Three. (50 points) Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn.
Three, using these ten points as vertices and the connected line segments as edges, we get a simple graph $G$ of order 10.
The following proves: The number of edges in graph $G$ does not exceed 15.
Let the vertices of graph $G$ be $v_{1}, v_{2}, \cdots, v_{10}$, with a total of $k$ edges, and use $\operatorname{deg}\left(v_{i}\right)$ to denote the degree of vertex $v_{i}$.

If $\operatorname{deg}\left(v_{i}\right) \leqslant 3$ for $i=1,2, \cdots, 10$, then $k=\frac{1}{2} \sum_{i=1}^{10} \operatorname{deg}\left(v_{i}\right) \leqslant \frac{1}{2} \times 10 \times 3=15$.

Assume there exists a vertex $v_{i}$ such that $\operatorname{deg}\left(v_{i}\right) \geqslant 4$. Without loss of generality, let $\operatorname{deg}\left(v_{1}\right)=n \geqslant 4$, and $v_{1}$ is adjacent to $v_{2}, v_{3}, \cdots, v_{n+1}$. Thus, there are no edges between $v_{2}, v_{3}, \cdots, v_{n+1}$, otherwise, a triangle would be formed. Therefore, there are exactly $n$ edges between $v_{1}, v_{2}, \cdots, v_{n+1}$.

For each $j(n+2 \leqslant j \leqslant 10)$, $v_{j}$ can be adjacent to at most one of $v_{2}, v_{3}$, $\cdots, v_{n+1}$ (otherwise, if $v_{j}$ is adjacent to $v_{s}$ and $v_{t}(2 \leqslant s<t \leqslant n+1)$, then $v_{1}, v_{s}, v_{j}, v_{t}$ would correspond to the four vertices of a spatial quadrilateral, which contradicts the given condition). Therefore, the number of edges between $v_{2}, v_{3}, \cdots, v_{n+1}$ and $v_{n+2}, v_{n+3}$, $\cdots, v_{10}$ is at most
$$
10-(n+1)=9-n \text {. }
$$

Among the $9-n$ vertices $v_{n+2}, v_{n+3}, \cdots, v_{n}$, since there are no triangles, by Turán's theorem, there are at most $\left[\frac{(9-n)^{2}}{4}\right]$ edges. Therefore, the number of edges in graph $G$ is
$$
\begin{array}{l}
k \leqslant n+(9-n)+\left[\frac{(9-n)^{2}}{4}\right] \\
=9+\left[\frac{(9-n)^{2}}{4}\right] \leqslant 9+\left[\frac{25}{4}\right]=15 .
\end{array}
$$

The graph in Figure 8 has 15 edges and meets the requirements. In summary, the maximum number of edges is 15.