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
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.
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.