Let parallels from $B,C$ to $AC,AB$ intersect at $P.$ Then $ABPC$ is a parallelogram and $\triangle PDE$ is isosceles, due to $\angle ADB=\angle AEC=90^{\circ}-\frac{_1}{^2}\angle BAC.$ Internal bisector of $\angle BAC$ cut $\overline{PB}$ at $Q.$ Then $\angle CAQ=\angle AQB=\angle BAQ$ implies that $\triangle BAQ$ is isosceles with apex $B$ $\Longrightarrow$ $BA=BQ=CF$ $\Longrightarrow$ $PF \parallel AQ$ $\Longrightarrow$ $PF$ bisects $\angle BPC$ internally, i.e. $PF$ is the P-angle bisector of the isosceles $\triangle PDE$ $\Longrightarrow$ $FD=FE.$

$\angle PAB=\angle APB$ Then $AB=PB=FC$ sice $PB\parallel FC$ and this implyes that $PFCB$ is parallelogram. Then $DF=BC$ and we must show that $BC=FE$. Whis obviously true because $\triangle BAC=\triangle FCE$ EDIT: 222 post

Let $M$ be the midpoint of the arc $\widehat{BAC}$ and $G$ the intersection of the circumcircle of $\triangle ABC$ and $BD$. Since $AB=FC$, $MB=MC$ and $\angle ABM = \angle FCM$ the triangles $ABM$ and $FCM$ are congruent. Hence $\angle MEC = 180^{\circ} - MAB = 180^{\circ} - MFC$ so $CFME$ is cyclic. Then $\angle FMC = \angle AMB = \angle ACB = \angle GBC = \angle GMC$ so $M$, $F$ and $G$ are collinear, and so $\angle ADG = \angle MAC = \angle MGB$. Together with $DG\perp AF$ this means $ADGF$ is an isosceles trapetz and hence cyclic. Thus $\angle FDA = \angle MGA = \angle MCA = \angle MEF$ so $DF=FE$.