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.

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.

The area K of a cyclic quadrilateral with sides a, b, c, d is given by Brahmagupta's formula^[9]: p.24

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

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 side lengths (regardless of sequence). This is another corollary to Bretschneider's formula. It can also be proved using calculus.^[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 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, angle A between sides a and d, and angle B between sides a and b can be expressed as^[9]: p.25

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

or

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

or^[9]: p.26

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

where θ is either angle between the diagonals. Provided A is not a right angle, the area can also be expressed as^[9]: p.26

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

Another formula is^[14]: p.83

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

where R is the radius of the circumcircle. As a direct consequence,^[15]

${\displaystyle K\leq 2R^{2}}$

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^{[9]: p.25,  [16][17]: p. 84}

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

so showing Ptolemy's theorem

${\displaystyle pq=ac+bd.}$

According to Ptolemy's second theorem,^{[9]: p.25,  [16]}

${\displaystyle {\frac {p}{q}}={\frac {ad+bc}{ab+cd}}}$

using the same notations as above.

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

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

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

Moreover,^{[18]: p.64, #1639}

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

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

If ABCD is a cyclic quadrilateral where AC meets BD at E, then^[19]

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

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}

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^[20]

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

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

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

The angle θ between the diagonals that is opposite sides a and c satisfies^[9]: p.26

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

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

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

where s is the semiperimeter.^[9]: p.31

Let ${\displaystyle B}$ denote the angle between sides ${\displaystyle a}$ and ${\displaystyle b}$, ${\displaystyle C}$ the angle between ${\displaystyle b}$ and ${\displaystyle c}$, and ${\displaystyle D}$ the angle between ${\displaystyle c}$ and ${\displaystyle d}$, then:^[21]

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

A cyclic quadrilateral with successive sides a, b, c, d and semiperimeter s has the circumradius (the radius of the circumcircle) given by^[16][22]

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

This was derived by the Indian mathematician 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

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

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.^{[23]: p.131,  [24]} These line segments are called the maltitudes,^[25] 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.^[24]

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.

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

• In a cyclic quadrilateral ABCD, the incenters M₁, M₂, M₃, M₄ (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.^[7]
• 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.^[26]

• 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.^[13]

A Brahmagupta quadrilateral^[27] 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:

${\displaystyle a=[t(u+v)+(1-uv)][u+v-t(1-uv)]}$

${\displaystyle b=(1+u^{2})(v-t)(1+tv)}$

${\displaystyle c=t(1+u^{2})(1+v^{2})}$

${\displaystyle d=(1+v^{2})(u-t)(1+tu)}$

${\displaystyle e=u(1+t^{2})(1+v^{2})}$

${\displaystyle f=v(1+t^{2})(1+u^{2})}$

${\displaystyle K=uv[2t(1-uv)-(u+v)(1-t^{2})][2(u+v)t+(1-uv)(1-t^{2})]}$

${\displaystyle 4R=(1+u^{2})(1+v^{2})(1+t^{2}).}$

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.^[30] One direction of this theorem was proved by Anders Johan Lexell in 1782.^[31] 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.^[32] Kiper et al.^[33] 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
• Cyclic polygon
• Power of a point
• Ptolemy's table of chords
• Robbins pentagon