Let (sinA)/(sinB) = ( ( A - C ))/( ( C - B )), where A, B, C are...

Let $\dfrac{sinA}{a} = \dfrac{sinB}{b} = \dfrac{sinC}{c} = k ⇒ sinA = ak, sinB = bk \text{ and } sinC = ck$

As $A + B + C = \pi$

$⇒ B + C = \pi - A, ⇒ \sin \left(B + C\right) = \sin \left(\pi - A\right) = sinA$

Similarly, $\sin \left(A + C\right) = sinB$

$⇒ \dfrac{\sin \left(B + C\right)}{\sin \left(A + C\right)} = \dfrac{\sin \left(A - C\right)}{\sin \left(C - B\right)}$

$⇒ \sin \left(B + C\right)\sin \left(C - B\right) = \sin \left(A + C\right)\sin \left(A - C\right)$

$⇒ \sin^{2}C - \sin^{2}B = \sin^{2}A - \sin^{2}C$

$⇒ {c}^{2} - {b}^{2} = {a}^{2} - {c}^{2} ⇒ 2{c}^{2} = {a}^{2} + {b}^{2}$

therefore we can say that ${b}^{2}, {c}^{2}, {a}^{2}$ are in A.P.