The three perpendicular bisectors of a triangle meet in a single point, called the circumcenter.

The vertices of a triangle are equidistant from the circumcenter.
Given:
, the perpendicular bisectors of 
To prove:
The perpendicular bisectors intersect in a point and that point is equidistant from the vertices.

The perpendicular bisectors of
intersect at point O.
Let us prove that point O lies on the perpendicular bisector of
and it is equidistant from A, B, and C.
Draw 
Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
So, OA = OC and OC = OB.
By the transitive property,
OA = OB.
Any point equidistant from the end points of a segment lies on its perpendicular bisector..
So, O is on the perpendicular bisector of
.
Since OA = OB = OC, point O is equidistant from A, B, and C.

This means that there is a circle having its center at the circumcenter and passing through all three vertices of the triangle. This circle is called the circumcircle.