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*.