
# 12.7: Axiom IV


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.