Given the square $A(0,0),B(2,0),C(2,2),D(0,2)$. Choose $A'(2,2\lambda)$, then the equation of $EF\ :\ y-\lambda=-\frac{1}{\lambda}(x-1)$ and $E(1+\lambda^{2},0),F(\ (1-\lambda)^{2},2)$. Point $D'(\frac{2(\lambda-1)^{2}}{\lambda^{2}+1},2+\frac{2\lambda(\lambda-1)^{2}}{\lambda^{2}+1})$. Equation of the line $A'D'\ :\ y-2\lambda=\frac{\lambda^{2}-1}{2\lambda}(x-2)$. Point $G(\frac{2-2\lambda}{\lambda+1},2)$. Inradius of $\triangle A'CG\ :\ r_{1}=\frac{2\lambda(1-\lambda)}{\lambda+1}$. Equation of $D'F\ :\ y-2=\frac{2\lambda}{1-\lambda^{2}}[x-(\lambda-1)^{2}]$. Inradius of $\triangle D'FG\ :\ r_{2}=\frac{\lambda(\lambda-1)^{2}}{\lambda+1}$. Equation of $A'E\ :\ y=\frac{2\lambda}{1-\lambda^{2}}[x-(\lambda^{2}+1)]$. Inradius of $\triangle A'BE\ :\ r_{3}=\lambda(1-\lambda)$. Then $r_{1}=r_{2}+r_{3}$.