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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.〔Michael John Bradley (2006). ''The Birth of Mathematics: Ancient Times to 1300''. Publisher Infobase Publishing. ISBN 0816054231. Page 70, 85.〕 It is named after the Indian mathematician Brahmagupta.〔Coxeter, H. S. M.; Greitzer, S. L.: ''Geometry Revisited''. Washington, DC: Math. Assoc. Amer., p. 59, 1967〕 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''. ==Proof== We need to prove that ''AF'' = ''FD''. We will prove that both ''AF'' and ''FD'' are in fact equal to ''FM''. To prove that ''AF'' = ''FM'', first note that the angles ''FAM'' and ''CBM'' are equal, because they are inscribed angles that intercept the same arc of the circle. Furthermore, the angles ''CBM'' and ''CME'' are both complementary to angle ''BCM'' (i.e., they add up to 90°), and are therefore equal. Finally, the angles ''CME'' and ''FMA'' are the same. Hence, ''AFM'' is an isosceles triangle, and thus the sides ''AF'' and ''FM'' are equal. The proof that ''FD'' = ''FM'' goes similarly: the angles ''FDM'', ''BCM'', ''BME'' and ''DMF'' are all equal, so ''DFM'' is an isosceles triangle, so ''FD'' = ''FM''. It follows that ''AF'' = ''FD'', as the theorem claims. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Brahmagupta theorem」の詳細全文を読む スポンサード リンク
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