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A quadrilateral is a polygon in Euclidean plane geometry with four edges (sides) and four vertices (corners). Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon and hexagon), and 4-gon (in analogy to ''n''-gons for arbitrary values of ''n''). A quadrilateral with vertices $A$, $B$, $C$ and $D$ is sometimes denoted as $\square ABCD$. The word "quadrilateral" is derived from the Latin words ''quadri'', a variant of four, and ''latus'', meaning "side". Quadrilaterals are either simple (not self-intersecting), or complex (self-intersecting, or 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 :$\angle A+\angle B+\angle C+\angle D=360^.$ This is a special case of the ''n''-gon interior angle sum formula: (''n'' − 2) × 180°. All non-self-crossing quadrilaterals tile the plane, by repeated rotation around the midpoints of their edges.

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.

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. They intersect at the "vertex centroid" of the quadrilateral (see below). The four maltitudes of a convex quadrilateral are the perpendiculars to a side—through the midpoint of the opposite side.

There are various general formulas for the area of a convex quadrilateral ''ABCD'' with sides .

Trigonometric formulas

The area can be expressed in trigonometric terms as :$K = \tfrac pq \cdot \sin \theta,$ where the lengths of the diagonals are and and the angle between them is . In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to $K=\tfracpq$ since is . The area can be also expressed in terms of bimedians as :$K = mn \cdot \sin \varphi,$ where the lengths of the bimedians are and and the angle between them is . Bretschneider's formula expresses the area in terms of the sides and two opposite angles: :$\begin K &= \sqrt \\ &= \sqrt \end$ where the sides in sequence are , , , , where is the semiperimeter, and and are two (in fact, any two) opposite angles. This reduces to Brahmagupta's formula for the area of a cyclic quadrilateral—when . Another area formula in terms of the sides and angles, with angle being between sides and , and being between sides and , is :$K = \tfracad \cdot \sin + \tfracbc \cdot \sin.$ In the case of a cyclic quadrilateral, the latter formula becomes $K = \tfrac\left(ad+bc\right)\sin.$ In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to $K=ab \cdot \sin.$ Alternatively, we can write the area in terms of the sides and the intersection angle of the diagonals, as long is not :Mitchell, Douglas W., "The area of a quadrilateral," ''Mathematical Gazette'' 93, July 2009, 306–309. :$K = \frac \cdot \left| a^2 + c^2 - b^2 - d^2 \.$ In the case of a parallelogram, the latter formula becomes $K = \tfrac|\tan \theta|\cdot \left| a^2 - b^2 \.$ Another area formula including the sides , , , is. :$K=\tfrac\sqrt\sin$ where is the distance between the midpoints of the diagonals, and is the angle between the bimedians. The last trigonometric area formula including the sides , , , and the angle (between and ) is: :$K=\tfracab\cdot\sin+\tfrac\sqrt ,$ which can also be used for the area of a concave quadrilateral (having the concave part opposite to angle ), by just changing the first sign to .

Non-trigonometric formulas

The following two formulas express the area in terms of the sides , , and , the semiperimeter , and the diagonals , : :$K = \sqrt,$ :$K = \tfrac \sqrt.$ The first reduces to Brahmagupta's formula in the cyclic quadrilateral case, since then . The area can also be expressed in terms of the bimedians , and the diagonals , : :$K=\tfrac\sqrt,$ :$K=\tfrac\sqrt.$ . In fact, any three of the four values , , , and suffice for determination of the area, since in any quadrilateral the four values are related by $p^2+q^2=2\left(m^2+n^2\right).$ The corresponding expressions are:Josefsson, Martin (2016) ‘100.31 Heron-like formulas for quadrilaterals’, ''The Mathematical Gazette'', 100 (549), pp. 505–508. :$K=\tfrac\sqrt,$ if the lengths of two bimedians and one diagonal are given, and :$K=\tfrac\sqrt,$ if the lengths of two diagonals and one bimedian are given.

Vector formulas

The area of a quadrilateral can be calculated using vectors. Let vectors and form the diagonals from to and from to . The area of the quadrilateral is then :$K = \tfrac |\mathbf\times\mathbf|,$ which is half the magnitude of the cross product of vectors and . In two-dimensional Euclidean space, expressing vector as a free vector in Cartesian space equal to and as , this can be rewritten as: :$K = \tfrac |x_1 y_2 - x_2 y_1|.$

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. 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).''

Lengths of the diagonals

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 :$p=\sqrt=\sqrt$ and :$q=\sqrt=\sqrt.$ Other, more symmetric formulas for the lengths of the diagonals, are :$p=\sqrt$ and :$q=\sqrt.$

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. 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 :$p^2q^2=a^2c^2+b^2d^2-2abcd\cos.$ 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 :$XY=\frac.$ In a convex quadrilateral ''ABCD'' with sides ''a'' = ''AB'', ''b'' = ''BC'', ''c'' = ''CD'', ''d'' = ''DA'', and where the diagonals intersect at ''E'', :$efgh\left(a+c+b+d\right)\left(a+c-b-d\right) = \left(agh+cef+beh+dfg\right)\left(agh+cef-beh-dfg\right)$ where ''e'' = ''AE'', ''f'' = ''BE'', ''g'' = ''CE'', and ''h'' = ''DE''. 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 by the Cayley-Menger determinant, as follows: :$\det \begin 0 & a^2 & p^2 & d^2 & 1 \\ a^2 & 0 & b^2 & q^2 & 1 \\ p^2 & b^2 & 0 & c^2 & 1 \\ d^2 & q^2 & c^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end = 0.$

Angle bisectors

The internal angle bisectors of a convex quadrilateral either form a cyclic quadrilateral (that is, the four intersection points of adjacent angle bisectors are concyclic) or they are concurrent. In the latter case the quadrilateral is a tangential quadrilateral. In quadrilateral ''ABCD'', if the angle bisectors of ''A'' and ''C'' meet on diagonal ''BD'', then the angle bisectors of ''B'' and ''D'' meet on diagonal ''AC''.

Bimedians

The bimedians of a quadrilateral are the line segments connecting the midpoints of the opposite sides. The intersection of the bimedians is the centroid of the vertices of the quadrilateral. 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. *A side of 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. *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.Altshiller-Court, Nathan, ''College Geometry'', Dover Publ., 2007. 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 :$m=\tfrac\sqrt$ where ''p'' and ''q'' are the length of the diagonals. The length of the bimedian that connects the midpoints of the sides ''b'' and ''d'' is :$n=\tfrac\sqrt.$ Hence :$\displaystyle p^2+q^2=2\left(m^2+n^2\right).$ This is also a corollary to the parallelogram law applied in the Varignon parallelogram. The lengths 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 :$m=\tfrac\sqrt$ and :$n=\tfrac\sqrt.$ 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:. * 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: :$\sin+\sin+\sin+\sin=4\sin\sin\sin$ and :$\frac=\frac.$ Also, :$\frac=\tan\tan\tan\tan.$ In the last two formulas, no angle is allowed to be a right angle, since tan 90° is 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 :$K\le \tfrac\left(a+c\right)\left(b+d\right)$ with equality only for a rectangle. :$K\le \tfrac\left(a^2+b^2+c^2+d^2\right)$ with equality only for a square. :$K\le \tfrac\left(p^2+q^2\right)$ with equality only if the diagonals are perpendicular and equal. :$K\le \tfrac\sqrt$ with equality only for a rectangle. From Bretschneider's formula it directly follows that the area of a quadrilateral satisfies :$K \le \sqrt$ 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. :$\displaystyle K\le \tfrac\sqrt$ Denoting the perimeter as ''L'', we have :$K\le \tfracL^2,$ with equality only in the case of a square. The area of a convex quadrilateral also satisfies :$K \le \tfracpq$ 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 :$K \leq \frac$ 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: :$K \leq \frac\left(ab+ac+ad+bc+bd+cd\right)- \frac\left(a^2+b^2+c^2+d^2\right)$ with equality only for a square.

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. 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. This follows directly from the quadrilateral identity $m^2+n^2=\tfrac\left(p^2+q^2\right).$

Sides

The sides ''a'', ''b'', ''c'', and ''d'' of any quadrilateral satisfy''Inequalities proposed in “Crux Mathematicorum”''

:$a^2+b^2+c^2 > \frac$ and :$a^4+b^4+c^4 \geq \frac.$

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 :$K\le \tfracL^2$ 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. Of all convex quadrilaterals with given diagonals, the orthodiagonal quadrilateral has the largest area. This is a direct consequence of the fact that the area of a convex quadrilateral satisfies :$K=\tfracpq\sin\le \tfracpq,$ 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.

Remarkable points and lines in a convex quadrilateral

*Let exterior squares be drawn on all sides of a quadrilateral. The segments connecting the centers of opposite squares are (a) equal in length, and (b) perpendicular. Thus these centers are the vertices of an orthodiagonal quadrilateral. This is called Van Aubel's theorem. *For any simple quadrilateral with given edge lengths, there is a cyclic quadrilateral with the same edge lengths.Peter, Thomas, "Maximizing the Area of a Quadrilateral", ''The College Mathematics Journal'', Vol. 34, No. 4 (September 2003), pp. 315–316. *The four smaller triangles formed by the diagonals and sides of a convex quadrilateral have the property that the product of the areas of two opposite triangles equals the product of the areas of the other two triangles.

Taxonomy

:. A hierarchical Taxonomy (general)|taxonomy of quadrilaterals is illustrated by the figure to the right. Lower classes are special cases of higher classes they are connected to. Note that "trapezoid" here is referring to the North American definition (the British equivalent is a trapezium). Inclusive definitions are used throughout.

260px|The (red) side edges of tetragonal disphenoid represent a regular zig-zag skew quadrilateral 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. Historically the term gauche quadrilateral was also used to mean a skew quadrilateral. 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.

References

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Quadrilaterals Formed by Perpendicular BisectorsProjective Collinearity
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Interactive Classification