Example 1 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. ${ }^{[1]}$
Let the graph that satisfies the conditions be $G(V, E)$.
First, we prove a lemma.
Lemma In any $n(n \leqslant 5)$-order subgraph $G^{\prime}$ of graph $G(V, E)$, there can be at most five edges.
Proof It suffices to prove the case when $n=5$.
If there exists a vertex $A$ in $G^{\prime}$ with a degree of 4, then no edges can be connected between any other two vertices. Therefore, the subgraph $G^{\prime}$ has exactly four edges;

If there exists a vertex $A$ in $G^{\prime}$ with a degree of 3, then the three vertices connected to $A$ cannot be connected to each other, and the remaining vertex can have at most one edge. Therefore, the subgraph $G^{\prime}$ has at most four edges;

If the degree of each vertex $A$ in $G^{\prime}$ is no more than 2, then the subgraph $G^{\prime}$ can have at most $\frac{5 \times 2}{2}=5$ edges.
The lemma is proved.
For the graph $G(V, E)$, let $A$ be one of the vertices in $G$ with the most edges, and let $d(A)=s$. Let the set of vertices connected to $A$ be $S$, and $|S|=s$. Then no edges can be connected between the vertices in set $S$. Let the set of vertices other than $A$ and the vertices in set $S$ be $T$.
If $s \geqslant 4$, then $|T|=9-s \leqslant 5$.
By the lemma, the vertices in set $T$ can have at most five edges.

Since each vertex in set $T$ can be connected to at most one vertex in set $S$, we have
$|E| \leqslant s+5+(9-s)=14$.
If $s \leqslant 3$, then $|E| \leqslant \frac{3 \times 10}{2}=15$.
Therefore, in either case, the graph $G(V, E)$ can have at most 15 edges.

The graph shown in Figure 1 has 15 edges and meets the requirements.

In conclusion, the maximum number of edges is 15.