If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$ 
If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
  show that 
  $$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$
where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c$ are sides of $\triangle ABC$.

Actually, I don't have any idea. Please someone help. It would be great if someone who answers include how he/she found where to begin. 

Hint:
$$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}=\frac{(b+c)+(c+a)+(a+b)}{(2k-1)+(2k)+(2k+1)}=\frac{a+b+c}{3k}$$
$$\frac{b+c}{2k-1}=\frac{a+b+c}{3k}=\frac{(a+b+c)-(b+c)}{(3k)-(2k-1)}$$
You can easily find $a:b:c$ and apply the sine formula.
