Ta có $\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}=\frac{ab}{cd}$
$cd\left( {{a}^{2}}+{{b}^{2}} \right)=ab\left( {{c}^{2}}+{{d}^{2}} \right)$
${{a}^{2}}cd+{{b}^{2}}cd={{c}^{2}}ab+{{d}^{2}}ab$
${{a}^{2}}cd+{{b}^{2}}cd-{{c}^{2}}ab-{{d}^{2}}ab=0$
$\left( {{a}^{2}}cd-{{c}^{2}}ab \right)+\left( {{b}^{2}}cd-{{d}^{2}}ab \right)=0$
$ac\left( ad-bc \right)+bd\left( bc-ad \right)=0$
$ac\left( ad-bc \right)-bd\left( ad-bc \right)=0$
$\left( ad-bc \right)\left( ac-bd \right)=0$
$\Rightarrow \left[ \begin{align} & ad-bc=0 \\ & ac-bd=0 \\ \end{align} \right.$
$\Rightarrow \left[ \begin{align} & ad=bc \\ & ac=bd \\ \end{align} \right.$
$\Rightarrow \left[ \begin{align} & \frac{a}{b}=\frac{c}{d} \\ & \frac{a}{b}=\frac{d}{c} \\ \end{align} \right.$

(a² + b²) / (c² + d²) = ab/cd

(a² + b²)cd = ab(c² + d²)

a²cd + b²cd = abc² + abd²

a²cd - abc² - abd² + b²cd = 0

ac(ad - bc) - bd(ad - bc) = 0

(ac - bd)(ad - bc) = 0

ac - bd = 0 hoặc ad - bc = 0

ac = bd hoặc ad = bc

a/b = d/c hoặc a/b = c/d (đpcm)