Since $AHCD$ is a parallelogram we have $\angle ADC = \angle CHA = 180^\circ - \beta$, which implies that $D$ lies on the circumference of the triangle $ABC$. Also, $\angle ACD = \angle HAC = 90^\circ - \gamma$.
Since $ABCD$ is cyclic we have $\angle ABD = \angle ACD$, and since lines $AG$ and $BD$ are parallel we have $\angle GAB = \angle ABD$.
Hence, $\angle BAG = 90^\circ - \gamma$.

Quadrilateral $AFA_1C$ is cyclic if and only if $\angle AFE = \angle ACA_1$, i.e. $\angle AFE = \gamma$. This is equivalent to $\angle FAE = 90^\circ - \gamma$.
Since $\angle GAE = 90^\circ - \gamma$, the above condition holds if and only if the triangles $AEF$ and $AEG$ are congruent, which is equivalent to $|EG| = |EF|$.
Since $F$ is the midpoint of the segment $\overline{A_1E}$, the given condition holds if and only if $F$ divides the segment $\overline{GA_1}$ in the ratio $2 : 1$, i.e. if and only if $|FG| = 2|FA_1|$. Since $\overline{GA_1}$ is a median in the triangle $BCG$, this holds if and only if $F$ is the centroid of that triangle.
This obviously holds if and only if the line $BF$ passes through the midpoint of the segment $\overline{CG}$.