Lines from the vertices of a unit square are drawn to the midpoints of the sides as shown in the figure below. What is the area of quadrilateral $ABCD$? Express your answer in simplest terms.

[asy]
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0,0)--(1,0.5));
draw((1,0)--(0.5,1));
draw((1,1)--(0,0.5));
draw((0,1)--(0.5,0));
label("$A$",(0.2,0.6),N);
label("$B$",(0.4,0.2),W);
label("$C$",(0.8,0.4),S);
label("$D$",(0.6,0.8),E);
[/asy]

$\text{(A) }\frac{\sqrt{2}}{9}\qquad\text{(B) }\frac{1}{4}\qquad\text{(C) }\frac{\sqrt{3}}{9}\qquad\text{(D) }\frac{\sqrt{8}}{8}\qquad\text{(E) }\frac{1}{5}$
1. **Label the vertices of the unit square**: Let the vertices of the unit square be \( (0,0) \), \( (1,0) \), \( (1,1) \), and \( (0,1) \). 

2. **Identify the midpoints of the sides**: The midpoints of the sides are \( (0.5,0) \), \( (1,0.5) \), \( (0.5,1) \), and \( (0,0.5) \).

3. **Determine the coordinates of points \( A, B, C, \) and \( D \)**: 
   - Point \( A \) is the intersection of the line from \( (0,0) \) to \( (1,0.5) \) and the line from \( (0,1) \) to \( (0.5,0) \).
   - Point \( B \) is the intersection of the line from \( (0,0) \) to \( (0.5,1) \) and the line from \( (1,0) \) to \( (0,0.5) \).
   - Point \( C \) is the intersection of the line from \( (1,0) \) to \( (0.5,1) \) and the line from \( (1,1) \) to \( (0,0.5) \).
   - Point \( D \) is the intersection of the line from \( (1,1) \) to \( (0.5,0) \) and the line from \( (0,1) \) to \( (1,0.5) \).

4. **Find the equations of the lines**:
   - The line from \( (0,0) \) to \( (1,0.5) \) has the equation \( y = 0.5x \).
   - The line from \( (0,1) \) to \( (0.5,0) \) has the equation \( y = 1 - 2x \).
   - The line from \( (0,0) \) to \( (0.5,1) \) has the equation \( y = 2x \).
   - The line from \( (1,0) \) to \( (0,0.5) \) has the equation \( y = 0.5 - 0.5x \).
   - The line from \( (1,0) \) to \( (0.5,1) \) has the equation \( y = 2x - 1 \).
   - The line from \( (1,1) \) to \( (0,0.5) \) has the equation \( y = 1.5 - 0.5x \).
   - The line from \( (1,1) \) to \( (0.5,0) \) has the equation \( y = 2 - 2x \).
   - The line from \( (0,1) \) to \( (1,0.5) \) has the equation \( y = 1 - 0.5x \).

5. **Find the intersection points**:
   - Point \( A \) is the intersection of \( y = 0.5x \) and \( y = 1 - 2x \):
     \[
     0.5x = 1 - 2x \implies 2.5x = 1 \implies x = \frac{2}{5}, \quad y = 0.5 \times \frac{2}{5} = \frac{1}{5}
     \]
     So, \( A = \left( \frac{2}{5}, \frac{1}{5} \right) \).

   - Point \( B \) is the intersection of \( y = 2x \) and \( y = 0.5 - 0.5x \):
     \[
     2x = 0.5 - 0.5x \implies 2.5x = 0.5 \implies x = \frac{1}{5}, \quad y = 2 \times \frac{1}{5} = \frac{2}{5}
     \]
     So, \( B = \left( \frac{1}{5}, \frac{2}{5} \right) \).

   - Point \( C \) is the intersection of \( y = 2x - 1 \) and \( y = 1.5 - 0.5x \):
     \[
     2x - 1 = 1.5 - 0.5x \implies 2.5x = 2.5 \implies x = 1, \quad y = 2 \times 1 - 1 = 1
     \]
     So, \( C = (1, 1) \).

   - Point \( D \) is the intersection of \( y = 2 - 2x \) and \( y = 1 - 0.5x \):
     \[
     2 - 2x = 1 - 0.5x \implies 1.5x = 1 \implies x = \frac{2}{3}, \quad y = 2 - 2 \times \frac{2}{3} = \frac{2}{3}
     \]
     So, \( D = \left( \frac{2}{3}, \frac{2}{3} \right) \).

6. **Calculate the area of quadrilateral \(ABCD\)** using the Shoelace Theorem:
   \[
   \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|
   \]
   Substituting the coordinates \( A = \left( \frac{2}{5}, \frac{1}{5} \right) \), \( B = \left( \frac{1}{5}, \frac{2}{5} \right) \), \( C = (1, 1) \), \( D = \left( \frac{2}{3}, \frac{2}{3} \right) \):
   \[
   \text{Area} = \frac{1}{2} \left| \frac{2}{5} \cdot \frac{2}{5} + \frac{1}{5} \cdot 1 + 1 \cdot \frac{2}{3} + \frac{2}{3} \cdot \frac{1}{5} - \left( \frac{1}{5} \cdot \frac{1}{5} + \frac{2}{5} \cdot 1 + 1 \cdot \frac{2}{3} + \frac{2}{3} \cdot \frac{2}{5} \right) \right|
   \]
   Simplifying:
   \[
   \text{Area} = \frac{1}{2} \left| \frac{4}{25} + \frac{1}{5} + \frac{2}{3} + \frac{2}{15} - \left( \frac{1}{25} + \frac{2}{5} + \frac{2}{3} + \frac{4}{15} \right) \right|
   \]
   Converting all terms to a common denominator of 75:
   \[
   \text{Area} = \frac{1}{2} \left| \frac{12}{75} + \frac{15}{75} + \frac{50}{75} + \frac{10}{75} - \left( \frac{3}{75} + \frac{30}{75} + \frac{50}{75} + \frac{20}{75} \right) \right|
   \]
   \[
   \text{Area} = \frac{1}{2} \left| \frac{87}{75} - \frac{103}{75} \right| = \frac{1}{2} \left| \frac{-16}{75} \right| = \frac{1}{2} \cdot \frac{16}{75} = \frac{8}{75}
   \]

The final answer is \(\boxed{\frac{1}{5}}\).