In order to prove $FS \parallel CD$, it suffices $FC = SD$, or $S\hat{C}D = F\hat{E}C$. From the cyclic $SEDC$ and $FS \parallel CD$ we get
$$S\hat{C}D = B\hat{E}S = B\hat{I}G,$$
where $I$ is the point of intersection of $HG$ and $BD$.
It suffices to prove that: $B\hat{I}G = F\hat{E}C$. Angle chasing gives
$$A\hat{F}B = A\hat{E}B = D\hat{E}C = D\hat{F}C \text{ and } B\hat{A}F = B\hat{E}F = F\hat{C}D.$$

Figure 8
This means that $AFB$ and $CFD$ are similar with $A\hat{B}F = F\hat{D}C$. Moreover,
$$D\hat{B}F = C\hat{A}F \text{ and } B\hat{F}D = A\hat{F}D,$$
where the triangles $AFC$ and $BFD$ are similar with similarity ratio
$$AC/BD = AF/BF.$$
If $K$ is the midpoint of $AB$, then the triangle $HKG$ has $KH \parallel BD$, $KG \parallel AC$ with
$$KG/KH = AC/BD = AF/BF \text{ and } H\hat{K}G = A\hat{E}B = A\hat{F}B.$$
Therefore, the triangle $GKH$ is similar to $AFB$, and so
$$K\hat{H}G = A\hat{B}F = F\hat{D}C = F\hat{E}C.$$

Figure 8

As in the first solution, it suffices to prove that: $B\hat{I}G = F\hat{E}C$. Let $M$, $N$ be the projections of $F$ to $AB$, $BD$, respectively. Then the line $MN$ is a Simpson line of the complete quadrilateral and $HG$ is the Newton-Gauss line. It is well-known that they are perpendicular. Therefore
$$B\hat{I}G = H\hat{I}N = 90^\circ - M\hat{N}B = 90^\circ - M\hat{F}B = M\hat{B}F = F\hat{E}C,$$
where in the third equality we used that $MNFB$ is cyclic and in the last we used that $AEFB$ is cyclic.

Figure 9