TITLE: Three homothetic centers are collinear
QUESTION [8 upvotes]: I am looking a proof for the problem as follows:

Let a convex hexagon, such that its principal diagonals are concurrent. For each side of the hexagon, extend the adjacent sides to their intersection, forming a triangle exterior to the given side. Then show that:  Three external (or internal) homothetic centers of three pair circumcircle of opposite triangles are collinear.

Please see the applet in Geogebra

PS: The line through the three external homothetic centers are perpendicular to the line through three internal homothetic centers.
Another applet: Three homothetic centers are collinear associated  a  circumscribed conic hexagon

REPLY [2 votes]: At least for the external homothetic centers, this follows from Monge's theorem, which actually tells you that "for any three circles in a plane, none of which is completely inside one of the others, the intersection points of each of the three pairs of external tangent lines are collinear".