Suppose we are given a finite plane connected graph $G$, whose edges are represented by lines. We can give a clockwise cyclic ordering to the edges adjacent to a vertex $v$ in the following way:
Take $r > 0$ to be small enough, then the circle around $v$ with radius $r$ will only intersect the edges adjacent to $v$. Thus the clockwise cyclic ordering of this circle will induce a cyclic ordering of the edges adjacent to $v$.
The following is somewhat in the spirit of half-edges:
Each edge $e = \{u,v\}$, we can orient to obtain the directed edges $(u,v)$ and $(v,u)$. To such directed edges we can define a succsessor, by setting $s(u,v) = (v,w)$, where $\{v,w\}$ should be the next edge after $\{u,v\}$ in the clockwise ordering of the edges adjacent to $v$.
Now assume we are given a face $f$ of $G$, and an edge $\{u,v\}$ on its boundrary, such that $f$ appears "on the left side" of the directed edge $(u,v)$. If we continuously take successors of $(u,v)$ we obtain a cycle of directed edges.
Now intuitivly, the corresponding set of undirected edges in $G$ should be the boundary of the face $f$, since by taking succsessors we "comlepetly walk around $f$". Is this true ? How can we proove this ?