Let $O$ - circumcenter $\Delta ACE$. Easy note that $BO$ - perpendicular bisector to $AC; DO$ - perpendicular bisector to $CE; FO$ - perpendicular bisector to $AE$. Note $\angle OAF = \angle OEF; \angle OAB = \angle OCB; \angle OCD = \angle OED$ from symmetry. Then $\angle BAO + \angle OAF = \angle BAF$ $= \angle DEF = \angle DEO + \angle OEF = \angle DEO + \angle OAF \Rightarrow \angle BAO = \angle DEO \Rightarrow \angle BCO = \angle DCO \Rightarrow CO$ - bisector $\angle BCD$. Similary $AO$ - bisector $\angle BAF, EO$ - bisector $\angle DEF$. Then all biscetors of angles $ABCDEF$ intersect at one point $O$ $\Rightarrow AD, DE, CF$ are concurrent from Brianchon's theorem.