How many kites are there such that all of its four vertices are vertices of a given regular icosagon ($20$-gon)?

$ 
\textbf{(A)}\ 105
\qquad\textbf{(B)}\ 100
\qquad\textbf{(C)}\ 95
\qquad\textbf{(D)}\ 90
\qquad\textbf{(E)}\ 85
$
1. **Identify the properties of a kite in a regular icosagon:**
   - A kite is a quadrilateral with two distinct pairs of adjacent sides that are equal.
   - In a regular icosagon (20-gon), a kite can be formed by selecting two pairs of vertices such that one pair forms a diagonal (diameter) and the other pair forms another diagonal that is perpendicular to the first.

2. **Count the number of diameters in the icosagon:**
   - A regular icosagon has 20 vertices.
   - Each vertex can be paired with the vertex directly opposite to it to form a diameter.
   - Therefore, the number of diameters in a 20-gon is:
     \[
     \frac{20}{2} = 10
     \]

3. **Count the number of ways to choose a perpendicular diagonal:**
   - For each diameter, there are 9 other vertices that can be chosen to form a perpendicular diagonal.
   - This is because the perpendicular diagonal must also pass through the center of the icosagon, and there are 9 other vertices on one side of the diameter.

4. **Calculate the total number of kites:**
   - For each of the 10 diameters, there are 9 ways to choose the perpendicular diagonal.
   - Therefore, the total number of kites is:
     \[
     10 \times 9 = 90
     \]

5. **Adjust for overcounting:**
   - Some kites are counted twice when both diagonals are diameters.
   - There are 5 such cases (since there are 10 diameters, and choosing any two of them forms a kite, but each pair is counted twice).
   - Therefore, we subtract these 5 cases from the total:
     \[
     90 - 5 = 85
     \]

Conclusion:
\[
\boxed{85}
\]