Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{b\cdot d}=k^2\) (1)
\(\Rightarrow\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{\left[k\left(b+d\right)\right]^2}{\left(b+d\right)^2}\)\(=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)