Denote the points of intersection of the line $EG$ with the lines $BD$ and $CF$ by

$P_1$ and $P_2$, respectively. We will prove that $P_1 \equiv P_2$. Denote the points of intersection of the lines $AB$ and $AC$ with the line $DF$ by $C_1$ and $B_1$, respectively. The segment $AA_1$ is the altitude and the bisector in the triangle $AB_1C_1$, hence $\triangle AB_1C_1$ is isosceles with the base $B_1C_1$ and $A_1B_1 = A_1C_1$. Without loss of generality, suppose that $\angle AA_1B > \angle AA_1C$. If $A_1B_1 \le AA_1$, the lines $BD$ and $CF$ intersect the ray $AE$, and if $A_1B_1 > AA_1$, these lines intersect the ray $AG$. Therefore, it is enough to prove the equality $AP_1 = AP_2$.

Suppose that $A_1B_1 \le AA_1$ (the case $A_1B_1 > AA_1$ can be considered similarly). Then $B_1F = A_1F - A_1B_1 = A_1D - A_1C_1 = DC_1$. From $DF \parallel EG$ follow the similarities $\triangle BC_1D \sim \triangle BAP_1$ and $\triangle CB_1F \sim \triangle CAP_2$, whence
$$AP_1 = \frac{AB \cdot DC_1}{BC_1} \quad \text{and} \quad AP_2 = \frac{AC \cdot B_1F}{CB_1}.$$

Therefore, it is enough to prove that $AB/BC_1 = AC/CB_1$. Let $C_2$ be the reflection of $C$ through the line $AA_1$. Since $\triangle AB_1C_1$ is isosceles, the point $C_2$ lies on the segment $AC_1$. From the equalities $\angle C_2A_1C_1 = \angle CA_1B_1 = \angle C_1A_1B$ it follows that the line $A_1C_1$ is the bisector in the triangle $A_1BC_2$. Finally,
$$\frac{CB_1}{BC_1} = \frac{C_1C_2}{BC_1} = \frac{A_1C_2}{A_1B} = \frac{A_1C}{A_1B} = \frac{AC}{AB}.$$