# Prove that $G,O,D$ are collinear.

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.

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