If $KL$ cut $CA,CB$ at $X,Y,$ then clearly $X$ defines $Y$ unambiguosly and $O$ is the circumcenter of $\triangle DEF$ $\Longleftrightarrow$ $F$ is on the incircle $(O).$ Thus we can restate the problem as follows: $F$ lies on $(O)$ (inside of $\triangle OAB$). $DF,EF$ cut $OB,OC$ at $L,K$ and $LK$ cuts $CA,CB$ at $X,Y.$ Then $XY$ bisects the perimeter of $\triangle ABC.$ Let $M$ be the midpoint of $AB$ (tangency point of $(O)$ with $AB$). Since $\angle KOL=120^{\circ},$ $\angle KFL=60^{\circ}$ and $\angle MFL=60^{\circ},$ it follows that $M \in \odot(FLOK)$ is the midpoint of its arc $KFL.$ Since $MOEA$ is cyclic, then $M$ is then the Miquel point of $KL$ WRT $\triangle OAE$ $\Longrightarrow$ $MKXA$ is cyclic $\Longrightarrow$ $\angle MXK=\angle MAO=30^{\circ}$ and similarly $\angle MYK=30^{\circ}.$ As a result $M$ is the midpoint of the arc $XY$ of $\odot(CXY).$ Thus by Ptolemy's theorem for $CXMY,$ we get: $CX+CY=\frac{XY}{MX} \cdot CM=\sqrt{3} \cdot CM=\tfrac{3}{2}AB \Longrightarrow 2 \cdot (CX+CY)=3 \cdot AB \Longrightarrow$ $CX+CY=AB+(AB-CX)+(AB-CY)=AB+AX+BY.$

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Okay, here is a more direct approach, working with the same previous notations. From the condition that $XY$ bisects the perimeter of $\triangle ABC,$ we deduce that $EX+DY=AX+BY$ and since $AE=DB$ $\Longrightarrow$ $AX=DY$ and $EX=BY$ $\Longrightarrow$ $X$ and $Y$ are then homologous points under the rotation $(M,120^{\circ})$ that swaps $AE$ and $DB.$ So this makes $\triangle MXY$ isosceles with $\angle MXY=\angle MYX=30^{\circ}$ $\Longrightarrow$ $AXKM$ is cyclic $\Longrightarrow$ $M$ is the Miquel point of $\triangle OAE$ WRT $KL$ $\Longrightarrow$ $MFKL$ is cyclic. But $\angle MKL=\angle MAX=60^{\circ}$ and similarly $\angle MLK=60^{\circ}$ $\Longrightarrow$ $\angle KFL=\angle KML=60^{\circ}$ $\Longrightarrow$ $F \in (O).$