View source for Cyclic quadrilateral

{{Use American English|date = February 2019}}

{{Short description|Quadrilateral whose vertices can all fall on a single circle}}

{{Use mdy dates|date = February 2019}}
[[Image:Cyclic quadrilateral.svg|thumb|right|Examples of cyclic quadrilaterals.]]
In [[Euclidean geometry]], a '''cyclic quadrilateral''' or '''inscribed quadrilateral''' is a [[quadrilateral]] whose [[vertex (geometry)|vertices]] all lie on a single [[circle]]. This circle is called the ''circumcircle'' or [[circumscribed circle]], and the vertices are said to be ''[[concyclic]]''. The center of the circle and its radius are called the ''circumcenter'' and the ''circumradius'' respectively. Other names for these quadrilaterals are '''concyclic quadrilateral''' and '''chordal quadrilateral''', the latter since the sides of the quadrilateral are [[Chord (geometry)|chords]] of the circumcircle. Usually the quadrilateral is assumed to be [[convex polygon|convex]], but there are also crossed cyclic quadrilaterals. The formulas and properties given below are valid in the convex case.

The word cyclic is from the [[Ancient Greek]] {{lang|grc|[[wiktionary:κύκλος#Ancient Greek|κύκλος]]}} (''kuklos'') which means "circle" or "wheel".

All [[triangle]]s have a [[Circumscribed circle|circumcircle]], but not all quadrilaterals do. An example of a quadrilateral that cannot be cyclic is a non-square [[rhombus]]. The section [[Cyclic quadrilateral#Characterizations|characterizations]] below states what [[necessary and sufficient condition]]s a quadrilateral must satisfy to have a circumcircle.

==Special cases==
Any [[Square (geometry)|square]], [[rectangle]], [[isosceles trapezoid]], or [[antiparallelogram]] is cyclic. A [[kite (geometry)|kite]] is cyclic [[if and only if]] it has two right angles. A [[bicentric quadrilateral]] is a cyclic quadrilateral that is also [[tangential quadrilateral|tangential]] and an [[Ex-tangential quadrilateral#Ex-bicentric quadrilateral|ex-bicentric quadrilateral]] is a cyclic quadrilateral that is also [[Ex-tangential quadrilateral|ex-tangential]]. A [[harmonic quadrilateral]] is a cyclic quadrilateral in which the product of the lengths of opposite sides are equal.

==Characterizations==
[[Image:Sehnenviereck2.PNG|thumb|right|A cyclic quadrilateral ABCD]]
A convex quadrilateral is cyclic [[if and only if]] the four [[perpendicular]] bisectors to the sides are [[Concurrent lines|concurrent]]. This common point is the [[circumcenter]].<ref name=Usiskin>{{citation |first1=Zalman |last1=Usiskin |first2=Jennifer |last2=Griffin |first3=David |last3=Witonsky |first4=Edwin |last4=Willmore |title=The Classification of Quadrilaterals: A Study of Definition |chapter=10. Cyclic quadrilaterals |chapterurl=https://books.google.com/books?id=ZkoUR5lRwdcC&pg=PA63 |year=2008 |publisher=IAP |isbn=978-1-59311-695-8 |pages=63–65 |series=Research in mathematics education}}</ref>

A convex quadrilateral {{math|''ABCD''}} is cyclic if and only if its opposite angles are [[supplementary angle|supplementary]], that is<ref name=Usiskin/>
<ref name=Fraivert>{{citation
|last1=Fraivert|first1=David|last2=Sigler|first2=Avi|last3=Stupel|first3=Moshe
|journal=International Journal of Mathematical Education in Science and Technology
|title=Necessary and sufficient properties for a cyclic quadrilateral
|url=https://doi.org/10.1080/0020739X.2019.1683772|year=2019}}</ref>
:<math>\alpha + \gamma = \beta + \delta = \pi \ (= 180^{\circ}).</math>

The direct theorem was Proposition 22 in Book 3 of [[Euclid]]'s [[Euclid's Elements|''Elements'']].<ref>{{citation |date=June 1997 |last=Joyce |first=D. E. |chapterurl=http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII22.html |title=Euclid's Elements |chapter=Book 3, Proposition 22 |publisher=Clark University|title-link=Euclid's Elements }}</ref> Equivalently, a convex quadrilateral is cyclic if and only if each [[exterior angle]] is equal to the opposite [[interior angle]].

Another [[necessary and sufficient condition]] for a convex quadrilateral {{math|''ABCD''}} to be cyclic is that an angle between a side and a [[diagonal]] is equal to the angle between the opposite side and the other diagonal.<ref name=Andreescu>{{citation |first1=Titu |last1=Andreescu |first2=Bogdan |last2=Enescu |title=Mathematical Olympiad Treasures |chapter=2.3 Cyclic quads |chapterurl=https://books.google.com/books?id=mwUHJpvLOPsC&pg=PA44 |year=2004 |publisher=Springer |isbn=978-0-8176-4305-8 |pages=44–46, 50 |mr=2025063}}</ref> That is, for example,
:<math>\angle ACB = \angle ADB.</math>

[[Ptolemy's theorem]] expresses the product of the lengths of the two diagonals {{math|''e''}} and {{math|''f''}} of a cyclic quadrilateral as equal to the sum of the products of opposite sides:<ref name=Durell/>{{rp|p.25}}<ref name=Fraivert/>
:<math>\displaystyle ef = ac + bd.</math>

The [[Theorem#Converse|converse]] is also true. That is, if this equation is satisfied in a convex quadrilateral, then a cyclic quadrilateral is formed.

In a convex quadrilateral {{math|''ABCD''}}, let {{math|''EFG''}} be the diagonal triangle of {{math|''ABCD''}} and let <math>\omega</math> be the nine-point circle of {{math|''EFG''}}.
{{math|''ABCD''}} is cyclic if and only if the point of intersection of the bimedians of {{math|''ABCD''}} belongs to the nine-point circle <math>\omega</math>.<ref>{{Cite journal|last=Fraivert|first=David|date=July 2019|title=New points that belong to the nine-point circle|journal=The Mathematical Gazette|volume=103|issue=557|pages=222–232|doi=10.1017/mag.2019.53}}</ref><ref>{{Cite journal|last=Fraivert|first=David|date=2018|title=New applications of method of complex numbers in the geometry of cyclic quadrilaterals|url=https://ijgeometry.com/wp-content/uploads/2018/04/5-16.pdf|journal=International Journal of Geometry|volume=7|issue=1|pages=5–16|via=}}</ref><ref name=Fraivert/>
[[File:Nine-point circle of diagonal triangle.png|thumb|''ABCD'' is a cyclic quadrilateral. ''EFG'' is the diagonal triangle of ''ABCD''. The point ''T'' of intersection of the bimedians of ''ABCD'' belongs to the nine-point circle of ''EFG''.]]

If two lines, one  containing segment {{math|''AC''}} and the other containing segment {{math|''BD''}}, intersect at {{math|''P''}}, then the four points {{math|''A''}}, {{math|''B''}}, {{math|''C''}}, {{math|''D''}} are concyclic if and only if<ref>{{citation |last=Bradley |first=Christopher J. |title=The Algebra of Geometry: Cartesian, Areal and Projective Co-Ordinates |publisher=Highperception |year=2007 |isbn=978-1906338008 |page=179 |oclc=213434422}}</ref>
:<math>\displaystyle AP\cdot PC = BP\cdot PD.</math>

The intersection {{math|''P''}} may be internal or external to the circle. In the former case, the cyclic quadrilateral is {{math|''ABCD''}}, and in the latter case, the cyclic quadrilateral is {{math|''ABDC''}}. When the intersection is internal, the equality states that the product of the segment lengths into which {{math|''P''}} divides one diagonal equals that of the other diagonal. This is known as the ''[[intersecting chords theorem]]'' since the diagonals of the cyclic quadrilateral are chords of the circumcircle.

Yet another characterization is that a convex quadrilateral {{math|''ABCD''}} is cyclic if and only if<ref>{{citation
|last=Hajja |first=Mowaffaq
|journal=Forum Geometricorum
|pages=103–6
|title=A condition for a circumscriptible quadrilateral to be cyclic
|url=http://forumgeom.fau.edu/FG2008volume8/FG200814.pdf
|volume=8 |year=2008}}</ref>
:<math>\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}=\tan{\frac{\beta}{2}}\tan{\frac{\delta}{2}}=1.</math>

==Area==
The [[area]] {{math|''K''}} of a cyclic quadrilateral with sides {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, {{math|''d''}} is given by [[Brahmagupta's formula]]<ref name=Durell/>{{rp|p.24}}
:<math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)} \,</math>

where {{math|''s''}}, the [[semiperimeter]], is {{math|''s'' {{=}} {{sfrac|1|2}}(''a'' + ''b'' + ''c'' + ''d'')}}. This is a [[corollary]] of [[Bretschneider's formula]] for the general quadrilateral, since opposite angles are supplementary in the cyclic case. If also {{math|''d'' {{=}} 0}}, the cyclic quadrilateral becomes a triangle and the formula is reduced to [[Heron's formula]].

The cyclic quadrilateral has [[Maxima and minima|maximal]] area among all quadrilaterals having the same sequence of side lengths. This is another corollary to Bretschneider's formula. It can also be proved using [[calculus]].<ref>{{citation |last=Peter |first=Thomas |title=Maximizing the area of a quadrilateral |journal=The College Mathematics Journal |volume=34 |issue=4 |date=September 2003 |pages=315–6 |jstor=3595770|doi=10.2307/3595770 }}</ref>

Four unequal lengths, each less than the sum of the other three, are the sides of each of three non-congruent cyclic quadrilaterals,<ref name=Coxeter/> which by Brahmagupta's formula all have the same area. Specifically, for sides {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, and {{math|''d''}}, side {{math|''a''}} could be opposite any of side {{math|''b''}}, side {{math|''c''}}, or side {{math|''d''}}.

The area of a cyclic quadrilateral with successive sides {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, {{math|''d''}} and angle {{math|''B''}} between sides {{math|''a''}} and {{math|''b''}} can be expressed as<ref name=Durell/>{{rp|p.25}}
:<math>K = \tfrac{1}{2}(ab+cd)\sin{B}</math>

or<ref name=Durell/>{{rp|p.26}}
:<math>K = \tfrac{1}{2}(ac+bd)\sin{\theta}</math>

where {{math|''θ''}} is either angle between the diagonals. Provided {{math|''A''}} is not a right angle, the area can also be expressed as<ref name=Durell/>{{rp|p.26}}
:<math>K = \tfrac{1}{4}(a^2-b^2-c^2+d^2)\tan{A}.</math>

Another formula is<ref>{{citation|last=Prasolov|first=Viktor|title=Problems in plane and solid geometry: v.1 Plane Geometry|url=http://students.imsa.edu/~tliu/Math/planegeo.pdf|access-date=November 6, 2011|archive-url=https://web.archive.org/web/20180921105112/http://students.imsa.edu/~tliu/Math/planegeo.pdf|archive-date=September 21, 2018|url-status=dead}}</ref>{{rp|p.83}}
:<math>\displaystyle K=2R^2\sin{A}\sin{B}\sin{\theta}</math>

where {{math|''R''}} is the radius of the [[Circumscribed circle|circumcircle]]. As a direct consequence,<ref name=Alsina>{{citation |first1=Claudi |last1=Alsina |first2=Roger |last2=Nelsen |title=When Less is More: Visualizing Basic Inequalities |chapter=4.3 Cyclic, tangential, and bicentric quadrilaterals |chapterurl=https://books.google.com/books?id=U1ovBsSRNscC&pg=PA64 |url=https://books.google.com/books?id=U1ovBsSRNscC |year=2009 |publisher=Mathematical Association of America |isbn=978-0-88385-342-9 |page=64}}</ref>

:<math>K\le 2R^2</math>
where there is equality if and only if the quadrilateral is a square.

==Diagonals==
In a cyclic quadrilateral with successive vertices {{math|''A''}}, {{math|''B''}}, {{math|''C''}}, {{math|''D''}} and sides {{math|''a'' {{=}} ''AB''}}, {{math|''b'' {{=}} ''BC''}}, {{math|''c'' {{=}} ''CD''}}, and {{math|''d'' {{=}} ''DA''}}, the lengths of the diagonals {{math|''p'' {{=}} ''AC''}} and {{math|''q'' {{=}} ''BD''}} can be expressed in terms of the sides as<ref name=Durell/>{{rp|p.25,}}<ref name=Alsina2>{{citation
|last1=Alsina |first1=Claudi |last2=Nelsen |first2=Roger B.
|journal=Forum Geometricorum
|pages=147–9
|title=On the diagonals of a cyclic quadrilateral
|url=http://forumgeom.fau.edu/FG2007volume7/FG200720.pdf |volume=7
|year=2007}}</ref><ref name=Johnson>Johnson, Roger A., ''Advanced Euclidean Geometry'', Dover Publ., 2007 (orig. 1929).</ref>{{rp|p. 84}}

:<math>p = \sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}</math> and <math>q = \sqrt{\frac{(ac+bd)(ab+cd)}{ad+bc}}</math>

so showing [[Ptolemy's theorem]]
:<math>pq = ac+bd.</math>

According to ''Ptolemy's second theorem'',<ref name=Durell/>{{rp|p.25,}}<ref name=Alsina2/>
:<math>\frac {p}{q}= \frac{ad+bc}{ab+cd}</math>

using the same notations as above.

For the sum of the diagonals we have the inequality<ref name=Crux>''Inequalities proposed in "[[Crux Mathematicorum]]"'', 2007,  [http://www.imomath.com/othercomp/Journ/ineq.pdf].</ref>{{rp|p.123,#2975}}

:<math>p+q\ge 2\sqrt{ac+bd}.</math>

Equality holds [[if and only if]] the diagonals have equal length, which can be proved using the [[AM-GM inequality]].

Moreover,<ref name=Crux/>{{rp|p.64,#1639}}

:<math>(p+q)^2 \leq (a+c)^2+(b+d)^2.</math>

In any convex quadrilateral, the two diagonals together partition the quadrilateral into four triangles; in a cyclic quadrilateral, opposite pairs of these four triangles are [[Similarity (geometry)|similar]] to each other.

If {{math|''M''}} and {{math|''N''}} are the midpoints of the diagonals {{math|''AC''}} and {{math|''BD''}}, then<ref>{{cite web |title=''ABCD'' is a cyclic quadrilateral. Let ''M'', ''N'' be midpoints of diagonals ''AC'', ''BD'' respectively...  |date=2010 |work=Art of Problem Solving |url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=350179}}</ref>
:<math>\frac{MN}{EF}=\frac{1}{2}\left |\frac{AC}{BD}-\frac{BD}{AC}\right|</math>

where {{math|''E''}} and {{math|''F''}} are the intersection points of the extensions of opposite sides.

If {{math|''ABCD''}} is a cyclic quadrilateral where {{math|''AC''}} meets {{math|''BD''}} at {{math|''E''}}, then<ref>[[A. Bogomolny]], An Identity in (Cyclic) Quadrilaterals, ''Interactive Mathematics Miscellany and Puzzles'',
[http://www.cut-the-knot.org/m/Geometry/SinesInQuadrilateral.shtml], Accessed 18 March 2014.</ref>

:<math> \frac{AE}{CE}=\frac{AB}{CB}\cdot\frac{AD}{CD}.</math>

A set of sides that can form a cyclic quadrilateral can be arranged in any of three distinct sequences each of which can form a cyclic quadrilateral of the same area in the same circumcircle (the areas being the same according to Brahmagupta's area formula). Any two of these cyclic quadrilaterals have one diagonal length in common.<ref name=Johnson/>{{rp|p. 84}}

==Angle formulas==
For a cyclic quadrilateral with successive sides {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, {{math|''d''}}, [[semiperimeter]] {{math|''s''}}, and angle {{math|''A''}} between sides {{math|''a''}} and {{math|''d''}}, the [[trigonometric functions]] of {{math|''A''}} are given by<ref>{{citation |last1=Siddons |first1=A. W. |first2=R. T. |last2=Hughes |title=Trigonometry |publisher=Cambridge University Press |year=1929 |page=202 |oclc=429528983}}</ref>
:<math>\cos A = \frac{a^2 + d^2 - b^2 - c^2}{2(ad + bc)},</math>
:<math>\sin A = \frac{2\sqrt{(s-a)(s-b)(s-c)(s-d)}}{(ad+bc)},</math>
:<math>\tan \frac{A}{2} = \sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}.</math>

The angle {{math|''θ''}} between the diagonals satisfies<ref name=Durell/>{{rp|p.26}}
:<math>\tan \frac{\theta}{2} = \sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}.</math>

If the extensions of opposite sides {{math|''a''}} and {{math|''c''}} intersect at an angle {{math|''φ''}}, then
:<math>\cos{\frac{\varphi}{2}}=\sqrt{\frac{(s-b)(s-d)(b+d)^2}{(ab+cd)(ad+bc)}}</math>

where {{math|''s''}} is the [[semiperimeter]].<ref name=Durell>{{citation |first1=C. V. |last1=Durell |first2=A. |last2=Robson |title=Advanced Trigonometry |url=https://books.google.com/books?id=3iYbExAsepEC |year=2003 |publisher=Courier Dover |isbn=978-0-486-43229-8 |origyear=1930}}</ref>{{rp|p.31}}

==Parameshvara's circumradius formula==
A cyclic quadrilateral with successive sides {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, {{math|''d''}} and [[semiperimeter]] {{math|''s''}} has the circumradius (the [[radius]] of the [[circumscribed circle|circumcircle]]) given by<ref name=Alsina2/><ref>{{citation |last=Hoehn |first=Larry |title=Circumradius of a cyclic quadrilateral |journal=Mathematical Gazette |volume=84 |issue=499 |date=March 2000 |pages=69–70  |jstor=3621477|doi=10.2307/3621477 }}</ref>
:<math>R=\frac{1}{4} \sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}.</math>

This was derived by the Indian mathematician Vatasseri [[Parameshvara]] in the 15th century.

Using [[Brahmagupta's formula]], Parameshvara's formula can be restated as
:<math>4KR=\sqrt{(ab+cd)(ac+bd)(ad+bc)}</math>

where {{math|''K''}} is the area of the cyclic quadrilateral.

==Anticenter and collinearities==
Four line segments, each [[perpendicular]] to one side of a cyclic quadrilateral and passing through the opposite side's [[midpoint]], are [[Concurrent lines|concurrent]].<ref name=Altshiller-Court>{{citation |first=Nathan |last=Altshiller-Court |title=College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle |year=2007 |publisher=Courier Dover |isbn=978-0-486-45805-2 |edition=2nd |origyear=1952 |oclc=78063045 |pages=131, 137–8}}</ref>{{rp|p.131;}}<ref name=Honsberger/> These line segments are called the ''maltitudes'',<ref>{{mathworld|title=Maltitude|urlname=Maltitude}}</ref> which is an abbreviation for midpoint altitude. Their common point is called the ''anticenter''. It has the property of being the reflection of the [[circumcenter]] in the [[Quadrilateral#Remarkable points and lines in a convex quadrilateral|"vertex centroid"]]. Thus in a cyclic quadrilateral, the circumcenter, the "vertex centroid", and the anticenter are [[Collinear points|collinear]].<ref name=Honsberger>{{citation |first=Ross |last=Honsberger |title=Episodes in Nineteenth and Twentieth Century Euclidean Geometry |chapterurl=https://books.google.com/books?id=6oduPgvOAhwC&pg=PA35 |year=1995 |publisher=Cambridge University Press |isbn=978-0-88385-639-0 |pages=35–39 |chapter=4.2 Cyclic quadrilaterals |series=New Mathematical Library |volume=37}}</ref>

If the diagonals of a cyclic quadrilateral intersect at {{math|''P''}}, and the [[midpoint]]s of the diagonals are {{math|''M''}} and {{math|''N''}}, then the anticenter of the quadrilateral is the [[orthocenter]] of [[triangle]] {{math|''MNP''}}.

==Other properties==
[[Image:Japanese theorem 2.svg|thumb|right|Japanese theorem]]

*In a cyclic quadrilateral {{math|''ABCD''}}, the [[Incircle and excircles of a triangle|incenter]]s ''M''<sub>1</sub>, ''M''<sub>2</sub>, ''M''<sub>3</sub>, ''M''<sub>4</sub> (see the figure to the right) in [[triangle]]s {{math|''DAB''}}, {{math|''ABC''}}, {{math|''BCD''}}, and {{math|''CDA''}} are the vertices of a [[rectangle]]. This is one of the theorems known as the [[Japanese theorem for cyclic quadrilaterals|Japanese theorem]]. The [[orthocenter]]s of the same four triangles are the vertices of a quadrilateral [[Congruence (geometry)|congruent]] to {{math|''ABCD''}}, and the [[centroid]]s in those four triangles are vertices of another cyclic quadrilateral.<ref name=Andreescu/>
*In a cyclic quadrilateral {{math|''ABCD''}} with circumcenter {{math|''O''}}, let {{math|''P''}} be the point where the diagonals {{math|''AC''}} and {{math|''BD''}} intersect. Then angle {{math|''APB''}} is the [[arithmetic mean]] of the angles {{math|''AOB''}} and {{math|''COD''}}. This is a direct consequence of the [[inscribed angle theorem]] and the [[exterior angle theorem]].
*There are no cyclic quadrilaterals with rational area and with unequal rational sides in either [[Arithmetic progression|arithmetic]] or [[geometric progression]].<ref name=Buchholz>{{citation
 | last1 = Buchholz | first1 = R. H.
 | last2 = MacDougall | first2 = J. A.
 | doi = 10.1017/S0004972700032883
 | issue = 2
 | journal = Bulletin of the Australian Mathematical Society
 | mr = 1680787
 | pages = 263–9
 | title = Heron quadrilaterals with sides in arithmetic or geometric progression
 | volume = 59
 | year = 1999}}</ref>

*If a cyclic quadrilateral has side lengths that form an [[arithmetic progression]] the quadrilateral is also [[Ex-tangential quadrilateral#Ex-bicentric quadrilateral|ex-bicentric]].
*If the opposite sides of a cyclic quadrilateral are extended to meet at {{math|''E''}} and {{math|''F''}}, then the internal [[Bisection#Angle bisector|angle bisectors]] of the angles at {{math|''E''}} and {{math|''F''}} are perpendicular.<ref name=Coxeter>{{citation |first1=Harold Scott MacDonald |last1=Coxeter |first2=Samuel L. |last2=Greitzer |title=Geometry Revisited |chapter=3.2 Cyclic Quadrangles; Brahmagupta's formula |chapterurl=https://books.google.com/books?id=VdAM58ksvcIC&pg=PA57 |year=1967 |publisher=Mathematical Association of America |isbn=978-0-88385-619-2 |pages=57, 60}}</ref>

==Brahmagupta quadrilaterals==
A '''Brahmagupta quadrilateral'''<ref>{{cite journal |last=Sastry |first=K.R.S. |title=Brahmagupta quadrilaterals |journal=Forum Geometricorum |volume=2 |pages=167–173 |year=2002 |url=http://forumgeom.fau.edu/FG2002volume2/FG200221.pdf }}</ref> is a cyclic quadrilateral with integer sides, integer diagonals, and integer area. All Brahmagupta quadrilaterals with sides {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, {{math|''d''}}, diagonals {{math|''e''}}, {{math|''f''}}, area {{math|''K''}}, and circumradius {{math|''R''}} can be obtained by [[clearing denominators]] from the following expressions involving rational parameters {{math|''t''}}, {{math|''u''}}, and {{math|''v''}}:

:<math>a=[t(u+v)+(1-uv)][u+v-t(1-uv)]</math>
:<math>b=(1+u^2)(v-t)(1+tv)</math>
:<math>c=t(1+u^2)(1+v^2)</math>
:<math>d=(1+v^2)(u-t)(1+tu)</math>
:<math>e=u(1+t^2)(1+v^2)</math>
:<math>f=v(1+t^2)(1+u^2)</math>
:<math>K=uv[2t(1-uv)-(u+v)(1-t^2)][2(u+v)t+(1-uv)(1-t^2)]</math>
:<math>4R=(1+u^2)(1+v^2)(1+t^2).</math>

==Orthodiagonal case==

===Circumradius and area===
For a cyclic quadrilateral that is also [[orthodiagonal quadrilateral|orthodiagonal]] (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths {{math|''p''<sub>1</sub>}} and {{math|''p''<sub>2</sub>}} and divides the other diagonal into segments of lengths {{math|''q''<sub>1</sub>}} and {{math|''q''<sub>2</sub>}}. Then<ref>{{citation
 |first1=Alfred S. |last1=Posamentier |first2=Charles T. |last2=Salkind
 |title=Challenging Problems in Geometry
 |chapterurl=https://books.google.com/books?id=ld-_Puq62mcC&pg=PA104
 |year=1970
 |publisher=Courier Dover
 |isbn=978-0-486-69154-1
 |pages=104–5
 |chapter=Solutions: 4-23 Prove that the sum of the squares of the measures of the segments made by two perpendicular chords is equal to the square of the measure of the diameter of the given circle.
 |edition=2nd}}</ref> (the first equality is Proposition 11 in [[Archimedes]]' ''[[Book of Lemmas]]'')
:<math> D^2=p_1^2+p_2^2+q_1^2+q_2^2=a^2+c^2=b^2+d^2 </math>

where {{math|''D''}} is the [[diameter]] of the [[circumcircle]].  This holds because the diagonals are perpendicular [[Circle#Chord|chords of a circle]]. These equations imply that the [[circumradius]] {{math|''R''}} can be expressed as
:<math> R=\tfrac{1}{2}\sqrt{p_1^2+p_2^2+q_1^2+q_2^2} </math>

or, in terms of the sides of the quadrilateral, as<ref name=Altshiller-Court/>
:<math> R=\tfrac{1}{2}\sqrt{a^2+c^2}=\tfrac{1}{2}\sqrt{b^2+d^2}. </math>

It also follows that<ref name=Altshiller-Court/>
:<math> a^2+b^2+c^2+d^2=8R^2. </math>

Thus, according to [[Quadrilateral#Diagonals|Euler's quadrilateral theorem]], the circumradius can be expressed in terms of the diagonals {{math|''p''}} and {{math|''q''}}, and the distance {{math|''x''}} between the midpoints of the diagonals as
:<math> R=\sqrt{\frac{p^2+q^2+4x^2}{8}}. </math>

A formula for the [[area]] {{math|''K''}} of a cyclic orthodiagonal quadrilateral in terms of the four sides is obtained directly when combining [[Cyclic quadrilateral#Diagonals|Ptolemy's theorem]] and the formula for the [[Orthodiagonal quadrilateral#Area|area of an orthodiagonal quadrilateral]]. The result is<ref>{{citation
 | last = Josefsson
 | first = Martin
 | journal = [[The Mathematical Gazette]]
 | pages = 213–224
 | title = Properties of Pythagorean quadrilaterals
 | volume = 100
 | issue = July
 | year = 2016| doi = 10.1017/mag.2016.57
 }}.</ref>{{rp|p.222}}
:<math> K=\tfrac{1}{2}(ac+bd). </math>

===Other properties===
*In a cyclic orthodiagonal quadrilateral, the [[Cyclic quadrilateral#Anticenter and collinearities|anticenter]] coincides with the point where the diagonals intersect.<ref name=Altshiller-Court/>
*[[Brahmagupta's theorem]] states that for a cyclic quadrilateral that is also [[Orthodiagonal quadrilateral|orthodiagonal]], the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side.<ref name=Altshiller-Court/>
*If a cyclic quadrilateral is also orthodiagonal, the distance from the [[circumcenter]] to any side equals half the length of the opposite side.<ref name=Altshiller-Court/>
*In a cyclic orthodiagonal quadrilateral, the distance between the midpoints of the diagonals equals the distance between the circumcenter and the point where the diagonals intersect.<ref name=Altshiller-Court/>

== Cyclic spherical quadrilaterals ==
In [[Spherical trigonometry|spherical geometry]], a spherical quadrilateral formed from four intersecting greater circles is cyclic if and only if the summations of the opposite angles are equal, i.e., α + γ = β + δ for consecutive angles α, β, γ, δ of the quadrilateral<ref>{{Cite journal|last=Wimmer|first=Lienhard|date=2011|title=Cyclic polygons in non-Euclidean geometry|url=|journal=Elemente der Mathematik|volume=66|issue=2|pages=74–82|doi=|pmid=|access-date=}}</ref>. One direction of this theorem was proved by I. A. Lexell in 1786. Lexell<ref>{{Cite journal|last=Lexell|first=A. J.|date=1786|title=De proprietatibus circulorum in superficie sphaerica descriptorum|url=|journal=Acta Acad. Sci. Petropol.|volume=6|issue=1|pages=58–103|doi=|pmid=|access-date=}}</ref> showed that in a spherical quadrilateral inscribed in a small circle of a sphere the sums of opposite angles are equal, and that in the circumscribed quadrilateral the sums of opposite sides are equal. The first of these theorems is the spherical analogue of a plane theorem, and the second theorem is its dual, that is, the result of interchanging great circles and their poles.<ref>{{Cite book|title=A History of Non-Euclidean Geometry - Springer|volume = 12|last=Rosenfeld|first=B. A.|doi=10.1007/978-1-4419-8680-1|series = Studies in the History of Mathematics and Physical Sciences|year = 1988|isbn = 978-1-4612-6449-1}}</ref> Kiper et al.<ref>{{Cite journal|last=Kiper|first=Gökhan|last2=Söylemez|first2=Eres|date=2012-05-01|title=Homothetic Jitterbug-like linkages|journal=Mechanism and Machine Theory|volume=51|pages=145–158|doi=10.1016/j.mechmachtheory.2011.11.014}}</ref> proved a converse of the theorem: If the summations of the opposite sides are equal in a spherical quadrilateral, then there exists an inscribing circle for this quadrilateral.

==See also==
*[[Butterfly theorem]]
*[[Circumcircle|Cyclic polygon]]
*[[Power of a point]]
*[[Ptolemy's table of chords]]
*[[Robbins pentagon]]

==References==
{{reflist}}

== Further reading ==
*D. Fraivert: [https://doi.org/10.1017/mag.2019.54 ''Pascal-points quadrilaterals inscribed in a cyclic quadrilateral'']

==External links==
*[http://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-area-cyclic-quadrilateral Derivation of Formula for the Area of Cyclic Quadrilateral]
*[http://www.cut-the-knot.org/Curriculum/Geometry/CyclicQuadrilateral.shtml Incenters in Cyclic Quadrilateral] at [[cut-the-knot]]
*[http://www.cut-the-knot.org/Curriculum/Geometry/Brahmagupta2.shtml Four Concurrent Lines in a Cyclic Quadrilateral] at [[cut-the-knot]]
*{{MathWorld |urlname=CyclicQuadrilateral |title=Cyclic quadrilateral}}
*[http://dynamicmathematicslearning.com/nine-point-quad.html Euler centre and maltitudes of cyclic quadrilateral] at [http://dynamicmathematicslearning.com/JavaGSPLinks.htm Dynamic Geometry Sketches], interactive dynamic geometry sketch.

[[Category:Quadrilaterals]]