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Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. It follows that any rhombus has the following properties: Opposite angles of a rhombus have equal measure. The two diagonals of a rhombus are perpendicular; that is, a rhombus is an orthodiagonal quadrilateral. Its diagonals bisect opposite ...
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]
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.
An orthodiagonal quadrilateral is a quadrilateral whose diagonals are perpendicular. These include the square, the rhombus, and the kite. By Brahmagupta's theorem, in an orthodiagonal quadrilateral that is also cyclic, a line through the midpoint of one side and through the intersection point of the diagonals is perpendicular to the opposite side.
The midpoints of the sides of any quadrilateral with perpendicular diagonals form a rectangle. A parallelogram with equal diagonals is a rectangle. The Japanese theorem for cyclic quadrilaterals [ 12 ] states that the incentres of the four triangles determined by the vertices of a cyclic quadrilateral taken three at a time form a rectangle.
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. [ 2 ] If an orthodiagonal quadrilateral is also cyclic, the distance from the circumcenter (the center of the circumscribed circle) to any side equals half the ...
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)
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).