Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$, respectively, such that $AE/ED = BF/FC$. Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$ respectively. Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through a common point.
Since this was an exercise for spiral similarity, I knew that this problem would use spiral similarity !! Please send your solutions too. This helps me a lot. Thanks in advance.
My Proof:Let $O$ be the spiral center of the spiral symmetry $S$ that sends $AD\rightarrow BC \implies S: A\rightarrow B $ and $S:D\rightarrow C$ . Then since $AE/ED = BF/FC$, we get that $S:E\rightarrow F$ .
Hence $S:AE \rightarrow BF $. Hence $O$ is the spiral center of the spiral similarity $S$, that sends $AE\rightarrow BF \implies O \in (SAE)$ and $O\in (SBF)$ as $AB\cap EF=S$
Similarly , $S:ED \rightarrow FC \implies O \in (TED)$ and $O\in (TFC)$ as $DC\cap EF=T$
Hence $(SAE)$, $(SBF)$, $(TCF)$, and $(TDE)$ pass through common point $O$ which is the spiral center of the spiral symmetry $S$ that sends $AD\rightarrow BC$.
Best Answer
Perhaps you could write which theorems you use, othervise it is perfect.
I suggest you to try this problem from IMO 2005 also:
Let $ABCD$ be a fixed convex quadrilateral with $BC=DA$ and $BC$ not parallel with $DA$. Let two variable points $E$ and $F$ lie of the sides $BC$ and $DA$, respectively and satisfy $BE=DF$. The lines $AC$ and $BD$ meet at $P$, the lines $BD$ and $EF$ meet at $Q$, the lines $EF$ and $AC$ meet at $R$.
Prove that the circumcircles of the triangles $PQR$, as $E$ and $F$ vary, have a common point other than $P$.