ^OCF                (Isosceles triangle)

∴ ∇BFO = ∇CFO                (2 sides and incl. angle equal)

∴ ^BFO = ^CFO

       = 90º                 (BC is a straight line)

∴ FO is the perpendicular bisector of BC 

Therefore the perpendicular bisectors of triangle meet at a single point which is equidistant from the vert