UpdateBot at 12:34, 18 March 2022

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[[Image:Brahmaguptra's theorem.svg|thumb|<math> \overline{BM}\perp\overline{AC},\overline{EF}\perp\overline{BC} </math> <math>\Rightarrow |\overline{AF}|=|\overline{FD}| </math>]]
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.<ref>Michael John Bradley (2006). ''The Birth of Mathematics: Ancient Times to 1300''. Publisher Infobase Publishing. {{ISBN|0816054231}}. Page 70, 85.</ref> It is named after the [[List of Indian mathematicians|Indian mathematician]] [[Brahmagupta]] (598-668).<ref>[[Harold Scott MacDonald Coxeter|Coxeter, H. S. M.]]; Greitzer, S. L.: ''Geometry Revisited''. Washington, DC: Math. Assoc. Amer., p. 59, 1967</ref>

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==
[[Image:Proof of Brahmagupta's theorem.svg|thumb|Proof of the theorem.]]
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 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.

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.

== See also==
* [[Brahmagupta's formula]] for the area of a cyclic quadrilateral

==References==
{{reflist}}

==External links==
{{ProofWiki|id=Brahmagupta Theorem|title=Brahmagupta theorem}}
* [http://www.cut-the-knot.org/Curriculum/Geometry/Brahmagupta.shtml Brahmagupta's Theorem] at [[cut-the-knot]]
*{{MathWorld|urlname=BrahmaguptasTheorem|title=Brahmagupta's theorem}}

[[Category:Brahmagupta]]
[[Category:Theorems about quadrilaterals and circles]]
[[Category:Articles containing proofs]]

Brahmagupta theorem - Revision history

UpdateBot at 12:34, 18 March 2022