On the condition $\frac{\sin x_1}{\sin(\alpha+x_2-x_1)} = \frac{\sin x_2}{\sin(\alpha+x_3-x_2)} = \cdots = \frac{\sin x_n}{\sin(\alpha+x_1-x_n)}$

Question

If we have $$\frac{\sin x_1}{\sin(\alpha+x_2-x_1)} = \frac{\sin x_2}{\sin(\alpha+x_3-x_2)} = \cdots = \frac{\sin x_n}{\sin(\alpha+x_1-x_n)}.$$

Here $0<x_i<\frac{\pi}{2} (i=1,2,\cdots,n), 0<\alpha<\frac{\pi}{2}$.

Is $x_1=x_2=\cdots=x_n$ true? Prove it or give a counter example.

I think it is true, but can not prove it.