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Li, H. Cyclic Quadrilateral. Encyclopedia. Available online: https://encyclopedia.pub/entry/31529 (accessed on 29 November 2023).

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Li, Handwiki. "Cyclic Quadrilateral." *Encyclopedia*. Web. 27 October, 2022.

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In Euclidean geometry, a cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose 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 chords of the circumcircle. Usually the quadrilateral is assumed to be 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 κύκλος (kuklos) which means "circle" or "wheel". All triangles have a circumcircle, but not all quadrilaterals do. An example of a quadrilateral that cannot be cyclic is a non-square rhombus. The section characterizations below states what necessary and sufficient conditions a quadrilateral must satisfy to have a circumcircle.

quadrilateral
quadrilaterals
chordal

Any square, rectangle, isosceles trapezoid, or antiparallelogram is cyclic. A kite is cyclic if and only if it has two right angles. 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.

A convex quadrilateral is cyclic if and only if the four perpendicular bisectors to the sides are concurrent. This common point is the circumcenter.^{[1]}

A convex quadrilateral *ABCD* is cyclic if and only if its opposite angles are supplementary, that is^{[1]} ^{[2]}

- [math]\displaystyle{ \alpha + \gamma = \beta + \delta = \pi \ (= 180^{\circ}). }[/math]

The direct theorem was Proposition 22 in Book 3 of Euclid's *Elements*.^{[3]} Equivalently, a convex quadrilateral is cyclic if and only if each exterior angle is equal to the opposite interior angle.

A convex quadrilateral *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.^{[4]} That is, for example,

- [math]\displaystyle{ \angle ACB = \angle ADB. }[/math]

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

(1) *ABCD* is a cyclic quadrilateral if and only if points *P* and *Q* are collinear with the center *O*, of circle [math]\displaystyle{ \omega }[/math].

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

If two lines, one containing segment *AC* and the other containing segment *BD*, intersect at *P*, then the four points *A*, *B*, *C*, *D* are concyclic if and only if^{[5]}

- [math]\displaystyle{ \displaystyle AP\cdot PC = BP\cdot PD. }[/math]

The intersection *P* may be internal or external to the circle. In the former case, the cyclic quadrilateral is *ABCD*, and in the latter case, the cyclic quadrilateral is *ABDC*. When the intersection is internal, the equality states that the product of the segment lengths into which *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.

Ptolemy's theorem expresses the product of the lengths of the two diagonals *e* and *f* of a cyclic quadrilateral as equal to the sum of the products of opposite sides:^{[6]}^{:p.25}^{[2]}

- [math]\displaystyle{ \displaystyle ef = ac + bd. }[/math]

The 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 *ABCD*, let *EFG* be the diagonal triangle of *ABCD* and let [math]\displaystyle{ \omega }[/math] be the nine-point circle of *EFG*. *ABCD* is cyclic if and only if the point of intersection of the bimedians of *ABCD* belongs to the nine-point circle [math]\displaystyle{ \omega }[/math].^{[2]}^{[7]}^{[8]}

A convex quadrilateral *ABCD* is cyclic if and only if^{[9]}

- [math]\displaystyle{ \tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}=\tan{\frac{\beta}{2}}\tan{\frac{\delta}{2}}=1. }[/math]

The area *K* of a cyclic quadrilateral with sides *a*, *b*, *c*, *d* is given by Brahmagupta's formula^{[6]}^{:p.24}

- [math]\displaystyle{ K=\sqrt{(s-a)(s-b)(s-c)(s-d)} \, }[/math]

where *s*, the semiperimeter, is *s* = 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 *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 sequence of side lengths. This is another corollary to Bretschneider's formula. It can also be proved using calculus.^{[10]}

Four unequal lengths, each less than the sum of the other three, are the sides of each of three non-congruent cyclic quadrilaterals,^{[11]} which by Brahmagupta's formula all have the same area. Specifically, for sides *a*, *b*, *c*, and *d*, side *a* could be opposite any of side *b*, side *c*, or side *d*.

The area of a cyclic quadrilateral with successive sides *a*, *b*, *c*, *d* and angle *B* between sides *a* and *b* can be expressed as^{[6]}^{:p.25}

- [math]\displaystyle{ K = \tfrac{1}{2}(ab+cd)\sin{B} }[/math]

or^{[6]}^{:p.26}

- [math]\displaystyle{ K = \tfrac{1}{2}(ac+bd)\sin{\theta} }[/math]

where *θ* is either angle between the diagonals. Provided *A* is not a right angle, the area can also be expressed as^{[6]}^{:p.26}

- [math]\displaystyle{ K = \tfrac{1}{4}(a^2-b^2-c^2+d^2)\tan{A}. }[/math]

Another formula is^{[12]}^{:p.83}

- [math]\displaystyle{ \displaystyle K=2R^2\sin{A}\sin{B}\sin{\theta} }[/math]

where *R* is the radius of the circumcircle. As a direct consequence,^{[13]}

- [math]\displaystyle{ K\le 2R^2 }[/math]

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

In a cyclic quadrilateral with successive vertices *A*, *B*, *C*, *D* and sides *a* = *AB*, *b* = *BC*, *c* = *CD*, and *d* = *DA*, the lengths of the diagonals *p* = *AC* and *q* = *BD* can be expressed in terms of the sides as^{[6]}^{:p.25,}^{[14]}^{[15]}^{:p. 84}

- [math]\displaystyle{ p = \sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}} }[/math] and [math]\displaystyle{ q = \sqrt{\frac{(ac+bd)(ab+cd)}{ad+bc}} }[/math]

so showing Ptolemy's theorem

- [math]\displaystyle{ pq = ac+bd. }[/math]

According to *Ptolemy's second theorem*,^{[6]}^{:p.25,}^{[14]}

- [math]\displaystyle{ \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^{[16]}^{:p.123,#2975}

- [math]\displaystyle{ 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,^{[16]}^{:p.64,#1639}

- [math]\displaystyle{ (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 similar to each other.

If *M* and *N* are the midpoints of the diagonals *AC* and *BD*, then^{[17]}

- [math]\displaystyle{ \frac{MN}{EF}=\frac{1}{2}\left |\frac{AC}{BD}-\frac{BD}{AC}\right| }[/math]

where *E* and *F* are the intersection points of the extensions of opposite sides.

If *ABCD* is a cyclic quadrilateral where *AC* meets *BD* at *E*, then^{[18]}

- [math]\displaystyle{ \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.^{[15]}^{:p. 84}

For a cyclic quadrilateral with successive sides *a*, *b*, *c*, *d*, semiperimeter *s*, and angle *A* between sides *a* and *d*, the trigonometric functions of *A* are given by^{[19]}

- [math]\displaystyle{ \cos A = \frac{a^2 + d^2 - b^2 - c^2}{2(ad + bc)}, }[/math]
- [math]\displaystyle{ \sin A = \frac{2\sqrt{(s-a)(s-b)(s-c)(s-d)}}{(ad+bc)}, }[/math]
- [math]\displaystyle{ \tan \frac{A}{2} = \sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}. }[/math]

The angle *θ* between the diagonals satisfies^{[6]}^{:p.26}

- [math]\displaystyle{ \tan \frac{\theta}{2} = \sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}. }[/math]

If the extensions of opposite sides *a* and *c* intersect at an angle *φ*, then

- [math]\displaystyle{ \cos{\frac{\varphi}{2}}=\sqrt{\frac{(s-b)(s-d)(b+d)^2}{(ab+cd)(ad+bc)}} }[/math]

where *s* is the semiperimeter.^{[6]}^{:p.31}

A cyclic quadrilateral with successive sides *a*, *b*, *c*, *d* and semiperimeter *s* has the circumradius (the radius of the circumcircle) given by^{[14]}^{[20]}

- [math]\displaystyle{ 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]\displaystyle{ 4KR=\sqrt{(ab+cd)(ac+bd)(ad+bc)} }[/math]

where *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.^{[21]}^{:p.131;}^{[22]} These line segments are called the *maltitudes*,^{[23]} 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.^{[22]}

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

- In a cyclic quadrilateral
*ABCD*, the incenters*M*_{1},*M*_{2},*M*_{3},*M*_{4}(see the figure to the right) in triangles*DAB*,*ABC*,*BCD*, and*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*ABCD*, and the centroids in those four triangles are vertices of another cyclic quadrilateral.^{[4]} - In a cyclic quadrilateral
*ABCD*with circumcenter*O*, let*P*be the point where the diagonals*AC*and*BD*intersect. Then angle*APB*is the arithmetic mean of the angles*AOB*and*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.
^{[24]}

- 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
*E*and*F*, then the internal angle bisectors of the angles at*E*and*F*are perpendicular.^{[11]}

A **Brahmagupta quadrilateral**^{[25]} is a cyclic quadrilateral with integer sides, integer diagonals, and integer area. All Brahmagupta quadrilaterals with sides *a*, *b*, *c*, *d*, diagonals *e*, *f*, area *K*, and circumradius *R* can be obtained by clearing denominators from the following expressions involving rational parameters *t*, *u*, and *v*:

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

For a cyclic quadrilateral that is also orthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths *p*_{1} and *p*_{2} and divides the other diagonal into segments of lengths *q*_{1} and *q*_{2}. Then^{[26]} (the first equality is Proposition 11 in Archimedes' *Book of Lemmas*)

- [math]\displaystyle{ D^2=p_1^2+p_2^2+q_1^2+q_2^2=a^2+c^2=b^2+d^2 }[/math]

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

- [math]\displaystyle{ 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^{[21]}

- [math]\displaystyle{ R=\tfrac{1}{2}\sqrt{a^2+c^2}=\tfrac{1}{2}\sqrt{b^2+d^2}. }[/math]

It also follows that^{[21]}

- [math]\displaystyle{ a^2+b^2+c^2+d^2=8R^2. }[/math]

Thus, according to Euler's quadrilateral theorem, the circumradius can be expressed in terms of the diagonals *p* and *q*, and the distance *x* between the midpoints of the diagonals as

- [math]\displaystyle{ R=\sqrt{\frac{p^2+q^2+4x^2}{8}}. }[/math]

A formula for the area *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^{[27]}^{:p.222}

- [math]\displaystyle{ K=\tfrac{1}{2}(ac+bd). }[/math]

- In a cyclic orthodiagonal quadrilateral, the anticenter coincides with the point where the diagonals intersect.
^{[21]} - 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.
^{[21]} - If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side.
^{[21]} - 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.
^{[21]}

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.^{[28]} One direction of this theorem was proved by I. A. Lexell in 1786. Lexell^{[29]} 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.^{[30]} Kiper et al.^{[31]} 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.

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