Editing Quadrilateral (section)

==Maximum and minimum properties==
Among all quadrilaterals with a given [[perimeter]], the one with the largest area is the [[Square (geometry)|square]]. This is called the ''[[isoperimetric inequality|isoperimetric theorem]] for quadrilaterals''. It is a direct consequence of the area inequality<ref name=Alsina/>{{rp|p.114}}
:<math>K\le \tfrac{1}{16}L^2</math>

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 [[Maxima and minima|maximum]] area is the [[cyclic quadrilateral]].<ref name=Peter/>

Of all convex quadrilaterals with given diagonals, the [[orthodiagonal quadrilateral]] has the largest area.<ref name=Alsina/>{{rp|p.119}} This is a direct consequence of the fact that the area of a convex quadrilateral satisfies
:<math>K=\tfrac{1}{2}pq\sin{\theta}\le \tfrac{1}{2}pq,</math>

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
:<math>AP+BP+CP+DP\ge AC+BD.</math>

From this inequality it follows that the point inside a quadrilateral that [[Maxima and minima|minimizes]] the sum of distances to the [[Vertex (geometry)|vertices]] is the intersection of the diagonals. Hence that point is the [[Fermat point]] of a convex quadrilateral.<ref name=autogenerated1>{{cite book |last1=Alsina |first1=Claudi |last2=Nelsen |first2=Roger |title=Charming Proofs : A Journey Into Elegant Mathematics |publisher=Mathematical Association of America |year=2010 |pages=114, 119, 120, 261 |isbn=978-0-88385-348-1 }}</ref>{{rp|p.120}}