Let $CF$ intersect $AB$ at $H$. It is clear that $\bigtriangleup BFC \cong \bigtriangleup BFH$, thus $F$ is the midpoint of $CH$. Combining this with the fact that $D$ is the midpoint of $AC$, we get $FD \parallel AH$. Claim: $GE \parallel BC$. Proof: $\frac{DG}{GB}=\frac{FD}{BH}=\frac{\frac{1}{2}AH}{BC}=\frac{AB-BC}{2BC}=\frac{AB+BC}{2BC}-1=\frac{AC}{2EC}-1=\frac{\frac{1}{2}AC-EC}{EC}=\frac{DE}{EC}$. Now let $DF$ intersect $BC$ at $M$. Since $DF$ is a midline of $\bigtriangleup ACH$, we have $DM$ is a midline of $\bigtriangleup ABC$, and this implies that $BM=MC$, which in turn implies that $DF$ bisects $GE$.

Let \( CF \cap AB = \{K\} \) and \( DF \cap BC = \{M\} \). Since \( BF \perp KC \) and \( BF \) is the angle bisector of \( \angle KBC \), we have that \( \triangle KBC \) is isosceles, i.e. \( BK = BC \). Also, \( F \) is the midpoint of \( KC \). Hence, \( DF \) is the midline for \( \triangle ACK \), i.e. \( DF \parallel AK \), from where it is clear that \( M \) is the midpoint of \( BC \). We will prove that \( GE \parallel BC \). It is sufficient to show that \[ \frac{BG}{GD} = \frac{CE}{ED}. \]From \( DF \parallel AK \) and \( DF = \frac{AK}{2} \), we have \[ \frac{BG}{GD} = \frac{BK}{DF} = \frac{2BK}{AK} \tag{1}. \]Also, \[ \frac{CE}{DE} = \frac{CD - DE}{DE} = \frac{CD}{DE} - 1 = \frac{AD}{DE} - 1 = \frac{AE - DE}{DE} - 1 = \frac{AE}{DE} - 2. \]\[ = \frac{AB}{DF} - 2 = \frac{AK + BK}{\frac{AK}{2}} - 2 = 2 + \frac{2BK}{AK} - 2 = \frac{2BK}{AK}. \tag{2} \]From (1) and (2), we have \[ \frac{BG}{GD} = \frac{CE}{ED}, \]so \( GE \parallel BC \). Since \( M \) is the midpoint of \( BC \), it follows that the segment \( DF \) bisects the segment \( GE \).