$c_1$ centered at $A$ passing through $B$.
$BB′$ is a diameter of $c_1$.
$T$ a random point in segment $BB′$.
$c_2$ centered at $B′$ passing through $T$.
$c_3$ centered at $B$ passing through $T$.
$c_4$ tangent externally to $c_2$ and $c_3$ and internally tangent to $c_1$
$F$ is center of $c_4$ and $H,I$ are tangency points.
It is clear to me that $Z = HI \cap AF$ is the second homothety center of $c_1$ and $c_4$ and I would like to prove that it also lies in that line perpendicular to $AB$ through $T$.
important related result that you probably should know: Show these three circles share their external common tangent lines
This seems to be a general result about soddy circles
Best Answer
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.