geometryprojective-geometry
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.
Do an inversion centered at $A$ with radius $AH$. The image of $P$ is $B$ and the image of $Q$ is $C$. Denote by $O'$ the image of $O$ under that inversion. Since $AH^2 = AO \cdot 2AO$, we get $AO' = 2AO$, that is, $AO'$ is the diameter of the circumcircle of $ABC$. Since $A, B, C$ and $O'$ lie on a circle passing through the center of inversion, the images of $B, C$ and $O'$ lie on a line. These images are $P, Q$ and $O$.
Let $R$ be the another point (different of $P$) where the $CP$ meets the circumcircle of $APQ$, let $F$ be the point where $BQ$ and $CP$ meets. Since $A$, $P$, $R$ and $Q$ lies in the same circumference it follows $$\measuredangle CRQ =180^{\circ}-\measuredangle PRQ=\measuredangle PAQ=\measuredangle BAC$$ Also, $\measuredangle BAC=\measuredangle PCB=\measuredangle RCB$ for construction, so $\measuredangle CRQ=\measuredangle RCB$ and $QR$ is parallel to $BC$. Then $\angle QRF=\angle FCB=\angle BAC=\angle QBC =\angle FQR$, i.e. $\triangle FBC$ and $\triangle FQR$ are isosceles. It follows that the bisector of $BC$ is perpendicular to $BC$ and passes by $F$, the same for the bisector of $QR$, so the bisector of $BC$ is the bisector of $QR$ too, and it is the line by the circumcircles of $\triangle ABC$ and $\triangle BPQ$
Best Answer
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