use an inversion circle to prove an identity for the arbelos and magic twin circles

Question

In the figure, we assume that everything is "how it looks." For instance, everything that looks tangent, is tangent, all the points that appears to be centers of the circles, are centers, and the line segment shown is vertical.

Also, assume circles $D$ and $G$ are orthogonal.

Let $r_G$ be the radius of circle $G$, let $r_B$ be the radius of circle $B$, and let $r_C$ be the radius of circle $C$.

Problem: We need to use circle $D$ as an inversion circle to prove that $$ r_G=\frac{r_Br_C}{r_B+r_C}. $$

I asked about this problem a few days ago and it was answered here using the Pythagorean Theorem, but without using the inversion circle. I was hoping that would give me an insight to using inversion, but sadly it did not.

Some observations.

We know that $\odot G$ is invariant under inversion since it's orthogonal to the inversion circle.

If we can prove that $\odot D$ passes through $E$, that would be sufficient for me to prove the rest.

Any ideas would be much appreciated. Thanks!

Answer 1

I will use the letters $\beta, \delta, \gamma$ for the circles with centers $B,D,G,$ respectively. The vertical line will be called $h.$

We want to prove that $\delta,$ which is centered at $D$ and orthogonal to $\gamma,$ passes through $E.$

Consider the inversion in circle $\delta.$

1. $\beta$ passes through the center of $\delta,$ therefore is inverted into a line, say $b.$ Necessarilly, $b$ is orthogonal to the line $DB.$
2. $\gamma$ is invariant, because is orthogonal to $\delta$
3. $\gamma$ is tangent to $\beta,$ thus the inverse of $\gamma$ is tangent to the inverse of $\beta$ (simply said, $\gamma$ is tangent to $b$)

As there is the only vertical line tangent to $\gamma,$ we have $b=h.$

Given that $h$ and $\beta$ are tangent at $E,$ the point $E$ lies on $\beta$ and on $b,$ which is the image od $\beta$ in the inversion. This says that $E$ is invariant, which completes the proof.