Quadrilateral

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

Convex quadrilaterals by symmetry, represented with a Hasse diagram.

In a convex quadrilateral all interior angles are less than 180°, and the two diagonals both lie inside the quadrilateral.

• Irregular quadrilateral (British English) or trapezium (North American English): no sides are parallel. (In British English, this was once called a trapezoid. For more, see Trapezoid § Trapezium vs Trapezoid)
• Trapezium (UK) or trapezoid (US): at least one pair of opposite sides are parallel. Trapezia (UK) and trapezoids (US) include parallelograms.
• Isosceles trapezium (UK) or isosceles trapezoid (US): one pair of opposite sides are parallel and the base angles are equal in measure. Alternative definitions are a quadrilateral with an axis of symmetry bisecting one pair of opposite sides, or a trapezoid with diagonals of equal length.
• Parallelogram: 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 include rhombi (including those rectangles called squares) and rhomboids (including those rectangles called oblongs). In other words, parallelograms include all rhombi and all rhomboids, and thus also include all rectangles.
• Rhombus, rhomb:[1] all four sides are of equal length (equilateral). An equivalent condition is that the diagonals perpendicularly bisect each other. Informally: "a pushed-over square" (but strictly including a square, too).
• Rhomboid: a parallelogram in which adjacent sides are of unequal lengths, and some angles are oblique (equiv., having no right angles). Informally: "a pushed-over oblong". Not all references agree, some define a rhomboid as a parallelogram that is not a rhombus.[4]
• Rectangle: all four angles are right angles (equiangular). An equivalent condition is that the diagonals bisect each other, and are equal in length. Rectangles include squares and oblongs. 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), and 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 (i.e., four equal sides and four equal angles).
• Oblong: longer than wide, or wider than long (i.e., a rectangle that is not a square).[5]
• Kite: two pairs of adjacent sides are of equal length. This implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. It also implies that the diagonals are perpendicular. Kites include rhombi.

• Tangential quadrilateral: the four sides are tangents to an inscribed circle. A convex quadrilateral is tangential if and only if opposite sides have equal sums.
• Tangential trapezoid: a trapezoid where the four sides are tangents to an inscribed circle.
• Cyclic quadrilateral: the four vertices lie on a circumscribed circle. A convex quadrilateral is cyclic if and only if opposite angles sum to 180°.
• Right kite: a kite with two opposite right angles. It is a type of cyclic quadrilateral.
• Harmonic quadrilateral: the products of the lengths of the opposing sides are equal. It is a type of cyclic quadrilateral.
• Bicentric quadrilateral: it is both tangential and cyclic.
• Orthodiagonal quadrilateral: the diagonals cross at right angles.
• Equidiagonal quadrilateral: the diagonals are of equal length.
• Ex-tangential quadrilateral: the four extensions of the sides are tangent to an excircle.
• 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.[6]
• A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square.[7]
• A diametric quadrilateral is a cyclic quadrilateral having one of its sides as a diameter of the circumcircle.[8]
• A Hjelmslev quadrilateral is a quadrilateral with two right angles at opposite vertices.[9]

In a concave quadrilateral, one interior angle is bigger than 180°, and one of the two diagonals lies outside the quadrilateral.

• A dart (or arrowhead) is a concave quadrilateral with bilateral symmetry like a kite, but where one interior angle is reflex. See Kite.

The area can be expressed in trigonometric terms as[14]

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

where the lengths of the diagonals are p and q and the angle between them is θ.[15] In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to ${\displaystyle K={\tfrac {pq}{2}}}$ since θ is 90°.

The area can be also expressed in terms of bimedians as[16]

${\displaystyle K=mn\sin \varphi ,}$

where the lengths of the bimedians are m and n and the angle between them is φ.

Bretschneider's formula[17][14] expresses the area in terms of the sides and two opposite angles:

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

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

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

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

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

Alternatively, we can write the area in terms of the sides and the intersection angle θ of the diagonals, as long θ is not 90°:[18]

${\displaystyle K={\tfrac {1}{4}}\left|\tan \theta \right|\cdot \left|a^{2}+c^{2}-b^{2}-d^{2}\right|.}$

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

Another area formula including the sides a, b, c, d is[16]

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

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

The last trigonometric area formula including the sides a, b, c, d and the angle α (between a and b) is:[19]

${\displaystyle K={\tfrac {1}{2}}ab\sin {\alpha }+{\tfrac {1}{4}}{\sqrt {4c^{2}d^{2}-(c^{2}+d^{2}-a^{2}-b^{2}+2ab\cos {\alpha })^{2}}},}$

which can also be used for the area of a concave quadrilateral (having the concave part opposite to angle α), by just changing the first sign + to -.

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

${\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)-{\tfrac {1}{4}}(ac+bd+pq)(ac+bd-pq)}},}$ [20]

${\displaystyle K={\tfrac {1}{4}}{\sqrt {4p^{2}q^{2}-\left(a^{2}+c^{2}-b^{2}-d^{2}\right)^{2}}}.}$ [21]

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:

${\displaystyle K={\tfrac {1}{2}}{\sqrt {(m+n+p)(m+n-p)(m+n+q)(m+n-q)}},}$ [22]

${\displaystyle K={\tfrac {1}{2}}{\sqrt {p^{2}q^{2}-(m^{2}-n^{2})^{2}}}.}$ [23]^{: Thm. 7}

In fact, any three of the four values m, n, p, and q suffice for determination of the area, since in any quadrilateral the four values are related by ${\displaystyle p^{2}+q^{2}=2(m^{2}+n^{2}).}$[24]^{: p. 126} The corresponding expressions are:[25]

${\displaystyle K={\tfrac {1}{2}}{\sqrt {[(m+n)^{2}-p^{2}]\cdot [p^{2}-(m-n)^{2}]}},}$

if the lengths of two bimedians and one diagonal are given, and[25]

${\displaystyle K={\tfrac {1}{4}}{\sqrt {[(p+q)^{2}-4m^{2}]\cdot [4m^{2}-(p-q)^{2}]}},}$

if the lengths of two diagonals and one bimedian are given.

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

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

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:

${\displaystyle K={\tfrac {1}{2}}|x_{1}y_{2}-x_{2}y_{1}|.}$

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.[26] The list applies to the most general cases, and excludes named subsets.

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).

The lengths of the diagonals in a convex quadrilateral ABCD can be calculated using the law of cosines on each triangle formed by one diagonal and two sides of the quadrilateral. Thus

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

and

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

Other, more symmetric formulas for the lengths of the diagonals, are[27]

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

and

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

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

${\displaystyle a^{2}+b^{2}+c^{2}+d^{2}=p^{2}+q^{2}+4x^{2}}$

where x is the distance between the midpoints of the diagonals.[24]^: 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[28]

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

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.

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[29]^: p.14

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

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

${\displaystyle efgh(a+c+b+d)(a+c-b-d)=(agh+cef+beh+dfg)(agh+cef-beh-dfg)}$

where e = AE, f = BE, g = CE, and h = DE.[30]

The shape and size of a convex quadrilateral are 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[14] by the Cayley-Menger determinant, as follows:

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

If a convex quadrilateral has the consecutive sides a, b, c, d and the diagonals p, q, then its area K satisfies[37]

${\displaystyle K\leq {\tfrac {1}{4}}(a+c)(b+d)}$ with equality only for a rectangle.

${\displaystyle K\leq {\tfrac {1}{4}}(a^{2}+b^{2}+c^{2}+d^{2})}$ with equality only for a square.

${\displaystyle K\leq {\tfrac {1}{4}}(p^{2}+q^{2})}$ with equality only if the diagonals are perpendicular and equal.

${\displaystyle K\leq {\tfrac {1}{2}}{\sqrt {(a^{2}+c^{2})(b^{2}+d^{2})}}}$ with equality only for a rectangle.[16]

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

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

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[38]

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

Denoting the perimeter as L, we have[38]^: p.114

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

with equality only in the case of a square.

The area of a convex quadrilateral also satisfies

${\displaystyle K\leq {\tfrac {1}{2}}pq}$

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

Let a, b, c, d be the lengths of the sides of a convex quadrilateral ABCD with the area K and diagonals AC = p, BD = q. Then[39]

${\displaystyle K\leq {\tfrac {1}{8}}(a^{2}+b^{2}+c^{2}+d^{2}+p^{2}+q^{2}+pq-ac-bd)}$ with equality only for a square.

Let a, b, c, d be the lengths of the sides of a convex quadrilateral ABCD with the area K, then the following inequality holds:[40]

${\displaystyle K\leq {\frac {1}{3+{\sqrt {3}}}}(ab+ac+ad+bc+bd+cd)-{\frac {1}{2(1+{\sqrt {3}})^{2}}}(a^{2}+b^{2}+c^{2}+d^{2})}$ with equality only for a square.

A corollary to Euler's quadrilateral theorem is the inequality

${\displaystyle a^{2}+b^{2}+c^{2}+d^{2}\geq p^{2}+q^{2}}$

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

${\displaystyle pq\leq ac+bd}$

where there is equality if and only if the quadrilateral is cyclic.[24]^{: 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

${\displaystyle pq\leq m^{2}+n^{2},}$

with equality holding if and only if the diagonals are equal.[41]^: Prop.1 This follows directly from the quadrilateral identity ${\displaystyle m^{2}+n^{2}={\tfrac {1}{2}}(p^{2}+q^{2}).}$

The sides a, b, c, and d of any quadrilateral satisfy[42]^{: p.228, #275}

${\displaystyle a^{2}+b^{2}+c^{2}>{\tfrac {1}{3}}d^{2}}$

and[42]^{: p.234, #466}

${\displaystyle a^{4}+b^{4}+c^{4}\geq {\tfrac {1}{27}}d^{4}.}$