A,B,C's can be difficult

$I\quad have\quad a\quad solution\quad with\quad using\quad just\quad this:\quad \frac { a-b }{ c } =\frac { a-c }{ b } \\ \frac { a-b }{ c } =\frac { a-c }{ b } \\ (a-b)b\quad =\quad (a-c)c\\ { b }^{ 2 }-{ c }^{ 2 }=ab-ac\\ (b-c)(b+c)=a(b-c)\\ \boxed { b+c=a } \\ \\ And\quad now:\\ \frac { a }{ a+b+c } =\frac { a }{ 2a } =0.5$

$Say ~ K= \dfrac{a-b}{c} = \dfrac{b+c}{a} = \dfrac{a-c}{b}\\We ~ use~ ratio~ property~k=\dfrac m n =\dfrac p q,~then ~ k=\dfrac{m+p}{n+q}\\Adding ~ First~ two~ K=\dfrac{a+c}{c+a}=1.~~Adding~ all ~three~K=\dfrac{2a}{a+b+c} \\\therefore~\dfrac{a}{a+b+c}=\dfrac K 2 = \color{#D61F06}{0.5}.$

I started this problem by giving values to two out of the three variables. Leaving the last as "x"

Example: Let a=2 b=1 and c=x

Which would give you: $\frac {2-1}{x} = \frac {1+x}{2}=2-x$

But we would only need one pair of expressions to solve for x namely: $\frac {1+x}{2}=2-x.$

Solving for x you would get x=1, which gives us a=2, b=1, and c=1.Which would be true for the conditions given.

Now, if you subtitute the three values into the expression given you would get: $\frac{2}{2+1+1}=\frac{1}{2}$

or in other words: 0.5

c=2a , b=-a

a/(a+b+c)=a/(a-a+2a)=1/2 or 0.5