Median points collinear: At least one outside triangle

Question

Suppose that ABC is a triangle and that $A'\in l_{BC}$, $B'\in l_{AC}$, and that $C'\in l_{AB}$. Prove that if $A', B', C'$ are collinear, then at least one of these points must be outside of the triangle.

I'm not sure how to prove that it is outside of the triangle. I know that one way to prove if an element is outside of the triangle is to see if all of the cross products have the same sign for their third component, but I don't think that can help here.

Answer 1

It is a direct consequence of Menelaus' theorem :

https://en.wikipedia.org/wiki/Menelaus'_theorem

Answer 2

Here is a proof that uses pigeonhole principle:

Assume that $A'$, $B'$ and $C'$ are collinear.

• If at least one of three vertices of $\triangle ABC$ is on line $A'B'$, then at least one of two other points must be outside of line $A'B'$ (since $A$, $B$ and $C$ can't be collinear). Therefore, line $A'B'$ can't intersect the segment between these two points.
• If none of the vertices of $\triangle ABC$ is on the line $A'B'$, then, by pigeonhole principle, at least two of them must lie on the same half plane. Therefore, line $A'B'$ cannot intersect segment between these two points in the same half plane.

Note: Above proof assumes vertices counts as outside of the triangle as per OP's comment.