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In geometry, Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has perpendicular diagonals), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. [1] It is named after the Indian mathematician Brahmagupta (598-668). [2]
Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.. The trigonometric adjustment in Bretschneider's formula for non-cyclicality of the quadrilateral can be rewritten non-trigonometrically in terms of the sides and the diagonals e and f to give [2] [3]
Labels used in proof concerning complete quadrilateral. It is a well-known theorem that the three midpoints of the diagonals of a complete quadrilateral are collinear. [2] There are several proofs of the result based on areas [2] or wedge products [3] or, as the following proof, on Menelaus's theorem, due to Hillyer and published in 1920. [4]
In Euclidean geometry, Brahmagupta's formula, named after the 7th century Indian mathematician, is used to find the area of any convex cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides. Its generalized version, Bretschneider's formula, can be used with non-cyclic quadrilateral.
Other names for these quadrilaterals are concyclic quadrilateral and chordal quadrilateral, the latter since the sides of the quadrilateral are chords of the circumcircle. Usually the quadrilateral is assumed to be convex, but there are also crossed cyclic quadrilaterals. The formulas and properties given below are valid in the convex case.
This is not a cyclic quadrilateral. The equality never holds here, and is unequal in the direction indicated by Ptolemy's inequality. The equation in Ptolemy's theorem is never true with non-cyclic quadrilaterals. Ptolemy's inequality is an extension of this fact, and it is a more general form of Ptolemy's theorem.
An arbitrary quadrilateral and its diagonals. Bases of similar triangles are parallel to the blue diagonal. Ditto for the red diagonal. The base pairs form a parallelogram with half the area of the quadrilateral, A q, as the sum of the areas of the four large triangles, A l is 2 A q (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, A s is a quarter of A ...
Follow the quadrilateral vertices in the same sequential direction and construct each square on the left hand side of each side of the given quadrilateral. The segments joining the centers of the squares constructed externally (or internally) to the quadrilateral over two opposite sides have been referred to as Van Aubel segments.
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