jmc

Adding the given equations, we get $$\frac{c(a+b)}{a+b}+\frac{a(b+c)}{b+c}+\frac{b(c+a)}{c+a}=1,$$which simplifies to $a+b+c=1.$

Subtracting the equations given in the problem, we get $$\frac{c(b-a)}{a+b}+\frac{a(c-b)}{b+c}+\frac{b(a-c)}{c+a}=19.$$Let $$\begin{array}{rl}u& =\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a},\\ v& =\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a},\end{array}$$so $u+v=3.$ Also, $$\begin{array}{rl}u-v& =\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\\ & =(a+b+c)\frac{a-b}{a+b}+(a+b+c)\frac{b-c}{b+c}+(a+b+c)\frac{c-a}{c+a}\\ & =a-b+\frac{c(a-b)}{a+b}+b-c+\frac{a(b-c)}{b+c}+c-a+\frac{b(c-a)}{c+a}\\ & =-19.\end{array}$$Subtracting the equations $u+v=3$ and $u-v=-19,$ we get $2v=22,$ so $v=\boxed{11}.$