$D$ lies on the smaller of arc $AB$ of the circumcircle of isoceles triangle $ABC$ at $A$. $E$ is a point on line segment $BC$ such that $AD \perp DE$. The perpendicular bisector $DE$ cuts $AC$ at point $F$. Prove that $EF \parallel AB$.
At first, I thought I needed to let $G$ be the intersection of $AE$ and the perpendicular bisector of $DE$ so $\triangle ADG$ is an isosceles triangle at $G \implies \widehat{EAD} = \widehat{GDA}$. (Because $\triangle ADE$ is a right-angled at $D$.)
Secondly, I assumed letting $H = AB \cap DF$ and proved that $H$ lies on the perpendicular bisector of $AD \implies \widehat{BAD} = \widehat{FDA}$. (But I couldn't.)
Therefore, $\widehat{EAD} - \widehat{BAD} = \widehat{GDA} - \widehat{FDA} \implies \widehat{EAB} = \widehat{GDF} \implies \widehat{EAB} = \widehat{AEF}$ (Because $G$ lies on the perpendicular of $DE$.)
We have that $AD \parallel FH \implies \widehat{CAD} = \widehat{CFH}$.
$ACBD$ is a cyclic quadrilateral $\implies \widehat{CAD} + \widehat{CBD} = 180^\circ$.
That means $\widehat{CFH} + \widehat{CBH} = 180^\circ \implies FCBH$ is a cyclic quadrilateral $\implies \widehat{FCB} = \widehat{DHF}$.
But $FH$ is the perpendicular bisector of $DE \implies \widehat{DHF} = \widehat{EHF}$.
$\triangle ABC$ is an isosceles triangle at $A \implies \widehat{ABC} = \widehat{ACB}$.
That means $\widehat{GBE} = \widehat{EHG} \implies GEBH$ is a cyclic quadrilateral $\implies \widehat{EBH} = \widehat{FGE}$.
But $FG$ is the perpendicular bisector of $ED \implies \widehat{FGE} = \widehat{FGD}$.
$ACBD$ is a cyclic quadrilateral $\implies \widehat{CAD} + \widehat{CBD} = 180^\circ$.
That means $\widehat{FAD} + \widehat{FGD} = 180^\circ \implies AFGD$ is a cyclic quadrilateral $\implies \widehat{DAG} = \widehat{DFG}$.
But $FG$ is the perpendicular bisector of $DE \implies \widehat{DFG} = \widehat{EFG}$.
But $AD \parallel FH \implies AB \parallel FE$.