My question is related to this question.
Given a Linear Program (LP) $max\{c^Tx: Ax \leq b, x \geq 0\}$ with $x,c \in \mathbb{R}^n$ and $A \in \mathbb{R}^{m \times n}$. The Simplex Method needs a feasible basic solution in order to start finding the optimal solution. If $x_1=x_2=\cdots=x_n=0$ is a feasible solution, then this is also a feasible basic solution since the non-negativity constraint guarantees a corner point in the polyhedron induced by $Ax \leq b$.
It gets more interesting when $x_1=\cdots=x_n=0$ is not a feasible solution. In order to find a feasible basic solution for the simplex to start, we can formulate a new linear program as follows
$$ \begin{aligned} & \underset{}{\min} && x_0 \\ & \text{subject to} & & \sum_{j=1}^n a_{ij}x_j -x_0 \leq b_i, \ \forall i \in \{1,\cdots,m\} \\ &&& x_j \geq 0,\ j \in \{1, \cdots, n\} \\ \end{aligned} $$
First we notice that if our original LP has a feasible solution, then our new LP has an optimal solution and vice versa.
Now, we start with $x_1=\cdots=x_n=0$ for our new LP and increase $x_0$ until the inequalities are satisfied. Then, we carry out the Simplex Method on the new LP using the found starting solution and if $x_0=0$ is indeed the optimal solution, then we have found the basic feasible solution as a starting point for our original LP.
This method is known as the "Two-Phase Simplex".
It is clear to me why this works if we assume that the initial increase of $x_0$ results in a feasible basic solution for the new LP. But why does it?
Question: Why does the process of increasing $x_0$ results in a basic feasible solution (a corner point of the corresponding polyhedron) for the new LP? And to be exact here: It is clear to me why this results in a feasible solution but not why it results in a basic feasible solution (a corner point of the corresponding polyhedron).
