9. Let $B$ be a point on the circle centred at $O$ with diameter $A C$ and let $D$ and $E$ be the circumcentres of the triangles $O A B$ and $O B C$ respectively. Given that $\sin \angle B O C=\frac{4}{5}$ and $A C=24$, find the area of the triangle $B D E$.
9. Answer: 45
Let $d=A C=24$. First, it is not difficult to see that $\angle D E B=\angle A C B$ and $\angle E D B=$ $\angle C A B$, so that the triangles $D B E$ and $A B C$ are similar.

Let $M$ and $N$ be the feet of the perpendiculars from $B$ onto $D E$ and $A C$ respectively. As $M$ is the midpoint of $O B$, we have $B M=\frac{d}{4}$. Also $B N=B O \sin \angle B O C=\frac{d}{2} \times \frac{4}{5}=\frac{2 d}{5}$. Therefore $D E=A C \times \frac{B M}{B N}=d \times \frac{5 d}{8 d}=\frac{5 d}{8}$. Thus the area of the triangle $B D E$ is $\frac{1}{2} \times B M \times D E=\frac{1}{2} \times \frac{d}{4} \times \frac{5 d}{8}=\frac{5 d^{2}}{64}$. Substituting $d=24$, the area of the triangle $B D E$ is 45.