\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Leftrightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)
\(\Leftrightarrow a\left(c-d\right)+b\left(c-d\right)=a\left(c+d\right)-b\left(c+d\right)\)
\(\Leftrightarrow ac-ad+bc-bd=ac+ad-bc+bd\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\rightarrowđpcm\)
Theo đề bài, ta có:
\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\) (*)
\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{2b}{2d}=\dfrac{b}{d}\) (**)
Từ (*) và (**), suy ra:
\(\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\) (đpcm)
