By J.N. Coldstream
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41: Given ∠(h, k)O and a point C inside it, for any points D on h and F on k, OC meets (DF ). 42: If a point B lies inside an angle ∠AOC, the angles ∠AOB, ∠BOC are adjacent. 10 Proof. B ∈ Int∠AOC =⇒ ∃D D ∈ OB & [ADC]. 20) lie on opposite sides of the line aOB . Together with the fact that the angles ∠AOB, ∠BOC share the side OB this means that ∠AOB, ∠BOC are adjacent. 12. In an angle ∠(k, m), adjacent to an angle ∠(h, k), the side m lies outside ∠(h, k). 13. If a point B lies inside an angle ∠AOC, neither the ray OC has any points inside or on the angle ∠AOB, nor the ray OA has any points inside or on ∠BOC.
E. a pair of parallel lines a, b), we define its interior, written Int ab, as the set of points lying on the same side of the line a as the line b and on the same side of the line b as the line a. 62 Equivalently, we could take some points A on a and B on b and define Int ab as the intersection aB ∩ bA . 16. If A ∈ a, B ∈ b, and a b then (AB) ⊂ Int ab. Furthermore, (AB) = PaAB ∩ Int ab. 2 Proof. 9. On the other hand, C ∈ aAB =⇒ C = A ∨ C = B ∨ [ABC] ∨ [ABC] ∨ [CAB]. From the definition of the interior of the strip ab it is evident that C ∈ Int ab contradicts all of these options except [ABC], which means that PaAB ∩ Int ab ⊂ (AB) .
4, which, of course, does not use the present lemma or any results following from it. 3 - see footnote accompanying that lemma. 48 Taking into account the following two facts lowers the number of cases to consider (cf. 6): If a point A ∈ {O} ∪ O Q precedes a point B ∈ a, then B ∈ OQ . If a point A precedes a point B ∈ OP ∪ {O}, then A ∈ OP . 49 Again, for brevity we shall usually leave out the word ”abstract” whenever there is no danger of confusion. e. (OA) = OA ∩ AO . 27: OA is complementary to OA The concept of a non-zero ordered abstract interval is intimately related to the concept of line order.