Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?


$\text{(A) } \frac23 \qquad \text{(B) } \frac {\sqrt3}{2} \qquad \text{(C) } \frac78 \qquad \text{(D) } \frac89 \qquad \text{(E) } \frac {1 + \sqrt3}{3}$
Let $O_{i}$ be the center of circle $i$ for all $i \in \{A,B,C,D\}$ and let $E$ be the tangent point of $B,C$. Since the radius of $D$ is the diameter of $A$, the radius of $D$ is $2$. 

Let the radius of $B,C$ be $r$ and let $O_{D}E = x$. If we connect $O_{A},O_{B},O_{C}$, we get an [isosceles triangle](https://artofproblemsolving.com/wiki/index.php/Isosceles_triangle) with lengths $1 + r, 2r$. 
Then right triangle $O_{D}O_{B}O_{E}$ has legs $r, x$ and [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) $2-r$. Solving for $x$, we get $x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}$. 
Also, right triangle $O_{A}O_{B}O_{E}$ has legs $r, 1+x$, and hypotenuse $1+r$. Solving, 
\begin{eqnarray*} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89  \end{eqnarray*}
So the answer is $\boxed{\mathrm{(D)}\ \frac{8}{9}}$.