14. In space, there are five points, no four of which are coplanar. If several line segments are drawn such that no tetrahedron exists in the graph, then the maximum number of triangles in the graph is $\qquad$.
14.4.

First, construct graph 6. It is easy to see that it meets the conditions and has exactly four triangles.

Now assume there exists some configuration where the number of triangles is no less than five.

If only two line segments are not connected, then these two line segments must have no common endpoints (as shown in graph 6), otherwise, a tetrahedron would exist. But there are only four triangles, which is a contradiction.

If at least three line segments are not connected, when one of these line segments serves as a side of three triangles, as shown in graph 7, there are only three triangles; when each line segment serves as a side of at most two triangles, then there are at most $\left[\frac{\left(\mathrm{C}_{5}^{2}-3\right) \times 2}{3}\right]=4$ triangles.