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Every kite is an orthodiagonal quadrilateral, meaning that its two diagonals are at right angles to each other. Moreover, one of the two diagonals (the symmetry axis) is the perpendicular bisector of the other, and is also the angle bisector of the two angles it meets. [1] Because of its symmetry, the other two angles of the kite must be equal.
Equivalently, a quadrilateral has equal diagonals if and only if it has perpendicular bimedians, and it has perpendicular diagonals if and only if it has equal bimedians. [7] Silvester (2006) gives further connections between equidiagonal and orthodiagonal quadrilaterals, via a generalization of van Aubel's theorem. [8]
A rhombus therefore has all of the properties of a parallelogram: for example, opposite sides are parallel; adjacent angles are supplementary; the two diagonals bisect one another; any line through the midpoint bisects the area; and the sum of the squares of the sides equals the sum of the squares of the diagonals (the parallelogram law).
In a convex quadrilateral, there is the following dual connection between the bimedians and the diagonals: [29] The two bimedians have equal length if and only if the two diagonals are perpendicular. The two bimedians are perpendicular if and only if the two diagonals have equal length.
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).
This is a list of two-dimensional geometric shapes in Euclidean and other geometries. For mathematical objects in more dimensions, see list of mathematical shapes. For a broader scope, see list of shapes.
The diagonals of an isosceles trapezoid have the same length; that is, every isosceles trapezoid is an equidiagonal quadrilateral. Moreover, the diagonals divide each other in the same proportions. As pictured, the diagonals AC and BD have the same length (AC = BD) and divide each other into segments of the same length (AE = DE and BE = CE).
For a cyclic quadrilateral that is also orthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths p 1 and p 2 and divides the other diagonal into segments of lengths q 1 and q 2. Then [28] (the first equality is Proposition 11 in Archimedes' Book of Lemmas)