[tex]\displaystyle\it\\\frac{a+2b}{2c+a}=\frac{2a+2c}{b}=\frac{a+2b+2c}{a+b}=\frac{4a+4b+4c}{2a+2b+2c}=2 \implies\\\frac{a+2b}{2c+a}=2 \Leftrightarrow a+2b=4c+2a\Leftrightarrow 2b=4c+a \Leftrightarrow a=2b-4c.\\\frac{2a+2c}{b}=2 \Leftrightarrow 2a+2c=2b \Leftrightarrow a+c=b\Leftrightarrow 2b-4c+c=b\Leftrightarrow b-3c=0\implies\\\boxed{\it b=3c},~de~unde~\boxed{\it a=6c-4c=2c}.\\deci,~raportul~\boxed{\it\boxed{\it\frac{a+c}{b}}=\frac{2c+c}{3c}=\frac{3c}{3c}=\boxed{\it1}}.[/tex]
