P is a bounded polyhedron in $\mathbb{R}^n$, $a$ a vector in $\mathbb{R}^n$, and $b$ some scalar. Define $$Q = {x \in P | a'x = b}$$. Show that every extreme point in Q is either an extreme point of P or a convex combination of two adjacent extreme points of P.
Intuitively, this makes complete sense to me. For example, given the 2D case, Q is simple a line within the polyhedron P. Thus, its two intersections with P are either on extreme points, or on the "boundary" of two extreme points, which can be written as a convex combination of two adjacent extreme points.
I have absolutely no idea how to approach this and prove it. Any suggestions would be greatly appreciated!
Suppose $x$ is an extreme point in $Q$. If this point coincides with an extreme point in $P$, there is nothing to show.
Suppose it is not an extreme point in $P$.
Since it's an extreme point in $Q$, it is also a basic feasible solution, which means that $n$ of the constraints are active in $x$.
One of these constraints is $a'x = b$. the other active constraints are $n-1$ of those constraints that are associated with the polyhedron $P$.
Two adjacent extreme points are two points that by definition share $n-1$ constraints. Therefore, $x$ lies in the segment that lies in between those two extreme points.