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Cyclic quadrilateral

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From Wikipedia, the free encyclopedia

Quadrilateral whose vertices lie on a circle

Special cases

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Anysquare,rectangle,isosceles trapezoid, orantiparallelogram is cyclic. Akite is cyclicif and only if it has two right angles – aright kite. Abicentric quadrilateral is a cyclic quadrilateral that is alsotangential and anex-bicentric quadrilateral is a cyclic quadrilateral that is alsoex-tangential. Aharmonic quadrilateral is a cyclic quadrilateral in which the product of the lengths of opposite sides are equal.

Characterizations

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Circumcenter

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A convex quadrilateral is cyclicif and only if the fourperpendicular bisectors to the sides areconcurrent. This common point is thecircumcenter.^[1]

Supplementary angles

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Proof without words using theinscribed angle theorem that opposite angles of a cyclic quadrilateral are supplementary:
2𝜃 + 2𝜙 = 360° ∴ 𝜃 + 𝜙 = 180°

A convex quadrilateralABCD is cyclic if and only if its opposite angles aresupplementary, that is^[1]^[2] $\alpha +\gamma =\beta +\delta =\pi \ {\text{radians}}\ (=180^{\circ }).$

The direct theorem was Proposition 22 in Book 3 ofEuclid'sElements.^[3] Equivalently, a convex quadrilateral is cyclic if and only if eachexterior angle is equal to the oppositeinterior angle.

In 1836Duncan Gregory generalized this result as follows: Given any convex cyclic2n-gon, then the two sums ofalternate interior angles are each equal to $(n-1)\pi$ .^[4] This result can be further generalized as follows: lfA₁A₂...A_2n(n > 1) is any cyclic2n-gon in which vertexA_i →A_i+k (vertexA_i is joined toA_i+k), then the two sums of alternate interior angles are each equal tomπ (wherem =n −k andk = 1, 2, 3, ... is the total turning).^[5]

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

${\dfrac {\tan {\frac {\alpha }{2}}+\tan {\frac {\gamma }{2}}}{1-\tan {\frac {\alpha }{2}}\tan {\frac {\gamma }{2}}}}={\dfrac {\tan {\frac {\beta }{2}}+\tan {\frac {\delta }{2}}}{1-\tan {\frac {\beta }{2}}\tan {\frac {\delta }{2}}}}=\infty .$

Which implies that^[6]

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

Angles between sides and diagonals

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A convex quadrilateralABCD is cyclic if and only if an angle between a side and adiagonal is equal to the angle between the opposite side and the other diagonal.^[7] That is, for example, $\angle ACB=\angle ADB.$

Pascal points

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*ABCD* is a cyclic quadrilateral.E is the point of intersection of the diagonals andF is the point of intersection of the extensions of sidesBC andAD. $\omega$ is a circle whose diameter is the segment,EF.P andQ are Pascal points formed by the circle $\omega$ . TrianglesFAB andFCD are similar.

{\displaystyle \omega } — *ABCD* is a cyclic quadrilateral.E is the point of intersection of the diagonals andF is the point of intersection of the extensions of sidesBC andAD. $\omega$ is a circle whose diameter is the segment,EF.P andQ are Pascal points formed by the circle $\omega$ . TrianglesFAB andFCD are similar.

Other necessary and sufficient conditions for a convex quadrilateralABCD to be cyclic are: letE be the point of intersection of the diagonals, letF be the intersection point of the extensions of the sidesAD andBC, let $\omega$ be a circle whose diameter is the segment,EF, and letP andQ bePascal points on sidesAB andCD formed by the circle $\omega$ .
(1)ABCD is a cyclic quadrilateral if and only if pointsP andQ are collinear with the centerO, of circle $\omega$ .
(2)ABCD is a cyclic quadrilateral if and only if pointsP andQ are the midpoints of sidesAB andCD.^[2]

Intersection of diagonals

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If two lines, one containing segmentAC and the other containing segmentBD, intersect atE, then the four pointsA,B,C,D are concyclic if and only if^[8] $\displaystyle AE\cdot EC=BE\cdot ED.$ The intersectionE may be internal or external to the circle. In the former case, the cyclic quadrilateral isABCD, and in the latter case, the cyclic quadrilateral isABDC. When the intersection is internal, the equality states that the product of the segment lengths into whichE divides one diagonal equals that of the other diagonal. This is known as theintersecting chords theorem since the diagonals of the cyclic quadrilateral are chords of the circumcircle.

Ptolemy's theorem

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Ptolemy's theorem expresses the product of the lengths of the two diagonalse andf of a cyclic quadrilateral as equal to the sum of the products of opposite sides:^[9]^: p.25^[2] $\displaystyle ef=ac+bd,$

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

Diagonal triangle

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In a convex quadrilateralABCD, letEFG be the diagonal triangle ofABCD and let $\omega$ be the nine-point circle ofEFG.ABCD is cyclic if and only if the point of intersection of the bimedians ofABCD belongs to the nine-point circle $\omega$ .^[10]^[11]^[2]

Area

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TheareaK of a cyclic quadrilateral with sidesa,b,c,d is given byBrahmagupta's formula^[9]^: p.24 $K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}\,$

wheres, thesemiperimeter, iss =⁠1/2⁠(a +b +c +d). This is acorollary ofBretschneider's formula for the general quadrilateral, since opposite angles are supplementary in the cyclic case. If alsod = 0, the cyclic quadrilateral becomes a triangle and the formula is reduced toHeron's formula.

The cyclic quadrilateral hasmaximal 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 usingcalculus.^[12]

Four unequal lengths, each less than the sum of the other three, are the sides of each of three non-congruent cyclic quadrilaterals,^[13] which by Brahmagupta's formula all have the same area. Specifically, for sidesa,b,c, andd, sidea could be opposite any of sideb, sidec, or sided.

The area of a cyclic quadrilateral with successive sidesa,b,c,d, angleA between sidesa andd, and angleB between sidesa andb can be expressed as^[9]^: p.25 $K={\tfrac {1}{2}}(ab+cd)\sin {B}$

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

or^[9]^: p.26 $K={\tfrac {1}{2}}(ac+bd)\sin {\theta }$

whereθ is either angle between the diagonals. ProvidedA is not a right angle, the area can also be expressed as^[9]^: p.26 $K={\tfrac {1}{4}}(a^{2}-b^{2}-c^{2}+d^{2})\tan {A}.$

Another formula is^[14]^: p.83 $\displaystyle K=2R^{2}\sin {A}\sin {B}\sin {\theta }$

whereR is the radius of thecircumcircle. As a direct consequence,^[15]

$K\leq 2R^{2}$ where there is equality if and only if the quadrilateral is a square.

Diagonals

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In a cyclic quadrilateral with successive verticesA,B,C,D and sidesa =AB,b =BC,c =CD, andd =DA, the lengths of the diagonalsp =AC andq =BD can be expressed in terms of the sides as^[9]^: p.25,^[16]^[17]^{: p. 84}

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

so showingPtolemy's theorem $pq=ac+bd.$

According toPtolemy's second theorem,^[9]^: p.25,^[16] ${\frac {p}{q}}={\frac {ad+bc}{ab+cd}}$

using the same notations as above.

For the sum of the diagonals we have the inequality^[18]^{: p.123, #2975}

$p+q\geq 2{\sqrt {ac+bd}}.$

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

Moreover,^[18]^{: p.64, #1639}

$(p+q)^{2}\leq (a+c)^{2}+(b+d)^{2}.$

In any convex quadrilateral, the two diagonals together partition the quadrilateral into four triangles; in a cyclic quadrilateral, opposite pairs of these four triangles aresimilar to each other.

IfABCD is a cyclic quadrilateral whereAC meetsBD atE, then^[19]

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

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.^[17]^{: p. 84}

Angle formulas

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For a cyclic quadrilateral with successive sidesa,b,c,d,semiperimeters, and angleA between sidesa andd, thetrigonometric functions ofA are given by^[20] $\cos A={\frac {a^{2}-b^{2}-c^{2}+d^{2}}{2(ad+bc)}},$ $\sin A={\frac {2{\sqrt {(s-a)(s-b)(s-c)(s-d)}}}{(ad+bc)}},$ $\tan {\frac {A}{2}}={\sqrt {\frac {(s-a)(s-d)}{(s-b)(s-c)}}}.$

The angleθ between the diagonals that is opposite sidesa andc satisfies^[9]^: p.26 $\tan {\frac {\theta }{2}}={\sqrt {\frac {(s-b)(s-d)}{(s-a)(s-c)}}}.$

If the extensions of opposite sidesa andc intersect at an angleφ, then $\cos {\frac {\varphi }{2}}={\sqrt {\frac {(s-b)(s-d)(b+d)^{2}}{(ab+cd)(ad+bc)}}}$

wheres is thesemiperimeter.^[9]^: p.31

A generalization ofMollweide's formula to cyclic quadrilaterals is given by the following two identities. Let $B {\displaystyle B}$ denote the angle between sides $a {\displaystyle a}$ and $b {\displaystyle b}$ , $C {\displaystyle C}$ the angle between $b {\displaystyle b}$ and $c {\displaystyle c}$ , and $D {\displaystyle D}$ the angle between $c {\displaystyle c}$ and $d {\displaystyle d}$ . If $E {\displaystyle E}$ is the point of intersection of the diagonals, denote $\angle {CED}=\theta$ , then:^[21]

${\begin{aligned}{\frac {a+c}{b+d}}&={\frac {\sin {\tfrac {1}{2}}(A+B)}{\cos {\tfrac {1}{2}}(C-D)}}\tan {\tfrac {1}{2}}\theta ,\\[10mu]{\frac {a-c}{b-d}}&={\frac {\cos {\tfrac {1}{2}}(A+B)}{\sin {\tfrac {1}{2}}(D-C)}}\cot {\tfrac {1}{2}}\theta .\end{aligned}}$

Moreover, a generalization of thelaw of tangents for cyclic quadrilaterals is:^[22]

${\frac {(a-c)(b-d)}{(a+c)(b+d)}}={\frac {\tan {\tfrac {1}{2}}(A-B)}{\tan {\tfrac {1}{2}}(A+B)}}.$

Parameshvara's circumradius formula

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A cyclic quadrilateral with successive sidesa,b,c,d andsemiperimeters has the circumradius (theradius of thecircumcircle) given by^[16]^[23] $R={\frac {1}{4}}{\sqrt {\frac {(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}}.$

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

UsingBrahmagupta's formula, Parameshvara's formula can be restated as $4KR={\sqrt {(ab+cd)(ac+bd)(ad+bc)}}$

whereK is the area of the cyclic quadrilateral.

Anticenter and collinearities

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Four line segments, eachperpendicular to one side of a cyclic quadrilateral and passing through the opposite side'smidpoint, areconcurrent.^[24]^: p.131,^[25] These line segments are called themaltitudes,^[26] which is an abbreviation for midpoint altitude. Their common point is called theanticenter. It has the property of being the reflection of thecircumcenter in the"vertex centroid". Thus in a cyclic quadrilateral, the circumcenter, the "vertex centroid", and the anticenter arecollinear.^[25]

If the diagonals of a cyclic quadrilateral intersect atP, and themidpoints of the diagonals areM andN, then the anticenter of the quadrilateral is theorthocenter oftriangleMNP.

The anticenter of a cyclic quadrilateral is thePoncelet point of its vertices.

Other properties

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In a cyclic quadrilateralABCD, theincentersM₁,M₂,M₃,M₄ (see the figure to the right) intrianglesDAB,ABC,BCD, andCDA are the vertices of arectangle. This is one of the theorems known as theJapanese theorem. Theorthocenters of the same four triangles are the vertices of a quadrilateralcongruent toABCD, and thecentroids in those four triangles are vertices of another cyclic quadrilateral.^[7]
In a cyclic quadrilateralABCD with circumcenterO, letP be the point where the diagonalsAC andBD intersect. Then angleAPB is thearithmetic mean of the anglesAOB andCOD. This is a direct consequence of theinscribed angle theorem and theexterior angle theorem.
There are no cyclic quadrilaterals with rational area and with unequal rational sides in eitherarithmetic orgeometric progression.^[27]

If a cyclic quadrilateral has side lengths that form anarithmetic progression the quadrilateral is alsoex-bicentric.
If the opposite sides of a cyclic quadrilateral are extended to meet atE andF, then the internalangle bisectors of the angles atE andF are perpendicular.^[13]

Brahmagupta quadrilaterals

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ABrahmagupta quadrilateral^[28] is a cyclic quadrilateral with integer sides, integer diagonals, and integer area. It is aprimitive Brahmagupta quadrilateral if no smallergeometrically similar quadrilateral has all these values as integers. Quadrilaterals whose side lengths, diagonals, and areas are allrational numbers are calledrational Brahmagupta quadrilaterals in this article. Every primitive Brahmagupta quadrilateral is a rational Brahmagupta quadrilateral. Conversely, every rational Brahmagupta quadrilateral is geometrically similar to exactly one primitive Brahmagupta quadrilateral.

All primitive Brahmagupta quadrilaterals can be obtained from the following expressions involving rational parameterst,u, andv. The computed side lengthsa,b,c,d, diagonalse,f, areaK, and circumradiusR will be rational numbers. These can be scaled to produce a unique primitive Brahmagupta quadrilateral; note thatK is an area and will be scaled by the square of the value that multiplies the other quantities. The Brahmagupta quadrilateral will be non-self-intersecting and non-degenerate ifmin(u,v) >t > 0.

$a=[t(u+v)+(1-uv)][u+v-t(1-uv)]$ $b=(1+u^{2})(v-t)(1+tv)$ $c=t(1+u^{2})(1+v^{2})$ $d=(1+v^{2})(u-t)(1+tu)$ $e=u(1+t^{2})(1+v^{2})$ $f=v(1+t^{2})(1+u^{2})$ $K=uv[2t(1-uv)-(u+v)(1-t^{2})][2(u+v)t+(1-uv)(1-t^{2})]$ $R=(1+u^{2})(1+v^{2})(1+t^{2})/4.$

SeeConcyclic points § Integer area and side lengths for a different parameterization of all non-degenerate primitive Brahmagupta quadrilaterals, which depends upon rational numbers,0 <c₁ <c₂ <c₃.

Orthodiagonal case

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Circumradius and area

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For a cyclic quadrilateral that is alsoorthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengthsp₁ andp₂ and divides the other diagonal into segments of lengthsq₁ andq₂. Then^[29] (the first equality is Proposition 11 inArchimedes'Book of Lemmas) $D^{2}=p_{1}^{2}+p_{2}^{2}+q_{1}^{2}+q_{2}^{2}=a^{2}+c^{2}=b^{2}+d^{2}$

whereD is thediameter of thecircumcircle. This holds because the diagonals are perpendicularchords of a circle. These equations imply that thecircumradiusR can be expressed as $R={\tfrac {1}{2}}{\sqrt {p_{1}^{2}+p_{2}^{2}+q_{1}^{2}+q_{2}^{2}}}$

or, in terms of the sides of the quadrilateral, as^[24] $R={\tfrac {1}{2}}{\sqrt {a^{2}+c^{2}}}={\tfrac {1}{2}}{\sqrt {b^{2}+d^{2}}}.$

It also follows that^[24] $a^{2}+b^{2}+c^{2}+d^{2}=8R^{2}.$

Thus, according toEuler's quadrilateral theorem, the circumradius can be expressed in terms of the diagonalsp andq, and the distancex between the midpoints of the diagonals as $R={\sqrt {\frac {p^{2}+q^{2}+4x^{2}}{8}}}.$

A formula for theareaK of a cyclic orthodiagonal quadrilateral in terms of the four sides is obtained directly when combiningPtolemy's theorem and the formula for thearea of an orthodiagonal quadrilateral. The result is^[30]^: p.222 $K={\tfrac {1}{2}}(ac+bd).$

Other properties

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In a cyclic orthodiagonal quadrilateral, theanticenter coincides with the point where the diagonals intersect.^[24]
Brahmagupta's theorem states that for a cyclic quadrilateral that is alsoorthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side.^[24]
If a cyclic quadrilateral is also orthodiagonal, the distance from thecircumcenter to any side equals half the length of the opposite side.^[24]
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.^[24]

Cyclic spherical quadrilaterals

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Inspherical 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.^[31] One direction of this theorem was proved byAnders Johan Lexell in 1782.^[32] 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.^[33] Kiper et al.^[34] 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.

References

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