Show convex combination If I have a bounded set $F$ in $N$ dimensional space and another set $G$ where every element $g$ in $G$ has $h'g=c$ and also must exist in $F$. $H$ is a vector in the $N$ dimensional space and $c$ is any constant $1\times 1$ matrix (scalar). $h$ is a vector of appropriate dimension.
How can I prove every extreme point of $G$ lies on the boundary of $F$?
That is to say if $x$ and $y$ are extreme points in $F$ then $xλ + (1-λ)y = g$

I'll prove that every point of $G$ in the interior of $F$ is not an extreme point of $G$. I'll assume that $N>1$.
LEMMA. There is a vector $v\neq 0$ such that $h'v=0$.
Proof. Since $N>1$ there is a vector $u$ which is not a multiple of $h$. Let
$$v=u-\left({{h'u}\over{h'h}}\right)h$$
Then $h'v=0$. Since $u$ is not a multiple of $h$, $v\neq 0$.
Answer. Suppose $x \in G$ and $x$ is also in the interior of $F$. Let $v$ be as in the lemma above. Then there is a small enough $\lambda$ such that $(x \pm \lambda v) \in F$. But $h'(x \pm \lambda v)=h'x$ so $(x \pm \lambda v) \in G$. But,
$$x= {(x + \lambda v) \over 2} + {(x - \lambda v) \over 2}$$
So $x$ is not an extreme point of $G$.
