Example $4 . \triangle \mathrm{ABC}$ has three sides $\mathrm{a}>\mathrm{b}>\mathrm{c}$ forming an arithmetic sequence, and $b=|A C|=2$, find the locus of vertex $\mathrm{B}$.

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Analysis: As shown in Figure 4, establish a coordinate system with the line of side $\mathrm{AC}$ as the $\mathrm{x}$-axis and the perpendicular bisector of $\mathrm{AC}$ as the $\mathrm{y}$-axis. From the given information, we have $a+c=2b=4$, which means the sum of the distances from the moving point $\mathrm{B}$ to the two fixed points A $(-1,0)$ and C $(1,0)$ is a constant. Therefore, the trajectory of point $\mathrm{B}$ is an ellipse with foci at $\mathrm{A}$ and $\mathrm{C}$, and the center at the origin. The length of the major axis is 4, the focal distance $|A C|=2$, and the length of the minor semi-axis is $\sqrt{3}$. Thus, the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. Since $a>\mathrm{b}>\mathrm{c}$, the trajectory of $\mathrm{B}$ is only the part of the ellipse to the left of the $\mathrm{y}$-axis, excluding the points $(0, \sqrt{3})$, $(-2,0)$, and $(0,-\sqrt{3})$.