Let $K=DE \cap BC$ and $P=AO \cap BC$. Since $I$ is the center of spiral similarity which takes $DE$ to $BC$, so $K \in \odot (CEI),\odot (BDI)$. Then $$\measuredangle OYI=\measuredangle EYI=\measuredangle EKI=\measuredangle DKI=\measuredangle DZI=\measuredangle OZI$$$$\text{and}$$$$\measuredangle OFI=\measuredangle AFI=\measuredangle ACI=\measuredangle ECI=\measuredangle EYI$$These two equalities give that $F,O,I,Y,Z$ are concyclic. So it suffices to show $G$ also lies on this circle. Fix $F$ (which automatically fixes point $G$), and animate $O$ linearly on line $AF$. Since $(B,C;P,K)=-1$, and $P$ is fixed, so that means that point $K$ is also fixed. Also, $O \mapsto E$ is a projective map. And, as $CEIK$ is cyclic, so inversion at $C$ gives that $E \mapsto I$ is also a projective map (Since after inversion at $C$, point $E$ still moves projectively on a line, while $I$ is now simply related to $E$ by a perspectivity at a fixed point $K$ to a fixed line). Thus, it suffices to show that $I,O,F,G$ are concyclic for $3$ positions of $O$ (the fact that $3$ is sufficient follows from inversion at $F$, since then $O$ and $I$ both move with degree $1$ on fixed lines, while now we are simply supposed to show a collinearity). For $O=A$, we have $I=A$ making the result obvious. Now consider $O=BG \cap AF$. Then $E=BG \cap AC$ and $I=\odot (ABC) \cap \odot (CEK)$. We first show that that $KCGE$ is cyclic. Note that we have $$-1=(B,C;P,K) \overset{A}{=} (B,C;F,\Gamma) \cap AK) \Rightarrow U \in AK$$This gives that $$\measuredangle GKC=\measuredangle CKU=\measuredangle CBU+\measuredangle BUK=\measuredangle GBC+\measuredangle BUA=\measuredangle EBC+\measuredangle BCE=\measuredangle GEC$$This means that $I,C,G,E,K$ are concyclic. Thus, we have $$\measuredangle OGI=\measuredangle EGI=\measuredangle ECI=\measuredangle ACI=\measuredangle AFI=\measuredangle OFI$$giving the desired result. Similarly, the result holds for $O=CG \cap AF$. $\blacksquare$ EDIT: There are some minor typos in the problem statement given above. It should be $I=\odot (ADE) \cap \Gamma$ and not $I=\odot (ADR) \cap \Gamma$. Also, the last word should be "concyclic" and not "cocyclic". Kindly correct these

Already posted before

math_pi_rate wrote: Let $K=DE \cap BC$ and $P=AO \cap BC$. Since $I$ is the center of spiral similarity which takes $DE$ to $BC$, so $K \in \odot (CEI),\odot (BDI)$. Then $$\measuredangle OYI=\measuredangle EYI=\measuredangle EKI=\measuredangle DKI=\measuredangle DZI=\measuredangle OZI$$$$\text{and}$$$$\measuredangle OFI=\measuredangle AFI=\measuredangle ACI=\measuredangle ECI=\measuredangle EYI$$These two equalities give that $F,O,I,Y,Z$ are concyclic. So it suffices to show $G$ also lies on this circle. Fix $F$ (which automatically fixes point $G$), and animate $O$ linearly on line $AF$. Since $(B,C;P,K)=-1$, and $P$ is fixed, so that means that point $K$ is also fixed. Also, $O \mapsto E$ is a projective map. And, as $CEIK$ is cyclic, so inversion at $C$ gives that $E \mapsto I$ is also a projective map (Since after inversion at $C$, point $E$ still moves projectively on a line, while $I$ is now simply related to $E$ by a perspectivity at a fixed point $K$ to a fixed line). Thus, it suffices to show that $I,O,F,G$ are concyclic for $3$ positions of $O$ (the fact that $3$ is sufficient follows from inversion at $F$, since then $O$ and $I$ both move with degree $1$ on fixed lines, while now we are simply supposed to show a collinearity). For $O=A$, we have $I=A$ making the result obvious. Now consider $O=BG \cap AF$. Then $E=BG \cap AC$ and $I=\odot (ABC) \cap \odot (CEK)$. We first show that that $KCGE$ is cyclic. Note that we have $$-1=(B,C;P,K) \overset{A}{=} (B,C;F,\Gamma) \cap AK) \Rightarrow U \in AK$$This gives that $$\measuredangle GKC=\measuredangle CKU=\measuredangle CBU+\measuredangle BUK=\measuredangle GBC+\measuredangle BUA=\measuredangle EBC+\measuredangle BCE=\measuredangle GEC$$This means that $I,C,G,E,K$ are concyclic. Thus, we have $$\measuredangle OGI=\measuredangle EGI=\measuredangle ECI=\measuredangle ACI=\measuredangle AFI=\measuredangle OFI$$giving the desired result. Similarly, the result holds for $O=CG \cap AF$. $\blacksquare$ EDIT: There are some minor typos in the problem statement given above. It should be $I=\odot (ADE) \cap \Gamma$ and not $I=\odot (ADR) \cap \Gamma$. Also, the last word should be "concyclic" and not "cocyclic". Kindly correct these Thanks for the correction!

AlastorMoody wrote: Already posted before Thanks for pointing that out! I tried the search function but it didn't work ( Ah I just checked the original statement, and it is different from what appears on the test (yeah I don't like the modification either XD but the intention was to make it easier I recall).