Problem #PRU-100648

Problem

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 $\mathrm{\angle }ECB$ and $\mathrm{\angle }EBA$ are both right angles. 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 $|ED|=|EF|$.