In how many ways, can we draw $n-3$ diagonals of a $n$-gon with equal sides and equal angles such that:
$i)$ none of them intersect each other in the polygonal.
$ii)$ each of the produced triangles has at least one common side with the polygonal.
1. **Understanding the problem**: We need to draw \( n-3 \) diagonals in an \( n \)-gon such that no two diagonals intersect inside the polygon, and each triangle formed by these diagonals has at least one side that is a side of the \( n \)-gon.

2. **Triangulation of an \( n \)-gon**: A triangulation of an \( n \)-gon is a division of the polygon into \( n-2 \) triangles by drawing \( n-3 \) non-intersecting diagonals. Each triangle must share at least one side with the \( n \)-gon.

3. **Counting the number of triangulations**: The number of ways to triangulate an \( n \)-gon is given by the \((n-2)\)-th Catalan number, \( C_{n-2} \). The \( k \)-th Catalan number is given by:
   \[
   C_k = \frac{1}{k+1} \binom{2k}{k}
   \]
   For our problem, we need \( C_{n-2} \):
   \[
   C_{n-2} = \frac{1}{(n-2)+1} \binom{2(n-2)}{n-2} = \frac{1}{n-1} \binom{2n-4}{n-2}
   \]

4. **Verification with smaller \( n \)**:
   - For \( n = 3 \), there are no diagonals, and the number of ways is \( C_1 = 1 \).
   - For \( n = 4 \), there is one diagonal, and the number of ways is \( C_2 = 2 \).
   - For \( n = 5 \), there are two diagonals, and the number of ways is \( C_3 = 5 \).
   - For \( n = 6 \), there are three diagonals, and the number of ways is \( C_4 = 14 \).

5. **Conclusion**: The number of ways to draw \( n-3 \) diagonals in an \( n \)-gon such that no two diagonals intersect inside the polygon and each triangle formed has at least one side that is a side of the \( n \)-gon is given by the \((n-2)\)-th Catalan number:
   \[
   \boxed{\frac{1}{n-1} \binom{2n-4}{n-2}}
   \]