(8) Given a regular 200-gon $A_{1} A_{2} \cdots A_{200}$, connect the diagonals $A_{i} A_{i+9}(i=1$, $2, \cdots, 200)$, where $A_{i+200}=A_{i}(i=1,2, \cdots, 9)$, then these 200 diagonals have $\qquad$ different intersection points inside the regular 200-gon.
81600 Hint: For each diagonal $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, $A_{i+1}, A_{i+2}, \cdots, A_{i+8}$ each draw two diagonals intersecting with $A_{i} A_{i+9}$. Since each such intersection point is counted twice, the number of different intersection points inside the regular 200-gon for these 200 diagonals does not exceed $\frac{2 \times 8 \times 200}{2}=1600$ (points). Below we prove that no three such diagonals intersect at one point.

Since each diagonal is of equal length, these 200 diagonals are all tangent to a fixed circle. Because only two tangents can be drawn from an intersection point, it is impossible for three such diagonals to intersect at one point.