What a beautiful result! By Newton's Theorem, $AC,BD,EG,FH$ are concurrent. Let $T$ be the point of intersection. Let $O$ be the centre of $\omega$. We can use Pascal on $EEHFFG, HHEGGF$ to show that $D,T,B,S=EH\cap FG$ are concurrent. $IG, FJ, AC$ are concurrent iff $AC$ is the polar of $Y=IJ\cap FG$. Now $AC$ is the polar of $S$ so $IG,FJ,AC$ are concurrent iff $Y=S$ iff $IJ, FG, EH$ are concurrent. We can now use Pascal twice (find the hexagons yourself hahaha) to show that $DA\cap BC, AB\cap CD, S$ are collinear. It is now obvious due to the complete quadrilateral that $(D,B;T,S)=-1$. Take perspective through $X$ so $(J,I; U=AC\cap IJ, V=IJ\cap BD)=-1$. Using Pascal again (find the hexagons yourself) we get $HI%Error. "capEJ" is a bad command. , JH\cap IE, A$ are collinear so we get another complete quadrilateral so $(J,I;U,W=EH\cap IJ)=-1$. So $V=W$; thus $IJ,FG,EH$ are concurrent. Done!

I think this works: Proof 1: Construct the tangents to the incircle at $I$ and $J$. Then these two points intersect at a point $Y$ collinear with $A$ and $X$ (Notice the tangential quadrilateral with sides containing $AE$, $AH$, $YI$, $YJ$). By Newton's theorem on the tangential quadrilateral containing the tangents through $I$, $F$, $G$, $J$, lines $FJ$, $IG$, and $CY = AC$ are concurrent. $\Box$ Proof 2: By Pascal's Theorem, $X = EJ \cap IH$, $Y = EG \cap FH$, and $Z = IG \cap FJ$ are collinear. Also, $Y$ lies on $AC$ by Newton's theorem, and thus, $YX = AC$ and $Z\in AC$. $\Box$

Lemma: Let $P$ be the Pole of $AC$ wrt $\omega$, If a line through $P$ intersects $\omega$ at $I,J$, then $BI\cap DJ, BJ\cap DI\in AC$ Proof: WLOG $(B,I), (D,J)$ are on the same side of $AC$. Since $P=HE\cap FG$(The polar of A,C), and we can use Menelaus to show that $BD,HE,FG$ concur, hence $P\in BD$. By properties of poles and polar $(P,AC\cap IJ;I,J)=-1,(CP,CA;CB,CD)=-1$ $\Rightarrow (P,AC\cap BD;B,D)=-1$. Hence $DJ\cap BI,BJ\cap DI$ lies on the line formed by $AC\cap IJ, ,AC\cap BD$, which is $AC$. The converse of this lemma is also true: If $X$ lies on $AC$, and $BX, DX\cap $incircle$=I,J$, then $J,I,P$ are collinear. Proof: Let $PI\cap \omega J'$ again, by the lemma $BI\cap DJ'\in AC$ or $=X\Rightarrow J\cap J'$. Now the original problem should be obvious, since $IG\cap JF$lies on the polar of $P$, which is $AC$.

Just make projective transfomation which sends circle into circle and move some point to infinity.