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12.7: Axioma IV

Última actualización

30 oct 2022
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- 12.6: Axioma III
- 12.8: Axioma H-v

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El siguiente reclamo dice que Axioma IV se sostiene en el plano h.

Reclamación $\PageIndex{1}$

En el plano h, tenemos $\triangle_h P Q R \cong \triangle_h P' Q' R'$ si y solo si

$Q'P_h' = QP_h,$ $Q'R_h' -QR_h$ y $\measuredangle_h P'Q'R' = \pm \measuredangle PQR$ .

Prueba

Aplicando la observación principal, podemos asumir eso $Q$ y $Q'$ coincidir con el centro de lo absoluto; en particular $Q=Q'$ . En este caso

$\measuredangle P' Q R'=\measuredangle_h P' Q R'=\pm\measuredangle_h P Q R=\pm\measuredangle P Q R.$

Desde

$Q P_h=Q P'_h$ y $Q R_h=Q R'_h,$

Lema 12.3.2 implica que lo mismo vale para las distancias euclidianas; es decir,

$Q P=Q P'$ y $Q R=Q R'.$

Por SAS, hay un movimiento del plano euclidiano que se envía $Q$ a sí mismo, $P$ hacia $P'$ y $R$ hacia $R'$

Tenga en cuenta que el centro del absoluto está fijado por el movimiento correspondiente. De ello se deduce que este movimiento da también un movimiento del plano h; en particular, los triángulos h $\triangle_h P Q R$ y $\triangle_h P' Q R'$ son h-congruentes.

Reclamación12.7.1\PageIndex{1}

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Reclamación $\PageIndex{1}$