18.32 $\triangle A B C$ is a non-equilateral acute triangle, $O$ is a point inside $\triangle A B C$ and satisfies: $O A=O B=O C$. The symmetric points of $O$ with respect to $B C, C A, A B$ are $A^{\prime}, B^{\prime}$, $C^{\prime}$, respectively. Using any four points from $A, B, C, O, A^{\prime}, B^{\prime}, C^{\prime}$ to form a quadrilateral, the number of parallelograms among them is 
(A) 3.
(B) 6.
(C) 9.
(D) 12.
（2nd "Jinyun Cup" Junior High School Mathematics Invitational, 1985）
[Solution] From the given, $O$ is the circumcenter of $\triangle ABC$.
Also, from the given, among the quadrilaterals formed by the seven points in the figure, there are two types of parallelograms:
$\square A O B C^{\prime}, \square A O C B^{\prime}, \angle B O C A^{\prime}$, etc., a total of 3. It is also easy to know that the quadrilateral $B C B^{\prime} C^{\prime}$ is also a parallelogram, and there are 3 such figures (the other two are $\square A B A^{\prime} B^{\prime}, \square A C A^{\{\prime} C^{\prime}$). In addition, there are no other parallelograms. Therefore, the answer is $(B)$.