Five numbers are given in ascending order, which are the lengths of the sides of a quadrilateral (not crossed, but not necessarily convex, meaning a diagonal is not necessarily inside the polygon) and one of its diagonals $D$. These numbers are $3, 5, 7, 13$ and 19. What can be the length of the diagonal $D$?
We can rephrase the problem: let $a, b, c, d$ be the lengths of the sides of the quadrilateral, and $e$ be the length of the diagonal that separates, on one side, the sides of lengths $a$ and $b$, and on the other side, the sides of lengths $d$ and $e$. It is necessary and sufficient that the triplets $(a, b, e)$ and $(c, d, e)$ satisfy the triangle inequality.

If $e=3$ or $e=5$, the triplet containing 19 will not work because $19-3>13$ and $19-5>13$: the side of length 19 is too large.

If $e=13$, it is required simultaneously that $a+b>13$ and $c+d>19$. However, if $a$ or $b=19$, then $c+d \leq 5+7<13$, and similarly if $c$ or $d=19$. Therefore, we cannot have $e=13$. We verify that the same applies for
$e=19$.

Thus, only $e=7$ remains. In this case, we can have $a=3, b=5, c=13, d=19$ for example.