1. In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If the sizes of angles $A, B, C$ form a geometric sequence, and $b^{2}-a^{2}=a c$, then the radian measure of angle $B$ is equal to
$$
B=\frac{2 \pi}{7}
$$
1. 【Analysis and Solution】Solution 1: Since $A$, $B$, $C$ form a geometric sequence,
$\therefore$ we can assume $A=\frac{1}{q} B, C=q B$.
From $A+B+C=\pi$, we know $B=\frac{q \pi}{q^{2}+q+1}$.
Given $b^{2}-a^{2}=a c$.
Using the cosine rule $b^{2}=a^{2}+c^{2}-2 a c \cos B$,
we get $a c=c^{2}-a c \cos B$.
$$
\begin{array}{l}
\because c \neq 0, \\
\therefore a=c-2 a \cos B .
\end{array}
$$

Using the sine rule, we get $\sin A=\sin C-2 \sin A \cos B$,
thus $\sin A=\sin C-[\sin (A+B)+\sin (A-B)]$.
$$
\therefore \sin A=\sin (B-A) \text {. }
$$

Solving, we get $A=\frac{1}{2} B$,
$$
\therefore q=2, \quad B=\frac{2 \pi}{7} \text {. }
$$

Solution 2: Below is a purely plane geometry solution. Extend $CB$ to point $D$ such that $BD=AB=c$, and connect $AD$. Given $b^{2}-a^{2}=a c$, this means $AC^{2}=BC \cdot CD$, making it easy to see that $\triangle ABC \cong \triangle DAC$, hence $\angle ABC=\angle ADB+\angle DAB=2$ $\angle BAC$. The rest follows the original solution.