It is clear that triangle $DFE$ is similar to $ABC$. I tried using inequalities such as $|AD|+|AE|\gt a$ to get information about the sides but with little success. Since $|FB|=|FC|\Longrightarrow \angle FCB = \angle FBC$ I attempted to divide the triangle into isosceles triangles and continue from there but I wasn't able to get the asnwer.
Thank you for any help.
$D,E$ lie on the perpendicular bisectors of $AC, AB$ respectively.
Since $FD // AB$ and $FE//AC$, the perpendicular bisectors of $AC, AB$ are also perpendicular to $FE$ and $FD$, making them altitudes of $\triangle DEF$, and their intersection the orthocenter of $\triangle DEF$. This intersection is also the circumcenter of $\triangle ABC$.
The line joining the this intersection and the point $F$ is therefore also an altitude of $\triangle DEF$ with base $DE$, and is perpendicular to $BC$. The line passing through the circumenter of $\triangle ABC$ and perpendicular to $BC$ must be the perpendicular bisector of $BC$. Hence $|FB| = |FC|$.