Let $\frac{\sin \mathrm{A}}{\sin \mathrm{B}}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{C}-\mathrm{B})\text{'}}$ where A, B and C are angles of a ∆ABC. If the lengths of the sides opposite these angles are a b, and c respectively, then

Given, $\frac{\sin \mathrm{A}}{\sin \mathrm{B}}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{C}-\mathrm{B})}$

$\begin{array}{l}\Rightarrow \frac{\sin (\mathrm{B}+\mathrm{C})}{\sin (\mathrm{A}+\mathrm{C})}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{C}-\mathrm{B})}[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }]\\ \Rightarrow \sin (\mathrm{C}+\mathrm{B})\cdot \sin (\mathrm{C}-\mathrm{B})=\sin (\mathrm{A}+\mathrm{C})\cdot \sin (\mathrm{A}-\mathrm{C})\\ \Rightarrow {\sin }^2\mathrm{C}-{\sin }^2\mathrm{B}={\sin }^2\mathrm{A}-{\sin }^2\mathrm{C}\\ \Rightarrow 2{\sin }^2\mathrm{C}={\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}\\ \Rightarrow 2(2\mathrm{Rsin}\mathrm{C}{)}^2=(2\mathrm{Rsin}\mathrm{A}{)}^2+(2\mathrm{Rsin}\mathrm{B}{)}^2\\ \Rightarrow 2{\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2\left[\because \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{C}}{\sin \mathrm{C}}=2\mathrm{R}\right]\\ \left.\Rightarrow {\mathrm{a}}^2,{\mathrm{c}}^2,{\mathrm{b}}^2\text{ are in AP. }\right]\\ \Rightarrow {\mathrm{b}}^2,{\mathrm{c}}^2,{\mathrm{a}}^2\text{ are in AP }\end{array}$