Given triangle ABC, if D is an interior pt and E is an exterior pt, segment DE intersects triangle ABC?

Question

This is regarding neutral geometry I think. It seems to be obviously true but I struggle to prove it.

My 'proof' goes by the following: since $E\notin\mathrm{ int}(\triangle{ABC})$, there exist an angle in $\triangle{ABC}$ s.t. $E$ not in the interior of that angle.

So say $E \notin \mathrm{int}(\angle{ABC})$

Then either $E$ is on the opposite side of line $BA$ as $C$ or on opposite side of line $BC$ as $A$

So say $E$ is on the opposite side of line $BA$ as $C$

Notice $D$ is on the same side of line $BA$ as $C$ using plane separation property.

By plane separation, $D$ is on the opposite side of line $BA$ as $E$. This implies there exist a point $F \notin {D,E}$ s.t. $DE\cap BA = F$.

If $F$ is in segment $AB$, then we are done. Otherwise, the proof goes really long and I have no idea if it is correct nor going the right way. I really appreciate any help.

Answer 1

How to prove this theorem depends on what you're allowed to use, that is, on what axioms you are using and on what results you have already proved. Since we don't know where you're starting from, we find it hard to give you an answer. If you can get your hands on a copy of Prenowitz and Jordan, Basic Concepts of Geometry, you'll find everything you need in Chapter 13, Separation properties of angles and triangles.

Answer 2

Well, I haven't managed to obtain a hard copy of the book Gerry mentioned, but I managed to come up an elegant solution to this problem on my own. So I would like to share this as a courtesy to you guys' help.

All we need to do is to prove a similar problem: Given $\triangle{ABC}$, if $D \in\mathrm{ int}(\triangle{ABC})$, $E\neq D$ & $DE\cap\triangle{ABC}=\phi$, then $E\in\mathrm{int}(\triangle{ABC})$.

This similar problem is essentially the contrapositive of the original problem, but the proof for this is much easier. The outline of the proof is the following:

1. First prove $E\notin\overleftrightarrow{AB}\cup\overleftrightarrow{AC}\cup\overleftrightarrow{BC}$
2. Then prove $DE\cap(\overleftrightarrow{AB}\cup\overleftrightarrow{AC}\cup\overleftrightarrow{BC})=\phi$. Notice this would need the result obtained in $1$ by assuming $\exists F \in DE$ s.t. $F \in\overleftrightarrow{AB}\cup\overleftrightarrow{AC}\cup\overleftrightarrow{BC}$ and derive the contradiction.
3. Then $E\in \mathrm{int}(\triangle{ABC})$ can be proven using plane separation axiom/theorem.