Two adjacent incircles

Let $BD=x$ and $AD=y$ .

A triangle's inradius $r$ , semiperimeter $s$ and area $T$ are related by $T=rs$

Here we have $T_{ABD}=r_{ABD} s_{ABD}\;\;\;\text{and}\;\;\;T_{ADC}=r_{ADC} s_{ADC}$

Dividing, $\frac{T_{ABD}}{T_{ADC}}=\frac{r_{ABD} s_{ABD}}{r_{ADC} s_{ADC}}$

Since $\Delta ABD$ and $\Delta ADC$ have the same height, the ratio of their areas is the ratio of their bases; so $\frac{T_{ABD}}{T_{ADC}}=\frac{BD}{DC}=\frac{x}{1-x}$ . Also, we're told $r_{ABD}=2r_{ADC}$ , so that $\frac{x}{1-x}=2\frac{1+x+y}{2-x+y}$

We can solve this for $y$ to get $y=\frac{x^2+2x-2}{2-3x}$

The cosine rule in $\Delta ABD$ gives $y^2=x^2-x+1$

Combining these to eliminate $y$ and rearranging, we get $x(x-1)\left(8x^2-17x+8\right)=0$

Solving this we find four roots for $x$ ; three of these are outside the interval $(0,1)$ , and are discarded; the other is $x=\frac{1}{16}\left(17-\sqrt{33}\right) \approx \boxed{0.70346}$

Let the tangent point of the big circle on $AB$ and small circle on $AC$ be $E$ and $F$ respectively, the radius of the smaller circle be $r$ , and $\angle BAD = \theta$ . Then

$\begin{cases} AE + EB = 2r \cot \dfrac \theta 2 + 2 r \cot 30^\circ = 1 & ...(1) \\ AF + FC = r\cot \left(30^\circ - \dfrac \theta 2\right) + r \cot 30^\circ = 1 & ...(2) \end{cases}$

From $(1) = (2)$ :

$\begin{aligned} 2r \cot \frac \theta 2 + 2\sqrt 3 r & = r\tan \left(60^\circ + \dfrac \theta 2\right) +\sqrt 3 r & \small \blue{\text{Let }t = \tan \frac \theta 2.} \\ \frac 2t + \sqrt 3 & = \frac {\sqrt 3+t}{1-\sqrt 3t} \\ (2+\sqrt 3t)(1-\sqrt 3t) & = \sqrt 3 t + t^2 \\ 2 - \sqrt 3 t - 3t^2 & = \sqrt 3 t + t^2 \\ 4t^2 + 2\sqrt 3 t - 2 & = 0 \\ 2t^2 + \sqrt 3 t - 1 & = 0 \\ \implies t & = \frac {\sqrt{11}-\sqrt 3}4 \end{aligned}$

Let $BD = x$ . By sine rule :

$\begin{aligned} \frac {BD}{DC} &= \frac {\sin \theta}{\sin (60^\circ - \theta)} \\ \frac x{1-x} & = \frac {\sin \theta}{\frac {\sqrt 3}2\cos \theta - \frac 12 \sin \theta} \\ x (\sqrt 3 \cos \theta - \sin \theta) & = 2\sin \theta - 2x \sin \theta \\ x (\sqrt 3 \cos \theta + \sin \theta) & = 2\sin \theta \end{aligned}$

$\begin{aligned} \implies x & = \frac {2\sin \theta}{\sqrt 3 \cos \theta + \sin \theta} = \frac {4t}{\sqrt 3(1-t^2) + 2t} = \frac 4{\frac {\sqrt 3}t - \sqrt 3 t + 2} \\ & = \frac 4{\frac {4\sqrt 3}{\sqrt{11}-\sqrt 3}-\frac {\sqrt{33}-3}4+2} = \frac 4{\frac {2\sqrt{33}+6}4-\frac {\sqrt{33}-3}4+2} \\ & = \frac {16}{\sqrt{33}+17} = \frac {17-\sqrt{33}}{16} \approx \boxed{0.703} \end{aligned}$