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The diagonals cut the quadrilateral into four triangles of which one opposite pair have equal areas. [16]: Prop.5 The product of the areas of the two triangles formed by one diagonal equals the product of the areas of the two triangles formed by the other diagonal. [16]: Thm.6
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).
One diagonal crosses the midpoint of the other diagonal at a right angle, forming its perpendicular bisector. [9] (In the concave case, the line through one of the diagonals bisects the other.) One diagonal is a line of symmetry. It divides the quadrilateral into two congruent triangles that are mirror images of each other. [7]
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.
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.
If two lines parallel to sides of a parallelogram are constructed concurrent to a diagonal, then the parallelograms formed on opposite sides of that diagonal are equal in area. [8] The diagonals of a parallelogram divide it into four triangles of equal area.
More specifically, let A, B, C and D be four points on a circle such that the lines AC and BD are perpendicular. Denote the intersection of AC and BD by M. Drop the perpendicular from M to the line BC, calling the intersection E. Let F be the intersection of the line EM and the edge AD. Then, the theorem states that F is the midpoint AD.
The two diagonals and the two tangency chords are concurrent. [11] [10]: p.11 One way to see this is as a limiting case of Brianchon's theorem, which states that a hexagon all of whose sides are tangent to a single conic section has three diagonals that meet at a point. From a tangential quadrilateral, one can form a hexagon with two 180 ...