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Ptolemy's Theorem

Let a convex quadrilateral ABCD be inscribed in a circle. Then the sum of the products of the two pairs of opposite sides equals the product of its two diagonals. In other words,

  AD·BC + AB·CD = AC·BD

Remark

Ptolemy of Alexandria (~100-168) gave the name to the Ptolemy's Planetary theory which he described in his treatise Almagest. The book is mostly devoted to astronomy and trigonometry where, among many other things, he also gives the approximate value of π as 377/120 and proves the theorem that now bears his name. The name Almagest is actually a corruption of the Arabic rendition "Al Magiste" - The Greatest - of the Greek H Megisth Suntaxiz (E Megiste Syntaxis).

This classical theorem has been proved many times over. Following is the simplest proof I am aware of.

Proof

On the diagonal BD locate a point M such that angles ACB and MCD be equal. Since angles BAC and BDC subtend the same arc, they are equal. Therefore, triangles ABC and DMC are similar. Thus we get CD/MD = AC/AB, or AB·CD = AC·MD.

Now, angles BCM and ACD are also equal; so triangles BCM and ACD are similar which leads to BC/BM = AC/AD, or BC·AD = AC·BM. Summing up the two identities we obtain

  AB·CD + BC·AD = AC·MD + AC·BM = AC·BD

Remark

Ptolemy's theorem admits a useful generalization: for four points A, B, C, D, not necessarily concyclic,

  AB·CD + BC·AD ≥ AC·BD

which is known as Ptolemy's inequality.

The following problem is discussed in Honsberger, Mathematical Morsels, p172:

  Let A1A2A3 denote an equilateral triangle inscribed in a circle. For any point P on the circle, show that the two shorter segments among PA1, PA2, PA3 add up to the third one.

Solution

Let s denote the length of the side of the given triangle. By Ptolemey's Theorem we have

  s·PA1 = s·PA2 + s·PA3

Therefore,

  PA1 = PA2 + PA3

Remark

This result has an interesting generalization to the case of a regular 3n-gon inscribed in a circle: Of the 3n chords obtained by connecting a point P with vertices of the polygon, the sum of the 2n shortest ones equals the sum of the n longest.

The problem itself is sometimes attributed to Van Schooten (1615 – 1660), see, for example, The Changing Shape of Geometry, (C. Pritchard, Cambridge University Press, 2003), p. 184, where two additional proofs could be found.

Remark

Ptolemy's theorem is a powerful result. With its help we establish the Pythagorean Theorem. Combined with the Law of Sines, Ptolemy's theorem serves to prove the addition and subtraction formulas for the sine function. It has a short proof in complex numbers. The following generalization is sometimes attributed to the great 9th century Indian mathematician Mahavira (or Mahaviracharya, meaning Mahavira the Teacher).

In a cyclic quadrilateral ABCD, let a, b, c, d denote the lengths of sides AB, BC, CD, DA, and m, n the lengths of the diagonals BD and BC. Then Mahavira's result is expressed as

  m2 = (ab + cd)(ac + bd)/(ad + bc) and
n2 = (ac + bd)(ad + bc)/(ab + cd)

H. Eves gives a proof as a sequence of exercises in [Great Moments in Mathematics Before 1650, p. 108]:

Let t be the diameter of the circmcircle of ABCD and θ the angle between either diagonal and the perpendicular upon the other.

Then, (using triangle's formula ab = 2hR applied to DAB and DCB), we get

  mt·cosθ = ab + cd and
nt·cosθ = ad + bc

So m / n = (ab + cd) / (ad + bc) which is called Ptolemy's second theorem.

Also mn = ac + bd (Ptolemy relation)

Multiplying those last 2 equations, we get:

  m2 = (ab + cd)(ac + bd) / (ad + bc)

Dividing instead, we get:

  n2 = (ac + bd)(ad + bc) / (ab + cd).

Finally, we also get

  (t cosθ)2 = (ab + cd)(ad + bc) / (ac + bd).

And if the diagonals in the quadrilateral are orthogonal,

  t2 = (ab + cd)(ad + bc) / (ac + bd).

An additional derivation has been posted to the CTK Exchange. This one can be found in Advanced Trigonomentry by C. V. Durrell and A. Robson, 1930, p. 25. (The book is available in a 2003 Dover edition and on google's bookshelf.)

In the same notations as above, by the Cosine Rule

(1)m2= b2 + c2 - 2·bc·cos(C).

And also

 m2= a2 + d2 - 2·ad·cos(A)
  = a2 + d2 - 2·ad·cos(180° - C)
(2) = a2 + d2 + 2·ad·cos(C)

Multiply (1) by ad and (2) by bc, add them up and you will arrive at

  m2 = (ab + cd)·(ac + bd)/(ad + bc)

The expression for n2 is obtained by picking the other pair of triangles.

Copyright © 1996-2008 Alexander Bogomolny

30765752Page copy protected against web site content infringement by Copyscape


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