Let $ABCD$ be a square. Let $E, Z$ be points on the sides $AB, CD$ of the square respectively, such that $DE\parallel BZ$. Assume that the triangles $\triangle EAD, \triangle ZCB$ and the parallelogram $BEDZ$ have the same area. 

If the distance between the parallel lines $DE$ and $BZ$ is equal to $1$, determine the area of the square.
1. **Identify the given conditions and relationships:**
   - $ABCD$ is a square.
   - $E$ and $Z$ are points on sides $AB$ and $CD$ respectively.
   - $DE \parallel BZ$.
   - The areas of $\triangle EAD$, $\triangle ZCB$, and parallelogram $BEDZ$ are equal.
   - The distance between the parallel lines $DE$ and $BZ$ is $1$.

2. **Express the areas of the given shapes:**
   - Let the side length of the square be $s$.
   - The area of the square is $s^2$.
   - The area of parallelogram $BEDZ$ is $DE \times 1 = DE$.
   - The area of $\triangle EAD$ is $\frac{1}{2} \times AD \times DE = \frac{1}{2} \times s \times DE$.
   - The area of $\triangle ZCB$ is $\frac{1}{2} \times CB \times BZ = \frac{1}{2} \times s \times BZ$.

3. **Set up the equation for the areas:**
   - Since the areas of $\triangle EAD$, $\triangle ZCB$, and parallelogram $BEDZ$ are equal, we have:
     \[
     \text{Area of } \triangle EAD = \text{Area of } \triangle ZCB = \text{Area of parallelogram } BEDZ = DE
     \]
   - Therefore:
     \[
     \frac{1}{2} \times s \times DE = DE
     \]
   - Solving for $DE$:
     \[
     \frac{1}{2} s \times DE = DE \implies \frac{1}{2} s = 1 \implies s = 2
     \]

4. **Verify the area of the square:**
   - The side length of the square is $s = 2$.
   - The area of the square is:
     \[
     s^2 = 2^2 = 4
     \]

The final answer is $\boxed{4}$