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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.

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You might find it easier if you look at $\text{int}(\angle ADB)$, $\text{int}(\angle BDC)$ and $\text{int}(\angle CDA)$ and find which $E$ is in. –  Henry Feb 8 '12 at 9:01
    
hmm, but it seems more complicated going this way b/c then I will have to first prove the plane can be separated into 3 parts, namely the interior of the angles you mentioned, whereas my way only using the angles in the triangles. I guess I could do it when I have the spare time... –  Nal Ra Feb 10 '12 at 1:43
    
I am sorry. I didn't realize it at first, but when I was working on p.80 exer. 7.12 Geometry: Euclid & Beyond by Robin Hartshorne, I finally recognize what you are trying to say (which seems to involve easy application of crossbar theorem). Thank you very much for you assistance :) –  Nal Ra Feb 12 '12 at 12:52

2 Answers 2

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.

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Thank you Gerry. That book you mentioned indeed contains a theorem that is the same to my question. The axioms I am using are Hilbert axioms under neutral geometry so that book really fits. –  Nal Ra Feb 10 '12 at 1:36

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.
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