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. 
 A: 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.
A: 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. $\square$ 
A: \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.
