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12.7: Axiom IV

Last updated

Sep 5, 2021
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- 12.6: Axiom III
- 12.8: Axiom h-V

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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.

Claim 12.7.1\PageIndex{1}

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Claim $\PageIndex{1}$