Cyclic quadrilateral#Anticenter and collinearities

Any square, rectangle, isosceles trapezoid, or antiparallelogram is cyclic. A kite is cyclic if and only if it has two right angles – a right kite. A bicentric quadrilateral is a cyclic quadrilateral that is also tangential and an ex-bicentric quadrilateral is a cyclic quadrilateral that is also ex-tangential. A harmonic quadrilateral is a cyclic quadrilateral in which the product of the lengths of opposite sides are equal.

File:Sehnenviereck.svg

A convex quadrilateral is cyclic if and only if the four perpendicular bisectors to the sides are concurrent. This common point is the circumcenter.{{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 |chapter-url=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}}

File:cyclic_quadrilateral_supplementary_angles_visual_proof.svg using the inscribed angle theorem that opposite angles of a cyclic quadrilateral are supplementary:
2𝜃 + 2𝜙 = 360° ∴ 𝜃 + 𝜙 = 180°]]

A convex quadrilateral {{math|ABCD}} is cyclic if and only if its opposite angles are supplementary, that is{{citation

|last1=Fraivert|first1=David|last2=Sigler|first2=Avi|last3=Stupel|first3=Moshe

|journal=International Journal of Mathematical Education in Science and Technology

|volume=51|issue=6|pages=913–938|title=Necessary and sufficient properties for a cyclic quadrilateral

|year=2020|doi=10.1080/0020739X.2019.1683772|s2cid=209930435}}

:$\backslash alpha\; +\; \backslash gamma\; =\; \backslash beta\; +\; \backslash delta\; =\; \backslash pi\; \backslash \; \backslash text\{radians\}\backslash \; (=\; 180^\{\backslash circ\}).$

The direct theorem was Proposition 22 in Book 3 of Euclid's Elements.{{citation |date=June 1997 |last=Joyce |first=D. E. |chapter-url=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 }} Equivalently, a convex quadrilateral is cyclic if and only if each exterior angle is equal to the opposite interior angle.

In 1836 Duncan Gregory generalized this result as follows: Given any convex cyclic 2n-gon, then the two sums of alternate interior angles are each equal to $(n-1)\backslash pi$.{{citation|first=Duncan|last=Gregory|title=Geometrical Theorem|journal=Cambridge Mathematical Journal|volume=1|year=1836|pages=92}}. This result can be further generalized as follows: lf A1A2...A2n (n > 1) is any cyclic 2n-gon in which vertex Ai->Ai+k (vertex Ai is joined to Ai+k), then the two sums of alternate interior angles are each equal to m$\backslash pi$ (where m = n—k and k = 1, 2, 3, ... is the total turning).{{citation|first=Michael|last=De Villiers|title=A unifying generalization of Turnbull's theorem|journal=International Journal of Mathematical Education in Science and Technology|volume=24|year=1993|issue=2 |pages=191–196|doi=10.1080/0020739930240204}}.

Taking the stereographic projection (half-angle tangent) of each angle, this can be re-expressed,

$$\backslash dfrac$$

{ \tan{\frac{\alpha}{2}} + \tan{\frac{\gamma}{2}} }

{ 1 - \tan{\frac{\alpha}{2}} \tan{\frac{\gamma}{2}} }

= \dfrac

{ \tan{\frac{\beta}{2}} + \tan{\frac{\delta}{2}} }

{ 1 - \tan{\frac{\beta}{2}} \tan{\frac{\delta}{2}} }

= \infty.

Which implies that{{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

|access-date=August 16, 2011

|archive-date=November 26, 2019

|archive-url=https://web.archive.org/web/20191126135341/http://forumgeom.fau.edu/FG2008volume8/FG200814.pdf

|url-status=dead

}}

:$\backslash tan\{\backslash frac\{\backslash alpha\}\{2\}\}\; \backslash tan\{\backslash frac\{\backslash gamma\}\{2\}\}\; =\; \backslash tan\{\backslash frac\{\backslash beta\}\{2\}\}\{\backslash tan\; \backslash frac\{\backslash delta\}\{2\}\}\; =\; 1$

A convex quadrilateral {{math|ABCD}} is cyclic if and only if an angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal.{{citation |first1=Titu |last1=Andreescu |first2=Bogdan |last2=Enescu |title=Mathematical Olympiad Treasures |chapter=2.3 Cyclic quads |chapter-url=https://books.google.com/books?id=mwUHJpvLOPsC&pg=PA44 |year=2004 |publisher=Springer |isbn=978-0-8176-4305-8 |pages=[https://archive.org/details/mathematicalolym0000andr_q5g0/page/44 44–46, 50] |mr=2025063 |url=https://archive.org/details/mathematicalolym0000andr_q5g0/page/44 }} That is, for example,

:$\backslash angle\; ACB\; =\; \backslash angle\; ADB.$

[[File:Cyclic_quadrilateral_-_pascal_points.png|thumb|ABCD is a cyclic quadrilateral. E is the point of intersection of the diagonals and F is the point of intersection of the extensions of sides BC and AD. $\backslash omega$ is a circle whose diameter is the segment, EF. P and Q are Pascal points formed by the circle $\backslash omega$.

Triangles FAB and FCD are similar.]]

Other necessary and sufficient conditions for a convex quadrilateral {{math|ABCD}} to be cyclic are: let {{math|E}} be the point of intersection of the diagonals, let {{math|F}} be the intersection point of the extensions of the sides {{math|AD}} and {{math|BC}}, let $\backslash omega$ be a circle whose diameter is the segment, {{math|EF}}, and let {{math|P}} and {{math|Q}} be Pascal points on sides {{math|AB}} and {{math|CD}} formed by the circle $\backslash omega$.

(1) {{math|ABCD}} is a cyclic quadrilateral if and only if points {{math|P}} and {{math|Q}} are collinear with the center {{math|O}}, of circle $\backslash omega$.

(2) {{math|ABCD}} is a cyclic quadrilateral if and only if points {{math|P}} and {{math|Q}} are the midpoints of sides {{math|AB}} and {{math|CD}}.

If two lines, one containing segment {{math|AC}} and the other containing segment {{math|BD}}, intersect at {{math|E}}, then the four points {{math|A}}, {{math|B}}, {{math|C}}, {{math|D}} are concyclic if and only if{{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}}

:$\backslash displaystyle\; AE\backslash cdot\; EC\; =\; BE\backslash cdot\; ED.$

The intersection {{math|E}} 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|E}} 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.

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:{{rp|p.25}}

:$\backslash displaystyle\; ef\; =\; ac\; +\; bd,$

where a, b, c, d are the side lengths in order. The converse is also true. That is, if this equation is satisfied in a convex quadrilateral, then a cyclic quadrilateral is formed.

File:Nine-point circle of diagonal triangle.png

In a convex quadrilateral {{math|ABCD}}, let {{math|EFG}} be the diagonal triangle of {{math|ABCD}} and let $\backslash omega$ 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 $\backslash omega$.{{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}}{{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}}

The area {{math|K}} of a cyclic quadrilateral with sides {{math|a}}, {{math|b}}, {{math|c}}, {{math|d}} is given by Brahmagupta's formula{{rp|p.24}}

:$K=\backslash sqrt\{(s-a)(s-b)(s-c)(s-d)\}\; \backslash ,$

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 maximal area among all quadrilaterals having the same side lengths (regardless of sequence). This is another corollary to Bretschneider's formula. It can also be proved using calculus.{{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 }}

Four unequal lengths, each less than the sum of the other three, are the sides of each of three non-congruent cyclic quadrilaterals, 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}}, angle {{math|A}} between sides {{math|a}} and {{math|d}}, and angle {{math|B}} between sides {{math|a}} and {{math|b}} can be expressed as{{rp|p.25}}

:$K\; =\; \backslash tfrac\{1\}\{2\}(ab+cd)\backslash sin\{B\}$

or

:$K\; =\; \backslash tfrac\{1\}\{2\}(ad+bc)\backslash sin\{A\}$

or{{rp|p.26}}

:$K\; =\; \backslash tfrac\{1\}\{2\}(ac+bd)\backslash sin\{\backslash theta\}$

where {{math|θ}} is either angle between the diagonals. Provided {{math|A}} is not a right angle, the area can also be expressed as{{rp|p.26}}

:$K\; =\; \backslash tfrac\{1\}\{4\}(a^2-b^2-c^2+d^2)\backslash tan\{A\}.$

Another formula is{{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}}{{rp|p.83}}

:$\backslash displaystyle\; K=2R^2\backslash sin\{A\}\backslash sin\{B\}\backslash sin\{\backslash theta\}$

where {{math|R}} is the radius of the circumcircle. As a direct consequence,{{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 |chapter-url=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}}

:$K\backslash le\; 2R^2$

where there is equality if and only if the quadrilateral is a square.

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{{rp|p.25,}}{{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}}Johnson, Roger A., Advanced Euclidean Geometry, Dover Publ., 2007 (orig. 1929).{{rp|p. 84}}

:$p\; =\; \backslash sqrt\{\backslash frac\{(ac+bd)(ad+bc)\}\{ab+cd\}\}$ and $q\; =\; \backslash sqrt\{\backslash frac\{(ac+bd)(ab+cd)\}\{ad+bc\}\}$

so showing Ptolemy's theorem

:$pq\; =\; ac+bd.$

According to Ptolemy's second theorem,{{rp|p.25,}}

:$\backslash frac\; \{p\}\{q\}=\; \backslash frac\{ad+bc\}\{ab+cd\}$

using the same notations as above.

For the sum of the diagonals we have the inequalityInequalities proposed in "Crux Mathematicorum", 2007, [http://www.imomath.com/othercomp/Journ/ineq.pdf].{{rp|p.123,#2975}}

:$p+q\backslash ge\; 2\backslash sqrt\{ac+bd\}.$

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

Moreover,{{rp|p.64,#1639}}

:$(p+q)^2\; \backslash leq\; (a+c)^2+(b+d)^2.$

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 similar to each other.

If {{math|ABCD}} is a cyclic quadrilateral where {{math|AC}} meets {{math|BD}} at {{math|E}}, thenA. 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.

:$\backslash frac\{AE\}\{CE\}=\backslash frac\{AB\}\{CB\}\backslash cdot\backslash frac\{AD\}\{CD\}.$

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.{{rp|p. 84}}

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{{citation |last1=Siddons |first1=A. W. |first2=R. T. |last2=Hughes |title=Trigonometry |publisher=Cambridge University Press |year=1929 |page=202 |oclc=429528983}}

:$\backslash cos\; A\; =\; \backslash frac\{a^2-b^2-c^2+d^2\}\{2(ad+bc)\},$

:$\backslash sin\; A\; =\; \backslash frac\{2\backslash sqrt\{(s-a)(s-b)(s-c)(s-d)\}\}\{(ad+bc)\},$

:$\backslash tan\; \backslash frac\{A\}\{2\}\; =\; \backslash sqrt\{\backslash frac\{(s-a)(s-d)\}\{(s-b)(s-c)\}\}.$

The angle {{math|θ}} between the diagonals that is opposite sides {{math|a}} and {{math|c}} satisfies{{rp|p.26}}

:$\backslash tan\; \backslash frac\{\backslash theta\}\{2\}\; =\; \backslash sqrt\{\backslash frac\{(s-b)(s-d)\}\{(s-a)(s-c)\}\}.$

If the extensions of opposite sides {{math|a}} and {{math|c}} intersect at an angle {{math|φ}}, then

:$\backslash cos\{\backslash frac\{\backslash varphi\}\{2\}\}=\backslash sqrt\{\backslash frac\{(s-b)(s-d)(b+d)^2\}\{(ab+cd)(ad+bc)\}\}$

where {{math|s}} is the semiperimeter.{{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 |orig-year=1930}}{{rp|p.31}}

Let $B$ denote the angle between sides $a$ and $b$, $C$ the angle between $b$ and $c$, and $D$ the angle between $c$ and $d$, then:{{citation

|url=http://matinf.upit.ro/MATINF9_10/files/downloads/Revista_2022.pdf#page=19 |last=José García |first=Emmanuel Antonio |title=A generalization of Mollweide's formula (rather Newton's) |journal=Matinf |volume=5 |number=9–10 |year=2022 |pages=19–22 |access-date=29 December 2023}}

:$\backslash begin\{align\}$

\frac{a+c}{b+d}

&= \frac{\sin\tfrac12(A+B)}{\cos\tfrac12(C-D)}\tan\tfrac12\theta, \\[10mu]

\frac{a-c}{b-d}

&= \frac{\cos\tfrac12(A+B)}{\sin\tfrac12(D-C)}\cot\tfrac12\theta.

\end{align}

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 circumcircle) given by{{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 }}

:$R=\backslash frac\{1\}\{4\}\; \backslash sqrt\{\backslash frac\{(ab+cd)(ac+bd)(ad+bc)\}\{(s-a)(s-b)(s-c)(s-d)\}\}.$

This was derived by the Indian mathematician Vatasseri Parameshvara in the 15th century. (Note that the radius is invariant under the interchange of any side lengths.)

Using Brahmagupta's formula, Parameshvara's formula can be restated as

:$4KR=\backslash sqrt\{(ab+cd)(ac+bd)(ad+bc)\}$

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

Four line segments, each perpendicular to one side of a cyclic quadrilateral and passing through the opposite side's midpoint, are concurrent.{{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 |orig-year=1952 |oclc=78063045 |pages=131, 137–8}}{{rp|p.131;}} These line segments are called the maltitudes,{{mathworld|title=Maltitude|urlname=Maltitude}} 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 "vertex centroid". Thus in a cyclic quadrilateral, the circumcenter, the "vertex centroid", and the anticenter are collinear.{{citation |first=Ross |last=Honsberger |title=Episodes in Nineteenth and Twentieth Century Euclidean Geometry |chapter-url=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}}

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

The anticenter of a cyclic quadrilateral is the Poncelet point of its vertices.

Image:Japanese theorem 2.svg

• In a cyclic quadrilateral {{math|ABCD}}, the incenters M₁, M₂, M₃, M₄ (see the figure to the right) in triangles {{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. The orthocenters of the same four triangles are the vertices of a quadrilateral congruent to {{math|ABCD}}, and the centroids in those four triangles are vertices of another cyclic quadrilateral.
• 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 or geometric progression.{{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| doi-access = free

| hdl = 1959.13/803798

| hdl-access = free

}}

• If a cyclic quadrilateral has side lengths that form an arithmetic progression the quadrilateral is also ex-bicentric.
• If the opposite sides of a cyclic quadrilateral are extended to meet at {{math|E}} and {{math|F}}, then the internal angle bisectors of the angles at {{math|E}} and {{math|F}} are perpendicular.{{citation |first1=Harold Scott MacDonald |last1=Coxeter |first2=Samuel L. |last2=Greitzer |title=Geometry Revisited |chapter=3.2 Cyclic Quadrangles; Brahmagupta's formula |chapter-url=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}}

A Brahmagupta quadrilateral{{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 |access-date=April 25, 2012 |archive-date=April 22, 2018 |archive-url=https://web.archive.org/web/20180422180842/http://forumgeom.fau.edu/FG2002volume2/FG200221.pdf |url-status=dead }} 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}}:

:$a=[t(u+v)+(1-uv)][u+v-t(1-uv)]$

:$b=(1+u^2)(v-t)(1+tv)$

:$c=t(1+u^2)(1+v^2)$

:$d=(1+v^2)(u-t)(1+tu)$

:$e=u(1+t^2)(1+v^2)$

:$f=v(1+t^2)(1+u^2)$

:$K=uv[2t(1-uv)-(u+v)(1-t^2)][2(u+v)t+(1-uv)(1-t^2)]$

:$4R=(1+u^2)(1+v^2)(1+t^2).$

For a cyclic quadrilateral that is also orthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths {{math|p₁}} and {{math|p₂}} and divides the other diagonal into segments of lengths {{math|q₁}} and {{math|q₂}}. Then{{citation

|first1=Alfred S.

|last1=Posamentier

|first2=Charles T.

|last2=Salkind

|title=Challenging Problems in Geometry

|chapter-url=https://books.google.com/books?id=ld-_Puq62mcC&pg=PA104

|year=1970

|publisher=Courier Dover

|isbn=978-0-486-69154-1

|pages=[https://archive.org/details/challengingprobl0000posa/page/104 104–5]

|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

|url=https://archive.org/details/challengingprobl0000posa/page/104

}} (the first equality is Proposition 11 in Archimedes' Book of Lemmas)

:$D^2=p\_1^2+p\_2^2+q\_1^2+q\_2^2=a^2+c^2=b^2+d^2$

where {{math|D}} is the diameter of the circumcircle. This holds because the diagonals are perpendicular chords of a circle. These equations imply that the circumradius {{math|R}} can be expressed as

:$R=\backslash tfrac\{1\}\{2\}\backslash sqrt\{p\_1^2+p\_2^2+q\_1^2+q\_2^2\}$

or, in terms of the sides of the quadrilateral, as

:$R=\backslash tfrac\{1\}\{2\}\backslash sqrt\{a^2+c^2\}=\backslash tfrac\{1\}\{2\}\backslash sqrt\{b^2+d^2\}.$

It also follows that

:$a^2+b^2+c^2+d^2=8R^2.$

Thus, according to 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

:$R=\backslash sqrt\{\backslash frac\{p^2+q^2+4x^2\}\{8\}\}.$

A formula for the area {{math|K}} of a cyclic orthodiagonal quadrilateral in terms of the four sides is obtained directly when combining Ptolemy's theorem and the formula for the area of an orthodiagonal quadrilateral. The result is{{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

}}.{{rp|p.222}}

:$K=\backslash tfrac\{1\}\{2\}(ac+bd).$

• In a cyclic orthodiagonal quadrilateral, the anticenter coincides with the point where the diagonals intersect.
• Brahmagupta's theorem states that for a cyclic quadrilateral that is also orthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side.
• If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side.
• 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.

In 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.{{Cite journal|last=Wimmer|first=Lienhard|date=2011|title=Cyclic polygons in non-Euclidean geometry|journal=Elemente der Mathematik|volume=66|issue=2|pages=74–82|doi=10.4171/EM/173|doi-access=free}} One direction of this theorem was proved by Anders Johan Lexell in 1782.{{cite journal |last=Lexell |first=Anders Johan |author-link=Anders Johan Lexell |year=1786 |title=De proprietatibus circulorum in superficie sphaerica descriptorum |journal=Acta Academiae Scientiarum Imperialis Petropolitanae |volume=6: 1782 |issue=1 |pages=58–103, [https://archive.org/details/actaacademiaesci82impe/page/n481/mode/1up figures tab. 3] |url=https://archive.org/details/actaacademiaesci82impe/page/58/ }} Lexell 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.{{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}} Kiper et al.{{Cite journal|last1=Kiper|first1=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|hdl=11511/40967|hdl-access=free}} 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.

• Butterfly theorem
• Brahmagupta triangle
• Cyclic polygon
• Power of a point
• Ptolemy's table of chords
• Robbins pentagon

{{reflist}}

• D. Fraivert: [https://doi.org/10.1017/mag.2019.54 Pascal-points quadrilaterals inscribed in a cyclic quadrilateral]

• [https://web.archive.org/web/20110621021620/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}}

{{Polygons}}

Category:Types of quadrilaterals