Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)=>a=b.k;c=d.k

\(\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\)(1)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}\)\(=\dfrac{\left(b^2+d^2\right).k^2}{b^2+c^2}=k^2\)(2)

Từ 1 và 2 =>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)