10. Given five points $A, B, C, D, E$ on a plane, no three of which are collinear. By connecting these points with 4 lines, each point is an endpoint of at least one line segment. The number of different connection methods is $\qquad$ kinds.
10.135.

In Figure 5, 4 connection methods meet the requirements of the problem (the diagram only represents the connection form between points and lines, without considering the position of the points).

Case (1), based on the choice of the central point, there are 5 connection methods;

Case (2), it can be considered as the permutation of 5 points $A, B, C, D, E$, but one permutation and its reverse permutation are the same, and they are in a one-to-one correspondence, so there are $\frac{5!}{2}=60$ connection methods;

Case (3), first, the choice of the branching point has 5 options, secondly, the selection of the two points for the fork has $\mathrm{C}_{4}^{2}=6$ options, and finally, the order of the remaining two connected points is different, with 2! options. In total, there are $5 \times 6 \times 2=60$ methods;

Case (4), choosing three points to construct a triangle, there are $\mathrm{C}_{5}^{3}=$ 10 options.
In total, there are $5+60+60+10=135$ connection methods.