Editing Ptolemy's theorem

{{short description|Relates the 4 sides and 2 diagonals of a quadrilateral with vertices on a common circle}}
[[File:Ptolemy equality.svg|right|thumb|upright=1.25|Ptolemy's theorem is a relation among these lengths in a cyclic quadrilateral.<math>\definecolor{V}{RGB}{148,0,211} \definecolor{B}{RGB}{0,0,255} \definecolor{R}{RGB}{204,0,0}
{\color{V}AC}\cdot{\color{V}BD}={\color{B}AB}\cdot{\color{B}CD}+{\color{R}BC}\cdot{\color{R}AD}</math>]]

In [[Euclidean geometry]], '''Ptolemy's theorem''' is a relation between the four sides and two diagonals of a [[cyclic quadrilateral]] (a quadrilateral whose [[Vertex (geometry)#Of a polytope|vertices]] lie on a common circle). The theorem is named after the [[Roman Greece|Greek]] [[astronomer]] and [[mathematician]] [[Ptolemy]] (Claudius Ptolemaeus).<ref>C. Ptolemy, [[Almagest]], Book 1, Chapter 10.</ref> Ptolemy used the theorem as an aid to creating [[Ptolemy's table of chords|his table of chords]], a trigonometric table that he applied to astronomy.

If the vertices of the cyclic quadrilateral are ''A'', ''B'', ''C'', and ''D'' in order, then the theorem states that:

: <math>AC\cdot BD = AB\cdot CD+BC\cdot AD</math>

This relation may be verbally expressed as follows:

:''If a quadrilateral is [[Cyclic quadrilateral|cyclic]] 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, the [[Theorem#Converse|converse]] 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 a [[cyclic quadrilateral]].''

To appreciate the utility and general significance of Ptolemy’s Theorem, it is especially useful to study its main [[#Corollaries|Corollaries]].

== Corollaries on inscribed polygons ==
===Equilateral triangle===
[[Image:Ptolemy Equilateral.svg|right|thumb|Equilateral triangle]]

Ptolemy's Theorem yields as a corollary a theorem<ref name="wilson">[http://jwilson.coe.uga.edu/emt725/Ptolemy/Ptolemy.html Wilson, Jim. "Ptolemy's Theorem."] link verified 2009-04-08</ref> 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:

:<math> qs=ps+rs \Rightarrow q=p+r. </math>

This corollary has as an application an algorithm for computing minimal [[Steiner tree]]s whose topology is fixed, by repeatedly replacing pairs of leaves of the tree ''A'', ''B'' that should be connected to a [[Steiner point (computational geometry)|Steiner point]], by the third point ''C'' of their equilateral triangle. The unknown Steiner point must lie on arc ''AB'' of the circle, and this replacement ensures that, no matter where it is placed, the length of the tree remains unchanged.<ref>{{citation|title=The Shortest-Network Problem|first1=Marshall W.|last1=Bern|first2=Ronald L.|last2=Graham|author2-link=Ronald Graham|journal=[[Scientific American]]|volume=260|issue=1|date=January 1989|pages=84–89|doi=10.1038/scientificamerican0189-84 |jstor=24987111|bibcode=1989SciAm.260a..84B |url=https://mathweb.ucsd.edu/~ronspubs/89_01_shortest_network.pdf}}</ref>

===Square===

Any [[Square (geometry)|square]] 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 <math>a</math> then the length of the diagonal is equal to <math>a\sqrt{2}</math> according to the [[Pythagorean theorem]], and Ptolemy's relation obviously holds.

===Rectangle===
[[Image:Ptolemy Rectangle.svg|right|thumb|Pythagoras's theorem: ''"manifestum est"'': Copernicus]]
More generally, if the quadrilateral is a [[rectangle]] 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 d<sup>2</sup>, the right hand side of Ptolemy's relation is the sum ''a''<sup>2</sup>&nbsp;+&nbsp;''b''<sup>2</sup>.

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.''<ref>[http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1543droc.book.....C&db_key=AST&page_ind=36&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES De Revolutionibus Orbium Coelestium: Page 37]. See last two lines of this page. Copernicus refers to Ptolemy's theorem as [http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1543droc.book.....C&db_key=AST&page_ind=37&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES "Theorema Secundum".]</ref>

===Pentagon===
[[Image:Ptolemy Pentagon.svg|right|thumb|The [[golden ratio]] follows from this application of Ptolemy's theorem]]
A more interesting example is the relation between the length ''a'' of the side and the (common) length ''b'' of the 5 chords in a regular pentagon. By [[completing the square]], the relation yields the [[golden ratio]]:<ref>[http://aleph0.clarku.edu/~djoyce/java/elements/bookXIII/propXIII8.html Proposition 8] in Book XIII of [http://aleph0.clarku.edu/~djoyce/java/elements/Euclid.html Euclid's Elements] proves by similar triangles the same result: namely that length a (the side of the pentagon) divides length b (joining alternate vertices of the pentagon) in "mean and extreme ratio".</ref>
:<math>\begin{array}{rl}
         b \cdot b                  \,\;\;\qquad\quad\qquad =&\!\!\!\! a \! \cdot \! a + a \! \cdot \! b
\\       b^2  \;\;  - ab                        \quad\qquad =&\!\!     a^2
\\ \frac{b^2}{a^2} \;\; - \frac{ab}{a^2}       \;\;\;\qquad =&\!\!\!        \frac{        a^2 }{a^2}
\\ \left(\frac{b}{a}\right)^2 - \frac{b}{a} + \left(\frac{1}{2}\right)^2 =&\!\!     1 + \left(\frac{          1 }{  2}\right)^2
\\ \left(\frac{b}{a}         -                  \frac{1}{2}\right)^2     =&\!\!    \quad \frac{          5 }{  4}
\\  \frac{b}{a}         -                \frac{1}{2} \;\;\; =&\!\!\!\!  \pm \frac{    \sqrt{5}}{  2}
\\  \frac{b}{a} > 0 \, \Rightarrow \, \varphi = \frac{b}{a} =&\!\!\!\!      \frac{1 + \sqrt{5}}{  2}
\end{array}</math>

===Side of decagon===
[[Image:Ptolemy Pentagon2.svg|right|thumb|Side of the inscribed decagon]]
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 diameter ''d'' as one of its diagonals:

:<math>ad=2bc</math>

:<math>\Rightarrow ad=2\varphi ac</math> where <math>\varphi</math> is the golden ratio.
:<math>\Rightarrow c=\frac{d}{2\varphi}.</math><ref>And in analogous fashion [http://aleph0.clarku.edu/~djoyce/java/elements/bookXIII/propXIII9.html Proposition 9] in Book XIII of [http://aleph0.clarku.edu/~djoyce/java/elements/Euclid.html Euclid's Elements] proves by similar triangles that length c (the side of the decagon) divides the radius in "mean and extreme ratio".</ref>

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 <ref>An interesting article on the construction of a regular pentagon and determination of side length can be found at the following reference [http://www.cut-the-knot.org/pythagoras/pentagon.shtml]</ref> is thereafter calculated as
::<math>a = \frac {b} {\varphi} = b \left( \varphi - 1 \right).</math>

As [[Copernicus]] (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."''<ref>[http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1543droc.book.....C&db_key=AST&page_ind=36&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES De Revolutionibus Orbium Coelestium: Liber Primus: Theorema Primum]</ref>

==Proofs==
{{Further |Proofs of trigonometric identities}}

=== Visual proof ===

[[File:Animated visual proof of Ptolemy's theorem, based on Derrick & Herstein (2012).gif|thumb|Animated visual proof of Ptolemy's theorem, based on Derrick & Herstein (2012).]]
The animation here shows a visual demonstration of Ptolemy's theorem, based on Derrick & Herstein (2012).<ref>W. Derrick, J. Herstein (2012) Proof Without Words: Ptolemy's Theorem, The College Mathematics Journal, v.43, n.5, p. 386.</ref>

=== Proof by similarity of triangles ===
[[Image:Ptolemy's theorem.svg|thumb|Constructions for a proof of Ptolemy's theorem]]
Let ABCD be a [[cyclic quadrilateral]].
On the [[chord (geometry)|chord]] BC, the [[inscribed angle]]s ∠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 is [[Similarity (geometry)|similar]] 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.]]<ref>{{citation|title=Charming Proofs: A Journey Into Elegant Mathematics|volume=42|series=Dolciani Mathematical Expositions|first1=Claudi|last1=Alsina|first2=Roger B.|last2=Nelsen|publisher=[[Mathematical Association of America]]|year=2010|isbn=9780883853481|page=112|url=https://books.google.com/books?id=mIT5-BN_L0oC&pg=PA112}}.</ref>

The proof as written is only valid for [[simple polygon|simple]] cyclic quadrilaterals. If the quadrilateral is self-crossing then K will be located outside the line segment AC. But in this case, AK&minus;CK = ±AC, giving the expected result.

=== Proof by trigonometric identities ===
Let the inscribed angles subtended by <math>AB</math>, <math>BC</math> and <math>CD</math> be, respectively, <math>\alpha</math>, <math>\beta</math> and <math>\gamma</math>, and the radius of the circle be <math>R</math>, then we have <math>AB=2R\sin\alpha</math>, <math>BC=2R\sin\beta</math>, <math>CD=2R\sin\gamma</math>, <math>AD=2R\sin(180^\circ-(\alpha+\beta+\gamma))</math>, <math>AC=2R\sin(\alpha+\beta)</math> and <math>BD=2R\sin(\beta+\gamma)</math>, and the original equality to be proved is transformed to

:<math> \sin(\alpha+\beta)\sin(\beta+\gamma) = \sin\alpha\sin\gamma + \sin\beta \sin(\alpha + \beta+\gamma)</math>

from which the factor <math>4R^2</math> has disappeared by dividing both sides of the equation by it.

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

:<math>
\begin{align}
& \sin\alpha\sin\beta\cos\beta\cos\gamma + \sin\alpha\cos^2\beta\sin\gamma \\ + {} & \cos\alpha\sin^2\beta\cos\gamma+\cos\alpha\sin\beta\cos\beta\sin\gamma.
\end{align}
</math>

Q.E.D.

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

:<math> 
\begin{align}
\mathrm{Area}(ABCD) & = \frac{1}{2}  AC\cdot BD \cdot \sin(\alpha + \gamma); \\
\mathrm{Area}(ABCD') & = \frac{1}{2} AB\cdot AD'\cdot \sin(180^\circ - \alpha - \gamma) +
 \frac{1}{2} BC\cdot CD' \cdot \sin(\alpha + \gamma)\\ & = 
\frac{1}{2} (AB\cdot CD +  BC\cdot AD)\cdot \sin(\alpha + \gamma).
\end{align}
</math>
 
Q.E.D.

=== Proof by inversion ===
[[File:Ptolemy-crop.svg|thumb|Proof of Ptolemy's theorem via [[circle inversion]]]]

Choose an auxiliary circle <math> \Gamma </math> of radius <math> r </math> centered at D with respect to which the circumcircle of ABCD is [[Inversive geometry#Circle inversion|inverted]] into a line (see figure).
Then
<math> A'B' + B'C' = A'C'. </math>
Then <math> A'B', B'C' </math> and <math> A'C' </math> can be expressed as 
<math display="inline"> \frac{AB \cdot DB'}{DA} </math>, <math display="inline"> \frac{BC \cdot DB'}{DC} </math> and <math display="inline"> \frac{AC \cdot DC'}{DA} </math> respectively. Multiplying each term by <math display="inline"> \frac{DA \cdot DC}{DB'} </math> and using <math display="inline">\frac{DC'}{DB'} = \frac{DB}{DC} </math> 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.

=== Proof using complex numbers ===
Embed ABCD in the [[complex plane]] <math>\mathbb{C}</math> by identifying <math>A\mapsto z_A,\ldots,D\mapsto z_D</math> as four distinct [[complex number]]s <math>z_A,\ldots,z_D\in\mathbb{C}</math>. Define the [[cross-ratio]]
:<math>\zeta:=\frac{(z_A-z_B)(z_C-z_D)}{(z_A-z_D)(z_B-z_C)}\in\mathbb{C}_{\neq0}</math>.
Then 
:<math>
\begin{align}
\overline{AB}\cdot\overline{CD}+\overline{AD}\cdot\overline{BC}
& = \left|z_A-z_B\right|\left|z_C-z_D\right| + \left|z_A-z_D\right|\left|z_B-z_C\right| \\
& = \left|(z_A-z_B)(z_C-z_D)\right| + \left|(z_A-z_D)(z_B-z_C)\right| \\
& = \left(\left|\frac{(z_A-z_B)(z_C-z_D)}{(z_A-z_D)(z_B-z_C)}\right| + 1\right) \left|(z_A-z_D)(z_B-z_C)\right| \\
& = \left(\left|\zeta\right| +1\right) \left|(z_A-z_D)(z_B-z_C)\right| \\
& \geq  \left|(\zeta +1)(z_A-z_D)(z_B-z_C)\right| \\
& = \left|(z_A-z_B)(z_C-z_D)+(z_A-z_D)(z_B-z_C)\right| \\
& = \left|(z_A-z_C)(z_B-z_D)\right| \\
& = \left|z_A-z_C\right|\left|z_B-z_D\right| \\
& = \overline{AC}\cdot\overline{BD}
\end{align}
</math>
with equality if and only if the cross-ratio <math>\zeta</math> is a positive real number.  This proves [[Ptolemy's inequality]] generally, as it remains only to show that <math>z_A,\ldots,z_D</math> lie consecutively arranged
on a circle (possibly of infinite radius, i.e. a line) in <math>\mathbb{C}</math> if and only if <math>\zeta\in\mathbb{R}_{>0}</math>.

From the [[Complex number#Polar form|polar form]] of a complex number <math>z=\vert z\vert e^{i\arg(z)}</math>, it follows
:<math>
\begin{align}
\arg(\zeta) & = \arg\frac{(z_A-z_B)(z_C-z_D)}{(z_A-z_D)(z_B-z_C)} \\
& = \arg(z_A-z_B)+\arg(z_C-z_D)-\arg(z_A-z_D)-\arg(z_B-z_C) \pmod{2\pi} \\
& = \arg(z_A-z_B)+\arg(z_C-z_D)-\arg(z_A-z_D)-\arg(z_C-z_B) - \arg(-1) \pmod{2\pi} \\
& = - \left[\arg(z_C-z_B)-\arg(z_A-z_B)\right] - \left[\arg(z_A-z_D)-\arg(z_C-z_D)\right] -\arg(-1) \pmod{2\pi} \\
& = - \angle ABC - \angle CDA -\pi \pmod{2\pi}\\
& = 0
\end{align}
</math>
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 <math>\pi</math>.

Q.E.D.

Note that this proof is equivalently made by observing that the cyclicity of ABCD, i.e. the [[Angle#Supplementary angle|supplementarity]] <math>\angle ABC</math> and <math>\angle CDA</math>, is equivalent to the condition
:<math>\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]  \pmod{2\pi}</math>;
in particular there is a rotation of <math>\mathbb{C}</math> in which this <math>\arg</math> is 0 (i.e. all three products are positive real numbers), and by which Ptolemy's theorem 
:<math>\overline{AB}\cdot \overline{CD}+\overline{AD}\cdot\overline{BC} = \overline{AC}\cdot \overline{BD}</math>
is then directly established from the simple algebraic identity
:<math>(z_A-z_B)(z_C-z_D)+(z_A-z_D)(z_B-z_C)=(z_A-z_C)(z_B-z_D).</math>

==Corollaries==
[[File:Ptolemy theore trig2 proof.svg|thumb|<math>|S_1|=\sin(\theta_1)</math>]]
[[File:Ptolemy theore trig3.svg|thumb|Corollary 1: Pythagoras's theorem]]

In the case of a circle of unit diameter the sides <math>S_1,S_2,S_3,S_4</math> of any cyclic quadrilateral ABCD are numerically equal to the sines of the angles <math>\theta_1,\theta_2,\theta_3</math> and <math>\theta_4</math> which they subtend (see [[Law of sines]]). Similarly the diagonals are equal to the sine of the sum of whichever '''pair''' of angles they subtend. We may then write Ptolemy's Theorem in the following trigonometric form:

:<math>\sin\theta_1\sin\theta_3+\sin\theta_2\sin\theta_4=\sin(\theta_1+\theta_2)\sin(\theta_1+\theta_4)</math>

Applying certain conditions to the subtended angles <math>\theta_1,\theta_2,\theta_3</math> and <math>\theta_4</math> 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 <math>\theta_1+\theta_2+\theta_3+\theta_4=180^\circ</math>.

===Corollary 1. Pythagoras's theorem===
Let <math>\theta_1=\theta_3</math> and <math>\theta_2=\theta_4</math>. Then <math>\theta_1+\theta_2=\theta_3+\theta_4=9
0^\circ</math>
(since opposite angles of a cyclic quadrilateral are supplementary). Then:<ref>In [[De Revolutionibus Orbium Coelestium]], [[Copernicus]] does not refer to Pythagoras's theorem by name but uses the term 'Porism' – a word which in this particular context would appear to denote an observation on – or obvious consequence of – another existing theorem. The 'Porism' can be viewed on pages [http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1543droc.book.....C&db_key=AST&page_ind=36&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES 36 and 37] of DROC (Harvard electronic copy)</ref>

:<math>\sin\theta_1\sin\theta_3+\sin\theta_2\sin\theta_4=\sin(\theta_1+\theta_2)\sin(\theta_1+\theta_4) </math>

:<math> \sin^2\theta_1+\sin^2\theta_2=\sin^2(\theta_1+\theta_2) </math>

:<math> \sin^2\theta_1+\cos^2\theta_1=1 </math>

===Corollary 2. The law of cosines===
[[File:Ptolemy theore trig4.svg|thumb|Corollary 2: the law of cosines]]

Let <math>\theta_2=\theta_4</math>. 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 <math>2x</math> units where:
:<math>x=S_2\cos(\theta_2+\theta_3)</math>

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

:<math>\begin{array}{lcl}
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}</math>

The cosine rule for triangle ABC.

===Corollary 3. Compound angle sine (+)===
Let

: <math>\theta_1+\theta_2=\theta_3+\theta_4=90^\circ. </math>

Then

:<math>
\sin\theta_1\sin\theta_3+\sin\theta_2\sin\theta_4=\sin(\theta_3+\theta_2)\sin(\theta_3+\theta_4)
</math>

Therefore,

: <math> \cos\theta_2\sin\theta_3+\sin\theta_2\cos\theta_3=\sin(\theta_3+\theta_2)\times 1 </math>

Formula for compound angle sine (+).<ref name="cut_the_knot">{{cite web | url = http://www.cut-the-knot.org/proofs/sine_cosine.shtml | title = Sine, Cosine, and Ptolemy's Theorem }}</ref>

===Corollary 4. Compound angle sine (−) ===
Let <math>\theta_1=90^\circ</math>. Then <math>\theta_2+(\theta_3+\theta_4)=90^\circ</math>. Hence,

:<math>\sin\theta_1\sin\theta_3+\sin\theta_2\sin\theta_4=\sin(\theta_3+\theta_2)\sin(\theta_3+\theta_4) </math>

: <math> \sin\theta_3+\sin\theta_2\cos(\theta_2+\theta_3)=\sin(\theta_3+\theta_2)\cos\theta_2 </math>

: <math>\sin\theta_3=\sin(\theta_3+\theta_2)\cos\theta_2-\cos(\theta_2+\theta_3)\sin\theta_2 </math>

Formula for compound angle sine&nbsp;(−).<ref name="cut_the_knot" />

This derivation corresponds to the [http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1543droc.book.....C&db_key=AST&page_ind=38&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES Third Theorem]
as chronicled by [[Copernicus]] following [[Ptolemy]] in [[Almagest]]. 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.<ref>To understand the Third Theorem, compare the Copernican diagram shown on page 39 of the [http://ads.harvard.edu/books/1543droc.book/ Harvard copy] of De Revolutionibus to that for the derivation of sin(A-B) found in the above [http://www.cut-the-knot.org/proofs/sine_cosine.shtml cut-the-knot] web page</ref>

===Corollary 5. Compound angle cosine (+)===
This corollary is the core of the [http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1543droc.book.....C&db_key=AST&page_ind=39&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES Fifth Theorem] as chronicled by Copernicus following Ptolemy in Almagest.

Let <math>\theta_3=90^\circ</math>. Then <math>\theta_1+(\theta_2+\theta_4)=90^\circ </math>. Hence

:<math> \sin\theta_1\sin\theta_3+\sin\theta_2\sin\theta_4=\sin(\theta_3+\theta_2)\sin(\theta_3+\theta_4) </math>

: <math> \cos(\theta_2+\theta_4)+\sin\theta_2\sin\theta_4=\cos\theta_2\cos\theta_4 </math>

: <math>\cos(\theta_2+\theta_4)=\cos\theta_2\cos\theta_4-\sin\theta_2\sin\theta_4 </math>

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 the [http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1543droc.book.....C&db_key=AST&page_ind=37&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES Second 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 by [[Hipparchus]] 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 of [[Timocharis]] 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.

==Ptolemy's inequality==
{{main|Ptolemy's inequality}}
[[Image:Ptolemy Inequality.svg|thumb|This is ''not'' a cyclic quadrilateral. The equality never holds here, and is unequal in the direction indicated by Ptolemy's inequality.]]

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 quadrilateral ''ABCD'', then

: <math>\overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA} \ge \overline{AC}\cdot \overline{BD}</math>

where equality holds [[if and only if]] the quadrilateral is [[cyclic quadrilateral|cyclic]]. This special case is equivalent to Ptolemy's theorem.
{{Clear}}

== Related theorem about the ratio of the diagonals ==
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.<ref>Claudi Alsina, Roger B. Nelsen: ''Charming Proofs: A Journey Into Elegant Mathematics''. MAA, 2010, {{ISBN|9780883853481}}, pp. [https://books.google.com/books?id=mIT5-BN_L0oC&pg=PA112 112–113]</ref>
: <math>
\frac{AC}{BD}=\frac{AB \cdot DA + BC \cdot CD}{AB \cdot BC + DA \cdot CD}</math>

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

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

: <math> \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}
</math>

: <math>\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}
</math>

Equating, we obtain the announced formula.

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

: <math>
\begin{align}
AC^2 & =AC \cdot BD \cdot \frac{AC}{BD}=(AB \cdot CD + BC \cdot DA)\frac{AB \cdot DA + BC \cdot CD}{AB \cdot BC + DA \cdot CD} \\[8pt]
BD^2 & =\frac {AC \cdot BD}{\frac{AC}{BD}}=(AB \cdot CD + BC \cdot DA)\frac{AB \cdot BC + DA \cdot CD}{AB \cdot DA + BC \cdot CD}
\end{align}
</math>

== See also ==

* [[Casey's theorem]]
* [[Intersecting chords theorem]]
* [[Greek mathematics]]

==Notes==
{{reflist}}

==References==
* [[Harold Scott MacDonald Coxeter|Coxeter, H. S. M.]] and [[S. L. Greitzer]] (1967) "Ptolemy's Theorem and its Extensions." §2.6 in ''Geometry Revisited'', [[Mathematical Association of America]] pp.&nbsp;42–43.
* [[Copernicus]] (1543) [[De Revolutionibus Orbium Coelestium]], English translation found in ''On the Shoulders of Giants'' (2002) edited by [[Stephen Hawking]], [[Penguin Books]] {{isbn|0-14-101571-3}}
* Amarasinghe, G. W. I. S. (2013) [http://geometry-math-journal.ro/pdf/Volume2-Issue1/A%20Concise%20Elementary%20Proof%20for%20the%20Ptolemy.pdf A Concise Elementary Proof for the Ptolemy's Theorem], ''Global Journal of Advanced Research on Classical and Modern Geometries (GJARCMG)'' 2(1): 20–25 (pdf).

==External links==
* [http://www.mathalino.com/reviewer/derivation-formulas/derivation-proof-ptolemy-s-theorem-cyclic-quadrilateral Proof of Ptolemy's Theorem for Cyclic Quadrilateral]
* [http://www.mathpages.com/home/kmath099/kmath099.htm MathPages – On Ptolemy's Theorem]
* {{cite web|url=http://hypertextbook.com/eworld/chords/|title=Ptolemy's Table of Chords|first=Glenn|last=Elert|work=E-World|year=1994}}
* [http://www.cut-the-knot.org/proofs/ptolemy.shtml Ptolemy's Theorem] at [[cut-the-knot]]
* [http://www.cut-the-knot.org/proofs/sine_cosine.shtml Compound angle proof] at [[cut-the-knot]]
*[http://planetmath.org/encyclopedia/PtolemysTheorem.html Ptolemy's Theorem] {{Webarchive|url=https://web.archive.org/web/20110724093414/http://planetmath.org/encyclopedia/PtolemysTheorem.html |date=2011-07-24 }} on [[PlanetMath]]
*[http://mathworld.wolfram.com/PtolemyInequality.html Ptolemy Inequality] on [[MathWorld]]
*[http://ads.harvard.edu/books/1543droc.book/ De Revolutionibus Orbium Coelestium] at Harvard.
*[https://web.archive.org/web/20080705164103/http://www.atara.net/deep_secrets/index.html Deep Secrets: The Great Pyramid, the Golden Ratio and the Royal Cubit]
*''[http://demonstrations.wolfram.com/PtolemysTheorem/ Ptolemy's Theorem]'' by Jay Warendorff, [[The Wolfram Demonstrations Project]].
*[http://aleph0.clarku.edu/~djoyce/java/elements/bookXIII/bookXIII.html Book XIII] of [http://aleph0.clarku.edu/~djoyce/java/elements/Euclid.html Euclid's Elements]
* [https://www.youtube.com/watch?v=bJOuzqu3MUQ A Miraculous Proof (Ptolemy's Theorem)] by Zvezdelina Stankova, on Numberphile.

{{Ancient Greek mathematics}}

[[Category:Theorems about quadrilaterals and circles]]
[[Category:Ptolemy|Theorem]]
[[Category:Articles containing proofs]]
[[Category:Euclidean plane geometry]]
[[Category:Greek mathematics]]