Lời giải:
Theo bài ra ta có:
$\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$
$=\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}$
$=\dfrac{a+b}{c+d}+1=\dfrac{b+c}{d+a}+1$
$=\dfrac{a+b+c+d}{c+d}=\dfrac{a+b+c+d}{d+a}$
$(a+b+c+d)(\dfrac{1}{c+d} - \dfrac{1}{d+a})=0$
`=>` \(\left[ \begin{array}{l}a+b+c+d=0\\\dfrac{1}{c+d} - \dfrac{1}{d+a}=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}a+b+c+d=0\\\ a=c\end{array} \right.\)
Vậy ta có điều phải chứng minh.
