No one?

Source: Perú TST- 2014 (I live in Perú then I know this) I think that the problem isn't attractive, for mathlinker on the geometry forum. Here exist many user too pro, at geometry they can solve this problem easily. My solution We apply Ceva's theorem in $\triangle BFD$ so $ AB, FC$ and $DE$ are concurrent if and only \[ \frac{\sin{\angle ABF}}{\sin{\angle ABD}}.\frac{\sin{\angle CFD}}{\sin{\angle {CFB}}}.\frac{\sin{\angle EDB}}{\sin{\angle EDF}}=1 \]We apply Ceva's theorem in $\triangle BFD$ and point $A$ \[ \frac{\sin{\angle ABF}}{\sin{\angle ABD}}.\frac{\sin{\angle AFD}}{\sin{\angle {AFB}}}.\frac{\sin{\angle ADB}}{\sin{\angle ADF}}=1 \] $ AB, FC$ and $DE$ are concurrent if and only if \[ \frac{\sin{\angle AFD}}{\sin{\angle {AFB}}}.\frac{\sin{\angle ADB}}{\sin{\angle ADF}}=\frac{\sin{\angle CFD}}{\sin{\angle {CFB}}}.\frac{\sin{\angle EDB}}{\sin{\angle EDF}} \]if and only if \[\sin{\angle AFD}.\sin{\angle ADB}=\sin{\angle CFD}.\sin{\angle EDB}\]Easy angle-chasing $\angle AFD+\angle ADB=\angle CFD+\angle EDB$ i.e. $\angle CFD=\angle AFD \vee \angle CFD=\angle ADB$ So $ AB, FC$ and $DE$ are concurrent if and only if $\angle CFD=\angle ADB \Longleftrightarrow BF=BD$ Now if $BFED$ is cyclical then $AF \parallel BE \parallel CD$ so isn't possible Consider $K$ the symmetrical of $E$ WRT the perpendicular bisector of $FD$ and $AF \cap CD \equiv G$ Then $K$ and $G$ are isogonal conjugate WRT $\triangle BFD$ $ AF, BE$ and $CD$ are concurrent if only if $\angle FBE=\angle KBD$ Denote $O_1$ the center of $\odot (BFE)$ and $O_2$ the center of $\odot (BDK)$ is clear that $O_1 \neq O_2$ $\angle FBE=\angle KBD \Longleftrightarrow \triangle EFO_1 \cong \triangle KDO_2 \Longleftrightarrow O_1$ is symmetrical of $O_2$ WRT the perpendicular bisector of $FD$ i.e. the radical axis between $\odot (BFE)$ and $\odot (BDK)$ belongs the perpendicular bisector of $FD$ Then $\angle FBE=\angle KBD \Longleftrightarrow B$ belongs the perpendicular bisector of $FD$, if only if $BF=BD$ $ AF, BE$ and $CD$ are concurrent if only if $BF=BD$