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Special cases

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.

Characterizations

Circumcenter

A convex quadrilateral is cyclic if and only if the four perpendicular bisectors to the sides are concurrent. This common point is the circumcenter.

Supplementary angles

A convex quadrilateral is cyclic if and only if its opposite angles are supplementary, that is :$\alpha + \gamma = \beta + \delta = \pi \ \text\ \left(= 180^\right).$ The direct theorem was Proposition 22 in Book 3 of 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 2''n''-gon, then the two sums of ''alternate'' interior angles are each equal to (''n''-1)$\pi$. Taking the stereographic projection (half-angle tangent) of each angle, this can be re-expressed, :$\frac = \frac = \infty.$ Which implies that :$\tan \tan = \tan = 1$

Angles between sides and diagonals

A convex quadrilateral 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. That is, for example, :$\angle ACB = \angle ADB.$

Pascal Points

Another necessary and sufficient conditions for a convex quadrilateral to be cyclic are: let be the point of intersection of the diagonals, let be the intersection point of the extensions of the sides and , let $\omega$ be a circle whose diameter is the segment, , and let and be Pascal points on sides and formed by the circle $\omega$.
(1) is a cyclic quadrilateral if and only if points and are collinear with the center , of circle $\omega$.
(2) is a cyclic quadrilateral if and only if points and are the midpoints of sides and .

Intersection of diagonals

If two lines, one containing segment and the other containing segment , intersect at , then the four points , , , are concyclic if and only if :$\displaystyle AP\cdot PC = BP\cdot PD.$ The intersection may be internal or external to the circle. In the former case, the cyclic quadrilateral is , and in the latter case, the cyclic quadrilateral is . When the intersection is internal, the equality states that the product of the segment lengths into which 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

Ptolemy's theorem expresses the product of the lengths of the two diagonals and of a cyclic quadrilateral as equal to the sum of the products of opposite sides: :$\displaystyle ef = ac + bd$, 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.

Diagonal triangle

In a convex quadrilateral , let be the diagonal triangle of and let $\omega$ be the nine-point circle of . is cyclic if and only if the point of intersection of the bimedians of belongs to the nine-point circle $\omega$.

Area

The area of a cyclic quadrilateral with sides , , , is given by Brahmagupta's formula :$K=\sqrt \,$ where , the semiperimeter, is . This is a corollary of Bretschneider's formula for the general quadrilateral, since opposite angles are supplementary in the cyclic case. If also , 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. 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 , , , and , side could be opposite any of side , side , or side . The area of a cyclic quadrilateral with successive sides , , , and angle between sides and can be expressed as :$K = \tfrac\left(ab+cd\right)\sin$ or :$K = \tfrac\left(ac+bd\right)\sin$ where is either angle between the diagonals. Provided is not a right angle, the area can also be expressed as :$K = \tfrac\left(a^2-b^2-c^2+d^2\right)\tan.$ Another formula is :$\displaystyle K=2R^2\sin\sin\sin$ where is the radius of the circumcircle. As a direct consequence, :$K\le 2R^2$ where there is equality if and only if the quadrilateral is a square.

Diagonals

In a cyclic quadrilateral with successive vertices , , , and sides , , , and , the lengths of the diagonals and can be expressed in terms of the sides asJohnson, Roger A., ''Advanced Euclidean Geometry'', Dover Publ., 2007 (orig. 1929). :$p = \sqrt$ and $q = \sqrt$ so showing Ptolemy's theorem :$pq = ac+bd.$ According to ''Ptolemy's second theorem'', :$\frac = \frac$ using the same notations as above. For the sum of the diagonals we have the inequality''Inequalities proposed in "Crux Mathematicorum"'', 2007,

:$p+q\ge 2\sqrt.$ Equality holds if and only if the diagonals have equal length, which can be proved using the AM-GM inequality. Moreover, :$\left(p+q\right)^2 \leq \left(a+c\right)^2+\left(b+d\right)^2.$ 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 and are the midpoints of the diagonals and , then :$\frac=\frac\left |\frac-\frac\$ where and are the intersection points of the extensions of opposite sides. If is a cyclic quadrilateral where meets at , then :$\frac=\frac\cdot\frac.$ 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.

Angle formulas

For a cyclic quadrilateral with successive sides , , , , semiperimeter , and angle between sides and , the trigonometric functions of are given by :$\cos A = \frac,$ :$\sin A = \frac,$ :$\tan \frac = \sqrt.$ The angle between the diagonals satisfies :$\tan \frac = \sqrt.$ If the extensions of opposite sides and intersect at an angle , then :$\cos=\sqrt$ where is the semiperimeter.

A cyclic quadrilateral with successive sides , , , and semiperimeter has the circumradius (the radius of the circumcircle) given by :$R=\frac \sqrt.$ This was derived by the Indian mathematician Vatasseri Parameshvara in the 15th century. Using Brahmagupta's formula, Parameshvara's formula can be restated as :$4KR=\sqrt$ where 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. These line segments are called the ''maltitudes'', 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. If the diagonals of a cyclic quadrilateral intersect at , and the midpoints of the diagonals are and , then the anticenter of the quadrilateral is the orthocenter of triangle .

Other properties

Japanese theorem *In a cyclic quadrilateral , the incenters ''M''1, ''M''2, ''M''3, ''M''4 (see the figure to the right) in triangles , , , and 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 , and the centroids in those four triangles are vertices of another cyclic quadrilateral. *In a cyclic quadrilateral with circumcenter , let be the point where the diagonals and intersect. Then angle is the arithmetic mean of the angles and . 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. *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 and , then the internal angle bisectors of the angles at and are perpendicular.

A Brahmagupta quadrilateral is a cyclic quadrilateral with integer sides, integer diagonals, and integer area. All Brahmagupta quadrilaterals with sides , , , , diagonals , , area , and circumradius can be obtained by clearing denominators from the following expressions involving rational parameters , , and : :$a=\left(u+v\right)+\left(1-uv\right)u+v-t\left(1-uv\right)\right]$ :$b=\left(1+u^2\right)\left(v-t\right)\left(1+tv\right)$ :$c=t\left(1+u^2\right)\left(1+v^2\right)$ :$d=\left(1+v^2\right)\left(u-t\right)\left(1+tu\right)$ :$e=u\left(1+t^2\right)\left(1+v^2\right)$ :$f=v\left(1+t^2\right)\left(1+u^2\right)$ :$K=uvt\left(1-uv\right)-\left(u+v\right)\left(1-t^2\right)2\left(u+v\right)t+\left(1-uv\right)\left(1-t^2\right)\right]$ :$4R=\left(1+u^2\right)\left(1+v^2\right)\left(1+t^2\right).$

Orthodiagonal case

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 and and divides the other diagonal into segments of lengths and . Then (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 is the diameter of the circumcircle. This holds because the diagonals are perpendicular chords of a circle. These equations imply that the circumradius can be expressed as :$R=\tfrac\sqrt$ or, in terms of the sides of the quadrilateral, as :$R=\tfrac\sqrt=\tfrac\sqrt.$ 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 and , and the distance between the midpoints of the diagonals as :$R=\sqrt.$ A formula for the area 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 :$K=\tfrac\left(ac+bd\right).$

Other properties

*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. One direction of this theorem was proved by I. A. Lexell in 1786. 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. Kiper et al. 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

References