# Prove that $EF \parallel AB$.

$$D$$ 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 $$DE$$.)

Let the perpendicular bisector of $$DE$$ cuts $$AB$$ and $$BD$$ respectively at $$G$$ and $$H$$.

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 $$AD \parallel FH \implies AB \parallel FE$$.