$\angle CXD=\angle CAD=\angle EXF$ and $\angle CDX=\angle CAX=\angle EAX=\angle EFX \implies \triangle XCD\sim \triangle XEF$. So $X$ is center of spiral similarity that maps $CD$ to $EF$. Similarly, $Y$ is center of sipiral similarity that maps $CD$ to $FE$. If $R=CD\cap EF$, from properties of spiral similarity we get $RCXE, RDYE, CRYF$ and $DRXF$ are cyclic. From Radical Axis thm we get $XC,ER,DY$ are concurrent, so are $CY,DX,FR$. Let first concurrency point be $P$ and second one be $Q$. If $XY$ is diameter of $\Omega$, then $Q$ is orthocenter of $\triangle XYP \implies PQ\perp XY$. Since $PQ$ coincides with $EF$, the result follows.

sketch of my solution: $AX$, $BY$, and $FE$ concurrent at T ( from by radical axes). Then from pascal on $BYDAXC$ the intersection of $YD$ and $XC$ is located on the $EF$ (say $K$ to this point). Morever, from pascal on $BYCAXD$ yields that the intersection of $YC$ and $XD$ is on $EF$ line. ( say M to this point). From these claims we can say that $E$, $T$, $F$, $M$, and $K$ is collinear. So M is the orthocenter of $XYK$. Therefore $EF$ is perpendicular to $XY$