If 3 diagonals of an $n$-gon meet in one point that is not a vertex of the $n$-gon then these diagonals have 6 endpoints in total, implying $n \ge 6$. If $n = 6$ then the only way to leave the endpoints of every two diagonals to different sides of the third diagonal is connecting each vertex to the opposite one (Fig. 17), but the obtained diagonals meet in the center of the $n$-gon. Suppose that there exist 3 diagonals satisfying the conditions for $n = 7$. Let $A$ be the vertex that is not an endpoint of any of the diagonals. Let $B$ be a vertex next to $A$ and let $C$ be the other endpoint of the diagonal whose one endpoint is $B$. Two endpoints of diagonals must lie on the same side of $BC$ as $A$ and two endpoints must lie on the other side. There is only one possibility to connect these points with two intersecting diagonals (Fig. 18). As these diagonals are symmetric w.r.t. the perpendicular bisector of $AB$, their common point $P$ lies on the perpendicular bisector of $AB$. As $C$ also lies on the perpendicular bisector and $C \neq P$, diagonal $BC$ could pass through $P$ only if $B$ were also located on the perpendicular bisector of $BC$, which is not the case. Hence finding the required 3 diagonals is impossible for $n = 7$.

For $n = 8$, draw one diagonal from some vertex to the opposite vertex. Adding two shorter diagonals symmetrically w.r.t. the first diagonal, all three intersect in one point inside the polygon that is not its center. (Fig. 19).

Fig. 17

Fig. 18

Fig. 19