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) =(uw)(vz)/(uz)(vw)
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
uv = 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 iline.
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 iline.
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 iline. If three lie on a hyperbolic circle (an iline), then so does
the fourth since there is only one iline 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.

