Let $H$ be an orthocenter in triangle $ABC$, and $O$ be the center of the circumscribed circle. We reflect $O$ with respect to the line $BC$ - we obtain the point $O'$. Consider the point $X$, which divides the segment $OH$ into $2$ segments in a ratio of $2$ to $1$, starting from point $H$. Prove that $X$ is the intersection point of the medians of the triangle $\Delta AOO'$.
I proved that $D$ is the center of $BC$ and the point $X$ is the centroid for the triangle $\Delta ABC$. I know about Euler's line but I can't solve this question. Any help?
This is trivial.
Since $X$ divide $HO$ in ratio $2:1$ point $X$ is actually gravity center of $ABC$, so $X$ lies on $AD$ and divide it in ratio $2:1$.
But $AD$ is median for triangle $AOO'$ so $X$ is also gravity center for $AOO'$.