# ptolemy's theorem in hyperbolic geometry

 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 thereforea 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). 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. To proceed further, we need to look further at the relationship between the euclidean and hyperbolic properties of a figure.