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# 12.6: Axiom III

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Note that the first part of Axiom III follows directly from the definition of the h-angle measure defined on page . It remains to show that $$\measuredangle_h$$ satisfies the conditions Axiom IIIa, Axiom IIIb, and Axiom IIIc.

The following two claims say that $$\measuredangle_h$$ satisfies IIIa and IIIb.

Claim $$\PageIndex{1}$$

Given an h-half-line $$[O P)_h$$ and $$\alpha\in(-\pi,\pi]$$, there is a unique h-half-line $$[O Q)_h$$ such that $$\measuredangle_h P O Q= \alpha$$.

Claim $$\PageIndex{2}$$

For any h-points $$P$$, $$Q$$, and $$R$$ distinct from an h-point $$O$$, we have

$$\measuredangle_h P O Q+\measuredangle_h Q O R \equiv\measuredangle_h P O R.$$

Proof of $$\PageIndex{1}$$ and $$\PageIndex{2}$$

Applying the main observation, we may assume that $$O$$ is the center of the absolute. In this case, for any h-point $$P \ne O$$, the h-half-line $$[OP)_h$$ is the intersection of the Euclidean half-line $$[OP)$$ with h-plane. Hence the claim $$\PageIndex{1}$$ and Claim $$\PageIndex{2}$$ follow from the axioms Axiom IIIa and Axiom IIIb of the Euclidean plane.

Claim $$\PageIndex{3}$$

The function

$$\measuredangle_h\:(P,Q,R)\mapsto\measuredangle_h P Q R$$

is continuous at any triple of points $$(P,Q,R)$$ such that $$Q\ne P$$, $$Q\ne R$$, and $$\measuredangle_h P Q R\ne\pi$$.

Proof

Suppose that $$O$$ denotes the center of the absolute. We can assume that $$Q$$ is distinct from $$O$$.

Suppose that $$Z$$ denotes the inverse of $$Q$$ in the absolute; suppose that $$\Gamma$$ denotes the circle perpendicular to the absolute and centered at $$Z$$. According to Lemma 12.3.1, the point $$O$$ is the inverse of $$Q$$ in $$\Gamma$$.

Let $$P'$$ and $$R'$$ denote the inversions in $$\Gamma$$ of the points $$P$$ and $$R$$ respectively. Note that the point $$P'$$ is completely determined by the points $$Q$$ and $$P$$. Moreover, the map $$(Q,P)\mapsto P'$$ is continuous at any pair of points $$(Q,P)$$ such that $$Q\ne O$$. The same is true for the map $$(Q,R)\mapsto R'$$

According to the main observation

$$\measuredangle_h P Q R\equiv -\measuredangle_h P' O R'.$$

Since $$\measuredangle_h P' O R'=\measuredangle P' O R'$$ and the maps $$(Q,P)\mapsto P'$$, $$(Q,R)\mapsto R'$$ are continuous, the claim follows from the corresponding axiom of the Euclidean plane.