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 Greek, Ancient (to 1453);: [[wiktionary:κύκλος#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.
A convex quadrilateral is cyclic if and only if the four perpendicular bisectors to the sides are concurrent. This common point is the circumcenter.
A convex quadrilateral is cyclic if and only if its opposite angles are supplementary, that is
\alpha+\gamma=\beta+\delta=\pi radians (=180\circ).
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 2n-gon, then the two sums of alternate interior angles are each equal to (n-1)
\pi
\pi
Taking the stereographic projection (half-angle tangent) of each angle, this can be re-expressed,
| \tan{ | ||||||
| + |
| \gamma | |
| 2 |
Which implies that
| \tan{ | \alpha |
| 2 |
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,
\angleACB=\angleADB.
Other 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
\omega
\omega
If two lines, one containing segment and the other containing segment, intersect at, then the four points,,, are concyclic if and only if
\displaystyleAE ⋅ EC=BE ⋅ ED.
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:
\displaystyleef=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.
In a convex quadrilateral, let be the diagonal triangle of and let
\omega
\omega
The area of a cyclic quadrilateral with sides,,, is given by Brahmagupta's formula
K=\sqrt{(s-a)(s-b)(s-c)(s-d)}
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,,,, angle between sides and, and angle between sides and can be expressed as
K=\tfrac{1}{2}(ab+cd)\sin{B}
or
K=\tfrac{1}{2}(ad+bc)\sin{A}
or
K=\tfrac{1}{2}(ac+bd)\sin{\theta}
where is either angle between the diagonals. Provided is not a right angle, the area can also be expressed as
K=\tfrac{1}{4}(a2-b2-c2+d2)\tan{A}.
Another formula is
\displaystyleK=2R2\sin{A}\sin{B}\sin{\theta}
where is the radius of the circumcircle. As a direct consequence,
K\le2R2
In a cyclic quadrilateral with successive vertices,,, and sides,,, and, the lengths of the diagonals and can be expressed in terms of the sides as[5]
p=\sqrt{
| (ac+bd)(ad+bc) | |
| ab+cd |
q=\sqrt{
| (ac+bd)(ab+cd) | |
| ad+bc |
so showing Ptolemy's theorem
pq=ac+bd.
According to Ptolemy's second theorem,
| p | |
| q |
=
| ad+bc | |
| ab+cd |
using the same notations as above.
For the sum of the diagonals we have the inequality[6]
p+q\ge2\sqrt{ac+bd}.
Equality holds if and only if the diagonals have equal length, which can be proved using the AM-GM inequality.
Moreover,[6]
(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 is a cyclic quadrilateral where meets at, then[7]
| AE | = | |
| CE |
| AB | ⋅ | |
| CB |
| 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.[5]
For a cyclic quadrilateral with successive sides,,,, semiperimeter, and angle between sides and, the trigonometric functions of are given by
\cosA=
| a2-b2-c2+d2 | |
| 2(ad+bc) |
,
\sinA=
| 2\sqrt{(s-a)(s-b)(s-c)(s-d) | |
\tan
| A | |
| 2 |
=\sqrt{
| (s-a)(s-d) | |
| (s-b)(s-c) |
The angle between the diagonals that is opposite sides and satisfies
\tan
| \theta | |
| 2 |
=\sqrt{
| (s-b)(s-d) | |
| (s-a)(s-c) |
If the extensions of opposite sides and intersect at an angle, then
| \cos{ | \varphi |
| 2 |
where is the semiperimeter.
Let
B
a
b
C
b
c
D
c
d
| \begin{align} | a+c |
| b+d |
&=
| \sin\tfrac12(A+B) | |
| \cos\tfrac12(C-D) |
\tan\tfrac12\theta,\\[10mu]
| a-c | |
| b-d |
&=
| \cos\tfrac12(A+B) | |
| \sin\tfrac12(D-C) |
\cot\tfrac12\theta. \end{align}
A cyclic quadrilateral with successive sides,,, and semiperimeter has the circumradius (the radius of the circumcircle) given by
| R= | 1 | \sqrt{ |
| 4 |
| (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=\sqrt{(ab+cd)(ac+bd)(ad+bc)}
where 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. 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 .
The anticenter of a cyclic quadrilateral is the Poncelet point of its vertices.
A Brahmagupta quadrilateral[8] 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=[t(u+v)+(1-uv)][u+v-t(1-uv)]
b=(1+u2)(v-t)(1+tv)
c=t(1+u2)(1+v2)
d=(1+v2)(u-t)(1+tu)
e=u(1+t2)(1+v2)
f=v(1+t2)(1+u2)
K=uv[2t(1-uv)-(u+v)(1-t2)][2(u+v)t+(1-uv)(1-t2)]
4R=(1+u2)(1+v2)(1+t2).
For a cyclic quadrilateral that is also 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)
| 2=a | |
| D | |
| 2 |
2+c2=b2+d2
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
| 2} | |
| R=\tfrac{1}{2}\sqrt{p | |
| 2 |
or, in terms of the sides of the quadrilateral, as
R=\tfrac{1}{2}\sqrt{a2+c2}=\tfrac{1}{2}\sqrt{b2+d2}.
It also follows that
a2+b2+c2+d2=8R2.
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{ | p2+q2+4x2 |
| 8 |
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[9]
K=\tfrac{1}{2}(ac+bd).
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.[10] One direction of this theorem was proved by Anders Johan Lexell in 1782.[11] 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.[12] Kiper et al.[13] 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.