Midpoint of hypotenuse of right triangle is circumcenter.

Show that the midpoint of the hypotenuse of a right triangle is the circumcenter.
Proof #1: We have right triangle ABC. We have A at (0,0); B at (x,0); and C at (0,y) Define D as the mid point of the hypotenuse. Therefore it is at (x/2, y/2). Show AD = DB = DC AD^2 = (y/2)^2 + (x/2)^2 AD^2 = 1/4(x^2 + y^2) DB^2 = EB^2 + DE^2 DB^2 = (x - x/2)^2 + (y/2)^2 DB^2 = (x/2)^2 + (y/2)^2 DB^2 = 1/4 (x^2 + y^2) Since D is the midpoint, DB = CD. DB^2 = AD^2 Therefore DB = AD We have, DB = CD, therefore AD = CD QED ================================================= Proof #2: DE is the perpendicular bisector of AB. All points on the perpendicular bisector are equidistant from the endpoints of the segment. D is on the bisector... Therefore A and B are both equidistant from D. Therefore AD = BD = CD. QED.