$\frac{ \sin \angle EYQ}{ \sin \angle BYQ} = \frac{ED}{BC} \cdot \frac{CF}{AD}$ $\frac{ \sin \angle AZR}{ \sin \angle DZR} = \frac{AF}{DE} \cdot \frac{BE}{CF}$ $\frac{ \sin \angle CXP}{ \sin \angle FXP} = \frac{BC}{AF} \cdot \frac{AD}{BE}$ $\frac{ \sin \angle EYQ}{ \sin \angle BYQ} \cdot \frac{ \sin \angle AZR}{ \sin \angle DZR} \cdot \frac{ \sin \angle CXP}{ \sin \angle FXP} = 1$ $\Longrightarrow XP, YQ$ and $ZR$ concur at $O$. Furthermore $\measuredangle XOP=180^{ \circ}$

Ramanujan_1729 wrote: Six points $A, B, C, D, E, F$ are chosen on a circle anticlockwise. None of $AB, CD, EF$ is a diameter. Extended $AB$ and $DC$ meet at $Z, CD$ and $FE$ at $X, EF$ and $BA$ at $Y. AC$ and $BF$ meets at $P, CE$ and $BD$ at $Q$ and $AE$ and $DF$ at $R.$ If $O$ is the point of intersection of $YQ$ and $ZR,$ find the $\angle XOP.$ Can you give me the source of this problem ? My teacher said, it may be China contest problem but I am not sure.

This is BdMO National Higher Secondary 2013 P9

use pascals theorem and desargues theorem.then prove that x,o,p are collinear and angle xop is 180 degree