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.