Assume w.l.o.g. that the points marked on the circle are equally spaced and number the marked points counterclockwise with natural numbers $0$, $1$, $\ldots$, $7$. Consider a triangle with vertices marked at points $0$, $1$, $3$ and its $7$ copies obtained by rotating the original triangle counterclockwise by $\frac{1}{8}$, $\frac{2}{8}$, $\ldots$, $\frac{7}{8}$ of a full turn around the center of the circle (illustrated in Fig. 29 with different colors). These $8$ triangles do not share any sides because all sides of the original triangle have different lengths, and each rotation of a side with a specific length results in different segments. Therefore, it is possible to draw $8$ triangles under the given conditions.

Fig. 29

On the other hand, note that from each marked point, at most $7$ segments can be drawn to the remaining marked points. Each triangle uses either $0$ or $2$ of these segments. Thus, each marked point can be the endpoint of at most $6$ different triangle sides in total. Since we count each side twice (once at each endpoint), there can be at most $\frac{8 \cdot 6}{2}$, or $24$ different triangle sides. Since each triangle has $3$ sides, there can be at most $\frac{24}{3}$, or $8$ triangles.