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$?
[asy]
unitsize(15mm);
pair A=(-1,0),B=(2/3,8/9),C=(2/3,-8/9),D=(0,0);
draw(Circle(D,2));
draw(Circle(A,1));
draw(Circle(B,8/9));
draw(Circle(C,8/9));
label("\(A\)", A);
label("\(B\)", B);
label("\(C\)", C);
label("D", (-1.2,1.8));[/asy]
$ \textbf{(A)}\ \frac23 \qquad \textbf{(B)}\ \frac {\sqrt3}{2} \qquad \textbf{(C)}\ \frac78 \qquad \textbf{(D)}\ \frac89 \qquad \textbf{(E)}\ \frac {1 \plus{} \sqrt3}{3}$
1. Let the centers of circles \( A, B, C \) be \( X, Y, Z \) respectively. Let the center of the larger circle \( D \) be \( O \). Denote the radius of circles \( B \) and \( C \) with \( r \). Notice that \( XY = XZ = r + 1 \) and \( YZ = 2r \).

2. Since circle \( A \) passes through the center of circle \( D \), the radius of circle \( D \) is \( 2 \) (since the radius of circle \( A \) is \( 1 \) and it is tangent to the center of \( D \)).

3. Using the Pythagorean theorem in triangle \( XYZ \), we can find the distance from \( X \) to the line segment \( YZ \). Let \( Q \) be the foot of the perpendicular from \( X \) to \( YZ \). Then, \( XQ = \sqrt{(r+1)^2 - r^2} = \sqrt{2r + 1} \).

4. The distance from \( O \) to \( Q \) is \( OQ = \sqrt{2r + 1} - 1 \).

5. The distance from \( O \) to \( Y \) (or \( Z \)) is \( OY = OZ = 2 - r \).

6. Using the Pythagorean theorem in triangle \( OYQ \), we get:
   \[
   (\sqrt{2r + 1} - 1)^2 + r^2 = (2 - r)^2
   \]

7. Expanding and simplifying the equation:
   \[
   (\sqrt{2r + 1} - 1)^2 = 2r + 1 - 2\sqrt{2r + 1} + 1 = 2r + 2 - 2\sqrt{2r + 1}
   \]
   \[
   (2 - r)^2 = 4 - 4r + r^2
   \]
   \[
   2r + 2 - 2\sqrt{2r + 1} + r^2 = 4 - 4r + r^2
   \]
   \[
   2r + 2 - 2\sqrt{2r + 1} = 4 - 4r
   \]
   \[
   2\sqrt{2r + 1} = 6r - 2
   \]
   \[
   \sqrt{2r + 1} = 3r - 1
   \]

8. Squaring both sides:
   \[
   2r + 1 = (3r - 1)^2
   \]
   \[
   2r + 1 = 9r^2 - 6r + 1
   \]
   \[
   0 = 9r^2 - 8r
   \]
   \[
   0 = r(9r - 8)
   \]

9. Solving for \( r \):
   \[
   r = 0 \quad \text{or} \quad r = \frac{8}{9}
   \]

Since \( r = 0 \) is not a valid solution for the radius of circle \( B \), we have:
\[
r = \frac{8}{9}
\]

The final answer is \(\boxed{\frac{8}{9}}\)