\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{a+b+d}=\frac{d}{a+b+c}=\frac{a+b+c+d}{b+c+d+c+d+a+a+b+d+a+b+c}=\frac{\left(a+b+c+d\right)}{3\left(a+b+c+d\right)}=3\Rightarrow a=b=c=d\Rightarrow\frac{a+c}{b+d}+\frac{a+b}{c+d}+\frac{a+c}{b+d}+\frac{b+c}{a+d}=1+1+1+1=4\)