9. (12 points) As shown in the figure, quadrilateral $C D E F$ is a parallelogram. If the area of trapezoid $A B C D$ is 320, and the areas of triangle $A F H$ and triangle $G E B$ are 32 and 48, respectively, then the area of triangle $O C D$ is $\qquad$ .
【Solution】Solution: Let the height of the trapezoid be $h$, then the area of trapezoid $A B C D$ is $\frac{1}{2} \times(A B+C D) \times h=320$, and the area of parallelogram $C D E F$ is $C D \times h=240$.

Therefore, $(A B+C D): C D=640: 240=8: 3$, so $A B: C D=5: 3$. Draw $P N$ perpendicular to $C D$ at $P$, and perpendicular to $A B$ at $N$,

Since $C D / / A B$, triangles $O C D \sim$ triangles $O A B$, so $O P: O N=C D: A B=3: 5, O P: P N=3: 8$,

Therefore, the area of triangle $O C D$ is $\frac{1}{2} \times C D \times O P=\frac{1}{2} \times C D \times \frac{3}{8} \times P N=\frac{1}{2} \times \frac{3}{8} \times$ the area of parallelogram $C D E F=$ $\frac{1}{2} \times \frac{3}{8} \times 240=45$,

Thus, the answer is 45.