Given a polyhedron $P$ we have two vertices $x'$, $y'$ to be adjacent, if their convex hull is a one-dimensional face of the polyhedron. I want to show that this is equivalent to the fact, that there is a vector $c \in \mathbb{R}^n$, s.t. $x'$, $y'$ are the only two vertices of the polyhedron which maximize $c^tx$.
My line of thought for the one direction is to choose two facets $F_1,$ $F_2$ of the polyhedron who live in affine hyperplanes $H_1$, $H_2$, then choose two normal vectors $c_1$, $c_2$ on the hyperplanes and claim that $c:=\frac{1}{2} (c_1+c_2)$ does the job. I still have trouble with the details (esp. why those facets exist and why no third vertex can lie in the orthogonal complement of $c$).
For the other direction I have no clue at all which route to follow.
No, in general the average of normal vectors to the incident facets won't work (and in higher dimensions, an edge will be in more than two facets).
The definition of a face is that it is the intersection of a supporting hyperplane with $P$, yes? Take $c$ to be the normal vector to a supporting hyperplane of the one-dimensional face $[x', y']$.