Given $A(0,0),B(b,0),C(b+c,d),D(c,d)$.
We use $w=\sqrt{c^{2}+d^{2}}$.
Midpoint $O_{1}(s,r_{1})$ and radius $r_{1}$ with $s=\frac{r_{1}(c+w)}{d}$.
Midpoint $O_{2}(t,d-r_{2})$ and radius $r_{2}$ with $t=\frac{d(b+c)-r_{2}(c+w)}{d}$.
After a lot of calculations $r_{2}=-r_{1}+d \cdot \frac{b(c+w)+c^{2}+cw+d^{2}-d\sqrt{2b(c+w)}}{2c^{2}+2cw+d^{2}}$.
The line $O_{1}O_{2}\ :\ y-r_{1}=\frac{d-r_{1}-r_{2}}{t-s}(x-s)$ cuts the line $E(t,d)F(s,0)\ :\ y=\frac{d}{t-s}(x-s)$
in the point $K(\frac{tr_{1}+sr_{2}}{r_{1}+r_{2}},\frac{dr_{1}}{r_{1}+r_{2}})$.
Using $\triangle AFO_{1} \sim \triangle CEO_{2}\ :\ \frac{r_{1}}{r_{2}}=\frac{s}{b+c-t}$, this point $K$ lies on $AC\ :\ y=\frac{d}{b+c} \cdot x$.
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