Đặt $\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=k$ (1)
\Rightarrow a=k(b+c); b=k(a+c); c=k(a+b).
\Rightarrow: $\frac{a+b+c}{b+c+a+c+a+b}=\frac{k(b+c+a+c+a+b)}{2(a+b+c)}=k$ (2)
Từ (1) và (2)\Rightarrow $\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}$
Do đó: $A=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}$