Selection Examination

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

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Alternative solution.

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