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12.7: Axiom IV

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    58344
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    The following claim says that Axiom IV holds in the h-plane.

    Claim \(\PageIndex{1}\)

    In the h-plane, we have \(\triangle_h P Q R \cong \triangle_h P' Q' R'\) if and only if

    \(Q'P_h' = QP_h,\) \(Q'R_h' -QR_h\) and \(\measuredangle_h P'Q'R' = \pm \measuredangle PQR\).

    Proof

    Applying the main observation, we can assume that \(Q\) and \(Q'\) coincide with the center of the absolute; in particular \(Q=Q'\). In this case

    \(\measuredangle P' Q R'=\measuredangle_h P' Q R'=\pm\measuredangle_h P Q R=\pm\measuredangle P Q R.\)

    Since

    \(Q P_h=Q P'_h\) and \(Q R_h=Q R'_h,\)

    Lemma 12.3.2 implies that the same holds for the Euclidean distances; that is,

    \(Q P=Q P'\) and \(Q R=Q R'.\)

    By SAS, there is a motion of the Euclidean plane that sends \(Q\) to itself, \(P\) to \(P'\) and \(R\) to \(R'\)

    Note that the center of the absolute is fixed by the corresponding motion. It follows that this motion gives also a motion of the h-plane; in particular, the h-triangles \(\triangle_h P Q R\) and \(\triangle_h P' Q R'\) are h-congruent.


    12.7: Axiom IV is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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