Five, (15 points) Draw five lines $l_{1}, l_{2}$, $\cdots, l_{5}$ on a plane, such that no two lines are parallel and no three lines intersect at the same point.
(1) How many intersection points do these five lines have? How many intersection points are there on each line? How many line segments are there on these five lines?
(2) What is the maximum number of isosceles triangles that can be formed with these line segments as sides? Please briefly explain the reason and draw the corresponding figures.
(1) As shown in Figure 5.
There are 10 intersection points formed by five lines, with exactly four intersection points on each line. Since there are 6 line segments on each line, the five lines together have
30 line segments.
(2) The maximum number of isosceles triangles that can be formed with these line segments as sides is 10.
Let the intersection point of $l_{a}$ and $l_{b}$ be $A_{ab}$.
Consider the isosceles triangles with the base on $l_{1}$, where $l_{1}$ has 4 intersection points $A_{1k} (k=2,3,4,5)$.
If the base is $A_{12} A_{13}$, then
$$
\angle A_{23} A_{12} A_{13} = \angle A_{23} A_{13} A_{12}
$$

is an acute angle. In this case, the line segment $A_{12} A_{14}$ cannot be the base of another isosceles triangle; otherwise, it would imply $l_{3} \parallel l_{4}$ or $l_{2} \parallel l_{4}$, which contradicts the given conditions. Similarly, $A_{12} A_{15}$, $A_{13} A_{14}$, and $A_{13} A_{15}$ cannot be the bases of any isosceles triangles.

Furthermore, when $A_{12} A_{13}$ is the base, $A_{14} A_{15}$ can also be the base of an isosceles triangle. Thus, on each line, there are at most two bases of isosceles triangles, which are formed by two different line segments. Therefore, the number of isosceles triangles with bases formed by these line segments is at most $2 \times 5 = 10$.

In fact, such a configuration of 10 isosceles triangles is possible.
As shown in Figure 6, a regular pentagon.