9. (This question is worth 16 points) In $\triangle A B C$, $B C=a, C A=b, A B=c$. If $b$ is the geometric mean of $a$ and $c$, and $\sin A$ is the arithmetic mean of $\sin (B-A)$ and $\sin C$, find the value of $\cos B$.

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Solution: Since $b$ is the geometric mean of $a$ and $c$, there exists $q>0$ such that
$$
b=q a, c=q^{2} a.
$$

Since $\sin A$ is the arithmetic mean of $\sin (B-A)$ and $\sin C$, we have
$$
2 \sin A=\sin (B-A)+\sin C=\sin (B-A)+\sin (B+A)=2 \sin B \cos A.
$$
4 points
Combining the sine and cosine laws, we get
$$
\frac{a}{b}=\frac{\sin A}{\sin B}=\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c},
$$

which simplifies to $b^{2}+c^{2}-a^{2}=2 a c$.
8 points
Substituting (1) and simplifying, we find $q^{2}+q^{4}-1=2 q^{2}$, or $q^{4}=q^{2}+1$, so
$$
q^{2}=\frac{\sqrt{5}+1}{2}.
$$

Thus,
$$
\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 a c}=\frac{q^{4}+1-q^{2}}{2 q^{2}}=\frac{1}{q^{2}}=\frac{\sqrt{5}-1}{2}.
$$
16 points