Proposition 3.1 - Find the center of a given circle.

Proof
Let circle ABC be drawn. Let AB be drawn at random, and let AB be bisected by E such that ED is the perpendicular bisector of AB where E lies on circle ABC. Extend ED to create DF such that F lies on circle ABC and D,E, and F are collinear. Bisect FD at G.
Suppose another point H is the center. Now, connect H to A, E, and B, thus creating AH, EH, and BH. We have that AE=EB, EH=EB, and AG=BH (because they are radii). Thus, triangle AHE is congruent to triangle HEB. This means that <AEH=<HEB. But for this to be true, both <AEH and <HEB must be right angles because they are supplementary. But, since <AEH and <HEB are not right angles, then H can not be the center. Only G can be the center because <AEG=<GEB=right angles. Therefore, G is the center of the circle ABC.