Orthodiagonal quadrilateral

A 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 orthodiagonal quadrilaterals.[1]

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

A square is a limiting case of both a kite and a rhombus.

Orthodiagonal 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 = 4 case of the biggest little polygon problem. The square is one such quadrilateral, but there are infinitely many others. An orthodiagonal quadrilateral that is also equidiagonal is a midsquare quadrilateral because its Varignon parallelogram is a square. Its area can be expressed purely in terms of its sides.

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 [2][3]

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

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 a² + c² = b² + d² must be orthodiagonal.[4] This can be proved in a number of ways, including using the law of cosines, vectors, an indirect proof, and complex numbers.[5]

The diagonals of a convex quadrilateral are perpendicular if and only if the two bimedians have equal length.[5]

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

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

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 projections of the diagonal intersection onto the sides of the quadrilateral are the vertices of a cyclic quadrilateral.[5]

An orthodiagonal quadrilateral ABCD (in blue). The Varignon parallelogram (in green) formed by the midpoints of the edges of ABCD is a rectangle. Additionally, the four midpoints (grey) and the four feet of the maltitudes (red) are cocyclic on the 8-point-circle.

A convex quadrilateral is orthodiagonal if and only if its Varignon parallelogram (whose vertices are the midpoints of its sides) is a rectangle.[5] 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 maltitudes are eight concyclic points; the eight point circle. The center of this circle is the centroid of the quadrilateral. The quadrilateral formed by the feet of the maltitudes is called the principal orthic quadrilateral.[6]

A second 8-point circle can be constructed from an orthodiagonal quadrilateral ABCD (in blue). The lines perpendicular to each side through the intersection of the diagonals intersect the sides in 8 different points, which are all cocyclic.

If the 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 to the diagonals of ABCD.[5]

There are several metric characterizations regarding the four triangles formed by the diagonal intersection P and the vertices of a convex quadrilateral ABCD. Denote by m₁, m₂, m₃, m₄ the medians in triangles ABP, BCP, CDP, DAP from P to the sides AB, BC, CD, DA respectively. If R₁, R₂, R₃, R₄ and h₁, h₂, h₃, h₄ denote the radii of the circumcircles and the altitudes respectively of these triangles, then the quadrilateral ABCD is orthodiagonal if and only if any one of the following equalities holds:[5]

• ${\displaystyle m_{1}^{2}+m_{3}^{2}=m_{2}^{2}+m_{4}^{2}}$
• ${\displaystyle R_{1}^{2}+R_{3}^{2}=R_{2}^{2}+R_{4}^{2}}$
• ${\displaystyle {\frac {1}{h_{1}^{2}}}+{\frac {1}{h_{3}^{2}}}={\frac {1}{h_{2}^{2}}}+{\frac {1}{h_{4}^{2}}}}$

Furthermore, a 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.[5]

Comparison with a tangential quadrilateral

A few metric characterizations of tangential quadrilaterals and orthodiagonal quadrilaterals are very similar in appearance, as can be seen in this table.[5] The notations on the sides a, b, c, d, the circumradii R₁, R₂, R₃, R₄, and the altitudes h₁, 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 area K of an orthodiagonal quadrilateral equals one half the product of the lengths of the diagonals p and q:[7]

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

Conversely, any convex quadrilateral where the area can be calculated with this formula must be orthodiagonal.[5] 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 the angle formed by the diagonals do not uniquely determine the area.[3] 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 squares are erected outward on the sides of any quadrilateral (convex, concave, or crossed), then their centres (centroids) are the vertices of an orthodiagonal quadrilateral that is also equidiagonal (that is, having diagonals of equal length). This is called Van Aubel's theorem.
• Each side of an orthodiagonal quadrilateral has at least one common point with the Pascal points circle.[8]

${\displaystyle ABCD}$ is an orthodiagonal quadrilateral, ${\displaystyle P_{1}X_{1}Z_{1}Y_{1}}$ and ${\displaystyle P_{2}X_{2}Z_{2}Y_{2}}$ are rectangles whose sides are parallel to the diagonals of the quadrilateral.

${\displaystyle ABCD}$ is an orthodiagonal quadrilateral. ${\displaystyle P_{1}}$ and ${\displaystyle Q_{1}}$ are Pascal points formed by the circle ${\displaystyle \omega _{1}}$, ${\displaystyle \sigma _{P_{1}Q_{1}}}$ is Pascal-points circle which defines the rectangle ${\displaystyle P_{1}V_{1}Q_{1}W_{1}}$. ${\displaystyle P_{2}}$ and ${\displaystyle Q_{2}}$ are Pascal points formed by the circle ${\displaystyle \omega _{2}}$, ${\displaystyle \sigma _{P_{2}Q_{2}}}$ is Pascal-points circle which defines the rectangle ${\displaystyle P_{2}V_{2}Q_{2}W_{2}}$.

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.[11]