A segment \(AB\) is a base of an isosceles triangle \(ABC\). A line perpendicular to the segment \(AC\) was drawn through point \(A\) – this line crosses an extension of the segment \(BC\) at point \(D\). There is also a point \(E\) somewhere, such that angles \(\angle ECB\) and \(\angle EBA\) are both right. Point \(F\) is on the extension of the segment \(AB\), such that \(B\) is between \(A\) and \(F\). We also know that \(BF = AD\). Show that \(ED =EF\).