Pigeonhole principle: Five points on an orange

Question

Five points are drawn on the surface of an orange. Prove that it is possible to cut the orange in half in such a way that at least four of the points are on the same hemisphere. (Any points lying along the cut count as being on both hemispheres.)

Answer 1

Through any two points, there is a great circle that passes through those two points. Such a cut will split the other 3 pigeons — oh, I mean points — among 2 halves.

[You can now handle additional points being on the great circle on your own, I believe.]

Answer 2

Given a sphere with radius $1$.

Use the points

$$ A_+(1,0,0), \quad A_-(-1,0,0), \quad B_+(0,1,0), \quad B_-(0,-1,0), \quad C(0,0,1) $$

And you cannot make such a cut.

As for any great circle, $(A_+,A_-)$ and $(B_+,B_-)$ are never on the same hemisphere.

That is if we exclude points on both hemispheres as being cut through a point.