We begin by constructing a triangle ABC and it's midpoints G, H, and I:
Construct the line perpendicular to AB that runs through point H, line j:
By the definition of a perpendicular bisector, we know that any point on line j is equidistant from points A and B.
We now construct a line perpendicular to AC that runs through G, line k and mark the intersection of k and j point O:
We know that the lines j and k have an intersection because they perpendicular to lines that are not parallel.
So again by definition, any point on line k is equidistant from points A and C.
If we know look at the intersection of lines j and k, point O, we have a point that is on both line j and on line k. Therefore by our previously stated definitions, point O is equidistant from points A, B, and C. We can check this by constructing a circle with center O and radius of any vertex of the triangle. It should go through all three vertices of the triangle:
It appears our original conjecture was correct!
Lastly, again using our knowledge that any point on perpendicular bisector is equidistant from the endpoints of the segment it bisects, and the fact that point O is equidistant from points B and C, we can conclude that line l, that runs through I and O must be the perpendicular bisector of BC:
Final GSP Sketch
As we can see from our GSP sketch, it appears our proof was correct! (Which should'nt be a surprise: we proved it!)
Because we defined lines j and k had an intersection O, and the perpendicular bisector of BC, line l, runs through O and I, we can conclude that the perpendicular bisectors of a triangle are concurrent.