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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.2012-02-08
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    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...2012-02-10
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    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 :)2012-02-12

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