Edges and vertices4
Schläfli symbol{4} (for square)
Areavarious methods;
see below
Internal angle (degrees)90° (for square and rectangle)

Edges and vertices4
Schläfli symbol{4} (for square)
Areavarious methods;
see below
Internal angle (degrees)90° (for square and rectangle)

In Euclidean plane geometry, a quadrilateral is a polygon with four sides (or edges) and four vertices or corners. Sometimes, the term quadrangle is used, by analogy with triangle, and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on.

The origin of the word "quadrilateral" is the two Latin words quadri, a variant of four, and latus, meaning "side".

Quadrilaterals are simple (not self-intersecting) or complex (self-intersecting), also called crossed. Simple quadrilaterals are either convex or concave.

The interior angles of a simple (and planar) quadrilateral ABCD add up to 360 degrees of arc, that is

$\angle A+\angle B+\angle C+\angle D=360^{\circ}.$

This is a special case of the n-gon interior angle sum formula (n − 2) × 180°. In a crossed quadrilateral, the four interior angles on either side of the crossing add up to 720°.[1]

All convex quadrilaterals tile the plane by repeated rotation around the midpoints of their edges.

Euler diagram of some types of quadrilaterals. (UK) denotes British English and (US) denotes American English.

A parallelogram is a quadrilateral with two pairs of parallel sides. Equivalent conditions are that opposite sides are of equal length; that opposite angles are equal; or that the diagonals bisect each other. Parallelograms also include the square, rectangle, rhombus and rhomboid.

• Rhombus or rhomb: all four sides are of equal length. An equivalent condition is that the diagonals perpendicularly bisect each other. An informal description is "a pushed-over square" (including a square).
• Rhomboid: a parallelogram in which adjacent sides are of unequal lengths and angles are oblique (not right angles). Informally: "a pushed-over rectangle with no right angles."[2]
• Rectangle: all four angles are right angles. An equivalent condition is that the diagonals bisect each other and are equal in length. Informally: "a box or oblong" (including a square).
• Square (regular quadrilateral): all four sides are of equal length (equilateral), and all four angles are right angles. An equivalent condition is that opposite sides are parallel (a square is a parallelogram), that the diagonals perpendicularly bisect each other, and are of equal length. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (four equal sides and four equal angles).
• Oblong: a term sometimes used to denote a rectangle which has unequal adjacent sides (i.e. a rectangle that is not a square).[3]

• An equilic quadrilateral has two opposite equal sides that, when extended, meet at 60°.
• A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length.[4]
• A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square.[5]
• A geometric chevron (dart or arrowhead) is a concave quadrilateral with bilateral symmetry like a kite, but one interior angle is reflex.
• A self-intersecting quadrilateral is called variously a cross-quadrilateral, crossed quadrilateral, butterfly quadrilateral or bow-tie quadrilateral. A special case of crossed quadrilaterals are the antiparallelograms, crossed quadrilaterals in which (like a parallelogram) each pair of nonadjacent sides has equal length. The diagonals of a crossed or concave quadrilateral do not intersect inside the shape.
• A non-planar quadrilateral is called a skew quadrilateral. Formulas to compute its dihedral angles from the edge lengths and the angle between two adjacent edges were derived for work on the properties of molecules such as cyclobutane that contain a "puckered" ring of four atoms.[6] See skew polygon for more. Historically the term gauche quadrilateral was also used to mean a skew quadrilateral.[7] A skew quadrilateral together with its diagonals form a (possibly non-regular) tetrahedron, and conversely every skew quadrilateral comes from a tetrahedron where a pair of opposite edges is removed.

## Special line segments

The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices.

The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides.[8] They intersect at the "vertex centroid" of the quadrilateral (see Remarkable points below).

The four maltitudes of a convex quadrilateral are the perpendiculars to a side through the midpoint of the opposite side.[9] .

## Area of a convex quadrilateral

There are various general formulas for the area K of a convex quadrilateral.

### Trigonometric formulas

The area can be expressed in trigonometric terms as

$K = \tfrac{1}{2} pq \cdot \sin \theta,$

where the lengths of the diagonals are p and q and the angle between them is θ.[10] In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to $K=\tfrac{1}{2}pq$ since θ is 90°.

Bretschneider's formula[11] expresses the area in terms of the sides and two opposite angles:

\begin{align} K &= \sqrt{(s-a)(s-b)(s-c)(s-d) - \tfrac{1}{2} abcd \; [ 1 + \cos (A + C) ]} \\ &= \sqrt{(s-a)(s-b)(s-c)(s-d) - abcd \left[ \cos^2 \left( \tfrac{A + C}{2} \right) \right]} \end{align}

where the sides in sequence are a, b, c, d, where s is the semiperimeter, and A and C are two (in fact, any two) opposite angles. This reduces to Brahmagupta's formula for the area of a cyclic quadrilateral when A+C = 180°.

Another area formula in terms of the sides and angles, with angle C being between sides b and c, and A being between sides a and d, is

$K = \tfrac{1}{2}ad \cdot \sin{A} + \tfrac{1}{2}bc \cdot \sin{C}.$

In the case of a cyclic quadrilateral, the latter formula becomes $K = \tfrac{1}{2}(ad+bc)\sin{A}.$

In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to $K=ab \cdot \sin{A}.$

Alternatively, we can write the area in terms of the sides and the intersection angle θ of the diagonals, so long as this angle is not 90°:[12]

$K = \frac{|\tan \theta|}{4} \cdot \left| a^2 + c^2 - b^2 - d^2 \right|.$

In the case of a parallelogram, the latter formula becomes $K = \tfrac{1}{2}|\tan \theta|\cdot \left| a^2 - b^2 \right|.$

Another area formula including the sides a, b, c, d is[13]

$K=\tfrac{1}{4}\sqrt{(2(a^2+c^2)-4x^2)(2(b^2+d^2)-4x^2)}\sin{\varphi}$

where x is the distance between the midpoints of the diagonals and φ is the angle between the bimedians.

### Non-trigonometric formulas

The following two formulas expresses the area in terms of the sides a, b, c, d, the semiperimeter s, and the diagonals p, q:

$K = \sqrt{(s-a)(s-b)(s-c)(s-d) - \tfrac{1}{4}(ac+bd+pq)(ac+bd-pq)},$ [14]
$K = \frac{1}{4} \sqrt{4p^{2}q^{2}- \left( a^{2}+c^{2}-b^{2}-d^{2} \right) ^{2}}.$ [15]

The first reduces to Brahmagupta's formula in the cyclic quadrilateral case, since then pq = ac + bd.

The area can also be expressed in terms of the bimedians m, n and the diagonals p, q:

$K=\tfrac{1}{2}\sqrt{(m+n+p)(m+n-p)(m+n+q)(m+n-q)},$ [16]
$K=\tfrac{1}{2}\sqrt{p^2q^2-(m^2-n^2)^2}.$ [17]

### Vector formulas

The area of a quadrilateral ABCD can be calculated using vectors. Let vectors AC and BD form the diagonals from A to C and from B to D. The area of the quadrilateral is then

$K = \tfrac{1}{2} |\mathbf{AC}\times\mathbf{BD}|,$

which is half the magnitude of the cross product of vectors AC and BD. In two-dimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x1,y1) and BD as (x2,y2), this can be rewritten as:

$K = \tfrac{1}{2} |x_1 y_2 - x_2 y_1|.$

## Diagonals

### Properties of the diagonals in some quadrilaterals

In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals are perpendicular, and if their diagonals have equal length.[18] The list applies to the most general cases, and excludes named subsets.

TrapezoidNoSee note 1No
Isosceles trapezoidNoSee note 1Yes
ParallelogramYesNoNo
KiteSee note 2YesSee note 2
RectangleYesNoYes
RhombusYesYesNo
SquareYesYesYes

Note 1: The most general trapezoids and isosceles trapezoids do not have perpendicular diagonals, but there are infinite numbers of (non-similar) trapezoids and isosceles trapezoids that do have perpendicular diagonals and are not any other named quadrilateral.

Note 2: In a kite, one diagonal bisects the other. The most general kite has unequal diagonals, but there is an infinite number of (non-similar) kites in which the diagonals are equal in length (and the kites are not any other named quadrilateral).

### Length of the diagonals

The length of the diagonals in a convex quadrilateral ABCD can be calculated using the law of cosines. Thus

$p=\sqrt{a^2+b^2-2ab\cos{B}}=\sqrt{c^2+d^2-2cd\cos{D}}$

and

$q=\sqrt{a^2+d^2-2ad\cos{A}}=\sqrt{b^2+c^2-2bc\cos{C}}.$

Other, more symmetric formulas for the length of the diagonals, are[19]

$p=\sqrt{\frac{(ac+bd)(ad+bc)-2abcd(\cos{B}+\cos{D})}{ab+cd}}$

and

$q=\sqrt{\frac{(ab+cd)(ac+bd)-2abcd(\cos{A}+\cos{C})}{ad+bc}}.$

### Generalizations of the parallelogram law and Ptolemy's theorem

In any convex quadrilateral ABCD, the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals plus four times the square of the line segment connecting the midpoints of the diagonals. Thus

$a^2 + b^2 + c^2 + d^2 = p^2 + q^2 + 4x^2$

where x is the distance between the midpoints of the diagonals.[20]:p.126 This is sometimes known as Euler's quadrilateral theorem and is a generalization of the parallelogram law.

The German mathematician Carl Anton Bretschneider derived in 1842 the following generalization of Ptolemy's theorem, regarding the product of the diagonals in a convex quadrilateral[21]

$p^2q^2=a^2c^2+b^2d^2-2abcd\cos{(A+C)}.$

This relation can be considered to be a law of cosines for a quadrilateral. In a cyclic quadrilateral, where A + C = 180°, it reduces to pq = ac + bd. Since cos (A + C) ≥ −1, it also gives a proof of Ptolemy's inequality.

### Other metric relations

If X and Y are the feet of the normals from B and D to the diagonal AC = p in a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, then[22]:p.14

$XY=\frac{|a^2+c^2-b^2-d^2|}{2p}.$

In a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, and where the diagonals intersect at E,

$efgh(a+c+b+d)(a+c-b-d) = (agh+cef+beh+dfg)(agh+cef-beh-dfg)$

where e = AE, f = BE, g = CE, and h = DE.[23]

The shape of a convex quadrilateral is fully determined by the lengths of its sides in sequence and of one diagonal between two specified vertices. The two diagonals p, q and the four side lengths a, b, c, d of a quadrilateral are related[24] by the Cayley-Menger determinant, as follows:

$\det \begin{bmatrix} 0 & a^2 & p^2 & d^2 & 1 \\ a^2 & 0 & b^2 & q^2 & 1 \\ p^2 & b^2 & 0 & c^2 & 1 \\ d^2 & q^2 & c^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix} = 0.$

## Bimedians

The Varignon parallelogram EFGH

The bimedians of a quadrilateral are the line segments connecting the midpoints of the opposite sides. The midpoints of the sides of any quadrilateral (convex, concave or crossed) are the vertices of a parallelogram called the Varignon parallelogram. It has the following properties:

• Each pair of opposite sides of the Varignon parallelogram are parallel to a diagonal in the original quadrilateral.
• The length of a side in the Varignon parallelogram is half as long as the diagonal in the original quadrilateral it is parallel to.
• The area of the Varignon parallelogram equals half the area of the original quadrilateral. This is true in convex, concave and crossed quadrilaterals provided the area of the latter is defined to be the difference of the areas of the two triangles it is composed of.[25]
• The perimeter of the Varignon parallelogram equals the sum of the diagonals of the original quadrilateral.

The diagonals of the Varignon parallelogram are the bimedians of the original quadrilateral.

The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection.[20]:p.125

In a convex quadrilateral with sides a, b, c and d, the length of the bimedian that connects the midpoints of the sides a and c is

$m=\tfrac{1}{2}\sqrt{-a^2+b^2-c^2+d^2+p^2+q^2}$

where p and q are the length of the diagonals.[26] The length of the bimedian that connects the midpoints of the sides b and d is

$n=\tfrac{1}{2}\sqrt{a^2-b^2+c^2-d^2+p^2+q^2}.$

Hence[20]:p.126

$\displaystyle p^2+q^2=2(m^2+n^2).$

This is also a corollary to the parallelogram law applied in the Varignon parallelogram.

The length of the bimedians can also be expressed in terms of two opposite sides and the distance x between the midpoints of the diagonals. This is possible when using Euler's quadrilateral theorem in the above formulas. Whence[17]

$m=\tfrac{1}{2}\sqrt{2(b^2+d^2)-4x^2}$

and

$n=\tfrac{1}{2}\sqrt{2(a^2+c^2)-4x^2}.$

Note that the two opposite sides in these formulas are not the two that the bimedian connects.

In a convex quadrilateral, there is the following dual connection between the bimedians and the diagonals:[22]

• The two bimedians have equal length if and only if the two diagonals are perpendicular.
• The two bimedians are perpendicular if and only if the two diagonals have equal length.

## Trigonometric identities

The four angles of a simple quadrilateral ABCD satisfy the following identities:[27]

$\sin{A}+\sin{B}+\sin{C}+\sin{D}=4\sin{\frac{A+B}{2}}\sin{\frac{A+C}{2}}\sin{\frac{A+D}{2}}$

and

$\frac{\tan{A}\tan{B}-\tan{C}\tan{D}}{\tan{A}\tan{C}-\tan{B}\tan{D}}=\frac{\tan{(A+C)}}{\tan{(A+B)}}.$

Also,[28]

$\frac{\tan{A}+\tan{B}+\tan{C}+\tan{D}}{\cot{A}+\cot{B}+\cot{C}+\cot{D}}=\tan{A}\tan{B}\tan{C}\tan{D}.$

In the last two formulas, no angle is allowed to be a right angle, since then the tangent functions are not defined.

## Inequalities

### Area

If a convex quadrilateral has the consecutive sides a, b, c, d and the diagonals p, q, then its area K satisfies[29]

$K\le \tfrac{1}{4}(a+c)(b+d)$ with equality only for a rectangle.
$K\le \tfrac{1}{4}(a^2+b^2+c^2+d^2)$ with equality only for a square.
$K\le \tfrac{1}{4}(p^2+q^2)$ with equality only if the diagonals are perpendicular and equal.
$K\le \tfrac{1}{2}\sqrt{(a^2+c^2)(b^2+d^2)}$ with equality only for a rectangle.[13]

From Bretschneider's formula it directly follows that the area of a quadrilateral satisfies

$K \le \sqrt{(s-a)(s-b)(s-c)(s-d)}$

with equality if and only if the quadrilateral is cyclic or degenerate such that one side is equal to the sum of the other three (it has collapsed into a line segment, so the area is zero).

The area of any quadrilateral also satisfies the inequality[30]

$\displaystyle K\le \tfrac{1}{2}\sqrt[3]{(ab+cd)(ac+bd)(ad+bc)}.$

Denoting the perimeter as L, we have[30]:p.114

$K\le \tfrac{1}{16}L^2,$

with equality only in the case of a square.

The area of a convex quadrilateral also satisfies

$K \le \tfrac{1}{2}pq$

for diagonal lengths p and q, with equality if and only if the diagonals are perpendicular.

### Diagonals and bimedians

A corollary to Euler's quadrilateral theorem is the inequality

$a^2 + b^2 + c^2 + d^2 \ge p^2 + q^2$

where equality holds if and only if the quadrilateral is a parallelogram.

Euler also generalized Ptolemy's theorem, which is an equality in a cyclic quadrilateral, into an inequality for a convex quadrilateral. It states that

$pq \le ac + bd$

where there is equality if and only if the quadrilateral is cyclic.[20]:p.128–129 This is often called Ptolemy's inequality.

In any convex quadrilateral the bimedians m, n and the diagonals p, q are related by the inequality

$pq \leq m^2+n^2,$

with equality holding if and only if the diagonals are equal.[31]:Prop.1

### Sides

The sides a, b, c, and d of any quadrilateral satisfy[32]:p.228,#275

$a^2+b^2+c^2 > \frac{d^2}{3}$

and[32]:p.234,#466

$a^4+b^4+c^4 \geq \frac{d^4}{27}.$

## Maximum and minimum properties

Among all quadrilaterals with a given perimeter, the one with the largest area is the square. This is called the isoperimetric theorem for quadrilaterals. It is a direct consequence of the area inequality[30]:p.114

$K\le \tfrac{1}{16}L^2$

where K is the area of a convex quadrilateral with perimeter L. Equality holds if and only if the quadrilateral is a square. The dual theorem states that of all quadrilaterals with a given area, the square has the shortest perimeter.

The quadrilateral with given side lengths that has the maximum area is the cyclic quadrilateral.[33]

Of all convex quadrilaterals with given diagonals, the orthodiagonal quadrilateral has the largest area.[30]:p.119 This is a direct consequence of the fact that the area of a convex quadrilateral satisfies

$K=\tfrac{1}{2}pq\sin{\theta}\le \tfrac{1}{2}pq,$

where θ is the angle between the diagonals p and q. Equality holds if and only if θ = 90°.

If P is an interior point in a convex quadrilateral ABCD, then

$AP+BP+CP+DP\ge AC+BD.$

From this inequality it follows that the point inside a quadrilateral that minimizes the sum of distances to the vertices is the intersection of the diagonals. Hence that point is the Fermat point of a convex quadrilateral.[34]:p.120

## Remarkable points and lines in a convex quadrilateral

The centre of a quadrilateral can be defined in several different ways. The "vertex centroid" comes from considering the quadrilateral as being empty but having equal masses at its vertices. The "side centroid" comes from considering the sides to have constant mass per unit length. The usual centre, called just centroid (centre of area) comes from considering the surface of the quadrilateral as having constant density. These three points are in general not all the same point.[35]

The "vertex centroid" is the intersection of the two bimedians.[36] As with any polygon, the x and y coordinates of the vertex centroid are the arithmetic means of the x and y coordinates of the vertices.

The "area centroid" of quadrilateral ABCD can be constructed in the following way. Let Ga, Gb, Gc, Gd be the centroids of triangles BCD, ACD, ABD, ABC respectively. Then the "area centroid" is the intersection of the lines GaGc and GbGd.[37]

In a general convex quadrilateral ABCD, there are no natural analogies to the circumcenter and orthocenter of a triangle. But two such points can be constructed in the following way. Let Oa, Ob, Oc, Od be the circumcenters of triangles BCD, ACD, ABD, ABC respectively; and denote by Ha, Hb, Hc, Hd the orthocenters in the same triangles. Then the intersection of the lines OaOc and ObOd is called the quasicircumcenter; and the intersection of the lines HaHc and HbHd is called the quasiorthocenter of the convex quadrilateral.[37] These points can be used to define an Euler line of a quadrilateral. In a convex quadrilateral, the quasiorthocenter H, the "area centroid" G, and the quasicircumcenter O are collinear in this order, and HG = 2GO.[37]

There can also be defined a quasinine-point center E as the intersection of the lines EaEc and EbEd, where Ea, Eb, Ec, Ed are the nine-point centers of triangles BCD, ACD, ABD, ABC respectively. Then E is the midpoint of OH.[37]

Another remarkable line in a convex quadrilateral is the Newton line.

## Taxonomy

A taxonomy of quadrilaterals is illustrated by the following graph. Lower forms are special cases of higher forms. Note that "trapezoid" here is referring to the North American definition (the British equivalent is a trapezium), and "kite" excludes the concave kite (arrowhead or dart). Inclusive definitions are used throughout.

Taxonomic hierarchy of quadrilaterals. Lower forms are special cases of the higher forms they are connected to.

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