7. The sides of a triangle $a, b, c$ satisfy the equation $\frac{c^{2}}{a+b}+\frac{a^{2}}{b+c}=b$, then the size of the angle $B$ opposite the side of length $b$ is
7. $60^{\circ}$ Hint: The original expression can be transformed into $\left(\frac{c^{2}}{a+b}+c\right)+\left(\frac{a^{2}}{b+c}+a\right)=b+a+c$

So $\frac{c(a+b+c)}{a+b}+\frac{a(a+b+c)}{b+c}=a+b+c$
So $\frac{c}{a+b}+\frac{a}{b+c}=1$
So $c(b+c)+a(a+b)=(a+b)(b+c)$
So $c^{2}+a^{2}-a c=b^{2}$
By the cosine rule, $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{1}{2}$
So $B=60^{\circ}$