Explanation:[br][br]Let ABC be a triangle. Create perpendicular bisectors i, j, and k of sides AB, BC, and AC respectively. Let their point of intersection be called point F. A circle can be made with center F and points A, B, and C on the circumference.[br]This is because i,j, and k are perpendicular bisectors. Thus, AE=BE, BF=CF, and CG=AG. Additionally angle AED= angle DEB, angle DEF=angle DEC, and angle CGD=angle AGD (because perpendicular lines form right angles and right angles are congruent). Finally, ED=ED, FD=FD, and GD=GD. Thus by SAS, triangle AED=triangle BED, triangle BDF= triangle CDF, and triangle CGD= triangle AGD.[br]Since these are congruent, then their sides are too. So AD=BD, BD=CD, and CD=AD. Thus AD=BD=CD.[br]Thus these form the radii of circle ABC with center D. Therefore a circle has been circumscribed.//