we have that $ \sin (C-A)\sin (A+C)=\sin^2C-\sin^2A=\sinh\sin B$... this implies that $ \sin(C-A)=\sin A$, and it can be easily checked that this implies that $ C=2A$, and we're done...

campos wrote: we have that $ \sin (C-A)\sin (A+C)=\sin^2C-\sin^2A=\sinh\sin B$... this implies that $ \sin(C-A)=\sin A$, and it can be easily checked that this implies that $ C=2A$, and we're done... $\sin A\sin(A+C)=\sin^2C-\sin^2A$ is correct, rather. Not sure how you arrived at the correct answer.. We have that $ab=c^2-a^2=b^2-2ab\cos C\implies a=b-2a\cos C$ so $\sin A=\sin (A+C)-2\sin A\cos C=\cos A\sin C - \sin A\cos C=\sin (C-A)$. So either $C=2A$ or $C=\pi$. In the latter case we have $ab=a^2+b^2-a^2=b^2\implies a=b$ so again $C=\pi=2\cdot\frac{\pi}{2}=2A$.

$\sin(C-A)\sin(A+C) = \sin^2 C - \sin^2 A$ is correct, it's a well known identity. He just forgot to write $\sin A$ in $\sin^2 C - \sin^2 A = \sin B \sin A$ ( which comes from dividing by $4R^2$ the condition of the problem).