# Proof of Lemma 2 for D3

 Lemma 2 If C does not lie on the h-segment AB, then d(A,B) < d(A,C) + d(C,B). Proof Suppose that C does not lie on the h-segment AB. If d(A,B) < d(A,C) or d(A,B) < d(C,B) then the result clearly holds. After Lemma 1, this is the case if C lies on the h-line AB, but not between A and B, so we now assume that C is not on AB Let Ka be the h-circle with h-centre A, through C, and let Kb be the h-circle with h-centre B, through C. As d(A,B) ≥ d(A,C), Ka meets AB at A' between A and B. Since A' is on Ka, d(A,A') = d(A,C). Likewise, Kb meets AB at B' between A and B, and, as B' is on Kb, d(B',B) = d(C,B). Note that Ka and Kb meet at C. Since C is not on AB, they meet again on the other side of AB. Thus A' and B' are as shown in the (static) figure on the right. Applying the above results, and lemma 1, d(A,B) = d(A,B') + d(B',B)           = d(A,B') + d(C,B)           = (d(A,A') - d(A',B')) + d(C,B)           = (d(A,C) - d(A',B')) + d(C,B)           = d(A,C) - d(A',B') + d(C,B)           < d(A,C) + d(C,B) return to distance page