Let $C F \cap A B=\{K\}$ and $D F \cap B C=\{M\}$. Since $B F \perp K C$ and $B F$ is angle bisector of $\angle K B C$, we have that $\triangle K B C$ is isosceles i.e. $B K=B C$, also $F$ is midpoint of $K C$. Hence $D F$ is midline for $\triangle A C K$ i.e. $D F \| A K$, from where it is clear that $M$ is a midpoint of $B C$.

We will prove that $G E \| B C$. It is sufficient to show $\frac{B G}{G D}=\frac{C E}{E D}$. From $D F \| A K$ and $D F=\frac{A K}{2}$ we have

$$\frac{B G}{G D}=\frac{B K}{D F}=\frac{2 B K}{A K}$$

Also

$$\begin{gathered} \frac{C E}{D E}=\frac{C D-D E}{D E}=\frac{C D}{D E}-1=\frac{A D}{D E}-1=\frac{A E-D E}{D E}-1=\frac{A E}{D E}-2= \\ =\frac{A B}{D F}-2=\frac{A K+B K}{\frac{A K}{2}}-2=2+2 \frac{B K}{A K}-2=\frac{2 B K}{A K} \end{gathered}$$

From (1) and (2) we have $\frac{B G}{G D}=\frac{C E}{E D}$, so $G E \| B C$, as $M$ is the midpoint of $B C$, it follows that the segment $D F$, bisects the segment $G E$.