National Math Olympiad

Denote $\angle BAC=\alpha$, $\angle ADF=\phi$ 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 $\angle FDE=\phi$ 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 $$\begin{gathered} 360^{\circ }=\angle BAD+\angle ADE+\angle DEB+\angle EBA \\ =\alpha +2\phi +2\psi +\alpha . \end{gathered}$$ which implies $\alpha +\phi +\psi =180^{\circ }$.

The sum of the inner angles of a triangle is $180^{\circ }$, so $\angle DFA=180^{\circ }-\alpha -\phi =\psi$, $\angle BFE=180^{\circ }-\alpha -\psi =\phi$ and $\angle DFE=180^{\circ }-\phi -\psi =\alpha$. The triangles $AFD$, $FED$ and $BEF$ have all three inner angles in common, so they are similar. This implies that $$\frac{|AF|}{|FD|}=\frac{|FE|}{|ED|}\text{and}\frac{|FD|}{|ED|}=\frac{|BF|}{|EF|}.$$ or $$|AF|=\frac{|FE|\cdot |FD|}{|ED|}=|EF|\cdot \frac{|FD|}{|ED|}=|BF|.$$ We have shown that $F$ is the midpoint of the segment $AB$.