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. 

\begin{align} 
\text{If }\quad
\frac{b+c}{2k-1}&=\frac{c+a}{2k}=\frac{a+b}{2k+1}
\tag{1}\label{1}
,\\
\text{show that }\quad
\frac{\sin A}{k+1}&=\frac{\sin B}{k}=\frac{\sin C}{k-1}
\tag{2}\label{2}
.
\end{align}
From \eqref{1} we have
by 
the rules based on componendo and dividendo,
\begin{align} 
\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{2a}{2k+2}
=
\frac{a}{k+1}
\\
&=
\frac{(b+c)-(c+a)+(a+b)}{(2k-1)-(2k)+(2k+1)}
=
\frac{2b}{2k}
=
\frac{b}{k}
\\
&=
\frac{(b+c)+(c+a)-(a+b)}{(2k-1)+(2k)-(2k+1)}
=
\frac{2c}{2k-2}
=
\frac{c}{k-1}
.
\end{align}
Now we have
\begin{align}
\frac{a}{k+1}&= 
\frac{b}{k}=
\frac{c}{k-1}
\tag{3}\label{3}
.
\end{align}
Recall that by the sine rule
for $\triangle ABC$
with the radius of circumscribed circle $R$
\begin{align}
a&=2R\sin A,\quad
b=2R\sin B,\quad
c=2R\sin C,
\end{align}
so \eqref{3} becomes
\begin{align}
\frac{2R\sin A}{k+1}&= 
\frac{2R\sin B}{k}=
\frac{2R\sin C}{k-1}
\end{align}
and \eqref{2} follows.
