Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$. Choose $D(-\lambda,0)$, then $E(-\lambda,\frac{a(b-\lambda)}{b}),F(\frac{c \lambda}{b},\frac{a(b-\lambda)}{b}),G(\frac{c \lambda}{b},0)$. $DEFG$ is a square if $\lambda=\frac{ab}{a+b+c}$ and the side $=\frac{a(b+c)}{a+b+c}$. Choose a line, through $C$, with slope $m\ :\ y=m(x-c)$. The line, through $B$ and $\bot CH\ :\ y=-\frac{1}{m}(x+b)$, intersects the line $CH$ in the point $H(\frac{cm^{2}-b}{m^{2}+1},-\frac{m(b+c)}{m^{2}+1})$. The line, through $A$ and $\bot HI\ :\ y-a=mx$, intersects the line $HI$ in the point $I(-\frac{am+b}{m^{2}+1},\frac{a+bm}{m^{2}+1})$. The line, thtough $C$ and $\bot IJ\ :\ y=-\frac{1}{m}(x-c)$, intersects the line $IJ$ in the point $J(\frac{c-am}{m^{2}+1},\frac{a+cm}{m^{2}+1})$. $CHIJ$ is a square if $m=\frac{b+c-a}{c}$ and the side $=\frac{(b+c)c}{\sqrt{(b+c-a)^{2}+c^{2}}}$.