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.

\end{enumerate}

Solution: \begin{center}
  \begin{asy}
    size(8cm);
    defaultpen(fontsize(9pt));
    pair A = (-1,3);
    pair B = (-3,-2);
    pair C = (6,-2);
    pair D = (3,5);

    pair M = (A*C-B*D)/(A+C-B-D);

    pair E = 0.7*A+0.3*D;
    pair F = 0.7*B+0.3*C;
    pair S = extension(E, F, A, B);
    pair T = extension(C, D, E, F);

    draw(T--F);
    draw(A--B--C--D--cycle);
    draw(A--S);
    draw(T--D);

    draw(circumcircle(S, A, E), dotted+blue);
    draw(circumcircle(S, B, F), dotted+blue);
    draw(circumcircle(T, C, F), dotted+blue);
    draw(circumcircle(T, D, E), dotted+blue);

    pair P = extension(A, B, C, D);
    pair Q = extension(A, D, B, C);
    draw(A--Q--B, dashed);
    draw(S--P, dashed);

    draw(circumcircle(P, A, D), dotted+red);
    draw(circumcircle(P, B, C), dotted+red);
    draw(circumcircle(Q, A, B), dotted+red);
    draw(circumcircle(Q, C, D), dotted+red);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$D$", D, dir(D));
    dot("$M$", M, dir(M));
    dot("$E$", E, dir(E));
    dot("$F$", F, dir(F));
    dot("$S$", S, dir(135));
    dot("$T$", T, dir(T));
    dot("$P$", P, dir(P));
    dot("$Q$", Q, dir(Q));

    /* Source generated by TSQ */
  \end{asy}
\end{center}

Let $M$ be the Miquel point of $ABCD$.
Then $M$ is the center of a spiral similarity taking $AD$ to $BC$.
The condition guarantees that it also takes $E$ to $F$.
Hence, we see that $M$ is the center of a spiral similarity taking $\ol{AB}$ to $\ol{EF}$,
and consequently the circumcircles of $QAB$, $QEF$, $SAE$, $SBF$ concur at point $M$.

In other words, the Miquel point of $ABCD$ is also the Miquel point of $ABFE$.
Similarly, $M$ is also the Miquel point of $EDCF$, so all four circles concur at $M$.