[Math] Sangaku. How to draw those three circles with only a ruler and a compass

Question

I found in a book of Sangakus the following problem.

Let $R_b$, $R_g$ and $R_r$ the radiuses of the blue, green and red circles $C_b$, $C_g$ and $C_r$.

Prove that
$$\frac{1}{\sqrt{R_r}}=\frac{1}{\sqrt{R_b}}+\frac{1}{\sqrt{R_g}}\,.\quad (1)$$
And this I can do. But then

I would like to draw the figure myself with only a ruler and a compass.

I know it is possible as I can construct inverses, sums and square roots with only a ruler and a compass, but when I tried to draw the figure with a "simple" or "natural" construction, I failed.

Does someone have an idea of how to draw it "naturally"?

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EDIT

Answer to a comment of Amzoti:
To prove Relation (1) I first prove the relation:
$$AB^2=4R_gR_b\quad (2)$$

Relation (2) is a consequence of Pythagoras' theorem in the triangle $O_bO_gH$:
$$AB^2+(R_g-R_b)^2=(R_b+R_g)^2\,.$$
(It was the previous Sengaku in the book.)

We thus get the relations
$$
\begin{align}
AB^2 & =4R_gR_b\quad (2) \\
AC^2 & =4R_gR_r\quad (3) \\
BC^2 & =4R_bR_r\quad (4) \\
\end{align}
$$
where $A$, $B$, $C$ are the orthogonal projections of the centers of the circles $C_b$, $C_g$ and $C_r$ on the line $d$. The relation $AB^2=AC^2+BC^2+2AC.BC$ then yields, using Relations (2) to (4),
$$4R_gR_b=4R_gR_r+4R_bR_r+8\sqrt{R_bR_g}R_r$$
Divided by $4R_rR_gR_b$ this equation is
$$\frac{1}{R_r}=\Big(\frac{\sqrt{R_g}+\sqrt{R_b}}{\sqrt{R_bR_g}}\Big)^2$$
which is in fact Relation (1) squared.

Answer 1

I found another straightforward method which can be perfectly show the meaning!