In order to prove that these three centers are collinear, extend the segment that contains the circumcenter and the centroid to the altitude CG. We will call this point H. If we can show H to be the orthocenter of the triangle our proof will be complete. [br][br]Observe the two triangles shaded in blue. Notice segment GH is parallel to segment EF because both segments are perpendicular to segment AB. (We know they are perpendicular because CG is an altitude and EF is the perpendicular bisector of AB.) Thus, because HF is a transversal of two parallel lines, m∠CHD = m∠EFD by the alternate interior angles theorem.[br][br]Further, m∠CDH = m∠FDE by the vertical angles theorem. From this we can conclude the two blue triangles are similar by AA~. [br][br]Note that CD = 2 DE because D is the centroid. With this ratio, we know DH = 2 DF and CH = 2 FE because of similar triangles.[br][br]Next, we construct the median of the triangle, AI. This must pass through D since D is the centroid. [br][br]We also construct segment AH and FI.[br][br]Now observe the triangles in green. Again, AD = 2 DI because D is the centroid. Further m∠ADH = m∠FDI by the vertical angles theorem. Since we already know that HD = 2 DF, the two green triangles are similar by SAS~. [br][br]Thus m∠FIA = m∠HAD because corresponding angles of similar triangles are congruent. This further implies that AH is parallel to IF by the converse of the alternate interior angles theorem. [br][br]We know that FI⊥CB because F is the circumcenter. Thus, by substitution AH⊥CB. [br][br]Since the orthocenter is the point of concurrency of the altitudes, H must be the orthocenter because it is the point of intersection of altitude CG and altitude AH (when AH is extended to segment BC). [i]Q.E.D.[/i]