Just a trivial fact: every polyhedron that is used in linear programming is convex, that is $$Ax \leq b$$ is convex, for a matrix $$A$$ and a (column) vector $$b$$.

Proof: Take any $$x', x''$$ that satisfy the system of inequalities $$Ax \leq b$$. Then, for $$0 \leq \lambda \leq 1$$, we have that $$\lambda Ax' \leq \lambda b$$, that is $$A \lambda x' \leq \lambda b$$. Similarly, for $$x''$$, we have that $$A (1-\lambda) x' \leq (1-\lambda) b$$. Summing the inequalities, we get: $A[\lambda x' + (1-\lambda) x''] \leq [\lambda + (1-\lambda)] b = b,$ which means that $$\hat{x} = \lambda x' + (1-\lambda) x''$$ is again a solution of the original set of inequalities, thus concluding the argument.