\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\Rightarrow\left(\dfrac{a}{a+b}+\dfrac{c}{c+d}-1\right)+\left(\dfrac{b}{b+c}+\dfrac{d}{d+a}-1\right)=0\)
\(\dfrac{ac-bd}{\left(a+b\right)\left(c+d\right)}-\dfrac{ac-bd}{\left(b+c\right)\left(a+d\right)}=0\)
=>\(\left(ac-bd\right).\left(\dfrac{1}{\left(a+b\right)\left(c+d\right)}-\dfrac{1}{\left(a+d\right)\left(b+c\right)}\right)=0\)=>\(\left(ac-bd\right).\left(\dfrac{ab+ac+bd+dc-ac-ad-bc-bd}{\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)}\right)=0\)

=>(ac-bd).(ab+cd-ad-bc)=0
=>(ac-bd)(a-c)(b-d)
vì \(a\ne b\ne c\ne d\Rightarrow a-c\ne0;b-d\ne0\)
suy ra ac-bd suy ra ac.bd=(ac)²=(bd)²