Problem: In triangle $ABC$,
points $P$, $Q$, $R$ lie on sides $BC$, $CA$, $AB$, respectively.
Let $\omega_A$, $\omega_B$, $\omega_C$ denote the
circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively.
Given the fact that segment $AP$ intersects
$\omega_A$, $\omega_B$, $\omega_C$ again at $X$, $Y$, $Z$ respectively,
prove that $YX/XZ=BP/PC$.

Solution: Let $M$ be the concurrence point of $\omega_A$, $\omega_B$, $\omega_C$ (by Miquel's theorem).

\begin{center}
  \begin{asy}
    size(8cm);
    defaultpen(fontsize(9pt));
    pair A = dir(110);
    pair B = dir(210);
    pair C = dir(330);
    pair P = 0.4*B+0.6*C;
    pair Q = 0.4*C+0.6*A;
    pair R = 0.7*A+0.3*B;
    draw(B--A--C);

    draw(circumcircle(A, Q, R), blue);
    draw(circumcircle(B, R, P), blue);
    draw(circumcircle(C, P, Q), blue);

    pair O_A = circumcenter(A, Q, R);
    pair O_B = circumcenter(B, R, P);
    pair O_C = circumcenter(C, P, Q);

    pair M = -P+2*foot(P, O_B, O_C);
    pair X = -A+2*foot(O_A, A, P);
    pair Y = -P+2*foot(O_B, A, P);
    pair Z = -P+2*foot(O_C, A, P);

    draw(X--M--P, dotted);
    draw(R--Q);
    draw(A--Y);
    draw(Z--P);

    draw(B--C, red);
    draw(Y--Z, red);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$P$", P, dir(P));
    dot("$Q$", Q, dir(Q));
    dot("$R$", R, dir(R));
    dot("$M$", M, dir(-90));
    dot("$X$", X, dir(225));
    dot("$Y$", Y, dir(45));
    dot("$Z$", Z, dir(180));

    /* Source generated by TSQ
    A = dir 110
    B = dir 210
    C = dir 330
    P = 0.4*B+0.6*C
    Q = 0.4*C+0.6*A
    R = 0.7*A+0.3*B
    B--A--C

    circumcircle A Q R blue
    circumcircle B R P blue
    circumcircle C P Q blue

    O_A := circumcenter A Q R
    O_B := circumcenter B R P
    O_C := circumcenter C P Q

    M = -P+2*foot P O_B O_C R-90
    X = -A+2*foot O_A A P R225
    Y = -P+2*foot O_B A P R45
    Z = -P+2*foot O_C A P R180

    X--M--P dotted
    R--Q
    A--Y
    Z--P

    B--C red
    Y--Z red
    */
  \end{asy}
\end{center}


Then $M$ is the center of a spiral similarity sending
$\ol{YZ}$ to $\ol{BC}$.
So it suffices to show that this spiral similarity
also sends $X$ to $P$, but
\[ \dang MXY = \dang MXA = \dang MRA = \dang MRB = \dang MPB \]
so this follows.