Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Xét VT \(\frac{ac}{bd}=\frac{bkdk}{bd}=\frac{bdk^2}{bd}=k^2\left(1\right)\)
Xét VP \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1) và (2) ->Đpcm