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| [[Image:Brahmaguptra's theorem.svg|thumb|<math> \overline{BD}\perp\overline{AC},\overline{EF}\perp\overline{BC} </math> <math>\Rightarrow |\overline{AF}|=|\overline{FD}| </math>]] | | Hi, I am Consuelo Goll. It's not a typical thing but what she [http://Likesperforming.net/ likes performing] is heading to [http://Www.google.com/search?q=karaoke&btnI=lucky karaoke] and she's been performing it for fairly a while. For years he's been operating as a manufacturing and planning officer but he's currently applied for an additional one. Wisconsin is where we've been living for many years and I don't strategy on altering it. His spouse and he preserve a website. You might want to check it out: https://wiki.lovefm.us/index.php?title=7_Incredible_Nya_Internet_Casino_P%C3%A5_N%C3%A4tet_Transformations<br><br> |
| In [[geometry]], '''Brahmagupta's theorem''' states that if a [[cyclic quadrilateral]] is [[Orthodiagonal quadrilateral|orthodiagonal]] (that is, has [[perpendicular]] [[diagonals]]), then the perpendicular to a side from the point of intersection of the diagonals always [[Bisection|bisects]] the opposite side. It is named after the [[List of Indian mathematicians|Indian mathematician]] [[Brahmagupta]].
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| 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''.
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| ==Proof==
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| [[Image:Proof of Brahmagupta's theorem.svg|thumb|Proof of the theorem.]]
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| We need to prove that ''AF'' = ''FD''. We will prove that both ''AF'' and ''FD'' are in fact equal to ''FM''.
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| To prove that ''AF'' = ''FM'', first note that the angles ''FAM'' and ''CBM'' are equal, because they are [[inscribed angle]]s that intercept the same arc of the circle. Furthermore, the angles ''CBM'' and ''CME'' are both [[complementary angles|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.
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| 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.
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| == See also==
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| * [[Brahmagupta's formula]] for the area of a cyclic quadrilateral
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| ==References==
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| *[[Harold Scott MacDonald Coxeter|Coxeter, H. S. M.]]; Greitzer, S. L.: ''Geometry Revisited''. Washington, DC: Math. Assoc. Amer., p. 59, 1967
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| *{{MathWorld|urlname=BrahmaguptasTheorem|title=Brahmagupta's theorem}}
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| * [http://www.cut-the-knot.org/Curriculum/Geometry/Brahmagupta.shtml Brahmagupta's Theorem] at [[cut-the-knot]]
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| [[Category:Brahmagupta]]
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| [[Category:Quadrilaterals]]
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| [[Category:Circles]]
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| [[Category:Theorems in geometry]]
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| [[Category:Articles containing proofs]]
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