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[[File:Orthodiagonal quadrilateral.svg|thumb|240px|An orthodiagonal quadrilateral. According to the characterization of these quadrilaterals, the two red squares on two opposite sides of the quadrilateral have the same total area as the two blue squares on the other pair of opposite sides.]]
In [[Euclidean geometry]], an '''orthodiagonal quadrilateral''' is a [[quadrilateral]] in which the [[diagonal]]s cross at [[right angle]]s. In other words, it is a four-sided figure in which the [[line segment]]s between non-adjacent [[vertex (geometry)|vertices]] are [[orthogonal]] to each other.

==Special cases==

A [[kite (geometry)|kite]] is an orthodiagonal quadrilateral in which one diagonal is a line of symmetry. The kites are exactly the orthodiagonal quadrilaterals that contain a [[circle]] tangent to all four of their sides; that is, the kites are the [[tangential quadrilateral|tangential]] orthodiagonal quadrilaterals.<ref>{{citation
 | last = Josefsson | first = Martin
 | journal = Forum Geometricorum
 | pages = 119–130
 | title = Calculations concerning the tangent lengths and tangency chords of a tangential quadrilateral
 | url = http://forumgeom.fau.edu/FG2010volume10/FG201013.pdf
 | volume = 10
 | year = 2010}}.</ref>

A [[rhombus]] is an orthodiagonal quadrilateral with two pairs of parallel sides (that is, an orthodiagonal quadrilateral that is also a [[parallelogram]]).

A [[square (geometry)|square]] is a limiting case of both a kite and a rhombus.

Orthodiagonal and [[Equidiagonal quadrilateral|equidiagonal]] quadrilaterals in which the diagonals are at least as long as all of the quadrilateral's sides have the maximum area for their diameter among all quadrilaterals, solving the ''n''&nbsp;=&nbsp;4 case of the [[biggest little polygon]] problem. The square is one such quadrilateral, but there are infinitely many others.

==Characterizations==
For any orthodiagonal quadrilateral, the sum of the squares of two opposite sides equals that of the other two opposite sides: for successive sides ''a'', ''b'', ''c'', and ''d'', we have <ref name=Altshiller-Court>{{citation|last=Altshiller-Court|first=N.|title=College Geometry|publisher=Dover Publications|year=2007}}. Republication of second edition, 1952, Barnes & Noble, pp. 136-138.</ref><ref name=Mitchell>Mitchell, {{citation|last=Douglas|first=W.|title=The area of a quadrilateral|journal=Mathematical Gazette|volume=93|year=2009|pages=306–309}}.</ref>
:<math>\displaystyle a^2+c^2=b^2+d^2. </math>

This follows from the [[Pythagorean theorem]], by which either of these two sums of two squares can be expanded to equal the sum of the four squared distances from the quadrilateral's vertices to the point where the diagonals intersect. Conversely, any quadrilateral in which ''a''<sup>2</sup> + ''c''<sup>2</sup> = ''b''<sup>2</sup> + ''d''<sup>2</sup> must be orthodiagonal.<ref>{{citation|url=http://forumgeom.fau.edu/FG2009volume9/FG200919.pdf|title=Class preserving dissections of convex quadrilaterals|journal=Forum Geometricorum|volume=9|year=2009|pages=195–211|first1=Dan|last1=Ismailescu|first2=Adam|last2=Vojdany}}.</ref> This can be proved in a number of ways, including using the [[law of cosines]], [[Euclidean vector|vectors]], an [[Proof by contradiction|indirect proof]], and [[complex number]]s.<ref name=Josefsson2>{{citation
 | last = Josefsson | first = Martin
 | journal = Forum Geometricorum
 | pages = 13–25
 | title = Characterizations of Orthodiagonal Quadrilaterals
 | url = http://forumgeom.fau.edu/FG2012volume12/FG201202.pdf
 | volume = 12
 | year = 2012}}.</ref>

The diagonals of a convex quadrilateral are perpendicular [[if and only if]] the two [[Quadrilateral#Special line segments|bimedians]] have equal length.<ref name=Josefsson2/>

According to another characterization, the diagonals of a convex quadrilateral ''ABCD'' are perpendicular if and only if
:<math>\angle PAB + \angle PBA + \angle PCD + \angle PDC = \pi</math>

where ''P'' is the point of intersection of the diagonals. From this equation it follows almost immediately that the diagonals of a convex quadrilateral are perpendicular if and only if the [[Projection (mathematics)|projections]] of the diagonal intersection onto the sides of the quadrilateral are the vertices of a [[cyclic quadrilateral]].<ref name=Josefsson2/>

A convex quadrilateral is orthodiagonal if and only if the [[midpoint]]s of the sides are the vertices of a [[rectangle]].<ref name=Josefsson2/> A related characterization states that a convex quadrilateral is orthodiagonal if and only if the midpoints of the sides and the feet of the four [[Quadrilateral#Special line segments|maltitudes]] are eight [[concyclic points]]; the ''eight point circle''. The center of this circle is the [[Quadrilateral#Special line segments|centroid]] of the quadrilateral. The quadrilateral formed by the feet of the maltitudes is called the ''principal orthic quadrilateral''.<ref>{{citation
 | last1 = Mammana | first1 = Maria Flavia | last2 = Micale | first2 = Biagio | last3 = Pennisi | first3 = Mario
 | journal = Forum Geometricorum
 | pages = 109–119
 | title = The Droz-Farny Circles of a Convex Quadrilateral
 | url = http://forumgeom.fau.edu/FG2011volume11/FG201111.pdf
 | volume = 11
 | year = 2011}}.</ref>

If the [[Normal (geometry)|normals]] to the sides of a convex quadrilateral ''ABCD'' through the diagonal intersection intersect the opposite sides in ''R'', ''S'', ''T'', ''U'', and ''K'', ''L'', ''M'', ''N'' are the feet of these normals, then ''ABCD'' is orthodiagonal if and only if the eight points ''K'', ''L'', ''M'', ''N'', ''R'', ''S'', ''T'' and ''U'' are concyclic; the ''second eight point circle''. A related characterization states that a convex quadrilateral is orthodiagonal if and only if ''RSTU'' is a rectangle whose sides are [[Parallel (geometry)|parallel]] to the diagonals of ''ABCD''.<ref name=Josefsson2/>

There are several metric characterizations regarding the four [[triangle]]s formed by the diagonal intersection ''P'' and the vertices of a convex quadrilateral ''ABCD''. Denote by ''m''<sub>1</sub>, ''m''<sub>2</sub>, ''m''<sub>3</sub>, ''m''<sub>4</sub> the [[Median (geometry)|medians]] in triangles ''ABP'', ''BCP'', ''CDP'', ''DAP'' from ''P'' to the sides ''AB'', ''BC'', ''CD'', ''DA'' respectively. If ''R''<sub>1</sub>, ''R''<sub>2</sub>, ''R''<sub>3</sub>, ''R''<sub>4</sub> and ''h''<sub>1</sub>, ''h''<sub>2</sub>, ''h''<sub>3</sub>, ''h''<sub>4</sub> denote the [[radius|radii]] of the [[circumcircle]]s and the [[Altitude (triangle)|altitude]]s respectively of these triangles, then the quadrilateral ''ABCD'' is orthodiagonal if and only if any one of the following equalities holds:<ref name=Josefsson2/>
* <math>m_1^2+m_3^2=m_2^2+m_4^2</math>
* <math>R_1^2+R_3^2=R_2^2+R_4^2</math>
* <math>\frac{1}{h_1^2}+\frac{1}{h_3^2}=\frac{1}{h_2^2}+\frac{1}{h_4^2}</math>

Furthermore, a quadrilateral ''ABCD''  with intersection ''P'' of the diagonals is orthodiagonal if and only if the circumcenters of the triangles ''ABP'', ''BCP'', ''CDP'' and ''DAP'' are the midpoints of the sides of the quadrilateral.<ref name=Josefsson2/> 

===Comparison with a tangential quadrilateral===
A few metric characterizations of [[tangential quadrilateral]]s and orthodiagonal quadrilaterals are very similar in appearance, as can be seen in this table.<ref name=Josefsson2/> The notations on the sides ''a'', ''b'', ''c'', ''d'', the circumradii ''R''<sub>1</sub>, ''R''<sub>2</sub>, ''R''<sub>3</sub>, ''R''<sub>4</sub>, and the altitudes ''h''<sub>1</sub>, ''h''<sub>2</sub>, ''h''<sub>3</sub>, ''h''<sub>4</sub> are the same as above in both types of quadrilaterals.

{| class=wikitable
|-
! Tangential quadrilateral
! Orthodiagonal quadrilateral
|-
| align=center|<math>a+c=b+d</math>
| align=center|<math>a^2+c^2=b^2+d^2</math>
|-
| align=center|<math>R_1+R_3=R_2+R_4</math>
| align=center|<math>R_1^2+R_3^2=R_2^2+R_4^2</math>
|-
| align=center|<math>\frac{1}{h_1}+\frac{1}{h_3}=\frac{1}{h_2}+\frac{1}{h_4}</math>
| align=center|<math>\frac{1}{h_1^2}+\frac{1}{h_3^2}=\frac{1}{h_2^2}+\frac{1}{h_4^2}</math>
|}

==Area==
The area ''K'' of an orthodiagonal quadrilateral equals one half the product of the lengths of the diagonals ''p'' and ''q'':<ref>Harries, J. "Area of a quadrilateral," ''Mathematical Gazette'' 86, July 2002, 310–311.</ref>
:<math>K=\frac{p\cdot q}{2}.</math>

Conversely, any convex quadrilateral where the area can be calculated with this formula must be orthodiagonal.<ref name=Josefsson2/> The orthodiagonal quadrilateral has the biggest area of all convex quadrilaterals with given diagonals.

==Other properties==

*Orthodiagonal quadrilaterals are the only quadrilaterals for which the sides and the angle formed by the diagonals do not uniquely determine the area.<ref name=Mitchell/>  For example, two rhombi both having common side ''a'' (and, as for all rhombi, both having a right angle between the diagonals), but one having a smaller [[acute angle]] than the other, have different areas (the area of the former approaching zero as the acute angle approaches zero).

*If [[Square (geometry)|square]]s are erected outward on the sides of any [[quadrilateral]] (convex, concave, or crossed), then their [[Centre (geometry)|centre]]s ([[centroid]]s) are the vertices of an orthodiagonal quadrilateral with diagonals of equal length. This is called [[Van Aubel's theorem]].

==Properties of orthodiagonal quadrilaterals that are also cyclic==
===Circumradius and area===
For a [[Cyclic quadrilateral|cyclic]] orthodiagonal quadrilateral (one that can be [[inscribed]] in a [[circle]]), suppose the intersection of the diagonals divides one diagonal into segments of lengths ''p''<sub>1</sub> and ''p''<sub>2</sub> and divides the other diagonal into segments of lengths ''q''<sub>1</sub> and ''q''<sub>2</sub>. Then<ref>Posamentier, Alfred S., and Charles T. Salkind, ''Challenging Problems in Geometry'', Dover Publ., second edition, 1996:pp. 104–105, #4–23.</ref> (the first equality is Proposition 11 in [[Archimedes]] [[Book of Lemmas]])
:<math> D^2=p_1^2+p_2^2+q_1^2+q_2^2=a^2+c^2=b^2+d^2 </math>

where ''D'' is the [[diameter]] of the [[circumcircle]].  This holds because the diagonals are perpendicular [[Circle#Chord|chords of a circle]]. These equations yields that the [[circumradius]] ''R'' can be expressed as
:<math> R=\tfrac{1}{2}\sqrt{p_1^2+p_2^2+q_1^2+q_2^2} </math>

or, in terms of the sides of the quadrilateral, as
:<math> R=\tfrac{1}{2}\sqrt{a^2+c^2}=\tfrac{1}{2}\sqrt{b^2+d^2}. </math>

It also follows that
:<math> a^2+b^2+c^2+d^2=8R^2. </math>

Thus, according to [[Quadrilateral#Diagonals|Euler's quadrilateral theorem]], the circumradius can be expressed in terms of the diagonals ''p'' and ''q'', and the distance ''x'' between the midpoints of the diagonals as
:<math> R=\sqrt{\frac{p^2+q^2+4x^2}{8}}. </math>

A formula for the [[area]] ''K'' of a cyclic orthodiagonal quadrilateral in terms of the four sides is obtained directly when combining [[Cyclic quadrilateral#Diagonals|Ptolemy's theorem]] and the formula for the [[Orthodiagonal quadrilateral#Area|area of an orthodiagonal quadrilateral]]. The result is
:<math> K=\tfrac{1}{2}(ac+bd). </math>

===Other properties===
*In a cyclic orthodiagonal quadrilateral, the [[Cyclic quadrilateral#Other properties|anticenter]] coincides with the point where the diagonals intersect.<ref name=Altshiller-Court/>

*[[Brahmagupta's theorem]] states that for a cyclic orthodiagonal quadrilateral, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side.<ref name=Altshiller-Court/>

*If an orthodiagonal quadrilateral is also cyclic, the distance from the [[circumcenter]] (the center of the circumscribed circle) to any side equals half the length of the opposite side.<ref name=Altshiller-Court/>

*In a cyclic orthodiagonal quadrilateral, the distance between the midpoints of the diagonals equals the distance between the circumcenter and the point where the diagonals intersect.<ref name=Altshiller-Court/>

==References==

{{reflist}}

[[Category:Quadrilaterals]]