To show the center of homothety of the biggest and smallest circle lies in the common tangent over T

Let common tangent at $T$ meet $AF$ at $Y$ and let perpendicular to $AB$ through $F$ meet $AB$ at $L$. Then we calculate $y=LT$ by Pythagoras theorem: $$ B'F^2-B'L^2 = LF^2 =BF^2-BL^2$$ so $$ (b+c)^2-(b-y)^2 = (2a+b+c)^2-(2a+b+y)^2$$ and so we get $$y= {ac\over a+b}$$ so $${AY\over FY} = {AT\over LT} = {a\over y} = {a+b\over c}$$

On the other hand let $X$ be in $HI\cap AF$. Homothety $H_1$ at $H$ and coefficient ${b\over c}$ takes $F$ to $B'$ and homothety $H_2$ at $G$ and coefficient ${a+b\over b}$ takes $B'$ to $A$, so composition $H_2\circ H_1$ takes $F$ to $A$ and has center at $FA\cap GH =X$. This composition has coefficent $${a+b\over b}\cdot {b\over c} = {a+b\over c}$$ so $X$ divides $AF$ in the same ratio as $Y$ and thus $X=Y$ and we are done.