Perpendicular Bisectors

Erin Mueller

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.