$7 \cdot 36$ 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) $-\frac{1}{2}$.
(China Junior High School Mathematics League, 1993)
[Solution] From the given, we know $\angle A D B=\angle A F O=90^{\circ}$, and $\angle B A D=\angle O A F$, hence
$$
\angle A O F=\angle A B C .
$$

In Rt $\triangle O A F$ and Rt $\triangle B C F$, $A O=B C, \angle A O F=\angle F B C$, then $\triangle O A F \cong \triangle B C F$. Therefore, $B F=F O$.
That is, Rt $\triangle B F O$ is an isosceles right triangle.
Thus,
$$
\begin{array}{l}
\cos (\angle O B C+\angle O C B) \\
=-\cos [\pi-(\angle O B C+\angle O C B)] \\
=-\cos \angle B O C=-\frac{\sqrt{2}}{2} .
\end{array}
$$

Therefore, the answer is $(A)$.