Prove that the three perpendicular bisectors of the sides of a triangle are concurrent.
Below is a diagram of our situation.
In order for all of the perpendicular bisectors (GE, GF, & GD) to be concurrent, we must prove that GD is perpendicular to AB. This will prove that GE, GF, & GD all meet at point G.
Since we constructed point E as the midpoint of AC, AE is congruent to EC. We also know that angles AEG and CEG are both right angles from our construction and thus congruent. We can also say that segment EG is congruent onto itself.
This means that triangles AEG and CEG are congruent by SAS. Since congruent parts of congruent triangles are congruent, AG is congruent to CG.
We can prove that triangles CGF and BGF are congruent in a similar manner.
Since GF is a perpendicular bisector of CB, this means CF is congruent to BF. Again, GF is congruent to itself by the Reflective Property and angles CFG and BFG are congruent because GF is perpendicular to CB.
Now, triangles CGF and BGF are congruent by SAS.
And again, because congruent parts of congruent triangles are congruent, CG is congruent to BG. By the Transitive Property, if AG is congruent to CG and CG is congruent to BG, then AG is congruent to BG.
We also know that D is the midpoint of AB, we just have to prove that DG is also a perpendicular bisector, which means angles ADG and BDG are both right angles.
If AG is congruent to BG, then triangle AGB is isosceles and we can use the properties of isosceles triangles. Since CG bisects AB, CG is the altitude of triangle AGB and the altitude is perpendicular to the isosceles triangle. This means that DG is perpendicular to AB and thus is a perpendicular bisector.