Example 1 In $\triangle ABC$, 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 $b^{2}-a^{2}=ac$, then the radian measure of angle $B$ is equal to what?
(1985 National High School League Question)
As shown in Figure 3-5, draw $CD \parallel AB$ intersecting the circumcircle of $\triangle ABC$ at $D$, and connect $AD$. Then quadrilateral $ABCD$ is an isosceles trapezoid. By Ptolemy's theorem, we have $b^2 = a^2 + c \cdot CD$.
Given $b^2 = a^2 + c \cdot a$, then $CD = a$, thus
$AD = DC = CB$, which means $\overparen{ADC} = 2 \overparen{BC}$, or $\angle B = 2 \angle BAC$.
Furthermore, in $\triangle ABC$, the angles $A, B, C$ form a geometric sequence, so the common ratio $q = \frac{\angle B}{\angle A} = 2$, thus $\angle A + \angle B + \angle C = \angle A + 2 \angle A + 4 \angle A = 7 \angle A = \pi$, hence $\angle A = \frac{\pi}{7}, \angle B = \frac{2 \pi}{7}$ is the solution.