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

We are given the following relationship between the angles of a triangle $$ABC$$:

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

Also, in any triangle $$ABC$$, we know that:

• The sum of angles is $$A + B + C = \pi$$ (or 180°).
• From the Sine Rule: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$. This means that $$\sin A = \frac{a}{2R}$$, $$\sin B = \frac{b}{2R}$$, and so on.
• Since $$A + B + C = \pi$$:

• $$A = \pi - (B + C) \implies \sin A = \sin(\pi - (B + C)) = \mathbf{\sin(B + C)}$$
• $$B = \pi - (A + C) \implies \sin B = \sin(\pi - (A + C)) = \mathbf{\sin(A + C)}$$

Substituting these into the left side of the given equation:

$$\frac{\sin(B + C)}{\sin(A + C)} = \frac{\sin(A - C)}{\sin(C - B)}$$

Cross-multiply the terms:

$$\sin(B + C) \cdot \sin(C - B) = \sin(A + C) \cdot \sin(A - C)$$

We can use the standard trigonometric identity: $$\sin(X + Y)\sin(X - Y) = \sin^2 X - \sin^2 Y$$.

• Right Side: $$\sin(A + C) \sin(A - C) = \mathbf{\sin^2 A - \sin^2 C}$$
• Left Side: $$\sin(C + B) \sin(C - B) = \mathbf{\sin^2 C - \sin^2 B}$$

Now, set them equal:

$$\sin^2 C - \sin^2 B = \sin^2 A - \sin^2 C$$

Using the Sine Rule ($$\sin \theta \propto \text{side length}$$), we can replace the sine terms with their corresponding side lengths $$a, b,$$ and $$c$$:

$$c^2 - b^2 = a^2 - c^2$$

Rearranging the terms:

$$2c^2 = a^2 + b^2$$

$$c^2 - b^2 = a^2 - c^2$$

There for $$a^2,b^2\ and\ c^2$$ are in A.P.