Lemma 1 (Well known) : Let $\omega$ and $\omega'$ be two circles which intersect each other at points $A , B$.Then for two arbitrary points $X,Y$ in the plane the quadrilateral $XABY$ is cyclic , if and only if we have : $$\frac{P_{\omega}{(X)}}{P_{\omega}{(Y)}}=\frac{P_{\omega'}{(X)}}{P_{\omega'}{(Y)}}$$ Lemma 2 :Let $ABCD$ be a cyclic quadrilateral which its diagonals $AC$ and $BD$ intersect at point $R$ and also $P , Q$ are intersection points of $AD , BC$ and $AB , CD$ respectively. So if we name foot of altitude from $P$ to $QR$ as point $H$ ; then quadrilaterals $ABHR$ and $CHRD$ are cyclic. Proof : Let $S$ be the Miquel point of quadrilateral $ABCD$ , then since $\angle ASQ + \angle ASP=\angle ABC + \angle ADC =180$ , points $P , S , Q$ are collinear and if $O$ be the center of circumcircle of $ABCD$ , one can see that : $$\angle ASC=\angle ASP - \angle PSB=\angle ABC - \angle ADC=180 - 2\angle ADC=180-\angle AOC$$Thus quadrilaterals $SAOC$ and $SBOD$ are cyclic and by concurrency of radical axes of circles , lines $AC , BD , OS$ are concurrent at point $R$. As the result , since $PQ$ is polar of $R$ wrt the circumcircle of $ABCD$ , we have $OS \perp PQ$ and $SPHR$ is cyclic too. So one can see that : $$QS.QP=QA.QB=QR.QH$$And we're done. Let $O$ be the center of circle $\omega'$ ( circumcircle of quadrilateral $ABCD$ ) and $O'$ the center of $\omega$ , also let $KL$ be a diameter of $\omega$ such that $EL$ and $EK$ are perpendecular to $FK$ and $EL$ at $N , M$ respectively and suppose that $AC$ and $BD$ are tangent to $\omega$ at $X , Y$. So if $P$ be the second intersection point of circumcircles of cyclic quadrilaterals $EMFN$ and $O'XFY$ , then we have $\angle EPF=\angle O'PF=90$ and points $O' , E , P$ are collinear. Now suppose that $F_1$ and $F_2$ be two points on lines $FL$ and $FK$ such that we have : $$(L,M;F_1,F)=(K,N;F_2,F)=-1$$Then we'll show that points $F_1 , F_2 , E$ are collinear and for that , we can get : $$\frac{LF_1}{MF_1}.\frac{EM}{EL}=\frac{NF_2}{KF_2}.\frac{EK}{EN} \iff \frac{LF}{MF}.\frac{EM}{EL}=\frac{NF}{KF}.\frac{EK}{EN}$$$$\triangle LME \sim \triangle LNF , \triangle KEN \sim \triangle KFM \implies \frac{LF}{MF}.\frac{EM}{EL}=\frac{NF}{KF}.\frac{EK}{EN}=\frac{NF}{MF}$$Thus these points are collinear and $E$ lies on the polar of point $F$ wrt the circle $\omega$ and as the result , points $X , Y , E$ are collinear. So note that we have $\angle EAX = \angle EBY$ , we can get $\triangle EAX \sim \triangle EBY$ and let $AC$ and $BD$ intersect the circumcircle of $EMFN$ for second time at points $X' , Y'$ respectively , thus while $ \angle EX'A=\angle EY'B=90$ one can see that : $$\frac{AX'}{AX}=\frac{BY'}{BY} \implies \frac{P_{\odot EMFN}{(A)}}{P_{\odot EMFN}{(B)}}=\frac{P_{\odot O'XFY}{(A)}}{P_{\odot O'XFY}{(B)}}$$Thus by lemma 1 , quadrilateral $ABFP$ and similary $CFPD$ are cyclic and by concurrency of radical axes , $AB , FP , CD$ are concurrent at a point $Q$. So by lemma 2 and since $O$ is the orthocenter of triangle $\triangle EFQ$ , $P$ is the intersection point of $EO$ and $FQ$ and as the result , points $O' , E , P , O$ are collinear. So we're done.

Let $AB\cap CD=G$ and denote the mid-point of $PQ$ as $M$. We know $GF\perp EO$ from brocard thus it's enough to show $EM\perp GF$. Let the intersections of $AC,BD$ with $\omega$ be $Y,X$ respectively and denote $K=XY\cap PQ$. $FE$ is the polar of $K$ hence $(K,E;X,Y)=-1$ which implies $(FK,FE;FD,FC)=-1 \implies G\in FK$. On the other hand, we have $FE\perp KM$ and $KE\perp FM$ thus $E$ is the ortho-center of $\triangle FKM$ and $FK\perp EM$ as desired.

Let $X$,$Y$ denote the touch points of $\omega$ with $AC$ and $BD$ Since $EFPQ$ are orthogonal quadruples polar of $F$ written $(PQ)$ passes through $E$. (Can be seen easily by brocard on $PQ$ and foot of perpendicular from $E$ and $F$). Also $XY$ is the polar of $F$ $\Rightarrow XYE$ is collinear. Notice both line $EXY$ and angle bisector of angle $BEA$ are perpendicular to angle bisector of $AFD$ (the latter comes from simple angle chase). Thus these two lines must coincides. Denote $M$,$N$ as midpoints of $AC$ and $BD$. Let $T_1$,$T_2$ be the projection of $E$ onto $AC$ and $BD$ Notice $EACMXT_1\cong EBDNYT_2$ Let perpendicular from $X$ to $AC$ intersect $EO$ at $Q_1$, perpendicular from $Y$ to $BD$ intersect $EO$ at $Q_2$. If we show $Q_1=Q_2$ we are done because this is the center of $\omega$. But $\frac{EQ_1}{Q_1O}=\frac{T_1X}{XM}=\frac{T_2Y}{YN}=\frac{EQ_2}{Q_2O}\Rightarrow Q_1=Q_2$ Q.E.D.

Denote $(ABCD)$ by $\Omega$. Let $I$ be the center of $\omega$ and $\omega$ touches $AC, BD$ at $X,Y$, respectively. Let lines $AD$ meet $IY$ at $N$ and $BC$ meet $IX$ at $M$. Let $R$ (possibly $D$) and $S$ (possibly $C$) be the points on $\Omega$ such that $BR,AS$ are diameters. Lines $DR,CS$ meet at $T$. First of all, we will prove the following well-known lemma. Lemma Let $ABC$ be a triangle with orthocenter $H$ and $\Gamma$ be the circle with diameter $BC$. Let the tangents from $A$ touches $\Gamma$ at $P,Q$, then we have $P,Q,H$ are collinear. Proof: Let $D,E,F$ be the feet of three altitudes drawn from $A,B,C$, respectively and $M$ be the midpoint of $BC$. Note that $\angle MDA=\angle MPA=\angle MQA=90^\circ$, then $A,P,D,M,Q$ are concyclic, called circle $\gamma$. Since $AH\cdot HD=BH\cdot HE$, then $pow(H,\Gamma)=pow(H,\gamma)$. But $\Gamma$ and $\gamma$ intersect at $P,Q$ and hence $PQ$ is their radical axis. Therefore, $P,Q,H$ are collinear. Back to the problem, since $F$ is orthocenter of $\triangle EPQ$, then $E$ is an orthocenter of $\triangle FPQ$. By the lemma, $X,E,Y$ are collinear. Since $\angle AXE=\angle BYE$ and $\angle XAE=\angle YBE$, then $\triangle AXE\sim\triangle BYE$. Also, $\triangle XME\sim\triangle YNE$. Hence, $$\frac{AE}{BE}=\frac{XE}{YE}=\frac{ME}{NE}$$Therefore, $A,B,N$ and $M$ are concyclic. Since $\angle NME=\angle BAE=\angle ECD$, then $MN\parallel CD$. From $IM\parallel CT$ and $IN\parallel TD$, then $\triangle IMN\sim\triangle TCD$. Consider $$\frac{TD}{IN}=\frac{CD}{MN}=\frac{DE}{NE}$$Therefore, $\triangle INE\sim\triangle TDE$ and then $I,E,T$ are collinear. Finally, use Desargues' therorem with $(ADRBCS)$ we have $E=AD\cap BC, T=DR\cap CS$ and $O=RB\cap SA$ are collinear and hence $OE$ pass through $I$ as desired.

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Dreamdream wrote: Denote $(ABCD)$ by $\Omega$. Let $I$ be the center of $\omega$ and $\omega$ touches $AC, BD$ at $X,Y$, respectively. Let lines $AD$ meet $IY$ at $N$ and $BC$ meet $IX$ at $M$. Let $R$ (possibly $D$) and $S$ (possibly $C$) be the points on $\Omega$ such that $BR,AS$ are diameters. Lines $DR,CS$ meet at $T$. First of all, we will prove the following well-known lemma. Lemma Let $ABC$ be a triangle with orthocenter $H$ and $\Gamma$ be the circle with diameter $BC$. Let the tangents from $A$ touches $\Gamma$ at $P,Q$, then we have $P,Q,H$ are collinear. Proof: Let $D,E,F$ be the feet of three altitudes drawn from $A,B,C$, respectively and $M$ be the midpoint of $BC$. Note that $\angle MDA=\angle MPA=\angle MQA=90^\circ$, then $A,P,D,M,Q$ are concyclic, called circle $\gamma$. Since $AH\cdot HD=BH\cdot HE$, then $pow(H,\Gamma)=pow(H,\gamma)$. But $\Gamma$ and $\gamma$ intersect at $P,Q$ and hence $PQ$ is their radical axis. Therefore, $P,Q,H$ are collinear. Back to the problem, since $F$ is orthocenter of $\triangle EPQ$, then $E$ is an orthocenter of $\triangle FPQ$. By the lemma, $X,E,Y$ are collinear. Since $\angle AXE=\angle BYE$ and $\angle XAE=\angle YBE$, then $\triangle AXE\sim\triangle BYE$. Also, $\triangle XME\sim\triangle YNE$. Hence, $$\frac{AE}{BE}=\frac{XE}{YE}=\frac{ME}{NE}$$Therefore, $A,B,N$ and $M$ are concyclic. Since $\angle NME=\angle BAE=\angle ECD$, then $MN\parallel CD$. From $IM\parallel CT$ and $IN\parallel TD$, then $\triangle IMN\sim\triangle TCD$. Consider $$\frac{TD}{IN}=\frac{CD}{MN}=\frac{DE}{NE}$$Therefore, $\triangle INE\sim\triangle TDE$ and then $I,E,T$ are collinear. Finally, use Desargues' therorem with $(ADRBCS)$ we have $E=AD\cap BC, T=DR\cap CS$ and $O=RB\cap SA$ are collinear and hence $OE$ pass through $I$ as desired. Excuse me, did you mean by Pascal's theorem on $(ADRBCS)$?