# Three homothetic centers are collinear

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

• Out of curiosity, how are you coming up with all these facts? Are you just playing around with GeoGebra (or a similar tool) and noticing that such-and-such relation always holds? Jul 8, 2016 at 12:34
• @Wojowu When I construct Dao's theorem on six circumcenters, I found this property true with a cyclic hexagon with principal diagonals are concurrent . Later I generalization for arbitrary hexagon with principal diagonals are concurrent. See Geogebra: Jul 8, 2016 at 12:53
• While you do specify it ... are you sure this property is solely for a convex hexagon ? and not valid for a general case.
– ARi
Jan 15, 2017 at 18:13
• General case: Alway have three homothetic centers are collinear, and three other homothetic centers are collinear. But exactly three external homothetic center are collinear and three internal homothetic center are collinear we need convex hexagon Jan 16, 2017 at 7:26
• @ARi maybe the result true for generalization case Jan 16, 2017 at 7:29