By Belyaev O.

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Conversely, suppose A ≺ B and B ≺ C in the given direct order on a. 3. If A, B ∈ OP , C = O then (A ≺ B)a ⇒ (B ≺ A)OP ⇒ [ABO]. 1 If A, B ∈ OP , C ∈ OQ then [ABO] & [BOC] =⇒ [ABC]. 11. 1 For A ∈ OP , B, C ∈ OQ we have [AOB] & [OBC] =⇒ [ABC]. If A = O and B, C ∈ OQ , we have B ≺ C ⇒ [OBC]. 1. e. that the interval A1 An is divided into n − 1 intervals A1 A2 , A2 A3 , . . , An−1 An (by the points A2 , A3 , . . An−1 ). Then in any order (direct or inverse), defined on the line containing these points, we have either A1 ≺ A2 ≺ .

4 OB ⊂ Int∠AOC. 20. 88 ¯ By the definition of adjacency Proof. (See Fig. ) By definition of the interior, A ∈ Int∠(k, m) ⇒ Amk. 5 ¯ ¯ ¯ ⇒ A ∈ Ext∠(h, k). ✷ ∠(k, m) = adj(h, k) ⇒ hkm. 21. 1. If points B, C lie on one side of a line aOA , and OB = OC , either the ray OB lies inside the angle ∠AOC, or the ray OC lies inside the angle ∠AOB. 2. Furthermore, if a point E lies inside the angle ∠BOC, it lies on the same side of aOA as B and C. That is, Int∠BOC ⊂ (aOA )B = (aOA )C . 15 c Proof. 1. Denote OD ⇋ OA .

48: If points B, C lie on one side of aOA , and OB = OC , either OB lies inside ∠AOC, or OC lies inside ∠AOB. 22. If a ray l with the same initial point as rays h, k lies inside the angle ∠(h, k) formed by them, then the ray k lies inside the angle ∠(hc , l). Proof. 15 we have l ⊂ Int∠(h, k) ⇒ k ⊂ Ext∠(h, l) & lk ¯h & l = k ⇒ k ⊂ Int∠(hc , l). 23. If open intervals (AF ), (EB) meet in a point G and there are three points in the set {A, F, E, B} known not to colline, the ray EB lies inside the angle ∠AEF .