The internal angles of a regular pentagon have size $108^\circ$. Thus $\angle ABC = 108^\circ$, implying $\angle ABF = 72^\circ$ (Fig. 7). As $AB = AF$, we have $\angle AFB = 72^\circ$ and $\angle BAF = 36^\circ$. Since $FG = FB$, $AG = AB$ and $AF = AF$, the triangles $AFB$ and $AFG$ are equal, implying that $\angle FAG = 36^\circ$. As $\angle EAB = 108^\circ$, we obtain $\angle GAE = 2 \cdot 36^\circ + 108^\circ = 180^\circ$, implying that $E$, $A$ and $G$ are collinear.

Let the lines $AB$ and $EF$ meet at point $K$ (Fig. 8). As $\angle AFC = 180^\circ - \angle FCD$, we have $AF \parallel CD$. Consequently also $AF \parallel BE$, whence

Fig. 7

Fig. 8

similarity of triangles $BKE$ and $AKF$ implies $\frac{BK}{AK} = \frac{BE}{AF}$.
Let the lines $AB$ and $DG$ meet at point $K'$ (Fig. 9). Lines $AG$ and $AE$ coincide because $E$, $A$ and $G$ are collinear. Consequently, $BD \parallel AG$, whence similarity of triangles $BK'D$ and $AK'G$ implies $\frac{BK'}{AK'} = \frac{BD}{AG}$.
Since $BD = BE$ and $AF = AG$, we have $\frac{BK}{AK} = \frac{BK'}{AK'}$, implying $K = K'$.

The internal angles of a regular pentagon have size $108^\circ$. Thus $\angle ABC = 108^\circ$ (Fig. 10), implying $\angle ABF = 72^\circ$. As $AB = AF$, we have $\angle AFB = 72^\circ$. Since $FG = FB$, $AG = AB$ and $AF = AF$, the triangles $AFB$ and $AFG$ are equal, implying also $\angle AFG = 72^\circ$. Therefore $\angle CFG = \angle BFG = 2 \cdot 72^\circ = 144^\circ$, whereas $\angle FCA = \angle BCA = \frac{180^\circ - \angle ABC}{2} = 36^\circ = 180^\circ - 144^\circ$. Consequently, $GF \parallel CA$, implying also $GF \parallel DE$.
Let the lines $AB$ and $EF$ meet at point $K$ (Fig. 11). As $CE \parallel BA$, we obtain $\frac{BF}{KF} = \frac{BC}{KE}$. By assumptions, $BF = GF$ and $BC = DE$; hence also $\frac{GF}{KF} = \frac{DE}{KE}$. As $\angle KFG = \angle KED$ because of $GF \parallel DE$, the triangles $KFG$ and $KED$ are similar. Thus $\angle FKG = \angle EKD$. Consequently, points $G$, $K$ and $D$ are collinear, meaning that the line $DG$ passes through $K$.

Fig. 10

Fig. 11

Remark: The claim follows directly from Desargues's theorem.