Arbelos and circles of Archimedes proof using an inversion circle

Question

In the image above, let $d$ denote the diameter of circle $\omega_4$, and set $d_1=AB$ and $d_2=BC$. Assume everything that looks tangent is in fact tangent.

Prove:
$$\displaystyle d=\frac{d_1d_2}{d_1+d_2}$$

I'm supposed to use a circle of inversion centered at $A$, and orthogonal to $\odot\omega_4$ in my proof. However, I would welcome any proof at all, even an elementary one.

Thanks!

Answer 1

Let $r$ denote radii. The following equation can be established by applying the Pythagorean theorem to the right triangles XYD and XZD,

$$XD^2 = (r_1+r_4)^2 - (r_1-r_4)^2 = (r_3-r_4)^2-(2r_1-r_3-r_4)^2$$

which simplifies to $$r_3r_4=r_1(r_3-r_1)$$

Recognize $r_3=r_1+r_2$ to obtain

$$r_4=\frac{r_1r_2}{r_1+r_2}$$

which is equivalent to $\displaystyle d=\frac{d_1d_2}{d_1+d_2}$.