Take $G$ on $AC$ such that $FC=FG$ and $G\neq C$ (case $\angle FCA=90$ is similar). It's easy to prove that $FGSB$ is inscribed quadrilateral, and $\angle BFG=\angle EAC$. Then $AE/AC=BD/AC=BF/FC=BF/FG$ hence $\triangle BFG$ is similar to $\triangle EAC$ which gives $\angle AEC=\angle FBG=\angle FSC$, which is equivalent to thesis.

$\frac{sin\angle AFS}{sin\angle SFD}=\frac{sin\angle ECD}{sin\angle CED}$ the rest is obvious by chasing angle

I am the author of this problem, and this is the solution I proposed (I wonder if it's the official solution?) Let $M$, $N$ be the feet of the attitudes from $S$ on the $AB$ and $CD$, respectively. $SMFN$ is cyclic. Therefore, $\angle AFS = \angle MFS = \angle MNS$. It remains to prove that $\angle MNS = \angle ECD$. It will follow from the similarity $MSN \sim EDC$. Since $ABCD$ is cyclic, triangles $ABS$ and $DCS$ are similar. Therefore, their attitudes are in the same ratio as the respective bases, namely $SM / SN = AB / CD$. From the parallelogram $ABDE$ we have $AB = ED$, therefore $SM / SN = ED / CD$. Moreover, $\angle MSN = \angle EDC$ (angles with perpendicular rays). Now we have the desired $MSN \sim EDC$. I'm also interested in how many teams solved this problem?

Nice solution, djuro! I solved it using the result of Problem G4 from the IMO-Shortlist 2009: Let $P$, $Q$ be the midpoints of $BC$, $DA$, respecticely. Furthermore, define $K:=PQ\cap AC$, $L:=PQ\cap BD$ and $G:=EC\cap BD$. By the stated IMO-SL-problem, $FS$ is tangent to the circumcircle of triangle $\triangle{PQS}$, i.e. $\sphericalangle{FSP}=\sphericalangle{SQP}$. Furthermore, as triangles $\triangle{PCS}$ and $\triangle{QDS}$ are similar, $\sphericalangle{PSC}=\sphericalangle{DSQ}$. Therefore, $\sphericalangle{FSK}=\sphericalangle{FSP}+\sphericalangle{PSC}=\sphericalangle{SQP}+\sphericalangle{DSQ}=\sphericalangle{SLK}$. Note that in the parallelogram $ABDE$, the midpoint $Q$ of $AD$ is also midpoint of the other diagonal $BE$. As $P$ is the midpoint of $BC$, $\frac{BQ}{QE}=\frac{BP}{PC}=1$ and lines $PQ$ and $CE$ are parallel. In particular, we have $\sphericalangle{SGC}=\sphericalangle{SLK}$. Altogether, we get $\sphericalangle{ASF}=180^{\circ}-\sphericalangle{FSK}=180^{\circ}-\sphericalangle{SGC}=\sphericalangle{CGD}$. As $ABCD$ is cyclic, we also have $\sphericalangle{FAS}=\sphericalangle{BAC}=\sphericalangle{BDC}=\sphericalangle{GDC}$, implying that triangles $\triangle{FSA}$ and $\triangle{FGD}$ are similar and hence $\sphericalangle{SFA}=\sphericalangle{DCG}$, as required.

Here is another solution. Let $K$ and $L$ be the points of intersection of the lines $FS, \: ED$ and $AS, \: ED$ respectively. $|DC|=u \: , \: |CF|=v \: , \: |BF|=x \: , \: |AB|=y \: , \: |KD|=xk \: \Longrightarrow \:$ $|DL|=yk \: , \: |ED|=y \: , \: x(x+y)=v(u+v) \: , \: \frac{v}{u}=\frac{x+y}{yk}$ $\Longrightarrow \: ky=\frac{u(x+y)}{v}=\frac{u(u+v)}{x} \: \Longrightarrow \: u(u+v)=xk \cdot y$ Therefore $E,K,C,F$ are concyclic and $\angle DKF= \angle AFS =\angle ECD$

My solution is here. Let $T := BD \cap CE$ and take $X$ on $BD$ such that $CX \parallel DE$. Since $AB \parallel DE \parallel CX$, $\angle DCX = \angle F$. With the angles on arc $BC$, $\triangle DCX \sim \triangle AFC$ (1). Now $XT : TD = CT : TE = CS : SA$, and therefore $T$ maps to $S$ in the similarity (1). That is, $\triangle DCT \sim \triangle AFS$.

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