Given $n$ lines each NOT passing through the origin $(0,0)$, and thus they can be represented by the equations $a_ix + b_iy - 1 = 0, i=1,...,n$. We know a line always cuts the whole plane into two half planes, and this representation of lines ensures that any point in the same half plane as the origin $(0,0)$ always makes $a_ix + b_iy - 1 < 0$. We define the half plane of a line where the origin locates as the inner side, and the other half plane as the outer side of the line. See Question formulate-constraints-such-that-the-intersection-area-of-4-half-planes-formed-b for more details.
What we want is to find: Under which conditions we can guarantee that all these $n$ lines NOT passing through the origin $(0,0)$ can make $S : = \{(x,y): \bigwedge a_ix + b_iy - 1 < 0, i = 1,...,n\}$ an n-sided polygon which harbors the origin inside. (Of course, the origin is definitely inside $S$, by definition of $S$.)
Formally speaking, The hypothesis we are going to prove is:
"These $n$ lines NOT passing through the origin $(0,0)$ can form a n-sided polygon $S$ which harbors the origin inside.
$\Leftrightarrow$ The $n$ normal vectors $[a_i, b_i]^T, i=1,...,n$ of all $n$ lines, when anchored at the origin $(0,0)$ (we can show thanks to the representation of lines, all the normal vectors point to the outer side of the lines), should not fit within any half plane whose edge passes through the origin.
$\Leftrightarrow$ When anchoring all the normal vectors $[a_i, b_i]^T, i=1,...,n$ at the origin $(0,0)$, we get an open quadrilateral with vertices $(a_i, b_i), i=1,...,n$ which harbors the origin inside."
An example sketch of $n=4$ is attached for your understanding:
Now I don't think the hypothesis is correct.
Considering the following example, although all 4 normal vectors of 4 lines don't fit within any half plane, S formed by the 4 lines (the yellow colored area) is not an enclosed quadrilateral (it's a triangular area instead):