SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $KF$ meets $AB$ at $S$. We have known that $EP$, $AK$, $BF$ are concurrent at Gergonne's point, then $(SP,AB)=-1$. Let $Q$ be the projection of $P$ on $KF$ then $Q(SP,AB)=-1$, but $QS\perp QP$ so $QP$ is the angle bisector of $\angle AQB$. From that, we get $\angle BQM=\angle AQN$.



In the other hand, easy to realize that $PQMB$ and $PQNA$ are cyclic, so $$\angle BPM=\angle BQM=\angle AQN=\angle APN.$$ Combining with the truth that $AB$ is the tangent of $(I)$, we get $$\angle PTH=\angle BPM=\angle APN=\angle PHT,$$ or $PH=PT$. $\square$