Let $CH(S)$ be a convex hull of a finite set $S$ and denote the set of all the vertices of $CH(S)$ as $Vert(S)$. For a vertex $v \in Vert(S)$, it has an associated set $E(v)$ which is defined as $E(v)=$ { $x \in Vert(S)$ | $xv$ is an edge of $CH(S)$ }. 


My question is best stated using an example in 2-D: 


In this example, $S = ${ $v\_1, v\_2, v\_3, v\_4, v\_5 $} and $v\_1, v\_2, v\_3, v\_4, v\_5$ in clockwise order constitutes $Vert(S)$. There exists a line $L$ between $v\_1$ and $CH(S - ${ $v\_1$}$)$. If we link $v\_1$ with every other 4 points, we will obtain 4 new points $p\_2, p\_3, p\_4, p\_5$ on $L$, where $p\_i$ is the intersection of $L$ and the line $v\_1v\_i$. **You will find that { $p\_2, p\_5$ } $= Vert($ { $p\_2, p\_3, p\_4, p\_5$ } $)$**. 


Here comes my problem. Given a finite set $S$ in higher dimensional (i.e. >= 3) vector space, we can find $Vert(S) = $ { $v\_1, v\_2, \cdots, v\_h$ }. Now consider a point $v \in Vert(S)$. There exists a hyperplane between $v$ and $CH(S - ${$v$}$)$. If we link $v$ with every other points in $Vert(S)$, we will get a set $P$ containing $h - 1$ points on the hyperplane. 


Will the points in $P$ corresponding to $E(v)$ constitute $Vert(P)$? Or is there some related theorems?



Higher dimensional convex hull
I think the answer is yes. First observe that $CH(P) \subset CH(S)\cap H$: if $x\in CH(P)$ then $x$ is written as a convex combination of things which are convex combinations of vertices of $CH(S)$, so is a convex combination of vertices of $CH(S)$.


Then we note that $CH(S)\cap H$ has as its vertices the points in $P$ corresponding to $E(v)$ (in general this is going to be all of $E(v)$ are, which are all contained in $P$ of course so $CH(S)\cap H \subset CH(P)$ and the two sets are equal, and we know the vertices of $CH(S)\cap H$


To see that $CH(S) \cap H$ has the correct vertices, the faces of $CH(S) \cap H$ correspond to faces of $CH(S)$ with their dimension dropped by at most 1 (if the face happens to be parallel to the hyperplane, then it will not drop in dimension). The edges in $E(v)$ cannot be parallel to the hyperplane because that means the hyperplane does not separate $v$ and the corresponding vertex on the other side of the edge


