My solution: Since $ X, Y $ is the center of $ \odot (ABDE), \odot (CHDE) $, respectively , so $ XY $ is the perpendicular bisector of their radical axis $ DE $ . Q.E.D

Dear Mathlinkers, 1. (O) the circumcircle of ABC 2. C’ the antipole of C wrt (O) 3. the line COC’ // XY 4. a result coming from Nagel : OC perpendicular to DE (see: http://jl.ayme.pagesperso-orange.fr/Docs/Cinq%20theoremes%20de%20Christian%20von%20Nagel.pdf p. 21-22) and we are done… Sincerely Jean-Louis

Since $X$ is circumcenter of circle $(ABEF)$ we have $DX=EX$ and since $Y$ is circumcenter of circle $(CEHF)$ we have $DY=EY$ so $XDYE$ is kite and thus conclusion.

$XY$ passes through midpoint of $DE$ (Newton-Gauss line) of the complete quad $DCEHBA$, and $X$ being the center of a circle having $DE$ as chord, we are done. Best regards, sunken rock

Because $DE$ is perpendicular to $CO$ due to $CH$, $CO$ being isogonal, it suffices to show that $CO$ and $XY$ are parallel. But in complex numbers $C=c$, $O=0$, $X=\frac {a+b} {2}$, and $Y=c+\frac {a+b} {2}$, so in fact $COXY$ is clearly a parallelogram and we are done.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%201.pdf p. 61... Sincerely Jean-Louis