The Theorem of Ptolemy via complex numbers.

The Ptolemy-Euler Theorem [1 chap. 2.2] similar to [7 p66-67]
-------------------------

For any four complex numbers a, b, c, d, the following identity
is easy to verify:

  (a-b)(c-d) + (a-d)(b-c) = (a-c)(b-d)

By triangle inequality, we obtain

  |a-b||c-d| + |a-d||b-c| >= |a-c||b-d|

Let us investigate when the inequality becomes an equality.
In case of the triangle inequality,

  |z1 + z2| <= |z1| + |z2|,

equality holds if and only if z1/z2 is a positive real number
(provided z1 z2 != 0). Thus we are looking for a condition to
ensure that (a-b)(c-d) : (a-d)(b-c) is a positive real number. 
But
      (a-b)(c-d)
      ----------  is a positive real number
      (a-d)(b-c)

      a-b   c-b
  <=> --- / ---  is a negative real number
      a-d   c-d

           a-b   c-b
  <=> arg( --- / --- )  =  PI
	   a-d   c-d

           a-b          c-b
  <=> arg( --- ) - arg( --- )  =  PI  (mod 2 PI).
           a-d          c-d

It follows that a, b, c, d are cocyclic, i. e. a, b, c, d are 
on the same circle or line (see COCYCLIC) and a and c are on the 
opposite sides of the chord joining b and d, which results in the
alphabetical order (clockwise or counterclockwise).
We have proven the following.

Theorem (Ptolemy-Euler)
  For any four points A, B, C, D in the plane,

  |AB| |CD| + |BC| |DA| >= |AC| |BD|.

  Equality holds if and only if these four points are cocyclic
  (colinear) and are in alphabetical order (clockwise or counterclockwise).

The equality was discovered by C. Ptolemy (ca. 85-165), while the
general case was found over a thousand years later by L. Euler (1707-1783).

COCYCLIC:
  Suppose C and D are on opposite sides of a line AB.
  Then the points A, B, C, D are cocyclic if and only if

       angle(ACB) + angle(ADB) = PI.


Quadrangles and Ptolemy [4 p86-90, 233-234]
-----------------------
This is based on Gauss' lecture on 'imaginaere Groessen' [5].
The points of a planar quadrangle ABCD can be described by 
the complex numbers a, b, c, d. We associate a triangle FGH
in symmetrical form it in the following way.

  h  =  ab + dc,
  g  =  ac + db,
  f  =  ad + bc.

Then we have for the oriented sides of the triangle FGH

  h - g  =  (d - a)(c - b),
  f - h  =  (d - b)(a - c),
  g - f  =  (b - a)(d - c).

As (h - g) + (f - h) + (g - f) = 0 we have for the quadrangle the identity

  (d - a)(c - b) + (d - b)(a - c) + (b - a)(d - c) = 0.

[...]

For the angles of the triangle FGH we find

  angle(HFG) = angle(BDC) - angle(BAC) = angle(DBA) - angle(DCA)
  angle(FGH) = angle(CDA) - angle(CBA) = angle(DCB) - angle(DAB)
  angle(GHF) = angle(ADB) - angle(ACB) = angle(DAC) - angle(DBC).

If the points A, B, C, D are cocyclic (in this order) we will find

  angle(HFG) = 0,
  angle(FGH) = 0,
  angle(GHF) = PI.

Therefore F, H, G are colinear in this order, which means |GH| + |HF| = |GF|.
And this gives |AD| |BC| + |BD| |CA| = |CD| |AB|.


References:

[1] Liang-shin Hahn;
   Complex Numbers and Geometry,
   MAA 1994 (Spectrum Series), ISBN 0-88385-510-0

[2] A. I. Markuschewitsch;
   Komplexe Zahlen und konforme Abbildungen,
   4. Aufl., Berlin 1973

[3] I. M. Jaglom;
   Komplexe Zahlen und ihre Anwendungen in der Geometrie,
   Moskau, 1969

[4] Herbert Pieper;
   Die komplexen Zahlen - Theorie - Praxis - Geschichte,
   Deutscher Verlag der Wissenschaften, Berlin, 1988, 2. Aufl.

[5] Carl F. Gauss;
   imaginaere Groessen,
   Vorlesung 1840, G"ottingen,
   Gauss-Werke VIII, 331-334
   Part: Vierecksbetrachtungen

[6] August F. M"obius;
   Berichte der K"oniglichen S"achsischen Gesellschaft der
   Wissenschaften zu Leipzig, math.-phys. Klasse Jg. 1852, 41-54

[7] Ebbinghaus, Hermes, Hirzebruch, Koecher, Mainzer, Prestel, Remmert;
   Zahlen,
   Springer Verlag, 1983
   Satz des Ptolem"aus, p66-67

[8] The inverse Ptolemy;
   Crelle Journal,
   8 (1832) 320                      - problem
   10 p41, 11 p264-271, 13 p233-236  - solutions

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