Let $ABC$ be a isosceles triangle ($AC=BC$). Points $D$ are on $AC$ and $E$ is on $BC$...
Let $P$ be the point on $AC$ such that $FP$ is perpendicular to $AC$.
Let $Q$ be the point on $BC$ such that $FQ$ is perpendicular to $BC$.
Let $R$ be the point on $DE$ such that $FR$ is perpendicular to $DE$.
Since $FD$ is the angle bisector of $\angle ADE$, we have $|FP| = |FR|$. Similarly $|FQ| = |FR|$.
Thus we have $|FP| = |FQ|$. Also $\angle A = \angle B$ and $\angle APF = 90^\circ = \angle BQF$, therefore the two triangles $\triangle APF$ and $\triangle BQF$ are congruent, which implies that $|AF| = |BF|$.