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We begin with a weak form
ptolemy's theorem in hyperbolic geometry
Suppose that ABCD is a convex hyperbolic quadrilateral inscribed in a
hyperbolic circle. Then s(AC)s(BD) = s(AB)s(CD) + s(AD)s(BC).
A direct proof from the distances theorem.
Unlike in euclidean geometry, we cannot expect a simple converse. It is
possible that four points lie on a euclidean but not hyperbolic circle.
A proof of the strong form of ptolemy's theorem comes quickly from
ideas in inversive geometry. This is a natural way to proceed since
ptolemy's theorem is really an inversive result.
Recall that the modulus of the hyperbolic cross ratio which is defined by
(u,v,w,z) =(u-w)(v-z)/(u-z)(v-w)
is an inversive invariant. It is therefore
a
hyperbolic invariant (and a euclidean invariant). In other words, the
function C(u,v,w,z) = |(u,v,w,z)| is a hyperbolic invariant. From
the proof of the distances theorem, we have
|u-v| = s(UV)/c(OU)c(OV), and similarly for the other factors. Thus
the hyperbolic invariant
For four distinct points of the disk, C(A,B,C,D) = s(AC)s(BD)/s(AD)s(BC).
From inversive geometry
Ptolemy's Theorem in Inversive Geometry
If A,B,C,D are distinct points, then C(B,C,A,D)+ C(B,A,C,D) ≥ 1, with
equality if and only if A,B,C,D lie in this order on an i-line.
Using our formula for C, this gives the
strong form of ptolemy's theorem for hyperbolic geometry
If A,B,C,D are distinct points of the disk, then
s(AB)s(CD) + s(AD)s(BC) ≥ s(AC)s(BD)
with equality if and only if A,B,C,D lie in this order on an i-line.
Note that, the condition "four points A,B,C,D lie in this order on a circle"
is equivalent to the condition "ABCD is a convex quadrilateral".
We finally obtain the
converse of ptolemy's theorem for hyperbolic geometry
A convex hyperbolic quadrilateral ABCD has a hyperbolic circumcircle if
(a) three of the points lie on a hyperbolic circle, and
(b) s(AB)s(CD) + s(AD)s(BC) = s(AC)s(BD)
This follows at once from the strong form. Condition (b) shows that all four
lie on an i-line. If three lie on a hyperbolic circle (an i-line), then so does
the fourth since there is only one i-line through three distinct points.
We could replace (a) by the metric condition H(s(AB),s(AC),s(BC)) > 0
by version 2 of the circumcircles theorem. A nicer solution would be
to find a metric conditon symmetric in A,B,C,D.
As in euclidean geometry, there is a version for hyperbolic hexagons.
This is known as fuhrmann's theorem.
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