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]
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.
^Michael John Bradley (2006). The Birth of Mathematics: Ancient Times to 1300. Publisher Infobase Publishing. ISBN 0816054231. Page 70, 85.
^Coxeter, H. S. M.; Greitzer, S. L.: Geometry Revisited. Washington, DC: Math. Assoc. Amer., p. 59, 1967
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