# 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,

$AD\cdot BC + AB\cdot CD = AC\cdot BD.$

## 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 $\pi$ 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. (There is another simple proof of a recent vintage.)

## 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 $AB\cdot CD = AC\cdot MD.$

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

$AB\cdot CD + BC\cdot AD = AC\cdot MD + AC\cdot BM = AC\cdot BD$

## Remark

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

$AB\cdot CD + BC\cdot AD \ge AC\cdot BD$

which is known as *Ptolemy's inequality*.

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

Let $A_{1}A_{2}A_{3}$ denote an equilateral triangle inscribed in a circle. For any point $P$ on the circle, show that the two shorter segments among $PA_{1},$ $PA_{2},$ $PA_{3}$ 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

$s\cdot PA_{1}=s\cdot PA_{2}+s\cdot PA_{3}$

Therefore,

$PA_{1}=PA_{2}+PA_{3}.$

### 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

### Remark

Ptolemy's theorem is a powerful result. With its help we establish the Pythagorean Theorem and

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

$\displaystyle m^{2} = \frac{(ab + cd)(ac + bd)}{ad + bc}$ and

$\displaystyle n^{2} = \frac{(ac + bd)(ad + bc)}{ab + cd}.$

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 $\theta$ 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

$mt\cdot \cos \theta = ab + cd$ and

$nt\cdot \cos \theta = ad + bc.$

So $\displaystyle\frac{m}{n} = \frac{ab + cd}{ad + bc}$ which is called *Ptolemy's second theorem*.

Also $mn = ac + bd$ (Ptolemy relation)

Multiplying those last 2 equations, we get:

$\displaystyle m^{2} = \frac{(ab + cd)(ac + bd)}{ad + bc}.$

Dividing instead, we get:

$\displaystyle n^{2} = \frac{(ac + bd)(ad + bc)}{ab + cd}.$

Finally, we also get

$\displaystyle (t \cos \theta )^{2} = \frac{(ab + cd)(ad + bc)}{ac + bd}.$

And if the diagonals in the quadrilateral are orthogonal,

$\displaystyle t^{2} = \frac{(ab + cd)(ad + bc)}{ac + bd}.$

An additional derivation has been posted to the old 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)

$m^{2}= b^{2} + c^{2} - 2\cdot bc\cdot \cos (C).$

And also

(2)

$\begin{align} m^{2} &= a^{2} + d^{2} - 2\cdot ad\cdot \cos (A)\\ &= a^{2} + d^{2} - 2\cdot ad\cdot \cos (180^{\circ} - C)\\ &= a^{2} + d^{2} + 2\cdot ad\cdot \cos (C). \end{align}$

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

$\displaystyle m^{2} = \frac{(ab + cd)(ac + bd)}{ad + bc}.$

The expression for $n^{2}$ is obtained by picking the other pair of triangles.

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