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 (5sided), hexagon (6sided) 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 selfintersecting) or complex (selfintersecting), also called crossed. Simple quadrilaterals are either convex or concave.
The interior angles of a simple 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 ngon interior angle sum formula (n 2) 180 . In a crossed quadrilateral, the 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.
Convex quadrilaterals parallelograms
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. Equivalent conditions are that opposite sides are parallel and opposite angles are equal, or that the diagonals perpendicularly bisect each other. An informal description is "a pushedover square" (including a square).

Rhomboid: a parallelogram in which adjacent sides are of unequal lengths and angles are oblique (not right angles). Informally: "a pushedover rectangle with no right angles."

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).
Convex quadrilaterals other
More quadrilaterals
 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.^{[2]}
 A geometric chevron (dart or arrowhead) is a concave quadrilateral with bilateral symmetry like a kite, but one interior angle is reflex.
 A selfintersecting quadrilateral is called variously a crossquadrilateral, crossed quadrilateral, butterfly quadrilateral or bowtie 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 nonplanar 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.^{[3]}
International Journal of Quantum Chemistry, 106 (1) 215227, 2006. See skew polygon for more. Historically the term gauche quadrilateral was also used to mean a skew quadrilateral.^{[4]}
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.^{[5]} 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.^{[6]}
Notations in metric formulas
In the metric formulas below, the following notations are used. A convex quadrilateral ABCD has the sides a = AB, b = BC, c = CD, and d = DA. The diagonals are p = AC and q = BD, and the angle between them is . The semiperimeter s is defined as

s=\tfrac{1}{2}(a+b+c+d).
Area of a convex quadrilateral
There are various general formulas for the area K of a convex quadrilateral.
Using trigonometry
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 .^{[7]} In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to K=\tfrac{1}{2}pq since is 90 .
Bretschneider's formula^{[8]} expresses the area in terms of the sides and two opposite angles:
 \begin{align} K &= \sqrt{(sa)(sb)(sc)(sd)  \tfrac{1}{2} abcd \; [ 1 + \cos (A + C) ]} \\ &= \sqrt{(sa)(sb)(sc)(sd)  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 of the diagonals, so long as this angle is not 90 :^{[9]}
 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.
Nontrigonometric 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{(sa)(sb)(sc)(sd)  \tfrac{1}{4}(ac+bd+pq)(ac+bdpq)}, ^{[10]}

K = \frac{1}{4} \sqrt{4p^{2}q^{2} \left( a^{2}+c^{2}b^{2}d^{2} \right) ^{2}}. ^{[11]}
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+np)(m+n+q)(m+nq)}, ^{[12]}

K=\tfrac{1}{2}\sqrt{p^2q^2(m^2n^2)^2}. ^{[13]}
Using vectors
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 the magnitude of the cross product of vectors AC and BD. In twodimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x_{1},y_{1}) and BD as (x_{2},y_{2}), this can be rewritten as:
 K = \tfrac{1}{2} x_1 y_2  x_2 y_1.
Area inequalities
If a convex quadrilateral has the consecutive sides a, b, c, d and the diagonals p, q, then its area K satisfies^{[14]}

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.
From Bretschneider's formula it directly follows that the area of a quadrilateral satisfies
 K \le \sqrt{(sa)(sb)(sc)(sd)}
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^{[15]}
 \displaystyle K\le \tfrac{1}{2}\sqrt[3]{(ab+cd)(ac+bd)(ad+bc)}.
Diagonals
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^22ab\cos{B}}=\sqrt{c^2+d^22cd\cos{D}}
and
 q=\sqrt{a^2+d^22ad\cos{A}}=\sqrt{b^2+c^22bc\cos{C}}.
Other, more symmetric formulas for the length of the diagonals, are^{[16]}
 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.^{[17]} This is sometimes known as Euler's quadrilateral theorem and is a generalization of the parallelogram law.
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.^{[17]} This is often called Ptolemy's inequality.
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^{[18]}
 p^2q^2=a^2c^2+b^2d^22abcd\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^{[19]}
 XY=\frac{a^2+c^2b^2d^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+cbd) = (agh+cef+beh+dfg)(agh+cefbehdfg)
where e = AE, f = BE, g = CE, and h = DE.^{[20]}
The shape of a simple quadrilateral is fully determined by the lengths of its sides and one diagonal. The two diagonals p, q and the four side lengths a, b, c, d of a quadrilateral are related^{[21]} by the CayleyMenger 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 midpoints of the sides of a quadrilateral are the vertices of a parallelogram called the Varignon parallelogram. The sides in this parallelogram are half the lengths of the diagonals of the original quadrilateral, the area of the Varignon parallelogram equals half the area of the original quadrilateral, and 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.^{[17]}
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^2c^2+d^2+p^2+q^2}
where p and q are the length of the diagonals.^{[22]} The length of the bimedian that connects the midpoints of the sides b and d is
 n=\tfrac{1}{2}\sqrt{a^2b^2+c^2d^2+p^2+q^2}.
Hence^{[17]}
 \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^{[13]}
 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 are the following dual connection between the bimedians and the diagonals:^{[19]}
 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:^{[23]}
 \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,^{[24]}
 \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.
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^{[15]}
 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.^{[25]}
Of all convex quadrilaterals with given diagonals, the orthodiagonal quadrilateral has the largest area.^{[15]} 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 is the angle between the diagonals p and q. Equality holds if and only if = 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.^{[15]}
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.^{[26]}
The "vertex centroid" is the intersection of the two bimedians.^{[27]} 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 G_{a}, G_{b}, G_{c}, G_{d} be the centroids of triangles BCD, ACD, ABD, ABC respectively. Then the "area centroid" is the intersection of the lines G_{a}G_{c} and G_{b}G_{d}.^{[28]}
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 O_{a}, O_{b}, O_{c}, O_{d} be the circumcenters of triangles BCD, ACD, ABD, ABC respectively; and denote by H_{a}, H_{b}, H_{c}, H_{d} the orthocenters in the same triangles. Then the intersection of the lines O_{a}O_{c} and O_{b}O_{d} is called the quasicircumcenter; and the intersection of the lines H_{a}H_{c} and H_{b}H_{d} is called the quasiorthocenter of the convex quadrilateral.^{[28]} 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.^{[28]}
There can also be defined a quasininepoint center E as the intersection of the lines E_{a}E_{c} and E_{b}E_{d}, where E_{a}, E_{b}, E_{c}, E_{d} are the ninepoint centers of triangles BCD, ACD, ABD, ABC respectively. Then E is the midpoint of OH.^{[28]}
Another remarkable line in a convex quadrilateral is the Newton line.
Other properties of convex quadrilaterals
Taxonomy
A taxonomy of quadrilaterals is illustrated by the following graph. Lower forms are special cases of higher forms. Note that "trapezium" here is referring to the British definition (the North American equivalent is a trapezoid), and "kite" excludes the concave kite (arrowhead or dart). Inclusive definitions are used throughout.

Taxonomy of quadrilaterals. Lower forms are special cases of higher forms.
See also
References
External links
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