Quadrilateral
Some types of quadrilaterals
Edges andvertices	4
Schläfli symbol	{4} (for square)
Area	various methods; see below
Internal angle (degrees)	90° (for square and rectangle)

Ingeometry aquadrilateral is a four-sidedpolygon, having fouredges (sides) and fourcorners (vertices). The word is derived from the Latin wordsquadri, a variant of four, andlatus, meaning "side". It is also called atetragon, derived from Greek "tetra" meaning "four" and "gon" meaning "corner" or "angle", in analogy to other polygons (e.g.pentagon). Since "gon" means "angle", it is analogously called aquadrangle, or 4-angle. A quadrilateral with vertices${\displaystyle A}$,${\displaystyle B}$,${\displaystyle C}$ and${\displaystyle D}$ is sometimes denoted as${\displaystyle ◻ABCD}$.^[1]

Quadrilaterals are eithersimple (not self-intersecting), orcomplex (self-intersecting, or crossed). Simple quadrilaterals are eitherconvex orconcave.

Theinterior angles of a simple (andplanar) quadrilateralABCD add up to 360degrees, that is^[1]

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

This is a special case of then-gon interior angle sum formula:S = (n − 2) × 180° (here, n=4).^[2]

All non-self-crossing quadrilateralstile the plane, by repeated rotation around the midpoints of their edges.^[3]

Any quadrilateral that is not self-intersecting is a simple quadrilateral.

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 atrapezoid. For more, seeTrapezoid § Trapezium vs Trapezoid.)
• Trapezium (UK) ortrapezoid (US): at least one pair of opposite sides areparallel. Trapezia (UK) and trapezoids (US) include parallelograms.
• Isosceles trapezium (UK) orisosceles trapezoid (US): one pair of opposite sides are parallel and the baseangles 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 areoblique (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 intocongruent 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 aretangents to aninscribed circle.
• Cyclic quadrilateral: the four vertices lie on acircumscribed 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: a cyclic quadrilateral such that the products of the lengths of the opposing sides are equal.
• Bicentric quadrilateral: it is both tangential and cyclic.
• Orthodiagonal quadrilateral: the diagonals cross atright angles.
• Equidiagonal quadrilateral: the diagonals are of equal length.
• Bisect-diagonal quadrilateral: one diagonal bisects the other into equal lengths. Every dart and kite is bisect-diagonal. When both diagonals bisect another, it's a parallelogram.
• Ex-tangential quadrilateral: the four extensions of the sides are tangent to anexcircle.
• Anequilic quadrilateral has two opposite equal sides that when extended, meet at 60°.
• AWatt quadrilateral is a quadrilateral with a pair of opposite sides of equal length.^[6]
• Aquadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square.^[7]
• Adiametric quadrilateral is a cyclic quadrilateral having one of its sides as a diameter of the circumcircle.^[8]
• AHjelmslev 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.

• Adart (or arrowhead) is aconcave quadrilateral with bilateral symmetry like a kite, but where one interior angle is reflex. SeeKite.

Aself-intersecting quadrilateral is called variously across-quadrilateral,crossed quadrilateral,butterfly quadrilateral orbow-tie quadrilateral. In a crossed quadrilateral, the four "interior" angles on either side of the crossing (twoacute and tworeflex, all on the left or all on the right as the figure is traced out) add up to 720°.^[10]

• Crossed trapezoid (US) or trapezium (Commonwealth):^[11] a crossed quadrilateral in which one pair of nonadjacent sides is parallel (like atrapezoid).
• Antiparallelogram: a crossed quadrilateral in which each pair of nonadjacent sides have equal lengths (like aparallelogram).
• Crossed rectangle: an antiparallelogram whose sides are two opposite sides and the two diagonals of arectangle, hence having one pair of parallel opposite sides.
• Crossed square: a special case of a crossed rectangle where two of the sides intersect at right angles.

The twodiagonals of a convex quadrilateral are theline segments that connect opposite vertices.

The twobimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides.^[12] They intersect at the "vertex centroid" of the quadrilateral (see§ Remarkable points and lines in a convex quadrilateral below).

The fourmaltitudes of a convex quadrilateral are the perpendiculars to a side—through the midpoint of the opposite side.^[13]

There are various general formulas for theareaK of a convex quadrilateralABCD with sidesa =AB,b =BC,c =CD andd =DA.

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

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

where the lengths of the diagonals arep andq and the angle between them isθ.^[15] In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to${\displaystyle K=\frac{pq}{2}}$  sinceθ is90°.

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

${\displaystyle K=mn\sin \phi ,}$ 

where the lengths of the bimedians arem andn and the angle between them isφ.

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

 

where the sides in sequence area,b,c,d, wheres is the semiperimeter, andA andC are two (in fact, any two) opposite angles. This reduces toBrahmagupta's formula for the area of a cyclic quadrilateral—whenA +C = 180°.

Another area formula in terms of the sides and angles, with angleC being between sidesb andc, andA being between sidesa andd, is

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

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

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 angleθ of the diagonals, as long asθ is not90°:^[18]

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

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

Another area formula including the sidesa,b,c,d is^[16]

${\displaystyle K=\frac{1}{2}\sqrt{((a^2+c^2)-2x^2)((b^2+d^2)-2x^2)}\sin \phi }$ 

wherex is the distance between the midpoints of the diagonals, andφ is the angle between thebimedians.

The last trigonometric area formula including the sidesa,b,c,d and the angleα (betweena andb) is:^[19]

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

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

The following two formulas express the area in terms of the sidesa,b,c andd, thesemiperimeters, and the diagonalsp,q:

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

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

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

The area can also be expressed in terms of the bimediansm,n and the diagonalsp,q:

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

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

In fact, any three of the four valuesm,n,p, andq 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]

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

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

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

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

The area of a quadrilateralABCD can be calculated usingvectors. Let vectorsAC andBD form the diagonals fromA toC and fromB toD. The area of the quadrilateral is then

${\displaystyle K=\frac{1}{2}|\mathbf{A}\mathbf{C}\times \mathbf{B}\mathbf{D}|,}$ 

which is half the magnitude of thecross product of vectorsAC andBD. In two-dimensional Euclidean space, expressing vectorAC as afree vector in Cartesian space equal to(x₁,y₁) andBD as(x₂,y₂), this can be rewritten as:

${\displaystyle K=\frac{1}{2}|x_1{y}_2-x_2{y}_1|.}$ 

In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals areperpendicular, and if their diagonals haveequal 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 quadrilateralABCD can be calculated using thelaw 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 quadrilateralABCD, 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

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

wherex is the distance between the midpoints of the diagonals.^{[24]: p.126} This is sometimes known asEuler's quadrilateral theorem and is a generalization of theparallelogram law.

The German mathematicianCarl Anton Bretschneider derived in 1842 the following generalization ofPtolemy'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 alaw of cosines for a quadrilateral. In acyclic quadrilateral, whereA +C = 180°, it reduces topq =ac +bd. Sincecos (A +C) ≥ −1, it also gives a proof of Ptolemy's inequality.

IfX andY are the feet of the normals fromB andD to the diagonalAC =p in a convex quadrilateralABCD with sidesa =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 quadrilateralABCD with sidesa =AB,b =BC,c =CD,d =DA, and where the diagonals intersect atE,

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

wheree =AE,f =BE,g =CE, andh =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 diagonalsp,q and the four side lengthsa,b,c,d of a quadrilateral are related^[14] by theCayley-Mengerdeterminant, as follows:

 

The internalangle bisectors of a convex quadrilateral either form acyclic quadrilateral^{[24]: p.127} (that is, the four intersection points of adjacent angle bisectors areconcyclic) or they areconcurrent. In the latter case the quadrilateral is atangential quadrilateral.

In quadrilateralABCD, if theangle bisectors ofA andC meet on diagonalBD, then the angle bisectors ofB andD meet on diagonalAC.^[31]

Thebimedians of a quadrilateral are the line segments connecting themidpoints of the opposite sides. The intersection of the bimedians is thecentroid of the vertices of the quadrilateral.^[14]

The midpoints of the sides of any quadrilateral (convex, concave or crossed) are the vertices of aparallelogram called theVarignon 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.^[32]
• Theperimeter 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 areconcurrent and are all bisected by their point of intersection.^{[24]: p.125}

In a convex quadrilateral with sidesa,b,c andd, the length of the bimedian that connects the midpoints of the sidesa andc is

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

wherep andq are the length of the diagonals.^[33] The length of the bimedian that connects the midpoints of the sidesb andd is

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

Hence^{[24]: p.126}

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

This is also acorollary to theparallelogram law applied in the Varignon parallelogram.

The lengths of the bimedians can also be expressed in terms of two opposite sides and the distancex between the midpoints of the diagonals. This is possible when using Euler's quadrilateral theorem in the above formulas. Whence^[23]

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

and

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

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

In a convex quadrilateral, there is the followingdual connection between the bimedians and the diagonals:^[29]

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

The four angles of a simple quadrilateralABCD satisfy the following identities:^[34]

${\displaystyle \sin A+\sin B+\sin C+\sin D=4\sin \frac{1}{2}(A+B)\sin \frac{1}{2}(A+C)\sin \frac{1}{2}(A+D)}$ 

and

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

Also,^[35]

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

In the last two formulas, no angle is allowed to be aright angle, since tan 90° is not defined.

Let${\displaystyle a}$ ,${\displaystyle b}$ ,${\displaystyle c}$ ,${\displaystyle d}$  be the sides of a convex quadrilateral,${\displaystyle s}$  is the semiperimeter, and${\displaystyle A}$  and${\displaystyle C}$  are opposite angles, then^[36]

${\displaystyle ad{\sin }^2\frac{1}{2}A+bc{\cos }^2\frac{1}{2}C=(s-a)(s-d)}$ 

and

${\displaystyle bc{\sin }^2\frac{1}{2}C+ad{\cos }^2\frac{1}{2}A=(s-b)(s-c)}$ .

We can use these identities to derive theBretschneider's Formula.

If a convex quadrilateral has the consecutive sidesa,b,c,d and the diagonalsp,q, then its areaK satisfies^[37]

${\displaystyle K\le \frac{1}{4}(a+c)(b+d)}$  with equality only for arectangle.

${\displaystyle K\le \frac{1}{4}(a^2+b^2+c^2+d^2)}$  with equality only for asquare.

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

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

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

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

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

Also,

${\displaystyle K\le \sqrt{abcd},}$ 

with equality for abicentric quadrilateral or a rectangle.

The area of any quadrilateral also satisfies the inequality^[38]

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

Denoting the perimeter asL, we have^{[38]: p.114}

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

with equality only in the case of a square.

The area of a convex quadrilateral also satisfies

${\displaystyle K\le \frac{1}{2}pq}$ 

for diagonal lengthsp andq, with equality if and only if the diagonals are perpendicular.

Leta,b,c,d be the lengths of the sides of a convex quadrilateralABCD with the areaK and diagonalsAC = p,BD = q. Then^[39]

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

Leta,b,c,d be the lengths of the sides of a convex quadrilateralABCD with the areaK, then the following inequality holds:^[40]

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

A corollary to Euler's quadrilateral theorem is the inequality

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

where equality holds if and only if the quadrilateral is aparallelogram.

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

${\displaystyle pq\le ac+bd}$ 

where there is equalityif and only if the quadrilateral is cyclic.^{[24]: p.128–129} This is often calledPtolemy's inequality.

In any convex quadrilateral the bimediansm, n and the diagonalsp, q are related by the inequality

${\displaystyle pq\le m^2+n^2,}$ 

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

The sidesa,b,c, andd of any quadrilateral satisfy^{[42]: p.228, #275}

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

and^{[42]: p.234, #466}

${\displaystyle a^4+b^4+c^4\ge \frac{1}{27}{d}^4.}$ 

Among all quadrilaterals with a givenperimeter, the one with the largest area is thesquare. This is called theisoperimetric theorem for quadrilaterals. It is a direct consequence of the area inequality^{[38]: p.114}

${\displaystyle K\le \frac{1}{16}{L}^2}$ 

whereK is the area of a convex quadrilateral with perimeterL. Equality holdsif 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 themaximum area is thecyclic quadrilateral.^[43]

Of all convex quadrilaterals with given diagonals, theorthodiagonal quadrilateral has the largest area.^{[38]: p.119} This is a direct consequence of the fact that the area of a convex quadrilateral satisfies

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

whereθ is the angle between the diagonalsp andq. Equality holds if and only ifθ = 90°.

IfP is an interior point in a convex quadrilateralABCD, then

${\displaystyle AP+BP+CP+DP\ge AC+BD.}$ 

From this inequality it follows that the point inside a quadrilateral thatminimizes the sum of distances to thevertices is the intersection of the diagonals. Hence that point is theFermat point of a convex quadrilateral.^{[44]: p.120}

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 justcentroid (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.^[45]

The "vertex centroid" is the intersection of the twobimedians.^[46] As with any polygon, thex andy coordinates of the vertex centroid are thearithmetic means of thex andy coordinates of the vertices.

The "area centroid" of quadrilateralABCD can be constructed in the following way. LetG_a,G_b,G_c,G_d be the centroids of trianglesBCD,ACD,ABD,ABC respectively. Then the "area centroid" is the intersection of the linesG_aG_c andG_bG_d.^[47]

In a general convex quadrilateralABCD, there are no natural analogies to thecircumcenter andorthocenter of atriangle. But two such points can be constructed in the following way. LetO_a,O_b,O_c,O_d be the circumcenters of trianglesBCD,ACD,ABD,ABC respectively; and denote byH_a,H_b,H_c,H_d the orthocenters in the same triangles. Then the intersection of the linesO_aO_c andO_bO_d is called thequasicircumcenter, and the intersection of the linesH_aH_c andH_bH_d is called thequasiorthocenter of the convex quadrilateral.^[47] These points can be used to define anEuler line of a quadrilateral. In a convex quadrilateral, the quasiorthocenterH, the "area centroid"G, and the quasicircumcenterO arecollinear in this order, andHG = 2GO.^[47]

There can also be defined aquasinine-point centerE as the intersection of the linesE_aE_c andE_bE_d, whereE_a,E_b,E_c,E_d are thenine-point centers of trianglesBCD,ACD,ABD,ABC respectively. ThenE is themidpoint ofOH.^[47]

Another remarkable line in a convex non-parallelogram quadrilateral is theNewton line, which connects the midpoints of the diagonals, the segment connecting these points being bisected by the vertex centroid. One more interesting line (in some sense dual to theNewton's one) is the line connecting the point of intersection of diagonals with the vertex centroid. The line is remarkable by the fact that it contains the (area) centroid. The vertex centroid divides the segment connecting the intersection of diagonals and the (area) centroid in the ratio 3:1.^[48]

For any quadrilateralABCD with pointsP andQ the intersections ofAD andBC andAB andCD, respectively, the circles(PAB), (PCD), (QAD), and(QBC) pass through a common pointM, called a Miquel point.^[49]

For a convex quadrilateralABCD in whichE is the point of intersection of the diagonals andF is the point of intersection of the extensions of sidesBC andAD, let ω be a circle throughE andF which meetsCB internally atM andDA internally atN. LetCA meet ω again atL and letDB meet ω again atK. Then, applyingPascal's theorem to the hexagonsEKNFML andEKMFNL inscribed in ω, there holds: the straight linesNK andML intersect at pointP that is located on the sideAB; the straight linesNL andKM intersect at pointQ that is located on the sideCD. PointsP andQ are called "Pascal points" formed by circle ω on sidesAB andCD.^[50][51][52]

• If exterior squares are drawn on all sides of a quadrilateral then the segments connecting thecenters of opposite squares are (a) equal in length, and (b)perpendicular. Thus these centers are the vertices of anorthodiagonal quadrilateral. This is calledVan Aubel's theorem.
• For any simple quadrilateral with given edge lengths, there is acyclic quadrilateral with the same edge lengths.^[43]
• 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.^[53]
• Let vectorsAC andBD form the diagonals fromA toC and fromB toD. The angle${\displaystyle \theta }$  at the intersection of the diagonals satisfies$${\displaystyle \cos \theta =\frac{b^2+d^2-a^2-c^2}{2pq},}$$  where${\displaystyle \theta }$  is the angle betweenAC andBD, and${\displaystyle p,q}$  are the diagonals of the quadrilateral.^[54]

A hierarchicaltaxonomy 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.

A non-planar quadrilateral is called askew 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 ascyclobutane that contain a "puckered" ring of four atoms.^[55] Historically the termgauche quadrilateral was also used to mean a skew quadrilateral.^[56] 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 oppositeedges is removed.

• Complete quadrangle
• Perpendicular bisector construction of a quadrilateral
• Saccheri quadrilateral
• Types of mesh § Quadrilateral
• Quadrangle (geography)
• Homography - Any quadrilateral can be transformed into another quadrilateral by a projective transformation (homography)

• "Quadrangle, complete",Encyclopedia of Mathematics,EMS Press, 2001 [1994]
• Quadrilaterals Formed by Perpendicular Bisectors,Projective Collinearity andInteractive Classification of Quadrilaterals fromcut-the-knot
• Definitions and examples of quadrilaterals andDefinition and properties of tetragons from Mathopenref
• A (dynamic) Hierarchical Quadrilateral Tree atDynamic Geometry Sketches
• An extended classification of quadrilateralsArchived 2019-12-30 at theWayback Machine atDynamic Math Learning HomepageArchived 2018-08-25 at theWayback Machine
• The role and function of a hierarchical classification of quadrilaterals by Michael de Villiers