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My friend give me some hint on this problem, i only able to find some clue but not able to finish the whole problem, he said i could use Miquel point theorem to finish the rest of it, how could i able to complete the solution by using Miquel point theorem?

Problem: Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC,$ respectively, such that $\frac{AE}{ED} = \frac{BF}{FC}.$ 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.

Partial solution:Let $X_{1337}$ be the point of intersection of the lines AD and BC. Let the circumcircles of triangles $X_{1337}AB$ and $X_{1337}CD$ intersect at a point $\Lambda$ apart from $X_{1337}$. Then, $\measuredangle\Lambda BX_{1337}=\measuredangle\Lambda AX_{1337}$, what is equivalent to $\measuredangle\Lambda BC=\measuredangle\Lambda AD$. Similarly, $\measuredangle\Lambda CB=\measuredangle\Lambda DA$. Hence, the triangles $\Lambda BC$ and $\Lambda AD$ are directly similar. The points F and E are corresponding points in these similar triangles, since they lie on the respective sides BC and AD and divide them in the same ratio $\frac{BF}{FC}=\frac{AE}{ED}$. As corresponding points in directly similar triangles form equal angles, this entails $\measuredangle\Lambda FB=\measuredangle\Lambda EA$. This is equivalent to $\measuredangle\Lambda FX_{1337}=\measuredangle\Lambda EX_{1337}$. Thus, the point $\Lambda$ lies on the circumcircle of triangle $X_{1337}EF$.


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The $1337$ looks completely out of place when the solution is taken out of context like this. (Source:… ; notice the thread title.) – darij grinberg Apr 14 '12 at 0:55
up vote 2 down vote accepted

Using the angle chase method (as you do) we can show that the Miquel point M of a quadrilateral ABCD is the center of a spiral similarity that maps side AD onto BC (and there is another spiral similarity that maps AB onto DC, but we need the first one in our problem). Since point E divides side AD in the same proportion as point F does with BC, we can observe that M happens to be the center of the same spiral similarity that maps side AE onto BF, and also side ED onto FC. This is because the center of a spiral similarity with given angle and ratio pair of (directed) segments is unique. Moreover, M is the Miquel point of both quadrilateral AEFB and EDCF for the said reason.

The details you can find in a brilliant note of Yufei Zhao here.

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