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If a cyclic quadrilateral has side lengths that form an arithmetic progression the quadrilateral is also ex-bicentric. If the opposite sides of a cyclic quadrilateral are extended to meet at E and F, then the internal angle bisectors of the angles at E and F are perpendicular. [13]
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]
Ptolemy's theorem is a relation among these lengths in a cyclic quadrilateral. = + In Euclidean geometry, Ptolemy's theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral (a quadrilateral whose vertices lie on a common circle).
It divides the quadrilateral into two congruent triangles that are mirror images of each other. [7] One diagonal bisects both of the angles at its two ends. [7] Kite quadrilaterals are named for the wind-blown, flying kites, which often have this shape [10] [11] and which are in turn named for a hovering bird and the sound it makes.
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.
The leftmost and rightmost vertices have right angles. In Euclidean geometry, a right kite is a kite (a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other) that can be inscribed in a circle. [1] That is, it is a kite with a circumcircle (i.e., a cyclic kite).
The two complete quadrilaterals have a shared diagonal, EF. N lies on the Newton–Gauss line of both quadrilaterals. N is equidistant from G and H, since it is the circumcenter of the cyclic quadrilateral EGFH. If triangles GMP, HMQ are congruent, and it will follow that M lies on the perpendicular bisector of the line HG.
The orange and green quadrilaterals are congruent; the blue one is not congruent to them. Congruence between the orange and green ones is established in that side BC corresponds to (in this case of congruence, equals in length) JK, CD corresponds to KL, DA corresponds to LI, and AB corresponds to IJ, while angle ∠C corresponds to (equals) angle ∠K, ∠D corresponds to ∠L, ∠A ...