12.4: Axiom I
Evidently, the h-plane contains at least two points. Therefore, to show that Axiom I holds in the h-plane, we need to show that the h-distance defined in 12.1 is a metric on h-plane; that is, the conditions (a) - (d) in Definition 1.3.1 hold for h-distance.
The following claim says that the h-distance meets the conditions (a) and (b)
Given the h-points \(P\) and \(Q\) , we have \(PQ_h \ge 0\) and \(PQ_h=0\) if and only if \(P=Q\) .
- Proof
-
According to Lemma 12.3.1 and the main observation ( Theorem 12.3.1 ), we may assume that \(Q\) is the center of the absolute. In this case
and therefore
Moreover, the equalities holds if and only if \(P=Q\) .
The following claim says that the h-distance meets the condition
For any h-points \(P\) and \(Q\) , we have \(PQ_h=QP_h\) .
- Proof
-
Let \(A\) and \(B\) be ideal points of \((PQ)_h\) and \(A,P,Q,B\) appear on the circline containing \((PQ)_h\) in the same order.
Then
\(\begin{array} {rcl} {PQ_h} & = & {\ln \dfrac{AQ \cdot BP}{QB \cdot PA} =} \\ {} & = & {=\ln \dfrac{BP \cdot AQ}{PA \cdot QB}=} \\ {} & = & {QP_h} \end{array}\)
The following claim shows, in particular, that the triangle inequality (which is Definition 1.3.1d ) holds for \(h\) -distance.
Given a triple of h-points \(P\) , \(Q\) , and \(R\) , we have
Moreover, the equality holds if and only if \(P\) , \(Q\) , and \(R\) lie on one h-line in the same order.
- Proof
-
Without loss of generality, we may assume that \(P\) is the center of the absolute and \(PQ_h \ge QR_h >0\) .
Suppose that \(\Delta\) denotes the h-circle with the center \(Q\) and h-radius \(\rho=QR_h\) . Let \(S\) and \(T\) be the points of intersection of \((PQ)\) and \(\Delta\) .
By Lemma 12.3.3 , \(PQ_h\z\ge QR_h\) . Therefore, we can assume that the points \(P\) , \(S\) , \(Q\) , and \(T\) appear on the h-line in the same order.
According to Lemma Lemma 12.3.4 , \(\Delta\) is a Euclidean circle; suppose that \(\hat Q\) denotes its Euclidean center. Note that \(\hat Q\) is the Euclidean midpoint of \([ST]\) .
By the Euclidean triangle inequality
\[PT = P\hat{Q}+\hat{Q} R \ge PR\]
and the equality holds if and only if \(T=R\) .
By Lemma Lemma 12.3.2 ,
\(\begin{array} {l} {PT_h = \ln \dfrac{1 + PT}{1 - PT},} \\ {PR_h = \ln \dfrac{1 + PR}{1 - PR}.} \end{array}\)
Since the function \(f(x)=\ln\frac{1+x}{1-x}\) is increasing for \(x\in[0,1)\) , inequality 12.4.1 implies
\(PT_h\ge PR_h\)
and the equality holds if and only if \(T=R\) .
Finally, applying Lemma 12.3.3 again, we get that
\(PT_h=PQ_h+QR_h.\)
Hence the claim follows.