The triangle $ A_0 B_0 C_0$ is the image of triangle $ ABC$ via symmetry with origin $ O$. Hence, their orthocenters are image of each other via this transformation. This means: $ O$ is the midpoint of $ HH_0$.

sinankaral53 wrote: Given an acute angled triangle $ ABC$ , $ O$ is the circumcenter and $ H$ is the orthocenter.Let $ A_1$,$ B_1$,$ C_1$ be the midpoints of the sides $ BC$,$ AC$ and $ AB$ respectively. Rays $ [HA_1$,$ [HB_1$,$ [HC_1$ cut the circumcircle of $ ABC$ at $ A_0$,$ B_0$ and $ C_0$ respectively.Prove that $ O$,$ H$ and $ H_0$ are collinear if $ H_0$ is the orthocenter of $ A_0B_0C_0$ Turkish solution

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Denote $ G$ the centroid of triangle $ ABC$. Consider the homothety with center $ G$ which turns $ A_1\mapsto A$, $ B_1\mapsto B$, $ C_1\mapsto C$. Therefore, it maps triangle $ A_1B_1C_1$ to triangle $ ABC$. But, according to a very well- known result, we get that $ H, O, G$ are collinear (The Euler line), further, $ \overline {GO} = -\frac {1}{2}\overline {GH}$. Hence it also turns $ O\mapsto H$, which means $ O$ is also the orthocenter of triangle $ A_1B_1C_1$. Now, consider the homothety with center $ H$, ratio $ 2$, which obviously turns $ A_1\mapsto A_0$, $ B_1\mapsto B_0$, $ C_1\mapsto C_0$. Therefore, triangle $ A_1B_1C_1$ is mapped into triangle $ A_0B_0C_0$, also. Hence it turns $ O\mapsto H_0$, which leads to the result that $ H,O,H_0$ are collinear. Our proof is completed .