{{Short description|Theorem on cyclic quadrilateral}}
[[Image:Brahmaguptra's theorem.svg|thumb|{{center|Diagram of Brahmagupta's theorem.}}
{{legend-line|solid red 2px|Cyclic quadrilateral}}
{{legend-line|solid black 2px|Perpendicular diagonals (concur at {{mvar|M}})}}
{{legend-line|solid blue 3px|Perpendicular of a side through {{mvar|M}}; bisects the opposite side per the theorem}}
]]

In [[geometry]], '''Brahmagupta's theorem''' states that if a [[cyclic quadrilateral]] is [[Orthodiagonal quadrilateral|orthodiagonal]] (that is, has [[diagonals]] that are [[perpendicular]]), 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 {{mvar|A}}, {{mvar|B}}, {{mvar|C}} and {{mvar|D}} be four points on a circle such that the lines {{mvar|AC}} and {{mvar|BD}} are perpendicular. Denote the intersection of {{mvar|AC}} and {{mvar|BD}} by {{mvar|M}}. Drop the perpendicular from {{mvar|M}} to the line {{mvar|BC}}, calling the intersection {{mvar|E}}. Let {{mvar|F}} be the intersection of the line {{mvar|EM}} and the edge {{mvar|AD}}. Then, the theorem states that {{mvar|F}} is the [[midpoint]] of {{mvar|AD}}.

==Proof==
[[Image:Proof of Brahmagupta's theorem.svg|thumb|Proof of the theorem]]
We need to prove that {{math|1=''AF'' = ''FD''}}. We will prove that both {{mvar|AF}} and {{mvar|FD}} are in fact equal to {{mvar|FM}}.

To prove that {{math|1=''AF'' = ''FM''}}, first note that the angles {{math|∠''FAM''}} and {{math|∠''CBM''}} are equal, because they are [[inscribed angle]]s that intercept the same [[circular arc|arc]] of the circle ({{mvar|CD}}). Furthermore, the angles {{math|∠''CBM''}} and {{math|∠''CME''}} are both [[complementary angles|complementary]] to angle {{math|∠''BCM''}} (i.e., they add up to 90°), and are therefore equal. Finally, the angles {{math|∠''CME''}} and {{math|∠''FMA''}} are the same. Hence, {{math|△''AFM''}} is an [[isosceles triangle]], and thus the sides {{mvar|AF}} and {{mvar|FM}} are equal.

The proof that {{math|1=''FD'' = ''FM''}} goes similarly: the angles {{math|∠''FDM''}}, {{math|∠''BCM''}}, {{math|∠''BME''}} and {{math|∠''DMF''}} are all equal, so {{math|△''DFM''}} is an isosceles triangle, so {{math|1=''FD'' = ''FM''}}. It follows that {{math|1=''AF'' = ''FD''}}, as the theorem claims.

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

==References==
{{reflist}}

==External links==
* [https://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]]