\(\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}\)
=>(a+b)(d+a)=(c+d)(b+c)
\(\Leftrightarrow ad+a^2+bd+ab=cb+c^2+db+dc\)
\(\Leftrightarrow ad+a^2+ab=cb+c^2+dc\)
\(\Leftrightarrow d\left(a-c\right)+\left(a+c\right)\left(a-c\right)+b\left(a-c\right)=0\)
\(\Leftrightarrow\left(a-c\right)\left(a+b+c+d\right)=0\)
=>a=c hoặc a+b+c+d=0(đpcm)