Search results
Results From The WOW.Com Content Network
Thus in a cyclic quadrilateral, the circumcenter, the "vertex centroid", and the anticenter are collinear. [24] If the diagonals of a cyclic quadrilateral intersect at P, and the midpoints of the diagonals are M and N, then the anticenter of the quadrilateral is the orthocenter of triangle MNP.
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.
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]
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.
[15] [16] The right kites are exactly the kites that are cyclic quadrilaterals, meaning that there is a circle that passes through all their vertices. [17] The cyclic quadrilaterals may equivalently defined as the quadrilaterals in which two opposite angles are supplementary (they add to 180°); if one pair is supplementary the other is as well ...
A convex quadrilateral is cyclic if and only if opposite angles sum to 180°. Right kite: a kite with two opposite right angles. It is a type of cyclic quadrilateral. Harmonic quadrilateral: a cyclic quadrilateral such that the products of the lengths of the opposing sides are equal. Bicentric quadrilateral: it is both tangential and cyclic.
Then the quadrilateral formed by M 1, M 2, M 3, M 4 is a rectangle. Proofs are given by Bogomolny [2] and Reyes. [1] This theorem may be extended to prove the Japanese theorem for cyclic polygons, according to which the sum of inradii of a triangulated cyclic polygon does not depend on how it is triangulated. The special case of the theorem for ...
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]