Given the square $ABCD\ :\ A(0,0),B(1,0),C(1,1),D(0,1)$. Choose $E(0,\lambda)$. $\triangle ABE$ has incenter $O(\frac{\lambda}{\lambda+w+1},\frac{\lambda}{\lambda+w+1})$ with $w=\sqrt{\lambda^{2}+1}$, radius $r=\frac{\lambda}{\lambda+w+1}$. $\triangle CDE$ has incenter $I(\frac{1-\lambda}{2-\lambda+v},1-x_{I})$ with $v=\sqrt{\lambda^{2}-2\lambda+2}$. The line $OI\ :\ y-r=\frac{1-x_{I}-r}{x_{I}-r}(x-r)$ intersects the y-axis in the point $T(0,\frac{r(2x_{I}-1)}{x_{I}-r})$. The line, through $T$ and tangent to the three incircles, has equation: $$y=\frac{-2x(v+1)(w+1)+\lambda^{4}+\lambda^{3}(v+w)+\lambda^{2}(vw-w-1)+\lambda(v+2)(w+1)}{2 \cdot [\lambda^{3}(v+w+2)-\lambda^{2}(v+2w+3)+\lambda(w+1)(2v+3)-(v+1)(w+1)]}$$.

parmenides51 wrote: Let $E$ be an arbitrary point on the side $BC$ of the square $ABCD$. Prove that the inscribed circles of triangles $ABE$, $CDE$, $ADE$ have a common tangent. Use Monge on the three circumcircles for an instakill.