ptolemy's theorem in inversive geometry

 In euclidean geometry, we have the following classical result Ptolemy's Theorem If the points A,B,C,D lie in this order on the circumference of a circle, then |AB||CD| + |AD||BC| = |AC||BD|. Some texts contain a better result, which hints at a connection with inversive geometry since it involves an object which is a line or circle, i.e. an i-line. Strong form of Ptolemy's Theorem For any four points, |AB||CD| + |AD||BC| ≥ |AC||BD|, with equality if and only if A,B,C,D lie in this order on a line or circle. Immediately we see, by dividing through by |AC||BD|, that each result involves only ratios, so is a theorem of similarity geometry. In fact, we will show that the theorem really belongs to inversive geometry, and it can be viewed as a generalisation of the Triangle Inequality in Similarity Geometry If A,B,C are any three points then |AB|/|AC| + |BC|/|AC| ≥ 1, with equality if and only if A,B,C lie in this order on a line. This is obtained from the standard euclidean triangle inequality |AB| + |BC| ≥ |AC|, by dividing through by |AC|. To say that A,B,C lie in this order on a line is just another way of saying that B lies between A and C ( or on the segment AC). Ptolemy's Theorem in Inversive Geometry If A,B,C,D are any four points, then |(B,C,A,D)| + |(B,A,C,D)| ≥ 1, with equality if and only if A,B,C,D lie in this order on an i-line. To recover the strong form of the euclidean theorem, we simply observe that |(B,C,A,D)| = |BA||CD|/|CA||BD| and |(B,A,C,D)| = |BC||AD|/|AC||BD|.