$20 \cdot 44$ Given that $H$ is the orthocenter of $\triangle ABC$, the circumradius of $\triangle ABC$ is $R$, and the circumradius of $\triangle BHC$ is $r$. Then the relationship between $R$ and $r$ is
(A) $R=r$.
(B) $R>r$.
(C) $Rr ; \triangle ABC$ is an obtuse triangle, $R$ $<r$.
(“Zu Chongzhi Cup” Junior High School Mathematics Invitational Competition, 1995)
[Solution] As shown in the figure, $\angle 1 + \angle 2 + \angle B H C = 180^{\circ}$,
$$
\begin{array}{l}
\text { and } \angle F C B + \angle 1 = 90^{\circ}, \angle E B C + \angle 2 = 90^{\circ}, \\
\text { also } \angle B A C + \angle A B C + \angle A C B = 180^{\circ}, \\
\text { thus } \angle B H C = 180^{\circ} - \angle B A C, \\
\text { in } \triangle A B C, B C = 2 R \sin A. \\
\text { in } \triangle B H C, B C = 2 r \sin \left(180^{\circ} - \angle B A C\right) \\
= 2 r \sin A, \\
\end{array}
$$

Therefore, $R = r$.
Hence, the answer is $(A)$.