Denote $\angle BAC = \alpha$, $\angle ADF = \varphi$ and $\angle FEB = \psi$. The triangle $ABC$ is isosceles with the apex at $C$, so $\angle CBA = \angle BAC = \alpha$. The segments $DF$ and $EF$ bisect the angles $\angle ADE$ and $\angle DEB$, so $\angle FDE = \varphi$ and $\angle DEF = \psi$.

The sum of the inner angles of a quadrilateral is equal to $360^\circ$. Hence, for the quadrilateral $ABED$ we have
$$360^\circ = \angle BAD + \angle ADE + \angle DEB + \angle EBA \\ = \alpha + 2\varphi + 2\psi + \alpha.$$
which implies $\alpha + \varphi + \psi = 180^\circ$.

The sum of the inner angles of a triangle is $180^\circ$, so $\angle DFA = 180^\circ - \alpha - \varphi = \psi$, $\angle BFE = 180^\circ - \alpha - \psi = \varphi$ and $\angle DFE = 180^\circ - \varphi - \psi = \alpha$. The triangles $AFD$, $FED$ and $BEF$ have all three inner angles in common, so they are similar. This implies that

$$\frac{|AF|}{|FD|} = \frac{|FE|}{|ED|} \quad \text{and} \quad \frac{|FD|}{|ED|} = \frac{|BF|}{|EF|}.$$
or
$$|AF| = \frac{|FE| \cdot |FD|}{|ED|} = |EF| \cdot \frac{|FD|}{|ED|} = |BF|.$$
We have shown that $F$ is the midpoint of the segment $AB$.