Point $K$ lies on the circumcircle of a rectangle $ABCD$. Line $CK$ intersects line segment $AD$ at point $M$ so that $AM:MD=2$. $O$ is the center the rectangle. Prove that the centroid of triangle $OKD$ belongs to the circumcircle of triangle $COD$. (Author: V. Shmarov)
Let $N$ be the midpoint of $DK$, $G$ is the centroid of $\triangle ODK$. Let $AK$ intersect $CD$ at $E$. Then $\angle EAD=\frac12\angle KOD=\angle NOD$, so $\triangle EAD\sim\triangle DON$. Thus $\angle MGD=\angle EMD=\angle DCO$, since $M$ is the orthocenter of $\triangle ACE$. Thus $GOCD$ is cyclic.