Có \(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a.b}{c.d}\) (1)
Có \(\dfrac{a}{c}=\dfrac{b}{d}\) => \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2\)
=> \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
ADTCDTSBN ta có
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\) (2)
Từ (1) và (2) =>\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a.b}{c.d}\) (đpcm)