SAUDI ARABIAN IMO Booklet 2023

Let $S$, $T$ be the intersection of the lines $KF$, $KE$ with $CA$, $AB$ and $P$, $Q$ the midpoint of $HT$, $HS$. Since $\angle HFT=\angle HMT=90^{\circ }$ so $HT$ is the diameter of $(HMF)$ and $P$ is the center of $(HMF)$. Similarly, $Q$ is the center of $(HNE)$.

Let $R$ be the intersection of $ST$, $EF$. Based on the basic properties of harmonic points, we have $$A(HR,EF)=-1.$$ On the other hand, if $EF$, $BC$ intersect at $R^{\prime }$, then $A(D{R}^{\prime },EF)=-1$, from which it follows that $R\equiv R^{\prime }$ and resulting in $R$ being the fixed point. Since $H$ is fixed, the midpoint $L$ of $HR$ is also a fixed point. On the other hand, $PL$, $QL$ are the medians of triangles $HTR$, $HSR$ such that $R$, $S$, $T$ are collinear, so $L$ belongs to $PQ$. So $PQ$ passes through the fixed $L$. $\square$