Michael Thomason's write-up for assignment four, problem twelve.
I will begin by constructing a triangle with two sides having their perpendicular bisectors drawn. They intersect at a point called the circumcenter, labeled O. I know they always intersect because they are perpendicular to lines with different slopes, so they too have different slopes and therefore must intersect. Click the drawing to open a GSP file; move the vertices of the triangle as you wish and note that the two perpendicular bisectors always intersect.
Claim: The perpendicular bisector of also passes also through O.
Proof: O lies on the perpendicular bisector of , so O is equidistant from A and C, i.e.. Likewise,. Substitute to see that , which implies that O lies on the perpendicular bisector of .
Therefore, the three perpendicular bisectors of the sides of a triangle are concurrent. Moreover, it has been shown that , i.e. that O is the circumcenter of triangle ABC. Click below to open a GSP file showing O and the circumcircle; move the vertices of the triangle as you wish and note that O is the center of the circumcircle of the triangle and that the three perpendicular bisectors are indeed concurrent at O.