Editing Quadrilateral (section)

==Inequalities==
===Area===
If a convex quadrilateral has the consecutive sides ''a'', ''b'', ''c'', ''d'' and the diagonals ''p'', ''q'', then its area ''K'' satisfies<ref>O. Bottema, ''Geometric Inequalities'', Wolters–Noordhoff Publishing, The Netherlands, 1969, pp. 129, 132.</ref>
:<math>K\le \tfrac{1}{4}(a+c)(b+d)</math> with equality only for a [[rectangle]].
:<math>K\le \tfrac{1}{4}(a^2+b^2+c^2+d^2)</math> with equality only for a [[square]].
:<math>K\le \tfrac{1}{4}(p^2+q^2)</math> with equality only if the diagonals are perpendicular and equal.
:<math>K\le \tfrac{1}{2}\sqrt{(a^2+c^2)(b^2+d^2)}</math> with equality only for a rectangle.<ref name=Josefsson4/>

From [[Bretschneider's formula]] it directly follows that the area of a quadrilateral satisfies
:<math>K \le \sqrt{(s-a)(s-b)(s-c)(s-d)}</math>

with equality [[if and only if]] the quadrilateral is [[cyclic quadrilateral|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).

Also,
:<math>K \leq \sqrt{abcd},</math>
with equality for a [[bicentric quadrilateral]] or a rectangle.

The area of any quadrilateral also satisfies the inequality<ref name=Alsina>{{citation|last1=Alsina|first1=Claudi|last2=Nelsen|first2=Roger|title=When Less is More: Visualizing Basic Inequalities|publisher=Mathematical Association of America|year=2009|page=68}}.</ref>
:<math>\displaystyle K\le \tfrac{1}{2}\sqrt[3]{(ab+cd)(ac+bd)(ad+bc)}.</math>

Denoting the perimeter as ''L'', we have<ref name=Alsina/>{{rp|p.114}}
:<math>K\le \tfrac{1}{16}L^2,</math>

with equality only in the case of a square.

The area of a convex quadrilateral also satisfies
:<math>K \le \tfrac{1}{2}pq</math>

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<ref>Dao Thanh Oai, Leonard Giugiuc, Problem 12033, American Mathematical Monthly, March 2018, p. 277</ref>

:<math> K \leq \tfrac18(a^2+b^2+c^2+d^2+p^2+q^2+pq-ac-bd) </math> 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:<ref>{{cite journal|author1=Leonard Mihai Giugiuc|author2=Dao Thanh Oai|author3=Kadir Altintas|title=An inequality related to the lengths and area of a convex quadrilateral|journal=International Journal of Geometry|volume=7|date=2018|pages=81–86|url=https://ijgeometry.com/wp-content/uploads/2018/04/81-86.pdf}}</ref>

:<math> 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) </math> with equality only for a square.

===Diagonals and bimedians===
A corollary to Euler's quadrilateral theorem is the inequality
:<math> a^2 + b^2 + c^2 + d^2 \ge p^2 + q^2 </math>

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

[[Leonhard Euler|Euler]] also generalized [[Ptolemy's theorem]], which is an equality in a [[cyclic quadrilateral]], into an inequality for a convex quadrilateral. It states that
:<math> pq \le ac + bd </math>

where there is equality [[if and only if]] the quadrilateral is cyclic.<ref name=Altshiller-Court/>{{rp|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
:<math>pq \leq m^2+n^2,</math>

with equality holding if and only if the diagonals are equal.<ref name=J2014>{{cite journal |last=Josefsson |first=Martin |title=Properties of equidiagonal quadrilaterals |journal=Forum Geometricorum |volume=14 |year=2014 |pages=129–144 |url=http://forumgeom.fau.edu/FG2014volume14/FG201412index.html |access-date=2014-08-28 |archive-date=2024-06-05 |archive-url=https://web.archive.org/web/20240605032351/https://forumgeom.fau.edu/FG2014volume14/FG201412index.html |url-status=dead }}</ref>{{rp|Prop.1}} This follows directly from the quadrilateral identity <math>m^2+n^2=\tfrac{1}{2}(p^2+q^2).</math>

===Sides===
The sides ''a'', ''b'', ''c'', and ''d'' of any quadrilateral satisfy<ref name=Crux>{{cite web|url=http://www.imomath.com/othercomp/Journ/ineq.pdf|title=Inequalities proposed in ''Crux Mathematicorum'' (from vol. 1, no. 1 to vol. 4, no. 2 known as "Eureka")|website=Imomath.com|access-date=March 1, 2022}}</ref>{{rp|p.228,#275}}
:<math>a^2+b^2+c^2 > \tfrac13 d^2</math>

and<ref name=Crux/>{{rp|p.234,#466}}
:<math>a^4+b^4+c^4 \geq \tfrac1{27} d^4.</math>