${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$ As A, B, C are angles of triangle. A + B + C = $\pi$ A = $\pi$ $-$ (B + C) ...... (1) Similarly sinB = sin(A + C) ..... (2) From (1) and (2) ${{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \over {\sin (C - B)}}$ $\sin (C + B).\sin (C - B) = \sin (A - C)\sin (A + C)$ ${\sin ^2}C - {\sin ^2}B = {\sin ^2}A - {\sin ^2}C$ $\because$ $\{ \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\}$ $2{\sin ^2}C = {\sin ^2}A + {\sin ^2}B$ By sine rule $2{c^2} = {a^2} + {b^2}$ $\Rightarrow$ b 2 , c 2 and a 2 are in A.P.