Let $\ell$ be a closed polyline on a plane, i.e. $$\ell = \bigcup_{i=1}^n A_iA_{i+1}$$ where $n\ge 3$ is an integer and $A_1, \ldots, A_n$ are points on a plane such that for every $i=1,2,\ldots, n$ points $A_i, A_{i+1}, A_{i+2}$ are not collinear (we assume that $A_{n+1}=A_1$ and $A_{n+2}=A_2$).
Show that there exists a convex polygon $B_1B_2\ldots B_n$ such that for every $i=1,2,\ldots,n$ $$B_iB_{i+1}=A_iA_{i+1}$$ (again, $B_{n+1}=B_1$).
$n$ sides $l_i=A_iA_{i+1}$ can form a convex polygon if and only if every side is less than the sum of the other sides. If we define $n$ vectors $\vec{l_i}=A_{i+1}-A_i$, then by definition $\sum_{i=0}^n\vec{l_i}=0$. It follows that for every $k$ between $1$ and $n$: $$ l_k=|\vec{l_k}|=\left|\sum_{i\ne k}\vec{l_i}\right|\le \sum_{i\ne k}|\vec{l_i}|=\sum_{i\ne k}{l_i}. $$ Notice that inequality is strict if vectors $\vec{l_i}$ are not all on the same line.