China Southeastern Mathematical Olympiad

Therefore, $\angle IGP=\angle IEP=90^{\circ }$, that is, $IG\perp PG$. Hence, points $P$, $S$ and $T$ are collinear.

Line $PST$ intersects $△ABC$. By Menelaus' Theorem, we have $$\frac{AS}{SC}\cdot \frac{CP}{PB}\cdot \frac{BT}{TA}=1.$$ Since $AS=AT$, $CS=CD$ and $BT=BD$, we have $$\frac{PC}{PB}\cdot \frac{BD}{CD}=1.\stackrel{◯}{1}$$ Let the extension of $BN$ intersect $PE$ at point $H$. Then line $BFH$ intersects $△PDE$. By Menelaus' Theorem, $$\frac{PH}{HE}\cdot \frac{EF}{FD}\cdot \frac{DB}{BP}=1.$$ Since $CF$ is parallel to $BE$, $\frac{EF}{FD}=\frac{PC}{CD}$, we have $$\frac{PH}{HE}\cdot \frac{PC}{CD}\cdot \frac{DB}{BP}=1.\stackrel{◯}{2}$$ By ① and ②, we have $PH=HE$. Hence, $P{H}^2=H{E}^2=HM\cdot HN$. Thus, we have $\frac{PH}{HM}=\frac{HN}{PH}$, $△PHN\sim △MHP$ and $\angle HPN=\angle HMP=\angle NEQ$. Further, since $\angle PEN=\angle EQN$, therefore $\angle ENP=\angle ENQ$. $\square$