[tex]\displaystyle\\\underline{\textnormal{Prima~parte}}:\\\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\implies \frac{a}{b+c}=2\Longleftrightarrow \boxed{2a=b+c}.\\\underline{\textnormal{Partea a doua}}:\\\textnormal{Dar, din prima parte ob\c{t}inem egalit\u{a}\c{t}ile:} \begin{cases} 2a=b+c\\ 2b=a+c \\ 2c=a+b\end{cases}\\\textnormal{sc\u{a}z\^{a}nd dou\u{a} c\^{a}te dou\u{a} egalit\u{a}\c{t}ile de mai sus, avem:}\begin{cases} 2(a-b)=b-a\\ 2(a-c)=c-a \end{cases}[/tex]
[tex]\displaystyle\\\textnormal{de unde}~a-b=b-a~\textnormal{iar}~a-c=c-a~\textnormal{de~unde}~\boxed{a=b=c}.[/tex]
