6. The perpendicular bisectors of the three sides of a triangle all meet at the center of the circumscribed circle.

Let O be the point where the perpendicular bisector of AB meets the perpendicular bisector of AC. Then by Theorem 7 about isosceles triangles, since O is on the perpendicular bisector of AB, its distance from A is the same as its distance from B. Since it is also on the perpendicular bisector of AC, its distance from A is the same as its distance from C. We conclude that the distance from O to B is the same as the distance from O to C since both are equal to the distance from O to A. By the converse of Theorem 7 about isosceles triangles, we conclude that O is on the perpendicular bisector of BC. Thus O is on the perpendicular bisectors of all three sides, and since it is equidistant from all three points of the triangle, all three points of the triangle will lie on the same circle centered about O.