Linear programming vertex proof I want to prove that $x$ is vertex of $\{x\in \mathbb{R}^n\ \colon Ax\leqslant b, \ x\geqslant 0\}$ iff $(x,b-Ax)$ is the vertex of $\{(x,u)\in \mathbb{R}^{n+m}\ \colon Ax+u=b, \ x\geqslant 0, \ u\geqslant 0\}$. Here $A$ is a $m\times n$ matrix. 
Definition is: $x \in X$ is called a vertex of a set $X$ if $x$ can't be represented as $x=\lambda y+(1-\lambda)z$, where $y,z\in X$, $y\neq z$ and $\lambda\in (0,1)$.
Any ideas?

Denote the first polyhedral as $P_1$ and the second polyhedral be $P_2$.
Part 1:
Suppose $x \in P_1$ is not a vertex of $P_1$, then $\exists y,z \in P_1$ such that $$x=\lambda y+(1-\lambda)z$$
check that $\begin{pmatrix} x \\ b-Ax \end{pmatrix}\in P_2$
and we have $$\begin{pmatrix} x \\ b-Ax \end{pmatrix}=\lambda \begin{pmatrix}y \\b-Ay \end{pmatrix}+(1-\lambda )\begin{pmatrix}z \\b-Az \end{pmatrix}$$
Notice that $y \in P_1 \implies \begin{pmatrix}y \\b-Ay \end{pmatrix} \in P_2$ and $z \in P_1 \implies \begin{pmatrix}z \\b-Az \end{pmatrix} \in P_2.$
Hence $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is not a vertex of $P_2$.
Part 2:
Conversely, if $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is not a vertex of $P_2$, then $\exists \begin{pmatrix} y \\ u_u \end{pmatrix}$, $\begin{pmatrix} z \\ u_z \end{pmatrix} \in P_2$.
From definition of $P_2$, we can deduce that $u_y=b-Ay$ and $u_z=b-Az$ and we have 
$$\begin{pmatrix} x \\ b-Ax \end{pmatrix}=\lambda \begin{pmatrix}y \\b-Ay \end{pmatrix}+(1-\lambda )\begin{pmatrix}z \\b-Az \end{pmatrix}$$
and we have $$x=\lambda y+(1-\lambda ) z$$
we have to check that $y,z \in P_1$, are you able to prove this?
If you can achieve this, you have proven that $x$ is not a vertex of $P_1$.
Summary:
$x \in P_1$ is not a vertex of $P_1$ if and of if $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is not a vertex of $P_2$ 
$x \in P_1$ is a vertex of $P_1$ if and of if $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is a vertex of $P_2$ 
