[asy][asy] size(12cm); pair G = dir(270); pair F = dir(165); pair E = dir(105); pair H = dir(30); pair I = (0,0); pair A = extension(E,E+(E-I)*dir(90),F,F+(F-I)*dir(90)); pair B = extension(G,G+(G-I)*dir(90),F,F+(F-I)*dir(90)); pair C = extension(G,G+(G-I)*dir(90),H,H+(H-I)*dir(90)); pair D = extension(E,E+(E-I)*dir(90),H,H+(H-I)*dir(90)); pair P = extension(A,C,B,D); pair O = circumcenter(A,B,C); pair X = extension(A,B,C,D); pair Y = extension(A,D,B,C); pair P = extension(E,G,F,H); draw(unitcircle); draw(E--G); draw(F--H); draw(A--B--C--D--A--cycle); draw(D--X--A--Y--B); draw(circumcircle(A,B,C)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); dot("$I$",I,dir(45)); dot("$O$",O,dir(O)); dot("$P$",P,dir(P)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); [/asy][/asy] It suffices to show that $I$ lies on the radical axis of $(OEG)$ and $(OFH)$. Let $P$ be the intersection of $EG$ and $FH$. We have $EP\cdot PG=FP\cdot PH$, thus $P$ is of equal power to the two circles, and lies on the radical axis. We now wish to show that $O, I, P$ are collinear. Let $X=AB\cap CD$ and $Y=AD\cap BC$. Since $P$ lies on the polar of $X$ wrt the incircle of $ABCD$ and $P$ also lies on the polar of $Y$ wrt the incircle of $ABCD$, thus $P$ is the pole of $XY$, which means that $IP\perp XY$. It is well-known that $AC\cap BD=P$ (say, by application of Brianchon's theorem on $AEDCGB$ and $AFBCHD$). Then by Brokard's theorem, $OP\perp XY$. We are done.

We solve this problem by claims. Let $X$ be the Miquel point of quadrilateral $ABCD$. Let $R$ be the intersection of $AC$ and $BD$. Let $K=AB\cap CD$ and $L=AD\cap BC$. Claim. $X$ is the inverse of $R$ wrt $(O)$. Proof. Note that $OACX$ and $OBDX$ are cyclic. This is angle chasing, \begin{align*} \measuredangle AOC&=2\measuredangle ABC\\&=\measuredangle ABC+ \measuredangle DLC+\measuredangle AKD+\measuredangle LDK\\&=\measuredangle DLC+\measuredangle AKD\\&=\measuredangle DXC+\measuredangle AXD\\&=\measuredangle AXC. \end{align*}Similarly $OBDX$ is cyclic. By inversion wrt $(O)$, we get that $R=AC\cap BD$ maps to $X=(AOC)\cap (BOD)$. The claim follows. $\square$ Claim. $R$ lies on $FH$ and $EG$. Proof. This is true due Brianchon's theorem on hexagons $AEBCGD$ and $DFABHC$. $\square$ Claim. $\dfrac{AE}{EB}=\dfrac{DG}{GC}$. Proof. We use Law of Sines and obtain that $$\frac{AE}{EB}= \frac{AR\cdot \sin{\angle ARE}}{BR\cdot \sin{\angle BRE}}=\frac{DR\cdot \sin{\angle CRG}}{CR\cdot \sin{\angle DRG}}=\frac{DG}{GC}.\quad \square$$ Claim. $X$ is the inverse of $R$ wrt $(I)$. Proof. As $X$ is the center of spiral similarity sending $AB$ to $DC$, it maps $E$ to $G$ by the last claim. Hence, $$\triangle AEX\sim\triangle DGX\implies \measuredangle KGX=\measuredangle DGX=\measuredangle AEX=\measuredangle KEX,$$which implies that $X$ lies on $(EIGK)$. Similarly $X$ lies on $(FIHL)$. By inversion wrt $(I)$, we get that $R=EG\cap FH$ maps to $X=(EIG)\cap (FIH)$. The claim follows. $\square$ We conclude that $O,I,R$ are collinear. Now by radical axis theorem on $(EFGH),(OEG)$ and $(OFH)$, we get that $R$ lies on the radical axis of $(OEG)$ and $(OFH)$. Therefore, $I$ also lies on the radical axis of $(OEG)$ and $(OFH)$. Hence, $OI$ is perpendicular to the line connecting the centers of $(OEG)$ and $(OFH)$. [asy][asy] import olympiad;import geometry; size(14cm);defaultpen(fontsize(10pt)); pair I,E,F,G,H,A,B,C,D,K,L,R,O,X; I=(0,0);E=dir(95.22);F=dir(201.72);G=dir(259.30);H=dir(332.80);path w=circumcircle(E,F,G); A=intersectionpoint(perpendicular(E,line(I,E)),perpendicular(F,line(I,F))); B=intersectionpoint(perpendicular(E,line(I,E)),perpendicular(H,line(I,H))); C=intersectionpoint(perpendicular(G,line(I,G)),perpendicular(H,line(I,H))); D=intersectionpoint(perpendicular(F,line(I,F)),perpendicular(G,line(I,G))); K=extension(A,B,C,D);L=extension(A,D,B,C);R=extension(A,C,B,D); O=circumcenter(A,B,C);X=extension(K,L,I,R); draw(A--B--C--D--cycle,orange);draw(circumcircle(A,B,C),heavyblue);draw(w,heavyblue); draw(A--K--D--L--C,orange);draw(K--L,orange);draw(O--X,heavygreen);draw(A--C,heavyred);draw(B--D,heavyred);draw(E--G,heavyred);draw(F--H,heavyred); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); dot("$I$",I,dir(90)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); dot("$R$",R,dir(R)); dot("$O$",O,dir(O)); dot("$X$",X,dir(X)); [/asy][/asy]

Hmm I think it's quiet famous of the following properties:Given a quadrilateral which has incenter and circumcenter (i)Its two diagonals and the two lines connecting the tangent point wrt incircle of the opposite side are concurrent at one point. (ii)That point lies on $OI$