We use barycentric coodinate wrt trangle $ABC$. We let $A=(1,0,0), B=(0,1,0), C=(0,0,1)$. We have $A'=(0:b:c)$, $B'$ and $C'$ are calculated cyclically. Take the circumcircle of $A'BC'$. The circle function is $-a^2yz-b^2zx-c^2xy+(\frac{c^2b}{a+b}x+\frac{c^2b}{a+b}z)(x+y+z)=0$. Plug in $B'=(a:0:c)$ to get $b^2ac=(\frac{c^2ab}{b+c}+\frac{a^2bc}{a+b})(a+c)$. which is equivalent to $\frac{b}{a+c}=\frac{a}{b+c}+\frac{c}{a+b}$.