$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$