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. 

Let $m$ be the common value of the initial fractions, so that
$$\begin{align}
a + b &= ( 2 k + 1 ) m \tag{1} \\
b + c &= ( 2 k - 1 ) m \tag{2}\\
c + a &= ( 2 k \phantom{+1\;\,}) m \tag{3}
\end{align}$$
We can combine these equations to get
$$\begin{align}
\phantom{-}(1)-(2)+(3):&\quad 2 a = 2(k+1)m \\
(1)+(2)-(3):&\quad 2b= 2(k\phantom{+1\;\,})m \\
-(1)+(2)+(3):&\quad 2c= 2(k-1)m
\end{align} \tag{4}$$
By the Law of Sines, 
$$\sin A : \sin B : \sin C = a : b : c = k + 1 : k : k - 1 \tag{5}$$
and the result follows. $\square$ 
