Distance between orthocenter and circumcenter. 
Let $O$ and $H$ be respectively the circumcenter and the orthocenter of triangle $ABC$. Let $a$, $b$ and $c$ denote the side lengths. We are given that $a^2+b^2+c^2=29$ and the circumradius is $R=9$. We need to find $OH^2$.
I know that there is formula $OH^2=9R^2-(a^2+b^2+c^2)$, but I cannot use it unless I prove it.

I tried placing ABC triangle in a Cartesian plane and found coordinates of $O$ and $H$, but the expressions are not nice and I didn't manage to simplify them to required result.
Maybe vectors can be used? Please, note that I cannot use sophisticated vector knowledge in the solution. Any suggestions or ideas would be appreciated.


since
$$\angle HAO=|B-C|.OA=R,AH=2R\cos{A}$$
use cosin theorem we have
\begin{align*}
OH^2&=AH^2+AO^2-2AH\cdot AO\cos{|B-C|}
=4R^2\cos^2{A}+R^2-4R^2\cos{A}\cos{(B-C)}\\
&=5R^2-4R^2\sin^2{A}+4R^2\cos{(B+C)}\cos{(B-C)}\\
&=9R^2-4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C})\\
&=9R^2-(a^2+b^2+c^2)
\end{align*}
