Equilateral Sangaku

Let AD =x then DF=2(√3)r
8-x+r/√(3) = x+2 √(3) r - r/√(3)
Apply co-sine rule in triangle ADC (angle ADC=120°)
(x+2 √(3) r)²+x²+x (x+2√(3) r)=64
r= √7 –√3

$\text{let r be the radius of the circle} \\ \Delta DEF \text{ is equilateral} \Rightarrow \left | E F \right | = 2 r\sqrt{3} \\ \left | H F \right | =\left | J E \right | = \frac{r}{\sqrt{3}} \\ \left | BJ \right | =\left | CH \right |=\left | CG \right | \\ \left | BG \right | =\left | BI \right | = \left | BJ \right |+\left | JE \right |+\left | EF \right |-\left | IF \right | \\ \left | BG \right | = \left | CG \right |+\frac{r}{\sqrt{3}}+2 r\sqrt{3}-\frac{r}{\sqrt{3}}=\left | CG \right |+2 r\sqrt{3}\\ \left | BC \right | =\left | BG \right | + \left | GC \right | \\ 8=\left | CG \right |+2 r\sqrt{3}+ \left | CG \right |\Rightarrow \left | CG \right |=4 - r\sqrt{3}\\ \left | BG \right |=4 + r\sqrt{3}\\ \beta =\angle {GBO}=\angle {IBO} \\ \alpha=\angle {HCO}=\angle {OCG} \\ \alpha + \beta =30° \\ \tan\left ( \alpha \right ) = \frac{r}{4 - r \sqrt{3}}\\ \tan\left ( \beta \right ) = \frac{r}{4 + r \sqrt{3}} \\ \tan\left ( \alpha + \beta \right ) = \frac{\tan\left ( \alpha \right ) + \tan\left ( \beta \right ) }{ 1 -\tan\left ( \alpha \right ) \tan\left ( \beta \right ) } \\ \tan\left ( 30° \right ) = \frac{\frac{r}{4 - r \sqrt{3}} + \frac{r}{4 + r \sqrt{3}} }{ 1 -\frac{r}{4 - r \sqrt{3}} \cdot \frac{r}{4 + r \sqrt{3}} } \\ \frac{1}{\sqrt{3}} = \frac{8r}{16 - 4r^2}\Rightarrow r^2 + 2 \sqrt{3} r - 4 =0 \\ r = -\sqrt{3} \pm \sqrt{7} \\ r = \sqrt{7} - \sqrt{3}$