In a triangle $\triangle ABC$ the cevian $AO$ and the incircles of the triangles $\triangle ABO$ and $\triangle AOC$ are traced. Then the two excircles opposite to the vertex A are drawn and with the tangent common to these two excircles a new triangle $\triangle ADE$ is formed. We Note $r$ and $R$ the radii of the red circles in the attached figure and $g$ and $G$ the radii of the green circles.
Prove that $$\frac 1r+\frac 1R=\frac 1g+\frac 1G$$
On
The result follows from this simpler proposition (whose notation does not match that of OP):
In $\triangle ABC$, the difference between the reciprocals of the inradius ($p$) and $A$-exradius ($q$) depends only on the distance ($h$) from $A$ to $\overleftrightarrow{BC}$.
Proof: If $P$ and $Q$ are points of tangency of incircle and excircle with the sides of $\angle A$, it is "known" (and/or readily shown) that $$|AP| = \tfrac12(-a+b+c) \qquad |AQ|=\tfrac12(a+b+c)$$ Thus, we have $$\frac{p}{\tfrac12(-a+b+c)}=\frac{q}{\tfrac12(a+b+c)}$$ which implies
$$\frac{1}{p}-\frac{1}{q}=\frac{q-p}{pq} =\frac{\tfrac12((a+b+c)-(-a+b+c))}{p\cdot \tfrac12 (a+b+c)} = \frac{a}{|\triangle ABC|} = \frac{2}{h}$$
$\square$
Let $S_{\Delta ABO}=S_1$, $S_{\Delta ACO}=S_2$, $AB=c$, $AC=b$, $BO=a_1$, $CO=a_2$ and $AO=m$.
Thus, $$r=r_{\Delta ABO}=\frac{2S_1}{c+m+a_1},$$ $$R=\frac{2S_2}{b+m-a_2},$$ $$g=r_{\Delta ACO}=\frac{2S_2}{b+m+a_2}$$ and $$G=\frac{2S_1}{c+m-a_1}.$$ Thus, we need to prove that: $$\frac{c+m+a_1}{S_1}+\frac{b+m-a_2}{S_2}=\frac{m+b+a_2}{S_2}+\frac{c+m-a_1}{S_1}$$ or $$c+m+a_1+\frac{S_1}{S_2}(b+m-a_2)=\frac{S_1}{S_2}(b+m+a_2)+c+m-a_1$$ or $$c+m+a_1+\frac{a_1}{a_2}(b+m-a_2)=\frac{a_1}{a_2}(b+m+a_2)+c+m-a_1,$$ which is obvious.