Equidiagonal quadrilateral

Examples of equidiagonal quadrilaterals include the isosceles trapezoids, rectangles and squares.

Among all quadrilaterals, the shape that has the greatest ratio of its perimeter to its diameter is an equidiagonal kite with angles π/3, 5π/12, 5π/6, and 5π/12.^[2]

A convex quadrilateral is equidiagonal if and only if its Varignon parallelogram, the parallelogram formed by the midpoints of its sides, is a rhombus. An equivalent condition is that the bimedians of the quadrilateral (the diagonals of the Varignon parallelogram) are perpendicular.^[3]

A convex quadrilateral with diagonal lengths ${\displaystyle p}$ and ${\displaystyle q}$ and bimedian lengths ${\displaystyle m}$ and ${\displaystyle n}$ is equidiagonal if and only if^{[4]: Prop.1}

${\displaystyle pq=m^{2}+n^{2}.}$

The area K of an equidiagonal quadrilateral can easily be calculated if the length of the bimedians m and n are known. A quadrilateral is equidiagonal if and only if^{[5]: p.19,   [4]: Cor.4}

${\displaystyle \displaystyle K=mn.}$

This is a direct consequence of the fact that the area of a convex quadrilateral is twice the area of its Varignon parallelogram and that the diagonals in this parallelogram are the bimedians of the quadrilateral. Using the formulas for the lengths of the bimedians, the area can also be expressed in terms of the sides a, b, c, d of the equidiagonal quadrilateral and the distance x between the midpoints of the diagonals as^[5]: p.19

${\displaystyle K={\tfrac {1}{4}}{\sqrt {(2(a^{2}+c^{2})-4x^{2})(2(b^{2}+d^{2})-4x^{2})}}.}$

Other area formulas may be obtained from setting p = q in the formulas for the area of a convex quadrilateral.