\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\\ \Leftrightarrow\left(a+b\right)\left(c-a\right)=\left(a-b\right)\left(c+a\right)\\ \Leftrightarrow ac-a^2+bc-ab=ac+a^2-bc-ab\\ \Leftrightarrow-2a^2=-2bc\\ \Leftrightarrow a^2=bc\)
Đặt \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}=k\), ta có \(a+b=k\left(a-b\right)\)
\(\Rightarrow a+b=ka-kb\Rightarrow a-ka=-kb-b\Leftrightarrow a\left(1-k\right)=b\left(-k-1\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{-k-1}{1-k}=\dfrac{-\left(k+1\right)}{-\left(k-1\right)}=\dfrac{k+1}{k-1}\Rightarrow\dfrac{a}{b}=\dfrac{k+1}{k-1}\)
\(c+a=k\left(c-a\right)\Rightarrow c+a=kc-ka\Rightarrow c-kc=-ka-a\)
\(\Rightarrow c\left(1-k\right)=a\left(-k-1\right)\Rightarrow\dfrac{c}{a}=\dfrac{-k-1}{1-k}=\dfrac{-\left(k+1\right)}{-\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\Rightarrow\dfrac{c}{a}=\dfrac{k+1}{k-1}\Rightarrow\dfrac{a}{b}=\dfrac{c}{a}=\dfrac{k+1}{k-1}\Rightarrow\dfrac{a}{b}=\dfrac{c}{a}\Rightarrow a^2=bc\)
