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
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.
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)
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.