Áp dụng TC DTSBN ta có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{\left(b+c\right)+\left(c+a\right)+\left(a+b\right)}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
Ta có :\(\frac{a}{b+c}=\frac{1}{2}\)
\(\Leftrightarrow2a=b+c\)
\(\Rightarrow a=\frac{b+c}{2}\) (dpcm)