$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1 \iff (a+b+c)(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b})=a+b+c \\\iff \dfrac{a^2}{b+c} + \dfrac{a(b+c)}{b+c} +\dfrac{b^2}{c+a}+ \dfrac{b(c+a)}{c+a} +\dfrac{c^2}{a+b} +\dfrac{c(a+b)}{a+b}=a+b+c \\\iff \dfrac{a^2}{b+c} + \dfrac{b^2}{c+a} +\dfrac{c^2}{a+b} + a+b+c= a+b+c \\\Rightarrow \dfrac{a^2}{b+c} + \dfrac{b^2}{c+a} +\dfrac{c^2}{a+b} =0$$