Let $T$ be a lattice triangle in $\Bbb R^2$ (i.e. the convex hull of three noncolinear points in $\Bbb Z^2$), and assume it has at least one interior lattice point. Is it always possible to find a point $p\in \operatorname{interior}(T)\cap \Bbb Z^2$ satisfying the following condition:

(*) $\operatorname{conv}(p,v)\cap \Bbb Z^2 = \{p,v\}$ for every vertex $v\in T$, where $\operatorname{conv}$ is convex hull.

For example: In this picture:

$p=D$ works, whereas in this picture:

$p=D$ does not work because the line segment $DC$ passes through $E$.

**Edit:** In light of fedja's counterexample (the triangle with vertices $(0,0)$, $(3,2)$, and $(1,3)$, and at their suggestion, I'm modifying the question to ask if (*) holds when $T$ contains "sufficiently many" points in the interior.