Let $T$ be the intersection point of $XY$and $FG$, $AK$ the altitude and $O$ the circumcenter of $(ABC)$. It is obvious that $AK$ is angle bisector in the isosceles triangle $EAD$. Also, $TD \perp FG$, as $XY$ is Simson line in cyclic $DEFG$. Hence $TYDG$, is cyclic and so we get that: , $\angle YTD = \angle YGD = \angle EGD = \angle EAD /2 = \angle KAD = \angle OAD$, where the last holds because $AK$, $AO$ are isogonal. Because, of the angle equality and using that $DT \parallel AO$ (as perpendiculars to $FG$), we conclude that $ TY \parallel AD => XY \parallel AD$, as needed.

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