Let $O$ be the circumcenter of $\triangle{ABC}$, and $O$ does not lies on $AB$, $BC$ or $CA$. Let $D$, $E$, and $F$ be circumcenters of $\triangle{OBC}$, $\triangle{OCA}$, and $\triangle{OAB}$, respectively. Let $G$ be the the circumcenter of $\triangle{OEF}$. Prove that $D$, $O$, and $G$ are collinear.🔑
Proof: Obviously, $\triangle{ABC}$ cannot be right, otherwise, $O$ will be on its hypotenuse, and the radius of one of the circumcircles of $\triangle{OBC}$, $\triangle{OCA}$, and $\triangle{OAB}$ will be infinite.
Case 1: If $\triangle{ABC}$ is acute, then $O$ is inside of $\triangle{ABC}$.
Draw line $OA$, $OB$, $OC$, $DE$, $EF$, $FD$. Additionally, $OA$ and $EF$ intersect a $P$, $OB$ and $FD$ at $Q$, and $OC$ and $DE$ at $R$.
Draw line $OD$, $OE$, $OF$, and $OG$. And $OD$ and $BC$ intersect at $S$, $OE$ and $AC$ intersect at $T$.
Because $OA$ is the common chord of circle centered at $E$ and $F$, $EF$ is perpendicular bisector of $OA$, so we have $OP=AP=\dfrac{1}{2}OA$, $OP\perp EF$. Similarly, we have $OQ=BQ=\dfrac{1}{2}OB$, $OQ\perp FD$, and $OR=CR=\dfrac{1}{2}OC$, $OR\perp DE$.
Additionally, because $O$ is the circumcenter of $\triangle{ABC}$, we have $OA=OB=OC$, therefore $O$ is the incenter of $\triangle{DEF}$.
Beacause $OP\perp EF$ and $OQ\perp FD$, we have $\angle{PFQ}+\angle{POQ}=180^\circ$, i.e. $\angle{PFQ}=180^\circ-\angle{POQ}=180^\circ-\angle{AOB}$.
Because $O$ is the circumcenter of acute $\triangle{ABC}$, we have $\angle{AOB}=2\angle{ACB}$. Therefore, $\angle{PFQ}=180^\circ-2\angle{ACB}$
Because $O$ is the incenter of $\triangle{DEF}$, we have $$\angle{EFO}=\dfrac{1}{2}\angle{PFO}=\dfrac{1}{2}\angle{PFQ}=\dfrac{1}{2}(180^\circ-2\angle{ACB})=90^\circ-\angle{ACB}$$
Because $O$, $E$ and $F$ are on the circle centered at $G$, $\triangle{OGE}$ is isosceles. Therefore $$\angle{EOG}=\angle{OEG}=\dfrac{1}{2}(180^\circ-\angle{EGO})=\dfrac{1}{2}(180^\circ-2\angle{EFO})$$ $$=90^\circ-\angle{EFO}=90^\circ-(90^\circ-\angle{ACB})=\angle{ACB}$$
Because $BC$ is the common chord of circles centered at $O$ and $D$, $OD\perp BC$. Therefore $\angle{OSC}=90^\circ$. Similarly, $\angle{OTC}=90^\circ$. Therefore, $\angle{ACB}+\angle{SOT}=180^\circ$. i.e. $\angle{ACB}+\angle{DOE}=180^\circ$.
Because $\angle{ACB}=\angle{EOG}$, we have $\angle{EOG}+\angle{DOE}=180^\circ$, i.e $\angle{DOG}=180^\circ$. Therefore, $D$, $O$, and $G$ are collinear.
Case 2: If $\triangle{ABC}$ is obtuse, assume $\angle{ACB}>90^\circ$, then $O$ is outside of $\triangle{ABC}$, on the exterior side of $AB$.
Draw line $OA$, $OB$, $OC$, $DE$, $EF$, $FD$. Extend $FE$ to intersect $OA$ at $E’$. Extend $FD$ to intersect $OB$ at $D’$, and $FE$ to intersect $OA$ at $E’$. Additionally, $OC$ and $DE$ intersect at $R$.
Draw line $OD$,$OE$, $OF$, and $OG$.
Because $O$ is the circumcenter of $\triangle{ABC}$, with $\angle{ACB}>90^\circ$, we have $$\angle{AOB}=\angle{AOB}+\angle{AOC}=2\angle{BAC}+2\angle{ABC}=2(\angle{BAC}+\angle{ABC})=2(180^\circ-\angle{ACB})=360^\circ-2\angle{ACB}$$
Because $OA$ is the common chord of circles centered at $E$ and $F$, $EF$ is perpendicular bisector of $OA$, so we have $OE’=AD’=\dfrac{1}{2}OA$, $OE’\perp E’F$, i.e. $OA\perp E’F$. Similarly, $OD’=BD’=\dfrac{1}{2}OB$, $OD’\perp D’F$; $OR=CR=\dfrac{1}{2}OC$, $OR\perp DE$.
Additionally, because $O$ is the circumcenter of $\triangle{ABC}$, we have $OA=OB=BC$. Therefore, $O$ is the excenter opposite vertex $F$ of $\triangle{DEF}$, with tangent points at $D’$, $E’$, and $R$.
Therefore, $EE’$ and $ER$ are tangent to circle centered at $O$, $EE’=ER$. Because $OE’=OR$, we have $\triangle{EOE’}\cong\triangle{EOR}$. Therefore $\angle{EOE’}=\angle{EOR}$. Similarly, we have $\angle{DOD’}=\angle{DOR}$
Therefore, $$\angle{DOE}=\angle{DOR}+\angle{EOR}=\dfrac{1}{2}\angle{E’OR}+\dfrac{1}{2}\angle{D’OR}=\dfrac{1}{2}(\angle{E’OR}+\angle{D’OR})=\dfrac{1}{2}\angle{D’O’E}=\dfrac{1}{2}{AOB}$$
Because $OD’=OE’$, $OD’\perp D’F$, $OE’\perp E’F$’, we have $\triangle{D’OF}\cong \triangle{E’OF}$. Thus, $\angle{D’FO}=\angle{E’FO}$ and $\angle{OD’F}=\angle{OE’F}$, i.e. $OF$ is angle bisector of both $\angle{D’FE’}$ and $\angle{AOB}$.
Because $O$, $E$, and $F$ are on the circle centered at $G$, $\triangle{EOG}$ is isosceles. Therefore $$\angle{GOE}=\angle{GEO}=\dfrac{1}{2}(180^\circ-\angle{OGE}) =\dfrac{1}{2}(180^\circ-2\angle{EFO})=90^\circ-\angle{EFO}=90^\circ-\angle{E’FO}$$
Because $OE’\perp E’F$’, $\triangle{OE’F}$ is right, therefore $\angle{E’OF}=90^\circ-\triangle{E’FO}$ Therefore $$\angle{GOE}=\angle{E’OF}$$
Because $\angle{E’FO}=\angle{AOF}$, and $OF$ is angle bisector of $\angle{AOB}$, we have $\angle{GOE}=\dfrac{1}{2}\angle{AOB}$. Thus, $\angle{DOE}=\angle{GOE}$. Therefore, $D$, $O$, and $G$ are collinear.
Combining Case 1 and 2, we proved that $D$, $O$, and $G$ are collinear, if $\triangle{ABC}$ is not right.