Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=k.b;c=k.d\)
Ta có: \(\dfrac{ab}{cd}=\dfrac{kb.b}{kd.d}=\dfrac{k.b^2}{k.d^2}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\dfrac{\left(k+1\right).b^2}{\left(k+1\right).d^2}=\dfrac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)