(2) => (3) i'll try to work on this tomorrow.
(3) => (1) Let D, the midpoint of AB, be the circumcenter of ABC. Let E be on CB such that DE is the perpendicular bisector of CB, and let F be on CA such that DF is the perpendicular bisector of CA. Then, CFDE forms a quadrilateral. In particular, since angles CDF and DEC forms right angles, then CFDE is a rectangle. therefore, angle C is a right angle.