Let's prove that $\angle YXL=\angle YKL$, so we will conclude the problem. To do this, observe that, since $BDXF$ is cyclic, it follows that: $\angle FXD=180°-\angle FBD \Rightarrow \angle FBD=\angle FXY$ and, from $AY//BC$, $\angle BCA=\angle CAY$, therefore $AFXY$ is cyclic. Now, see that, by construction, $X$ is the Miquel point relative to $D$, $E$ and $F$ in $\triangle ABC$, with this $AFXE$ will also be cyclic, and then $A, F, X, E$ and $Y$ all belong to the same circle. However, by the midline $FE$, we know that $FE//BC//AY$ and since $AFEY$ is cyclic, it must be an isosceles trapezoid, giving us $YE=AF$. But, for the same reason, we conclude that $FBDK$ and $EDCL$ are also isosceles trapezoids. And then $BF=KD=AF=YE$, but, from $\angle FBD=\angle AFE= \angle BDK=\angle FEY$, we have that $KDEY$ is a parallelogram. From this, it suffices to see that $\angle DCL=\angle CDE=\angle LXY=\angle DEK$, but, as $DE//YK$, $\angle YXL=\angle DEK=\angle EKY=\angle YKL $, as we wanted to show.

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Let $E'$ be the reflection of $E$ over $F$ and let $C'$ be the reflection of $C$ over $B$. Then $BE'=EA=CE=DL$, $BF=DK$, $\angle DBF=\angle KDB$ and $\angle E'BC'=\angle ECD=\angle CDL$, so $\angle FBE'=\angle KDL$. Hence $BE'F\equiv DLK$, so $KL=FE'=EF$, and therefore $FL$ and $KE$ share the same midpoint $M$ and $FK=EL$. Hence $M$ lies on the radical axis $DX$ of $(DBF)$ and $(DCE)$. As $AY\parallel FM\parallel BD$ and $F$ is the midpoint of $AB$, we get that $M$ is the midpoint of $DY$, so$$MX\cdot MY=MX\cdot MD=MK\cdot MF=MK\cdot ML,$$which means that $KXLY$ is cyclic, as desired.

Let $S=DX\cap FE$, notice that $S$ lies on the radical axis of $(FKDB)$ and $(ELCD) \implies SX*SD=SK*SF=SE*SL$. Now we have that the quarilaterals $FKDB$ and $ELCD$ are isosceles trapezoids therefore $DL=EC=EA$ and $BK=FD$. Also $\angle FDB=\angle FXY$ and $\angle ELD=\angle EXY \implies \angle FXE=\angle FXY+\angle EXY=\angle ABC+\angle ACB$ but $\angle FAE=180^{\circ}-\angle ABC-\angle ACB \implies AFXE$ is cyclic. Since $AY//BC//DE \implies \angle ACB=\angle YAE=\angle YXE \implies AXEY$ is cyclic so the points $A,X,Y, E, F$ are concylic which means $AYFE$ is an isosceles trapezoid so $DL=EC=EA=YF$ and also $\angle YAE=\angle YFE=\angle FLD \implies YF//LD \implies YFLD $ is a paralelogram so the diagonals bisect each other therefore $SY=SD$ and $SF=SL$. Finally using $SX*SD=SK*SF=SE*SL$ we get $SK*SL=SX*SY \implies KXLY$ is cyclic $\square$

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