InEuclidean geometry, anorthodiagonal quadrilateral is aquadrilateral in which thediagonals cross atright angles. In other words, it is a four-sided figure in which theline segments between non-adjacentvertices areorthogonal (perpendicular) to each other.

Akite is an orthodiagonal quadrilateral in which one diagonal is aline of symmetry. The kites are exactly the orthodiagonal quadrilaterals that contain acircletangent to all four of their sides; that is, the kites are thetangential orthodiagonal quadrilaterals.^[1]

Arhombus is an orthodiagonal quadrilateral with two pairs ofparallel sides (that is, an orthodiagonal quadrilateral that is also aparallelogram).

Asquare is a limiting case of both a kite and a rhombus.

Orthodiagonal quadrilaterals that are alsoequidiagonal quadrilaterals are calledmidsquare quadrilaterals.^[2]

For any orthodiagonal quadrilateral, the sum of the squares of two opposite sides equals that of the other two opposite sides: for successive sidesa,b,c, andd, we have^[3][4]

${\displaystyle {\displaystyle a^2+c^2=b^2+d^2.}}$

This follows from thePythagorean 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 whicha² +c² =b² +d² must be orthodiagonal.^[5]This can beproved in a number of ways, including using thelaw of cosines,vectors, anindirect proof, andcomplex numbers.^[6]

The diagonals of aconvex quadrilateral are perpendicularif and only if the twobimedians have equal length.^[6]

According to another characterization, the diagonals of a convex quadrilateralABCD are perpendicular if and only if

${\displaystyle \mathrm{\angle }PAB+\mathrm{\angle }PBA+\mathrm{\angle }PCD+\mathrm{\angle }PDC=\pi }$

whereP 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 theprojections of the diagonal intersection onto the sides of the quadrilateral are the vertices of acyclic quadrilateral.^[6]

A convex quadrilateral is orthodiagonal if and only if itsVarignon parallelogram (whose vertices are themidpoints of its sides) is arectangle.^[6] A related characterization states that a convex quadrilateral is orthodiagonal if and only if the midpoints of the sides and the feet of the fourmaltitudes are eightconcyclic points; theeight point circle. The center of this circle is thecentroid of the quadrilateral. The quadrilateral formed by the feet of the maltitudes is called theprincipal orthic quadrilateral.^[7]

If thenormals to the sides of a convex quadrilateralABCD through the diagonal intersection intersect the opposite sides inR,S,T,U, andK,L,M,N are the feet of these normals, thenABCD is orthodiagonal if and only if the eight pointsK,L,M,N,R,S,T andU are concyclic; thesecond eight point circle. A related characterization states that a convex quadrilateral is orthodiagonal if and only ifRSTU is a rectangle whose sides are parallel to the diagonals ofABCD.^[6]

There are several metric characterizations regarding the fourtriangles formed by the diagonal intersectionP and the vertices of a convex quadrilateralABCD. Denote bym₁,m₂,m₃,m₄ themedians in trianglesABP,BCP,CDP,DAP fromP to the sidesAB,BC,CD,DA respectively. IfR₁,R₂,R₃,R₄ andh₁,h₂,h₃,h₄ denote theradii of thecircumcircles and thealtitudes respectively of these triangles, then the quadrilateralABCD is orthodiagonal if and only if any one of the following equalities holds:^[6]

• ${\displaystyle m_1^2+m_3^2=m_2^2+m_4^2}$
• ${\displaystyle R_1^2+R_3^2=R_2^2+R_4^2}$
• ${\displaystyle \frac{1}{h_1^2}+\frac{1}{h_3^2}=\frac{1}{h_2^2}+\frac{1}{h_4^2}}$

Furthermore, a quadrilateralABCD with intersectionP of the diagonals is orthodiagonal if and only if thecircumcenters of the trianglesABP,BCP,CDP andDAP are the midpoints of the sides of the quadrilateral.^[6]

A few metric characterizations oftangential quadrilaterals and orthodiagonal quadrilaterals are very similar in appearance, as can be seen in this table.^[6] The notations on the sidesa,b,c,d, thecircumradiiR₁,R₂,R₃,R₄, and the altitudesh₁,h₂,h₃,h₄ are the same as above in both types of quadrilaterals.

Tangential quadrilateral	Orthodiagonal quadrilateral
${\displaystyle a+c=b+d}$	${\displaystyle a^2+c^2=b^2+d^2}$
${\displaystyle R_1+R_3=R_2+R_4}$	${\displaystyle R_1^2+R_3^2=R_2^2+R_4^2}$
${\displaystyle \frac{1}{h_1}+\frac{1}{h_3}=\frac{1}{h_2}+\frac{1}{h_4}}$	${\displaystyle \frac{1}{h_1^2}+\frac{1}{h_3^2}=\frac{1}{h_2^2}+\frac{1}{h_4^2}}$

The areaK of an orthodiagonal quadrilateral equals one half the product of the lengths of the diagonalsp andq:^[8]

${\displaystyle K=\frac{pq}{2}.}$

Conversely, any convex quadrilateral where the area can be calculated with this formula must be orthodiagonal.^[6] The orthodiagonal quadrilateral has the biggest area of all convex quadrilaterals with given diagonals.

• Orthodiagonal quadrilaterals are the only quadrilaterals for which the sides and theangle formed by the diagonals do not uniquely determine the area.^[4] For example, two rhombi both having common sidea (and, as for all rhombi, both having a right angle between the diagonals), but one having a smalleracute angle than the other, have different areas (the area of the former approaching zero as the acute angle approaches zero).
• If squares are erected outward on the sides of any quadrilateral (convex, concave, or crossed), then theircentres (centroids) are the vertices of an orthodiagonal quadrilateral that is alsoequidiagonal (that is, having diagonals of equal length). This is calledVan Aubel's theorem.
• Each side of an orthodiagonal quadrilateral has at least one common point with thePascal points circle.^[9]

For acyclic orthodiagonal quadrilateral (one that can beinscribed in a circle), suppose the intersection of the diagonals divides one diagonal into segments of lengthsp₁ andp₂ and divides the other diagonal into segments of lengthsq₁ andq₂. Then^[10] (the first equality is Proposition 11 inArchimedes'Book of Lemmas)

${\displaystyle D^2=p_1^2+p_2^2+q_1^2+q_2^2=a^2+c^2=b^2+d^2}$

whereD is thediameter of the circumcircle. This holds because the diagonals are perpendicularchords of a circle. These equations yield the circumradius expression

${\displaystyle R=\frac{1}{2}\sqrt{p_1^2+p_2^2+q_1^2+q_2^2}}$

or, in terms of the sides of the quadrilateral, as^[3]

${\displaystyle R=\frac{1}{2}\sqrt{a^2+c^2}=\frac{1}{2}\sqrt{b^2+d^2}.}$

It also follows that^[3]

${\displaystyle a^2+b^2+c^2+d^2=8R^2.}$

Thus, according toEuler's quadrilateral theorem, the circumradius can be expressed in terms of the diagonalsp andq, and the distancex between the midpoints of the diagonals as

${\displaystyle R=\sqrt{\frac{p^2+q^2+4x^2}{8}}.}$

A formula for the areaK of a cyclic orthodiagonal quadrilateral in terms of the four sides is obtained directly when combiningPtolemy's theorem and the formula for thearea of an orthodiagonal quadrilateral. The result is^{[11]: p.222}

${\displaystyle K=\frac{1}{2}(ac+bd).}$

• In a cyclic orthodiagonal quadrilateral, theanticenter coincides with the point where the diagonals intersect.^[3]
• 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.^[3]
• If an orthodiagonal quadrilateral is also cyclic, the distance from thecircumcenter (the center of the circumscribed circle) to any side equals half the length of the opposite side.^[3]
• 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.^[3]

For every orthodiagonal quadrilateral, we can inscribe two infinite sets of rectangles:

(i) a set of rectangles whose sides are parallel to the diagonals of the quadrilateral

(ii) a set of rectangles defined by Pascal-points circles.^[12]

• A Van Aubel like property of an Orthodiagonal Quadrilateral atDynamic Geometry Sketches, interactive geometry sketches.

• Types of quadrilaterals

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