Английская Википедия:Cyclic quadrilateral

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Файл:Cyclic quadrilateral.svg

Examples of cyclic quadrilaterals

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 Шаблон:Lang (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.

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 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.

Characterizations

Файл:Sehnenviereck.svg

A cyclic quadrilateral ABCD

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 Шаблон:Math is cyclic if and only if its opposite angles are supplementary, that is

<math>\alpha + \gamma = \beta + \delta = \pi \ \text{radians}\ (= 180^{\circ}).</math>

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)<math>\pi</math>. This result can be further generalized as follows: lf A1A2...A2n (n > 1) is any cyclic 2n-gon in which vertex Ai->Ai+k (vertex Ai is joined to Ai+k), then the two sums of alternate interior angles are each equal to m<math>\pi</math> (where m = n—k and k = 1, 2, 3, ... is the total turning).

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

<math>\frac

{ \tan{\frac{\alpha}{2}} + \tan{\frac{\gamma}{2}} } { 1 - \tan{\frac{\alpha}{2}} \tan{\frac{\gamma}{2}} } = \frac { \tan{\frac{\beta}{2}} + \tan{\frac{\delta}{2}} } { 1 - \tan{\frac{\beta}{2}} \tan{\frac{\delta}{2}} } = \infty.</math>

Which implies that

<math>\tan{\frac{\alpha}{2}} \tan{\frac{\gamma}{2}} = \tan{\frac{\beta}{2}}{\tan \frac{\delta}{2}} = 1</math>

Angles between sides and diagonals

A convex quadrilateral Шаблон:Math 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,

<math>\angle ACB = \angle ADB.</math>

Pascal points

Файл:Cyclic quadrilateral - pascal points.png

ABCD is a cyclic quadrilateral. E is the point of intersection of the diagonals and F is the point of intersection of the extensions of sides BC and AD. <math>\omega</math> is a circle whose diameter is the segment, EF. P and Q are Pascal points formed by the circle <math>\omega</math>. Triangles FAB and FCD are similar.

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

Intersection of diagonals

If two lines, one containing segment Шаблон:Math and the other containing segment Шаблон:Math, intersect at Шаблон:Math, then the four points Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math are concyclic if and only if

<math>\displaystyle AE\cdot EC = BE\cdot ED.</math>

The intersection Шаблон:Math may be internal or external to the circle. In the former case, the cyclic quadrilateral is Шаблон:Math, and in the latter case, the cyclic quadrilateral is Шаблон:Math. When the intersection is internal, the equality states that the product of the segment lengths into which Шаблон:Math 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 Шаблон:Math and Шаблон:Math of a cyclic quadrilateral as equal to the sum of the products of opposite sides:Шаблон:Rp

<math>\displaystyle ef = ac + bd,</math>

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

Файл:Nine-point circle of diagonal triangle.png

ABCD is a cyclic quadrilateral. EFG is the diagonal triangle of ABCD. The point T of intersection of the bimedians of ABCD belongs to the nine-point circle of EFG.

In a convex quadrilateral Шаблон:Math, let Шаблон:Math be the diagonal triangle of Шаблон:Math and let <math>\omega</math> be the nine-point circle of Шаблон:Math. Шаблон:Math is cyclic if and only if the point of intersection of the bimedians of Шаблон:Math belongs to the nine-point circle <math>\omega</math>.

Area

The area Шаблон:Math of a cyclic quadrilateral with sides Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math is given by Brahmagupta's formulaШаблон:Rp

<math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)} \,</math>

where Шаблон:Math, the semiperimeter, is Шаблон:Math. This is a corollary of Bretschneider's formula for the general quadrilateral, since opposite angles are supplementary in the cyclic case. If also Шаблон:Math, 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 Шаблон:Math, Шаблон:Math, Шаблон:Math, and Шаблон:Math, side Шаблон:Math could be opposite any of side Шаблон:Math, side Шаблон:Math, or side Шаблон:Math.

The area of a cyclic quadrilateral with successive sides Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math, angle Шаблон:Math between sides Шаблон:Math and Шаблон:Math, and angle Шаблон:Math between sides Шаблон:Math and Шаблон:Math can be expressed asШаблон:Rp

<math>K = \tfrac{1}{2}(ab+cd)\sin{B}</math>

or

<math>K = \tfrac{1}{2}(ad+bc)\sin{A}</math>

orШаблон:Rp

<math>K = \tfrac{1}{2}(ac+bd)\sin{\theta}</math>

where Шаблон:Math is either angle between the diagonals. Provided Шаблон:Math is not a right angle, the area can also be expressed asШаблон:Rp

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

Another formula isШаблон:Rp

<math>\displaystyle K=2R^2\sin{A}\sin{B}\sin{\theta}</math>

where Шаблон:Math is the radius of the circumcircle. As a direct consequence,

<math>K\le 2R^2</math>

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

Diagonals

In a cyclic quadrilateral with successive vertices Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math and sides Шаблон:Math, Шаблон:Math, Шаблон:Math, and Шаблон:Math, the lengths of the diagonals Шаблон:Math and Шаблон:Math can be expressed in terms of the sides asШаблон:RpШаблон:Rp

<math>p = \sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}</math> and <math>q = \sqrt{\frac{(ac+bd)(ab+cd)}{ad+bc}}</math>

so showing Ptolemy's theorem

<math>pq = ac+bd.</math>

According to Ptolemy's second theorem,Шаблон:Rp

<math>\frac {p}{q}= \frac{ad+bc}{ab+cd}</math>

using the same notations as above.

For the sum of the diagonals we have the inequalityШаблон:Rp

<math>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,Шаблон:Rp

<math>(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 Шаблон:Math and Шаблон:Math are the midpoints of the diagonals Шаблон:Math and Шаблон:Math, then

<math>\frac{MN}{EF}=\frac{1}{2}\left |\frac{AC}{BD}-\frac{BD}{AC}\right|</math>

where Шаблон:Math and Шаблон:Math are the intersection points of the extensions of opposite sides.

If Шаблон:Math is a cyclic quadrilateral where Шаблон:Math meets Шаблон:Math at Шаблон:Math, then

<math> \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.Шаблон:Rp

Angle formulas

For a cyclic quadrilateral with successive sides Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math, semiperimeter Шаблон:Math, and angle Шаблон:Math between sides Шаблон:Math and Шаблон:Math, the trigonometric functions of Шаблон:Math are given by

<math>\cos A = \frac{a^2-b^2-c^2+d^2}{2(ad+bc)},</math>

<math>\sin A = \frac{2\sqrt{(s-a)(s-b)(s-c)(s-d)}}{(ad+bc)},</math>

<math>\tan \frac{A}{2} = \sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}.</math>

The angle Шаблон:Math between the diagonals that is opposite sides Шаблон:Math and Шаблон:Math satisfiesШаблон:Rp

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

If the extensions of opposite sides Шаблон:Math and Шаблон:Math intersect at an angle Шаблон:Math, then

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

where Шаблон:Math is the semiperimeter.Шаблон:Rp

Let <math>B</math> denote the angle between sides <math>a</math> and <math>b</math>, <math>C</math> the angle between <math>b</math> and <math>c</math>, and <math>D</math> the angle between <math>c</math> and <math>d</math>, then:

<math>\begin{align}

\frac{a+c}{b+d} &= \frac{\sin\tfrac12(A+B)}{\cos\tfrac12(C-D)}\tan\tfrac12\theta, \\[10mu] \frac{a-c}{b-d} &= \frac{\cos\tfrac12(A+B)}{\sin\tfrac12(D-C)}\cot\tfrac12\theta. \end{align}</math>

Parameshvara's circumradius formula

A cyclic quadrilateral with successive sides Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math and semiperimeter Шаблон:Math has the circumradius (the radius of the circumcircle) given by

<math>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. (Note that the radius is invariant under the interchange of any side lengths.)

Using Brahmagupta's formula, Parameshvara's formula can be restated as

<math>4KR=\sqrt{(ab+cd)(ac+bd)(ad+bc)}</math>

where Шаблон:Math 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.Шаблон:Rp 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 Шаблон:Math, and the midpoints of the diagonals are Шаблон:Math and Шаблон:Math, then the anticenter of the quadrilateral is the orthocenter of triangle Шаблон:Math.

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

Other properties

Файл:Japanese theorem 2.svg

Japanese theorem

• In a cyclic quadrilateral Шаблон:Math, the incenters M₁, M₂, M₃, M₄ (see the figure to the right) in triangles Шаблон:Math, Шаблон:Math, Шаблон:Math, and Шаблон:Math 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 Шаблон:Math, and the centroids in those four triangles are vertices of another cyclic quadrilateral.
• In a cyclic quadrilateral Шаблон:Math with circumcenter Шаблон:Math, let Шаблон:Math be the point where the diagonals Шаблон:Math and Шаблон:Math intersect. Then angle Шаблон:Math is the arithmetic mean of the angles Шаблон:Math and Шаблон:Math. 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 Шаблон:Math and Шаблон:Math, then the internal angle bisectors of the angles at Шаблон:Math and Шаблон:Math are perpendicular.

Brahmagupta quadrilaterals

A Brahmagupta quadrilateral is a cyclic quadrilateral with integer sides, integer diagonals, and integer area. All Brahmagupta quadrilaterals with sides Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math, diagonals Шаблон:Math, Шаблон:Math, area Шаблон:Math, and circumradius Шаблон:Math can be obtained by clearing denominators from the following expressions involving rational parameters Шаблон:Math, Шаблон:Math, and Шаблон:Math:

<math>a=[t(u+v)+(1-uv)][u+v-t(1-uv)]</math>

<math>b=(1+u^2)(v-t)(1+tv)</math>

<math>c=t(1+u^2)(1+v^2)</math>

<math>d=(1+v^2)(u-t)(1+tu)</math>

<math>e=u(1+t^2)(1+v^2)</math>

<math>f=v(1+t^2)(1+u^2)</math>

<math>K=uv[2t(1-uv)-(u+v)(1-t^2)][2(u+v)t+(1-uv)(1-t^2)]</math>

<math>4R=(1+u^2)(1+v^2)(1+t^2).</math>

Orthodiagonal case

Circumradius and area

For a cyclic quadrilateral that is also orthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths Шаблон:Math and Шаблон:Math and divides the other diagonal into segments of lengths Шаблон:Math and Шаблон:Math. Then (the first equality is Proposition 11 in Archimedes' Book of Lemmas)

<math> D^2=p_1^2+p_2^2+q_1^2+q_2^2=a^2+c^2=b^2+d^2 </math>

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

<math> 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

<math> R=\tfrac{1}{2}\sqrt{a^2+c^2}=\tfrac{1}{2}\sqrt{b^2+d^2}. </math>

It also follows that

<math> 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 Шаблон:Math and Шаблон:Math, and the distance Шаблон:Math between the midpoints of the diagonals as

<math> R=\sqrt{\frac{p^2+q^2+4x^2}{8}}. </math>

A formula for the area Шаблон:Math 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 isШаблон:Rp

<math> K=\tfrac{1}{2}(ac+bd). </math>

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.

Cyclic spherical quadrilaterals

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 Anders Johan Lexell in 1782. 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.

See also

• Butterfly theorem
• Cyclic polygon
• Power of a point
• Ptolemy's table of chords
• Robbins pentagon

Further reading

• D. Fraivert: Pascal-points quadrilaterals inscribed in a cyclic quadrilateral

External links

• Derivation of Formula for the Area of Cyclic Quadrilateral
• Incenters in Cyclic Quadrilateral at cut-the-knot
• Four Concurrent Lines in a Cyclic Quadrilateral at cut-the-knot
• Шаблон:MathWorld