Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k,c=d.k\)
Ta có:
\(\frac{a.c}{b.d}=\frac{b.k.d.k}{b.d}=k^2\) (1)
\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(b.k+d.k\right)^2}{\left(b+d\right)^2}=\frac{\left[k.\left(b+d\right)\right]^2}{\left(b+d\right)^2}=k^2\) (2)
Từ (1) và (2) suy ra \(\frac{a.c}{b.d}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)