If you have a point at the boundary, finding one in the interior is pretty trivial. The wording of the first post says that he would be satisfied with that.

Maybe I missed it, but I don't think a better solution has been suggested so far... What you suggested returns the same answer if you flip one of the inequalities, so it can't possibly return something in the interior in both cases.

I don't have inequalities in my solution. What I suggested disregards direction because it already assumes that the feasible region of the half spaces defines a convex polyhedron. If we'd flip one of the inequalities, the feasible region would be empty.

No, that's not how it works. Imagine you have a bunch of half-spaces that already define a convex polyhedron. Now take a plane that cuts that convex polyhedron through the middle, and either add one half-space or the other to the list. In both situations the feasible region is not empty, and your proposed solution will return the same answer.