Given a parallelogram with area $1$ and we will construct lines where this lines connect a vertex with a midpoint of the side no adjacent to this vertex; with the $8$ lines formed we have  a octagon inside of the parallelogram. Determine the area of this octagon
1. **Understanding the Problem:**
   - We are given a parallelogram with an area of 1.
   - We need to construct lines connecting each vertex of the parallelogram to the midpoint of the side not adjacent to this vertex.
   - These lines will form an octagon inside the parallelogram.
   - We need to determine the area of this octagon.

2. **Constructing the Lines:**
   - Consider a parallelogram \(ABCD\) with vertices \(A, B, C, D\).
   - Let \(M, N, O, P\) be the midpoints of sides \(BC, CD, DA, AB\) respectively.
   - Draw lines \(AM, BN, CO, DP\). These lines will intersect inside the parallelogram forming an octagon.

3. **Analyzing the Octagon:**
   - The octagon is formed by the intersection of the lines connecting vertices to midpoints.
   - Each of these lines divides the parallelogram into two regions, one of which is a triangle and the other a quadrilateral.

4. **Calculating the Area:**
   - Each of the lines divides the parallelogram into two regions of equal area.
   - Since there are 4 such lines, they divide the parallelogram into 8 smaller regions.
   - The octagon is formed by excluding the 4 triangles at the corners of the parallelogram.
   - Each of these triangles has an area of \(\frac{1}{8}\) of the parallelogram's area.

5. **Area of the Octagon:**
   - The total area of the parallelogram is 1.
   - The area of each of the 4 triangles is \(\frac{1}{8}\).
   - The total area of the 4 triangles is \(4 \times \frac{1}{8} = \frac{1}{2}\).
   - Therefore, the area of the octagon is the total area of the parallelogram minus the area of the 4 triangles:
     \[
     \text{Area of the octagon} = 1 - \frac{1}{2} = \frac{1}{2}
     \]

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