Is something wrong with this solution? We claim that the answer is $n = 4$. This works by taking a square and drawing one diagonal. For other $n$, we claim that every diagonal drawn in the triangulation must be a diameter of the circle it is circumscribed in. Indeed, suppose that $AB$ is a diagonal of the triangulation, and that $\triangle ABC, \triangle ABD$ are two triangles in the triangulation. Then we have that either $C$ is the reflection of $D$ over $AB$ or $ACBD$ is a parallelogram. In either case, $\angle ACB = \angle ADB$. Since $\angle ACB + \angle ADB = 180$, this implies that $\angle ACB = \angle ADB = 90$ and $AB$ is a diameter. Hence, since any two diameters intersect, at most one diagonal is drawn. This gives $n \le 4$, hence the answer. $\square$