Iranian Mathematical Olympiad

By Pappus's hexagon theorem, $F$, $A$ and $E$ are collinear. Therefore $$\frac{EA}{AF}=\frac{EN}{NM}=\frac{EK}{BK}\Rightarrow AK\parallel BF\Rightarrow \angle BFK=\angle AKF=\angle AKE=\angle FBK$$ Thus $KF=KB$ and similarly $KE=KC$, so $\angle EKF\sim \angle CKB$. Hence $\angle FEK=\angle FCB=\angle FKP$, so $PN$ is tangent to the circumcircle of triangle $EKF$. The lines $NP$ and $EF$ meet at $T$, and $TQ$ is the second tangent from $T$ to the circumcircle of triangle $EKF$.



In triangle $EKF$, $A$ is the foot of the angle bisector $\angle BAC$ to $BC$, so $TA=TK$. Also $H$ is the reflection of $A$ with respect to $NP$, hence $TH=TA=TK=TQ$, so $QAKH$ is a cyclic quadrilateral. In triangle $EKF$, $MK$ is the symmedian, hence passes through $Q$. The lines $MK$ and $EF$ meet at $J$. Thus $JE\cdot JF=JQ\cdot JK$, so $J$ lies on the radical axis of the circumcircle of triangles $AKH$ and $EFH$, hence the result follows.