Geometry question on triangle Let ABC be an isosceles triangle with AB = AC and let Γ denote its circumcircle. A point D is on the arc AB of Γ not containing C and a point E is on the arc AC of Γ not containing B such that AD = CE.   how can I prove that BE is parallel to AD.?



Note $|AB|=|D'E'|$.
Let $|AF|=|AD'|$, then $\angle AD'B=\angle AFC$.
Rotate $A\to D'$ and so $\angle AD'B=\angle D'AE'$.
$|D'B|=|AE'|$ so $AD' \parallel BE'$.
