In the figure, determine $x$ as a function of $R$. For reference: In the figure, determine $x$ as a function of $R$.

My progress: I don't know and I can affirm that $\angle FGD$ is $90^o$ ...I think you would need to demonstrate ...but I can't
$\triangle FDB: (R+r)^2 = R^2+(2R-2r)^2\\R^2+2Rr+r^2=R^2+4R^2-4Rr+4r^2\\6Rr-3r^2 = 4Rr\\6R-3r=4R \implies \boxed{r = \frac{2R}{3}}\\\triangle FGD: (R+r)^2 = (x+r)^2+(x+R)^2\\R^2+\frac{4R^2}{3} + \frac{4R^2}{9} = x^2+\frac{4xR}{3}+\frac{4R^2}{9}+x^2+2xR+R^2\\\frac{4R^2}{3} =2x^2+\frac{10xR}{3}\\\frac{4R^2}{3} = \frac{6x^2+10xR}{3}\implies 4R^2 = 6x^2+10xR \implies\\ 3x^2+5xR - 2R^2 = 0\\\therefore \boxed{\color{red}x = \frac{R}{3}}$


If the radius of the other half circle is $r$, you already showed that $ \small \displaystyle r = \frac{2R}{3}$.
Now applying Descartes' theorem,
$ \small \displaystyle \left(- \frac{1}{2R} + \frac{1}{R} + \frac{1}{2R/3} + \frac{1}{x} \right)^2 = 2 \left(\frac{1}{4R^2} + \frac{1}{R^2} + \frac{1}{(2R/3)^2} + \frac{1}{x^2}\right)$
Simplifying, $ \small \displaystyle \left(\frac{2}{R} + \frac{1}{x} \right)^2 = \frac{7}{R^2} + \frac{2}{x^2}$
Solving, $ \small \displaystyle x = \frac{R}{3}$
