In rectangle $ ADEH$, points $ B$ and $ C$ trisect $ \overline{AD}$, and points $ G$ and $ F$ trisect $ \overline{HE}$.  In addition, $ AH \equal{} AC \equal{} 2.$  What is the area of quadrilateral $ WXYZ$ shown in the figure?

[asy]defaultpen(linewidth(0.7));pointpen=black; pathpen=black;
size(7cm);
pair A,B,C,D,E,F,G,H,W,X,Y,Z;
A=(0,2); B=(1,2); C=(2,2); D=(3,2);
H=(0,0); G=(1,0); F=(2,0); E=(3,0);
D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,NE); D('F',F,NE); D('G',G,NW); D('H',H,NW);
D(A--F); D(B--E); D(D--G); D(C--H);
Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);
D('W',W,N); D('X',X,plain.E); D('Y',Y,S); D('Z',Z,plain.W);
D(A--D--E--H--cycle);[/asy]

$ \textbf{(A) } \frac 12 \qquad \textbf{(B) } \frac {\sqrt {2}}2\qquad \textbf{(C) } \frac {\sqrt {3}}2 \qquad \textbf{(D) } \frac {2\sqrt {2}}3 \qquad \textbf{(E) } \frac {2\sqrt {3}}3$
1. **Identify the dimensions of the rectangle:**
   - Given \( AH = AC = 2 \), and since \( AD \) is trisected by points \( B \) and \( C \), we have \( AD = 3 \).
   - Therefore, the dimensions of the rectangle \( ADEH \) are \( 3 \times 2 \).

2. **Determine the coordinates of key points:**
   - Let \( A = (0, 2) \), \( D = (3, 2) \), \( H = (0, 0) \), and \( E = (3, 0) \).
   - Points \( B \) and \( C \) trisect \( \overline{AD} \), so \( B = (1, 2) \) and \( C = (2, 2) \).
   - Points \( G \) and \( F \) trisect \( \overline{HE} \), so \( G = (1, 0) \) and \( F = (2, 0) \).

3. **Find the equations of the lines forming quadrilateral \( WXYZ \):**
   - Line \( AF \) passes through \( A = (0, 2) \) and \( F = (2, 0) \). Its slope is \( \frac{0-2}{2-0} = -1 \), so the equation is \( y = -x + 2 \).
   - Line \( BE \) passes through \( B = (1, 2) \) and \( E = (3, 0) \). Its slope is \( \frac{0-2}{3-1} = -1 \), so the equation is \( y = -x + 3 \).
   - Line \( DG \) passes through \( D = (3, 2) \) and \( G = (1, 0) \). Its slope is \( \frac{0-2}{1-3} = 1 \), so the equation is \( y = x - 1 \).
   - Line \( CH \) passes through \( C = (2, 2) \) and \( H = (0, 0) \). Its slope is \( \frac{0-2}{0-2} = 1 \), so the equation is \( y = x \).

4. **Find the intersection points \( W, X, Y, Z \):**
   - \( W \) is the intersection of \( BE \) and \( CH \):
     \[
     \begin{cases}
     y = -x + 3 \\
     y = x
     \end{cases}
     \]
     Solving, \( x = 1.5 \), \( y = 1.5 \). So, \( W = (1.5, 1.5) \).

   - \( X \) is the intersection of \( BE \) and \( DG \):
     \[
     \begin{cases}
     y = -x + 3 \\
     y = x - 1
     \end{cases}
     \]
     Solving, \( x = 2 \), \( y = 1 \). So, \( X = (2, 1) \).

   - \( Y \) is the intersection of \( AF \) and \( DG \):
     \[
     \begin{cases}
     y = -x + 2 \\
     y = x - 1
     \end{cases}
     \]
     Solving, \( x = 1.5 \), \( y = 0.5 \). So, \( Y = (1.5, 0.5) \).

   - \( Z \) is the intersection of \( AF \) and \( CH \):
     \[
     \begin{cases}
     y = -x + 2 \\
     y = x
     \end{cases}
     \]
     Solving, \( x = 1 \), \( y = 1 \). So, \( Z = (1, 1) \).

5. **Calculate the area of quadrilateral \( WXYZ \):**
   - Using the Shoelace Theorem for the vertices \( W(1.5, 1.5) \), \( X(2, 1) \), \( Y(1.5, 0.5) \), \( Z(1, 1) \):
     \[
     \text{Area} = \frac{1}{2} \left| 1.5 \cdot 1 + 2 \cdot 0.5 + 1.5 \cdot 1 + 1 \cdot 1.5 - (1.5 \cdot 2 + 1 \cdot 1.5 + 0.5 \cdot 1 + 1 \cdot 1.5) \right|
     \]
     \[
     = \frac{1}{2} \left| 1.5 + 1 + 1.5 + 1.5 - (3 + 1.5 + 0.5 + 1.5) \right|
     \]
     \[
     = \frac{1}{2} \left| 5.5 - 6.5 \right| = \frac{1}{2} \left| -1 \right| = \frac{1}{2}
     \]

The final answer is \(\boxed{\frac{1}{2}}\)