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.
, the perpendicular bisectors of and .
The perpendicular bisectors intersect in a point and that point is equidistant from the vertices.
The perpendicular bisectors of and intersect at point .
Let us prove that point lies on the perpendicular bisector of and it is equidistant from , and .
Draw and .
Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
So, and .
By the transitive property,
Any point equidistant from the end points of a segment lies on its perpendicular bisector.
So, is on the perpendicular bisector of .
Since , point is equidistant from , and .
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 .