Problem 1402

Solution:
Let $CF \cap AB = \{K\}$ and $DF \cap BC = \{M\}$. Since $BF \perp KC$ and $BF$ is angle bisector of $\text{KBC}$, we have that $\triangle KBC$ is isosceles, i.e. $BK = BC$, also $F$ is midpoint of $KC$. Hence $DF$ is midline for $\triangle ACK$, i.e. $DF \parallel AK$, from where it is clear that $M$ is a midpoint of $BC$.

We will prove that $GE \parallel BC$. It is sufficient to show $\frac{BG}{GD} = \frac{CE}{ED}$. From $DF \parallel AK$ and $DF = \frac{AK}{2}$ we have
$$\frac{BG}{GD} = \frac{BK}{DF} = \frac{2BK}{AK}$$
Also
$$\begin{gathered} \frac{CE}{DE} = \frac{CD - DE}{DE} = \frac{CD}{DE} - 1 = \frac{AD}{DE} - 1 = \frac{AE - DE}{DE} - 1 = \frac{AE}{DE} - 2 = \\ = \frac{AB}{DF} - 2 = \frac{AK + BK}{\frac{AK}{2}} - 2 = 2 + 2\frac{BK}{AK} - 2 = \frac{2BK}{AK} \end{gathered}$$
From (1) and (2) we have $\frac{BG}{GD} = \frac{CE}{ED}$, so $GE \parallel BC$, as $M$ is the midpoint of $BC$, it follows that the segment $DF$ bisects the segment $GE$.