8. Given six points in space $A, B, C, D, E, F$ with no four points being coplanar. Then the maximum number of line segments that can be drawn so that no tetrahedron exists in the graph is $\qquad$.
8. 12 .

Notice that, among six points, at most $\mathrm{C}_{6}^{2}=15$ line segments can be connected, forming $\mathrm{C}_{6}^{4}=15$ tetrahedra, with each line segment corresponding to $\mathrm{C}_{4}^{2}=6$ tetrahedra. If the number of removed edges is less than 3, there will still be at least one tetrahedron. Therefore, to meet the requirement, the maximum number of line segments that can be connected is 12.

Furthermore, when connecting $A C, A D, A E, A F, B C, B D, B E, B F, C D, D E, C F, E F$, the requirement is satisfied, thus 12 line segments can be connected.