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).
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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.
proof
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|.
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