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 $YX/XZ=BP/PC$.