8. Given 5 points $A, B, C, D, E$ on a plane, no three of which are collinear. By connecting these points with 4 line segments, each point being an endpoint of at least one line segment, how many different connection methods are there?
8. Solution As shown in the figure, four connection methods meet the requirements of the problem.
(The figure only represents the connection form between points and lines, without considering the position of the points)
(1) There is one central point connected to the other 4 points, with a total of 5 connection methods;
(2) It can be regarded as the permutation of 5 points $A, B, C, D, E$, but a permutation and its reverse permutation are the same connection method, and the two are in one-to-one correspondence, so there are a total of $\frac{5!}{2}=$ 60 connection methods;
(3) First, the selection of the branching point has 5 ways, then the selection of the two fork points has $C_{4}^{2}=6$ ways, and finally, the order of the remaining two connected points is different, with $2!$ ways, for a total of $5 \times 6 \times 2=60$ connection methods;
(4) Select 3 points to construct a triangle, with $C_{5}^{3}=10$ ways.
In total, there are $5+60+60+10=135$ connection methods.