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Let $ABCD$ be a convex quadrilateral with $\angle DAB=\angle ABC=\angle BCD$. Let $G$ and $O$ be the centroid and circumcenter of $\triangle ABC$ respectively. Prove that $G,O,D$ are collinear.

Attempt: Let the line through $A\|BC$ intersect $DC$ at $E$ and the line through $C\|AB$ intersect $AD$ at $F$. Now after some trivial angle chasing we get $B,A,E,F,C$ are concyclic.

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Based on your notations and what you have: $$B,A,E,F,C~{\rm are~concyclic~on~the~circumcircle~}(ABC)~{\rm with~center~}O,$$ $$AB\parallel FC~{\rm and~}AE\parallel BC.$$ Let $H$ be the orthocenter of triangle $ABC$. Let $C'$ be the alternative intersection of $CH$ with $(ABC)$. Let $A'$ be the alternative intersection of $AH$ with $(ABC)$. Then $AA'$ (resp. $CC'$) being orthogonal transversal of parallel lines $AE$ and $BC$ (resp. $AB$ and $FC$), one has $$\angle A'AE=90^\circ~({\rm resp.}~\angle C'CF=90^\circ).$$ It follows that both $A'E$ and $C'F$ are diameters of $(ABC)$, so $A'E\cap C'F=O$. Now the six points $C',A,E,A',C,F$ on circle $(ABC)$ satisfy $$CC'\cap AA'=H,C'F\cap EA'=O,AF\cap EC=D.$$ So by Pascal's theorem, $H,O,D$ are collinear. Since $HO$ coincides with the Euler line $GO$, the same is true for $G,O,D$. QED

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