What is the non-trivial, general solution of these equal ratios? Provide non-trivial solution of the following:

$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$$

$a=?, b=?, c=?$
The solution should be general.

Besides to the good answers posted for this question (this answer and this one), I think finding eigenvalues and eigenvectors of such system will help you to find the solution.
$$\begin{array}{l}\frac{a}{{b + c}} = \frac{b}{{c + a}} = \frac{c}{{a + b}} = t\\\left[ {\begin{array}{*{20}{c}}1&{ - t}&{ - t}\\{ - t}&1&{ - t}\\{ - t}&{ - t}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\\\left[ C \right]\left\{ X \right\} = \left\{ 0 \right\}\\\det \left[ {C - \lambda I} \right] = 0\\\det \left[ {\begin{array}{*{20}{c}}{1 - \lambda }&{ - t}&{ - t}\\{ - t}&{1 - \lambda }&{ - t}\\{ - t}&{ - t}&{1 - \lambda }\end{array}} \right] = 0\\\left( {1 - \lambda } \right)\left\{ {{{\left( {1 - \lambda } \right)}^2} - {{\left( { - t} \right)}^2}} \right\} = 0\\\lambda  = 1,1 - t,1 + t\\\begin{array}{*{20}{c}}{\lambda  = 1}\\{\lambda  = 1 - t}\\{\lambda  = 1 + t}\end{array}\left| {\begin{array}{*{20}{c}}{ + \left( {1 - \lambda } \right)a - \left( t \right)b - \left( t \right)c = 0}\\{ - \left( t \right)a + \left( {1 - \lambda } \right)b - \left( t \right)c = 0}\\{ - \left( t \right)a - \left( t \right)b + \left( {1 - \lambda } \right)c = 0}\end{array}} \right\rangle \left\{ {\left. {\begin{array}{*{20}{c}}{a = b = c = 0}\\{a = b = c = 0}\\{a + b + c = 0}\end{array}} \right\rangle } \right.\left. {\underline {\, 
 {a + b + c = 0} \,}}\! \right| \end{array}$$
As it seems the answer is compatible with the other answers to this question. 
