Suppose there are matrix $A\in\mathbb{R}^{n \times m}$ and vector $b\in\mathbb{R}^n$. Consider a non-empty polyhedron $P = \{Ax \leq b\} $. Then, there exists a vector $\bar{x}\in P $ such that $A^\prime \bar{x} = b^\prime$ for some nonsingular $p \times p$ submatrix $A^\prime$ of $A$ and a subvector $b^\prime$ of $b$.
This comes from the polyhedral theory, but I have no idea where I start proving the above statement.
Here is my thoughts: since $P$ is not empty, there is a vector $\bar{x}\in P$. This means there is a subsystem $A^\prime x=B^\prime$ with nonsingular square matrix $A$ and the solution of this system is $\bar{x}$. Any comment please?