InEuclidean geometry,Ptolemy's theorem is a relation between the four sides and two diagonals of acyclic quadrilateral (a quadrilateral whosevertices lie on a common circle). The theorem is named after theGreekastronomer andmathematicianPtolemy (Claudius Ptolemaeus).^[1] Ptolemy used the theorem as an aid to creatinghis table of chords, a trigonometric table that he applied to astronomy.

If the vertices of the cyclic quadrilateral areA,B,C, andD in order, then the theorem states that:

${\displaystyle AC\cdot BD=AB\cdot CD+BC\cdot AD}$

This relation may be verbally expressed as follows:

If a quadrilateral iscyclic then the product of the lengths of its diagonals is equal to the sum of the products of the lengths of the pairs of opposite sides.

Moreover, theconverse of Ptolemy's theorem is also true:

In a quadrilateral, if the sum of the products of the lengths of its two pairs of opposite sides is equal to the product of the lengths of its diagonals, then the quadrilateral can be inscribed in a circle i.e. it is acyclic quadrilateral.

To appreciate the utility and general significance of Ptolemy’s Theorem, it is especially useful to study its mainCorollaries.

Ptolemy's Theorem yields as a corollary a theorem^[2] regarding an equilateral triangle inscribed in a circle.

Given An equilateral triangle inscribed on a circle, and a point on the circle.

The distance from the point to the most distant vertex of the triangle is the sum of the distances from the point to the two nearer vertices.

Proof: Follows immediately from Ptolemy's theorem:

${\displaystyle qs=ps+rs\Rightarrow q=p+r.}$ 

This corollary has as an application an algorithm for computing minimalSteiner trees whose topology is fixed, by repeatedly replacing pairs of leaves of the treeA,B that should be connected to aSteiner point, by the third pointC of their equilateral triangle. The unknown Steiner point must lie on arcAB of the circle, and this replacement ensures that, no matter where it is placed, the length of the tree remains unchanged.^[3]

Anysquare can be inscribed in a circle whose center is the center of the square. If the common length of its four sides is equal to${\displaystyle a}$  then the length of the diagonal is equal to${\displaystyle a\sqrt{2}}$  according to thePythagorean theorem, and Ptolemy's relation obviously holds.

More generally, if the quadrilateral is arectangle with sides a and b and diagonal d then Ptolemy's theorem reduces to the Pythagorean theorem. In this case the center of the circle coincides with the point of intersection of the diagonals. The product of the diagonals is then d², the right hand side of Ptolemy's relation is the suma² + b².

Copernicus – who used Ptolemy's theorem extensively in his trigonometrical work – refers to this result as a 'Porism' or self-evident corollary:

Furthermore it is clear (manifestum est) that when the chord subtending an arc has been given, that chord too can be found which subtends the rest of the semicircle.^[4]

A more interesting example is the relation between the lengtha of the side and the (common) lengthb of the 5 chords in a regular pentagon. Bycompleting the square, the relation yields thegolden ratio:^[5]

 

If now diameter AF is drawn bisecting DC so that DF and CF are sides c of an inscribed decagon, Ptolemy's Theorem can again be applied – this time to cyclic quadrilateral ADFC with diameterd as one of its diagonals:

${\displaystyle ad=2bc}$ 

${\displaystyle \Rightarrow ad=2\phi ac}$  where${\displaystyle \phi }$  is the golden ratio.

${\displaystyle \Rightarrow c=\frac{d}{2\phi }.}$ ^[6]

whence the side of the inscribed decagon is obtained in terms of the circle diameter. Pythagoras's theorem applied to right triangle AFD then yields "b" in terms of the diameter and "a" the side of the pentagon^[7] is thereafter calculated as

${\displaystyle a=\frac{b}{\phi }=b\left(\phi -1\right).}$ 

AsCopernicus (following Ptolemy) wrote,

"The diameter of a circle being given, the sides of the triangle, tetragon, pentagon, hexagon and decagon, which the same circle circumscribes, are also given."^[8]

The animation here shows a visual demonstration of Ptolemy's theorem, based on Derrick & Herstein (2012).^[9]

Let ABCD be acyclic quadrilateral.On thechord BC, theinscribed angles ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB.Construct K on AC such that ∠ABK = ∠CBD; since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.

Now, by common angles △ABK issimilar to △DBC, and likewise △ABD is similar to △KBC.Thus AK/AB = CD/BD, and CK/BC = DA/BD;equivalently, AK⋅BD = AB⋅CD, and CK⋅BD = BC⋅DA.By adding two equalities we have AK⋅BD + CK⋅BD = AB⋅CD + BC⋅DA, and factorizing this gives (AK+CK)·BD = AB⋅CD + BC⋅DA.But AK+CK = AC, so AC⋅BD = AB⋅CD + BC⋅DA,Q.E.D.^[10]

The proof as written is only valid forsimple cyclic quadrilaterals. If the quadrilateral is self-crossing then K will be located outside the line segment AC. But in this case, AK−CK = ±AC, giving the expected result.

Let the inscribed angles subtended by${\displaystyle AB}$ ,${\displaystyle BC}$  and${\displaystyle CD}$  be, respectively,${\displaystyle \alpha }$ ,${\displaystyle \beta }$  and${\displaystyle \gamma }$ , and the radius of the circle be${\displaystyle R}$ , then we have${\displaystyle AB=2R\sin \alpha }$ ,${\displaystyle BC=2R\sin \beta }$ ,${\displaystyle CD=2R\sin \gamma }$ ,${\displaystyle AD=2R\sin ({180}^{\circ }-(\alpha +\beta +\gamma ))}$ ,${\displaystyle AC=2R\sin (\alpha +\beta )}$  and${\displaystyle BD=2R\sin (\beta +\gamma )}$ , and the original equality to be proved is transformed to

${\displaystyle \sin (\alpha +\beta )\sin (\beta +\gamma )=\sin \alpha \sin \gamma +\sin \beta \sin (\alpha +\beta +\gamma )}$ 

from which the factor${\displaystyle 4R^2}$  has disappeared by dividing both sides of the equation by it.

Now by using the sum formulae,${\displaystyle \sin (x+y)=\sin x\cos y+\cos x\sin y}$  and${\displaystyle \cos (x+y)=\cos x\cos y-\sin x\sin y}$ , it is trivial to show that both sides of the above equation are equal to

 

Q.E.D.

Here is another, perhaps more transparent, proof using rudimentary trigonometry.Define a new quadrilateral${\displaystyle ABC{D}^{\prime }}$  inscribed in the same circle, where${\displaystyle A,B,C}$  are the sameas in${\displaystyle ABCD}$ , and${\displaystyle D^{\prime }}$  located at a new point on the same circle, defined by${\displaystyle |\overline{A{D}^{\prime }}|=|\overline{CD}|}$ ,${\displaystyle |\overline{C{D}^{\prime }}|=|\overline{AD}|}$ . (Picture triangle${\displaystyle ACD}$  flipped, so that vertex${\displaystyle C}$  moves to vertex${\displaystyle A}$  and vertex${\displaystyle A}$  moves to vertex${\displaystyle C}$ . Vertex${\displaystyle D}$  will now be located at a new point D’ on the circle.)Then,${\displaystyle ABC{D}^{\prime }}$  has the same edges lengths, and consequently the same inscribed angles subtended by the corresponding edges, as${\displaystyle ABCD}$ , only in a different order. That is,${\displaystyle \alpha }$ ,${\displaystyle \beta }$  and${\displaystyle \gamma }$ , for, respectively,${\displaystyle AB,BC}$  and${\displaystyle A{D}^{\prime }}$ .Also,${\displaystyle ABCD}$  and${\displaystyle ABC{D}^{\prime }}$  have the same area. Then,

 

Q.E.D.

Choose an auxiliary circle${\displaystyle \mathrm{\Gamma }}$  of radius${\displaystyle r}$  centered at D with respect to which the circumcircle of ABCD isinverted into a line (see figure).Then${\displaystyle A^{\prime }{B}^{\prime }+B^{\prime }{C}^{\prime }=A^{\prime }{C}^{\prime }.}$ Then${\displaystyle A^{\prime }{B}^{\prime },B^{\prime }{C}^{\prime }}$  and${\displaystyle A^{\prime }{C}^{\prime }}$  can be expressed as$\frac{AB\cdot D{B}^{\prime }}{DA}$ ,$\frac{BC\cdot D{B}^{\prime }}{DC}$  and$\frac{AC\cdot D{C}^{\prime }}{DA}$  respectively. Multiplying each term by$\frac{DA\cdot DC}{D{B}^{\prime }}$  and using$\frac{D{C}^{\prime }}{D{B}^{\prime }}=\frac{DB}{DC}$  yields Ptolemy's equality.

Q.E.D.

Note that if the quadrilateral is not cyclic then A', B' and C' form a triangle and hence A'B'+B'C' > A'C', giving us a very simple proof of Ptolemy's Inequality which is presented below.

Embed ABCD in thecomplex plane${\displaystyle \mathbb{C}}$  by identifying${\displaystyle A\mapsto z_A,\dots ,D\mapsto z_D}$  as four distinctcomplex numbers${\displaystyle z_A,\dots ,z_D\in \mathbb{C}}$ . Define thecross-ratio

${\displaystyle \zeta :=\frac{(z_A-z_B)(z_C-z_D)}{(z_A-z_D)(z_B-z_C)}\in {\mathbb{C}}_{\ne 0}}$ .

Then

 

with equality if and only if the cross-ratio${\displaystyle \zeta }$  is a positive real number. This provesPtolemy's inequality generally, as it remains only to show that${\displaystyle z_A,\dots ,z_D}$  lie consecutively arrangedon a circle (possibly of infinite radius, i.e. a line) in${\displaystyle \mathbb{C}}$  if and only if${\displaystyle \zeta \in {\mathbb{R}}_{>0}}$ .

From thepolar form of a complex number${\displaystyle z=|z|e^{i\arg (z)}}$ , it follows

 

with the last equality holding if and only if ABCD is cyclic, since a quadrilateral is cyclic if and only if opposite angles sum to${\displaystyle \pi }$ .

Q.E.D.

Note that this proof is equivalently made by observing that the cyclicity of ABCD, i.e. thesupplementarity${\displaystyle \mathrm{\angle }ABC}$  and${\displaystyle \mathrm{\angle }CDA}$ , is equivalent to the condition

${\displaystyle \arg \left[(z_A-z_B)(z_C-z_D)\right]=\arg \left[(z_A-z_D)(z_B-z_C)\right]=\arg \left[(z_A-z_C)(z_B-z_D)\right](\mathrm{mod}2\pi )}$ ;

in particular there is a rotation of${\displaystyle \mathbb{C}}$  in which this${\displaystyle \arg }$  is 0 (i.e. all three products are positive real numbers), and by which Ptolemy's theorem

${\displaystyle \overline{AB}\cdot \overline{CD}+\overline{AD}\cdot \overline{BC}=\overline{AC}\cdot \overline{BD}}$ 

is then directly established from the simple algebraic identity

${\displaystyle (z_A-z_B)(z_C-z_D)+(z_A-z_D)(z_B-z_C)=(z_A-z_C)(z_B-z_D).}$ 

In the case of a circle of unit diameter the sides${\displaystyle S_1,S_2,S_3,S_4}$  of any cyclic quadrilateral ABCD are numerically equal to the sines of the angles${\displaystyle {\theta }_1,{\theta }_2,{\theta }_3}$  and${\displaystyle {\theta }_4}$  which they subtend (seeLaw of sines). Similarly the diagonals are equal to the sine of the sum of whicheverpair of angles they subtend. We may then write Ptolemy's Theorem in the following trigonometric form:

${\displaystyle \sin {\theta }_1\sin {\theta }_3+\sin {\theta }_2\sin {\theta }_4=\sin ({\theta }_1+{\theta }_2)\sin ({\theta }_1+{\theta }_4)}$ 

Applying certain conditions to the subtended angles${\displaystyle {\theta }_1,{\theta }_2,{\theta }_3}$  and${\displaystyle {\theta }_4}$  it is possible to derive a number of important corollaries using the above as our starting point. In what follows it is important to bear in mind that the sum of angles${\displaystyle {\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4={180}^{\circ }}$ .

Let${\displaystyle {\theta }_1={\theta }_3}$  and${\displaystyle {\theta }_2={\theta }_4}$ . Then${\displaystyle {\theta }_1+{\theta }_2={\theta }_3+{\theta }_4={90}^{\circ }}$ (since opposite angles of a cyclic quadrilateral are supplementary). Then:^[11]

${\displaystyle \sin {\theta }_1\sin {\theta }_3+\sin {\theta }_2\sin {\theta }_4=\sin ({\theta }_1+{\theta }_2)\sin ({\theta }_1+{\theta }_4)}$ 

${\displaystyle {\sin }^2{\theta }_1+{\sin }^2{\theta }_2={\sin }^2({\theta }_1+{\theta }_2)}$ 

${\displaystyle {\sin }^2{\theta }_1+{\cos }^2{\theta }_1=1}$ 

Let${\displaystyle {\theta }_2={\theta }_4}$ . The rectangle of corollary 1 is now a symmetrical trapezium with equal diagonals and a pair of equal sides. The parallel sides differ in length by${\displaystyle 2x}$  units where:

${\displaystyle x=S_2\cos ({\theta }_2+{\theta }_3)}$ 

It will be easier in this case to revert to the standard statement of Ptolemy's theorem:

${\displaystyle \begin{array}{l}{S}_1{S}_3+S_2{S}_4=\overline{AC}\cdot \overline{BD}\\ \Rightarrow S_1{S}_3+{S_2}^2={\overline{AC}}^2\\ \Rightarrow S_1[S_1-2S_2\cos ({\theta }_2+{\theta }_3)]+{S_2}^2={\overline{AC}}^2\\ \Rightarrow {S_1}^2+{S_2}^2-2S_1{S}_2\cos ({\theta }_2+{\theta }_3)={\overline{AC}}^2\end{array}}$ 

The cosine rule for triangle ABC.

Let

${\displaystyle {\theta }_1+{\theta }_2={\theta }_3+{\theta }_4={90}^{\circ }.}$ 

Then

${\displaystyle \sin {\theta }_1\sin {\theta }_3+\sin {\theta }_2\sin {\theta }_4=\sin ({\theta }_3+{\theta }_2)\sin ({\theta }_3+{\theta }_4)}$ 

Therefore,

${\displaystyle \cos {\theta }_2\sin {\theta }_3+\sin {\theta }_2\cos {\theta }_3=\sin ({\theta }_3+{\theta }_2)\times 1}$ 

Formula for compound angle sine (+).^[12]

Let${\displaystyle {\theta }_1={90}^{\circ }}$ . Then${\displaystyle {\theta }_2+({\theta }_3+{\theta }_4)={90}^{\circ }}$ . Hence,

${\displaystyle \sin {\theta }_1\sin {\theta }_3+\sin {\theta }_2\sin {\theta }_4=\sin ({\theta }_3+{\theta }_2)\sin ({\theta }_3+{\theta }_4)}$ 

${\displaystyle \sin {\theta }_3+\sin {\theta }_2\cos ({\theta }_2+{\theta }_3)=\sin ({\theta }_3+{\theta }_2)\cos {\theta }_2}$ 

${\displaystyle \sin {\theta }_3=\sin ({\theta }_3+{\theta }_2)\cos {\theta }_2-\cos ({\theta }_2+{\theta }_3)\sin {\theta }_2}$ 

Formula for compound angle sine (−).^[12]

This derivation corresponds to theThird Theoremas chronicled byCopernicus followingPtolemy inAlmagest. In particular if the sides of a pentagon (subtending 36° at the circumference) and of a hexagon (subtending 30° at the circumference) are given, a chord subtending 6° may be calculated. This was a critical step in the ancient method of calculating tables of chords.^[13]

This corollary is the core of theFifth Theorem as chronicled by Copernicus following Ptolemy in Almagest.

Let${\displaystyle {\theta }_3={90}^{\circ }}$ . Then${\displaystyle {\theta }_1+({\theta }_2+{\theta }_4)={90}^{\circ }}$ . Hence

${\displaystyle \sin {\theta }_1\sin {\theta }_3+\sin {\theta }_2\sin {\theta }_4=\sin ({\theta }_3+{\theta }_2)\sin ({\theta }_3+{\theta }_4)}$ 

${\displaystyle \cos ({\theta }_2+{\theta }_4)+\sin {\theta }_2\sin {\theta }_4=\cos {\theta }_2\cos {\theta }_4}$ 

${\displaystyle \cos ({\theta }_2+{\theta }_4)=\cos {\theta }_2\cos {\theta }_4-\sin {\theta }_2\sin {\theta }_4}$ 

Formula for compound angle cosine (+)

Despite lacking the dexterity of our modern trigonometric notation, it should be clear from the above corollaries that in Ptolemy's theorem (or more simply theSecond Theorem) the ancient world had at its disposal an extremely flexible and powerful trigonometric tool which enabled the cognoscenti of those times to draw up accurate tables of chords (corresponding to tables of sines) and to use these in their attempts to understand and map the cosmos as they saw it. Since tables of chords were drawn up byHipparchus three centuries before Ptolemy, we must assume he knew of the 'Second Theorem' and its derivatives. Following the trail of ancient astronomers, history records the star catalogue ofTimocharis of Alexandria. If, as seems likely, the compilation of such catalogues required an understanding of the 'Second Theorem', then the true origins of the latter disappear thereafter into the mists of antiquity; but it cannot be unreasonable to presume that the astronomers, architects and construction engineers of ancient Egypt may have had some knowledge of it.

The equation in Ptolemy's theorem is never true with non-cyclic quadrilaterals.Ptolemy's inequality is an extension of this fact, and it is a more general form of Ptolemy's theorem. It states that, given a quadrilateralABCD, then

${\displaystyle \overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA}\ge \overline{AC}\cdot \overline{BD}}$ 

where equality holdsif and only if the quadrilateral iscyclic. This special case is equivalent to Ptolemy's theorem.

Ptolemy's theorem gives the product of the diagonals (of a cyclic quadrilateral) knowing the sides. The following theorem yields the same for the ratio of the diagonals.^[14]

${\displaystyle \frac{AC}{BD}=\frac{AB\cdot DA+BC\cdot CD}{AB\cdot BC+DA\cdot CD}}$ 

Proof: It is known that the area of a triangle${\displaystyle ABC}$  inscribed in a circle of radius${\displaystyle R}$  is:${\displaystyle \mathcal{A}=\frac{AB\cdot BC\cdot CA}{4R}}$ 

Writing the area of the quadrilateral as sum of two triangles sharing the same circumscribing circle, we obtain two relations for each decomposition.

${\displaystyle {\mathcal{A}}_{\text{tot}}=\frac{AB\cdot BC\cdot CA}{4R}+\frac{CD\cdot DA\cdot AC}{4R}=\frac{AC\cdot (AB\cdot BC+CD\cdot DA)}{4R}}$ 

${\displaystyle {\mathcal{A}}_{\text{tot}}=\frac{AB\cdot BD\cdot DA}{4R}+\frac{BC\cdot CD\cdot DB}{4R}=\frac{BD\cdot (AB\cdot DA+BC\cdot CD)}{4R}}$ 

Equating, we obtain the announced formula.

Consequence: Knowing both the product and the ratio of the diagonals, we deduce their immediate expressions:

 

• Casey's theorem
• Intersecting chords theorem
• Greek mathematics

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• Compound angle proof atcut-the-knot
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