Let $ABCD=M$, $WXYZ=P$ and $EFGH=Q$. We know that $AI\perp WZ$ then $WZ\parallel EH$, likewise $WX\parallel EF$,....... It is easy to see that $P$ and $Q$ are cyclical. $\angle HAI=\angle HDI=\angle GCI=90$ so $HAID$ and $GCID$ are cyclical. For cyclical sayings and also for $ZDY$ isosceles with bisector $DI$, we have $\angle AHI=\angle ADI=\angle IDC=\angle IGC$ (1), the equality of the first and last angle implies that $H,I,F$ and $G,I,E$ are collinear. Let (1)$=\phi$ then $\angle ADC=2\phi$ and $\angle ABC=180-2\phi$. We know that $BXIW$ is cyclical (<BWI=<BXI=90) then $\angle XIW=2\phi$ and $\angle WYX=\phi$, that is, $\angle WYX=\angle CGI\rightarrow WY\parallel EG$. We conclude that $\Delta WXY\sim\Delta EFG$ as well as $\Delta WZY\sim\Delta EHG$ both similarities have center of homothecy at the intersection of $EW$ with $GY$ (WY and EG are common sides of the respective similar triangles)

ricarlos wrote: Let $ABCD=M$, $WXYZ=P$ and $EFGH=Q$. We know that $AI\perp WZ$ then $WZ\parallel EH$, likewise $WX\parallel EF$,....... It is easy to see that $P$ and $Q$ are cyclical. $\angle HAI=\angle HDI=\angle GCI=90$ so $HAID$ and $GCID$ are cyclical. For cyclical sayings and also for $ZDY$ isosceles with bisector $DI$, we have $\angle AHI=\angle ADI=\angle IDC=\angle IGC$ (1), the equality of the first and last angle implies that $H,I,F$ and $G,I,E$ are collinear. Let (1)$=\phi$ then $\angle ADC=2\phi$ and $\angle ABC=180-2\phi$. We know that $BXIW$ is cyclical (<BWI=<BXI=90) then $\angle XIW=2\phi$ and $\angle WYX=\phi$, that is, $\angle WYX=\angle CGI\rightarrow WY\parallel EG$. We conclude that $\Delta WXY\sim\Delta EFG$ as well as $\Delta WZY\sim\Delta EHG$ both similarities have center of homothecy at the intersection of $EW$ with $GY$ (WY and EG are common sides of the respective similar triangles) I might be missing something but didn't u prove the statement for quadrilaterals and not general polygons?

Beautiful Problem!

n-gon problems are always super cool, i liked this one. For a n-gon $X$ let $X_1, X_2,..., X_n$ be its vertices, and in this case let $P_i$ lie in $M_iM_{i+1}$ and $M_{i+1}$ lies in $Q_iQ_{i+1}$. Let $(X)$ be the circumcircle of n-gon $X$ in case it existed. Since $(M)$ exists we have that $I$ has an isogonal conjugate in $Q$ (because of pedal circle), call it $I'$. Claim 1: $I'$ is the center of $(Q)$ (so $Q$ cyclic) Proof: By angle chase using the isogonal conjugates we have: $$\angle Q_{i+1}Q_iI'=\angle Q_{i-1}Q_iI=\angle M_iM_{i+1}I=\angle M_{i+2}M_{i+1}I=\angle Q_{i+2}Q_{i+1}I=\angle Q_iQ_{i+1}I'$$Therefore $I'Q_i=I'Q_{i+1}$ and by repeating this we get $Q$ cyclic and $(Q)$ has center $I$ as claimed. Claim 2: $Q_{i-1}M_iM_{i+1}Q_{i+1}$ is cyclic Proof: By angle chase using all the 90 degress which give cyclics we get: $$\angle Q_iQ_{i-1}Q_{i+1}=\frac{\angle Q_iI'Q_{i+1}}{2}=90-\angle Q_{i+1}Q_iI'=90-\angle M_iQ_iI=\angle Q_iIM_i=\angle Q_iM_{i+1}M_i$$Which is enough to verify the claim. Finishing: Note that $I$ is center of $(P)$ as well, now trivially since $Q_iQ_{i+1} \perp M_{i+1}I$ and $P_iM_{i+1}P_{i+1}I$ is a cyclic kite we have $P_iP_{i+1} \perp IM_{i+1}$ so $Q_iQ_{i+1} \parallel P_iP_{i+1}$, notice from the previous angle chase we get: $$\angle Q_{i+1}Q_iI'=\angle M_iM_{i+1}I=\angle P_iP_{i+1}I \implies \triangle Q_iI'Q_{i+1} \sim \triangle P_iIP_{i+1}$$And note that if we let $r_Q, r_P$ the radiuses of $(Q), (P)$ respectively then $\frac{r_P}{r_Q}$ is the similarity ratio for all similarities. Therefore all is left to prove is that $I'Q_i \parallel IP_i$ but this is trivial as by Claim 2 and anti-parallels we get $Q_iI' \perp M_iM_{i+1}$ but also $IP_i \perp M_iM_{i+1}$ is given, therefore with this parallel repeated cyclically we have that $P \cup I, Q \cup I'$ are homothetic thus we are done .