Can you prove or disprove the following claim:
Claim. A convex hexagon $ABCDEF$ is circumscribed about an ellipse. Let $G$ be the point of concurrency of hexagon's principal diagonals , and let the points $O_1$ , $O_2$ , $O_3$ , $O_4$ , $O_5$ , $O_6$ be the circumcenters of $\triangle ABG$ , $\triangle BCG$ , $\triangle CDG$ , $\triangle DEG$ , $\triangle EFG$ and $\triangle FAG$ , respectively . Then, the points $O_1$ , $O_2$ , $O_3$ , $O_4$ , $O_5$ , $O_6$ lie on a new common ellipse.
GeoGebra applet that demonstrates this claim can be found here.
Consider a hexagon $O_1O_2O_3O_4O_5O_6$ . My idea is to apply Braikenridge–Maclaurin theorem on it in order to prove that points $O_1$ , $O_2$ , $O_3$ , $O_4$ , $O_5$ , $O_6$ are conelliptic. But, to do that I need to show that three intersection points of the three pairs of lines through opposite sides of a hexagon $O_1O_2O_3O_4O_5O_6$ lie on a line l. How this can be done?