SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $F$ as the intersection of $AD$, $BC$. Suppose that $FK$ intersects $AB$, $CD$ at $S$, $T$ respectively. Since the harmonic points of complete quadrilateral, we have $(E,T,D,C)=-1$. Then $ET\cdot EH=ED\cdot EC$. Similarly, $ES\cdot EG=EA\cdot EB$. But $EA\cdot EB=ED\cdot EC$ then $ES\cdot EG=ET\cdot EH$ which implies that $G,H,T,S$ are concyclic.



Denote $Q^{\prime }$ as the midpoint of $FK$ then $Q^{\prime }$, $G$, $H$ belong to the Gauss line of the complete quadrilateral $AFBKCD$. We shall prove that $Q^{\prime }$ belongs to the radical axis of $(O),(I)$. It is easy to see that $EK$ is the antipole of $F$ then denote $EK\cap (O)=\{X,Y\}$ then $FX$, $FY$ are the tangent lines of $(O)$. Let $U,V$ be midpoints of $FX$, $FY$. We consider the power of point to circle $(O)$ and degenerate circle $F$. $${\mathcal{P}}_{U/(O)}=U{X}^2,{\mathcal{P}}_{U/(F)}=U{F}^2$$ but $UX=UF$ then ${\mathcal{P}}_{U/(O)}={\mathcal{P}}_{U/(F)}$. This implies that $U$ belongs to the radical axis of $(O),(F)$. Similarly with the point $V$ then $UV$ is the radical axis of $(O),(F)$. But $Q^{\prime }\in UV$ which is the midline of triangle $FXY$, then $${\mathcal{P}}_{Q^{\prime }/(F)}={\mathcal{P}}_{Q^{\prime }/(O)}\iff {\mathcal{P}}_{Q^{\prime }/(O)}=Q^{\prime }{F}^2=Q^{\prime }{K}^2=Q^{\prime }S\cdot Q^{\prime }T=Q^{\prime }G\cdot Q^{\prime }H={\mathcal{P}}_{Q^{\prime }/(I)}.$$ Then $Q^{\prime }$ belongs to the radical axis of $(I),(O)$ which means $Q^{\prime }\in MN$ or $Q^{\prime }\equiv Q$. Thus $Q{K}^2={\mathcal{P}}_{Q/(I)}$ implies that $QK$ is the tangent line of $(I)$. Hence, $QK\perp KI$. Using the Brocard's theorem, we have $K$ is the orthocenter of triangle $OEF$ then $EO\perp FK$. Combining all these results, we get $EO\parallel IK$.

2) Two lines pass through $Q$ and intersect $(I)$ at $M,N$ and $G,H$ then $P$ is the intersection of $MG$, $NH$ which means $P$ belongs to the antipole of $Q$. Note that $QK$ is tangent to $(I)$ which means $K$ also belong to the antipole of the pole $Q$ respect to circle $(I)$. Then $PK$ is the antipole of $Q$ respect to the circle $(I)$. This implies that $PK\perp IQ$.