In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$$
$$\dfrac {\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+b^2-c^2}{2ab}}{\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+c^2-b^2}{2ac}}=\dfrac {b}{c}$$
On simplification,
$$\dfrac {ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3}{ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3}=\dfrac {b}{c}$$

Alternate approach:
Cross multiply
$\cos A \sin C + \sin 2C = \sin B \cos A + \sin 2B$
$\cos A (\sin C - \sin B) - (\sin 2B - \sin 2C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} - 2* \sin (B-C) \cos(B+C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2* \sin (B-C) \cos A=0$
$2\cos A [\sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2 \sin \frac{C-B}{2} \cos \frac{B-C}{2}] = 0$
$2\cos A *\sin \frac{B-C}{2}[\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2}]=0$ 
IF:
$\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2} =0 \implies \tan \frac B2 \tan \frac C2 = -\frac 13$ and $\frac B2<90 $ and $\frac C2 < 90$ 
