9. In $\triangle A B C, D$ is a point on $B C$ such that $A D$ bisects $\angle B A C$. If $A C=2, B D=2$ and $D C=1$, find $\cos \angle A B C$.
(1 mark)
在 $\triangle A B C$ 中, $D$ 是 $B C$ 上的一點, 使得 $A D$ 平分 $\angle B A C$ 。若 $A C=2 、 B D=2$而 $D C=1$, 求 $\cos \angle A B C$ 。
9. $\frac{7}{8}$
9. Produce $A D$ to $E$ so that $A B / / C E$. Then $\angle C A E=\angle B A D=$ $\angle C E A$ and so $C E=C A=2$. As $\triangle A B D \sim \triangle E C D$, we have $\frac{A B}{E C}=\frac{B D}{C D}$, or $\frac{A B}{2}=\frac{2}{1}$, which gives $A B=4$. Applying cosine law in $\triangle A B C$, we have $2^{2}=4^{2}+3^{2}-2(4)(3) \cos \angle A B C$ and hence $\cos \angle A B C=\frac{7}{8}$.