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Lippmann's axiom and Lebesgue's axiom are equivalent to the Lotschnittaxiom

机译:Lippmann的Axiom和Lebesgue的公理相当于Lotschnittaxiom

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We prove that both Lippmann's axiom of 1906, stating that for any circle there exists a triangle circumscribing it, and Lebesgue's axiom of 1936, stating that for every quadrilateral there exists a triangle containing it, are equivalent,with respect to Hilbert's plane absolute geometry, to Bachmann's Lotschnittaxiom, which states that perpendiculars raised on the two legs of a right angle meet. We also show that, in the presence of the Circle Axiom, the statement "There is an angle such that the perpendiculars raised on its legs at equal distances from the vertex meet" is equivalent to the negation of Hilbert's hyperbolic parallel postulate.
机译:我们证明了1906年的李普曼的公理,说明任何圆圈都存在三角形,而1936年的Lebesgue的公理,陈述,对于每个四边形,存在含有它的三角形,相当于希尔伯特的平面绝对几何形状, 到Bachmann的Lotschnittaxiom,它指出垂直于直角相遇的两条腿升起。 我们还表明,在圆形公理的存在下,声明“存在一定角度,使得在其腿上沿着顶点相遇的距离上升的垂直”相当于希尔伯特的双曲线平行假设的否定。

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