a2 = bc
=> a . a = b . c
=> \(\frac{a}{c}=\frac{b}{a}=\frac{a+b}{a+c}=\frac{a-b}{c-a}\Rightarrow\frac{a+b}{a-b}=\frac{a+c}{c-a}\)
Ta có: \(\frac{a+b}{a-b}=\frac{a+c}{c-a}\Rightarrow\left(a+b\right)\left(c-a\right)=\left(a+c\right)\left(a-b\right)\)
\(\Rightarrow ac-a^2+bc-ab=a^2-ab+ac-bc\)
\(\Rightarrow ac-bc+bc-ab-bc+ab-ac+bc=0\)
\(\Rightarrow0=0\) (luôn đúng)
Vậy đpcm
Theo giải thiết ta được :
\(a^2=bc\Rightarrow a.a=c.b\)
\(\Rightarrow\frac{a}{d}=\frac{b}{a}=\frac{a+b}{a+d}=\frac{a-b}{a-d}\Rightarrow\frac{a+b}{a-b}=\frac{a+c}{c-a}\)