Question

$a\left(\cos B\cos C+\cos A\right)=b\left(\cos C\cos A+\cos B\right)=c\left(\cos A\cos B+\cos C\right)$

Answer 1

Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Consider:
$$\begin{gathered} a\left(\cos B\cos C+\cos A\right) \\ =k\sin A\left(\cos B\cos C+\cos A\right) \\ =k\left(\sin A\cos B\cos C+\cos A\sin A\right) \\ =k\left[\frac{1}{2}\cos C\left\{\sin \left(A+B\right)+\sin \left(A-B\right)\right\}+\sin A\cos A\right] \\ =k\left[\frac{1}{2}\left\{\sin \left(A+B\right)\cos C+\sin \left(A-B\right)\cos C\right\}+\sin A\cos A\right] \\ =k\left[\frac{1}{2}\left\{\frac{1}{2}\left[\sin \left(A+B+C\right)+\sin \left(A+B-C\right)+\sin \left(A-B+C\right)+\sin \left(A-B-C\right)\right]\right\}+\sin A\cos A\right] \\ =k\left[\frac{1}{4}\left\{\mathrm{sin\pi }+\sin \left(\mathrm{\pi }-2\mathrm{C}\right)+\sin \left(\mathrm{\pi }-2\mathrm{B}\right)-\sin \left(\mathrm{\pi }-2\mathrm{A}\right)\right\}+\frac{\sin 2\mathrm{A}}{2}\right]\left(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right) \\ =\frac{k}{4}\left(\sin 2C+\sin 2B+\sin 2A\right)....\left(1\right) \\ \mathrm{and} \\ b\left(\cos A\cos C+\cos B\right) \\ =k\left(\sin B\cos A\cos C+\mathrm{sinBcos}B\right) \\ =k\left[\frac{1}{2}\cos A\left\{\sin \left(B+C\right)+\sin \left(B-C\right)\right\}+\frac{\sin 2B}{2}\right] \\ =k\left(\frac{1}{2}\left(\sin \left(B+C\right)\cos A+\sin \left(B-C\right)\cos A\right)+\frac{\sin 2B}{2}\right) \\ =k\left(\frac{1}{4}\left(\sin \left(B+C+A\right)+\sin \left(B+C-A\right)+\sin \left(B-C+A\right)+\sin \left(B-C-A\right)\right)+\frac{\sin 2B}{2}\right) \\ =\frac{k}{4}\left(\mathrm{sin\pi }+\sin \left(\mathrm{\pi }-2\mathrm{A}\right)+\sin \left(\mathrm{\pi }-2\mathrm{C}\right)-\sin \left(\mathrm{\pi }-2\mathrm{B}\right)+\frac{\sin 2\mathrm{B}}{2}\right)\left(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right) \\ =\frac{k}{4}\left(\sin 2A+\sin 2C+\sin 2B\right)...\left(2\right) \\ \mathrm{Similarly}, \\ c\left(\cos A\cos B+\cos C\right)=\frac{k}{4}\left(\sin 2A+\sin 2B+\sin 2C\right)...\left(3\right) \end{gathered}$$


From (1), (2) and (3), we get:

$a\left(\cos B\cos C+\cos A\right)=b\left(\cos C\cos A+\cos B\right)=c\left(\cos A\cos B+\cos C\right)$

Hence proved.