Euclid II-1. (pg 51)

Explanation:[br]Create circle ABC. Connect AB. Draw a perpendicular bisector of AB such that D is the midpoint of AB and EC is perpendicular to AB and D lies on EC. Then create the midpoint of EC, labeled pt F.[br]F can either be the center of the circle or not. If it is, end proof.[br]If it is not, let G be the center. Join GA, GD, and GB.[br]AD=DB because D is the midpoint. GD=GD, and AG=BG because they are both radii. Thus by SSS, triangle ADG is congruent to triangle BDG. Thus angle ADG= angle BDG. Because these two angles form a straight line (AB) and equal each other, they must both be right angles. [br]But, angle ADF is also right because of the given conditions earlier where EC is perpendicular to AB. Thus angle ADF= angle ADG. This is a contradiction. Thus G is NOT the center of the circle and F therefore is.

Information: Euclid II-1. (pg 51)