6. In $\triangle A B C$, $\angle A$ is an obtuse angle, $O$ is the orthocenter, $A O=B C$,
then the value of $\cos (\angle O B C+\angle O C B)$ is
(A) $-\frac{\sqrt{2}}{2}$;
(B) $\frac{\sqrt{2}}{2}$;
(C) $\frac{\sqrt{3}}{2}$;
(D) $-\frac{1}{2}$.
6. (A)

Let the altitudes $AD$, $BE$, and $CF$ of $\triangle ABC$ intersect at point $O$. Since $\triangle ABC$ is an obtuse triangle, its orthocenter $O$ is outside $\triangle ABC$ (as shown in the figure).
$\because B, D, F, O$ are concyclic,

thus $\angle 1=\angle 2$.
Given $OA=BC$,
we know that Rt $\triangle OAF \cong \text{Rt} \triangle BCF$,
$\therefore OF=BF$.

Therefore, $\angle BOF=45^{\circ}$.
$$
\begin{array}{l} 
\text { and } \angle OBC + \angle OCB \\
=180^{\circ}-\angle BOF= 135^{\circ} . \\
\therefore \quad \cos (\angle OBC + \angle OCB)=\cos 135 \\
=-\frac{\sqrt{2}}{2} .
\end{array}
$$