Consider a matrix $A_{m\times n}$ and $b_{n\times 1}$. Let the space be $\mathbb R^n$. Let $S:=\{ x\in\mathbb R^n:Ax\le b, x\ge 0\}$ and $T:=\{x\in\mathbb R^n:Ax=b, x\ge 0\}$.
I want to show that if $\bar x$ is an extreme point$^1$ for $T$ then it is also an extreme point for $S$. But I have no idea how to proceed. Any help?
Edit:
I just realized that I can use the result (that we've proven) that if a constraint of $S$ is binding at $\bar x$, and if $\bar x$ is a strict convex combination of $a, b\in S$, then the same constraint is also binding at $a$ and $b$. Since all the constraints of $S$ are binding for any point in $T$, we have that $a, b\in S$, leading a contradiction.
Am I correct?
$^1x_0$ is said to be an extreme point in a convex set $A$ iff $x_0\in A$ and it cannot be represented as a strict convex combination of two distinct points in $A$.