Chinese Mathematical Olympiad

Proof Draw lines $AC$, $AD$, $AE$, $AF$, $DF$. From $\angle ADB=\angle AFB$, $\angle ACB=\angle AEF$ and the assumption $CD=EF$; one obtains $△ACD\cong △AEF$. So we have $AD=AF$, $\angle ADC=\angle AFE$ and $\angle ADF=\angle AFD$. Then $\angle ABC=\angle AFD=\angle ADF=\angle ABF$; $AB$ is the bisector of $\angle CBF$. Draw lines $CM$, $FN$. Since $M$ is the middle point of arc $PB$, $CM$ is the bisector of $\angle DCF$, and $FN$ is the bisector of $\angle CFB$. Then $BA$, $CM$, $FN$ have a common intersection, say, at $I$. In the circles ${\Gamma }_1$ and ${\Gamma }_2$, one has $CI\times IM=AI\times IB$, $AI\times IB=NI\times IF$, according to the theorem of power with respect to circles. So $NI\times IF=CI\times IM$, and $C$, $F$, $M$, $N$ are concyclic.