# A Protogeometric Nature Goddess from Knossos by J.N. Coldstream

March 9, 2017 | | By admin |

By J.N. Coldstream

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If B ∈ (−∞, O), then B ≺ O ≺ A, which contradicts the condition A ∼ B. If B ∈ (O, +∞), then C ≺ O ≺ B, which contradicts the condition B ∼ C. In both cases the assumption of nonequivalence of A and C leads to a contradiction. Therefore, the required condition A ∼ C is fulfilled. In the second case, assuming that A and C are not equivalent, we find that the line a passing through none of the points A, B, and C intersects the segment [AC] at some interior point O. Then due to Pasch’s axiom A12 it should intersect one of the segments [AB] or [BC] at an interior point.

An . 5) i=1 the segments in the right hand side of this equality intersecting only by their ending points. 5). Assume that B ∈ [Aq , Aq+1 ]. We advance by one the numbers of the points Aq+1 , . . , An : Aq+1 → Aq+2 , . . , An → An+1 . Then assign B = Aq+1 and get the required monotonic sequence A1 , . . , An+1 . 4) is fulfilled we need to advance the numbers in the whole sequence A1 , . . , An : A1 → A2 , . . , An → An+1 . § 4. DIRECTIONS. VECTORS ON A STRAIGHT LINE. 37 Then we assign B = A1 and as a result we obtain the required monotonic sequence of points A1 , .

In order to prove that the point B lies in the interior of the segment [AD] it is sufficient to exchange the notations of the points A with D and B with C. Thereafter the rest is to use the first proposition, which is already proved, and then return to the initial notations. 2. Let A, B, C, and D be a group of four points. If the point C lies in the interior of the segment [AD] and if the point B lies in the interior of the segment [AC], then B is in the interior of [AD] and C is in the interior of [BD].