$O$ is the circumcenter of triangle $ABC$, whereas $G$ is the centroid and $H$ is the orthocenter. $R$ denotes the circumradius. How can I prove the following relations:
- $OH^2=9R^2-(a^2+b^2+c^2)$.
- $OG=\frac13 OH=2 HG$
I honestly don't have any ideas. Any hints would appreciated :)
Assuming second, you can prove your first result by vectors. Take origin at circumcentre. hence, $\vec {OH}=\vec a+\vec b+\vec c$ and $\vec {OG}=\vec{GH}/3$
$|\vec a+\vec b+\vec c|^2=\vec{OH}.\vec{OH}$
Note that $|\vec a|=R$ and same with other two and angles between them is twice of angle subtended at vertex(remember theorem from circles?)
$OH^2=R^2(3+2(\sum\cos 2A))=R^2(1-8\prod \cos A)$. You should be able to simplify now.
If you want a proof for centroid one, google Ceva's theorem.