- Brahmagupta theorem
**Brahmagupta's theorem**is a result ingeometry . It states that if acyclic quadrilateral hasperpendicular diagonals, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. It is named after the Indian mathematicianBrahmagupta .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 in the middle of "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 angle s 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°). Finally, the angles "CME" and "FMA" are the same. Hence, "AFM" is anisosceles 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.

**External links***MathWorld|urlname=BrahmaguptasTheorem|title=Brahmagupta's theorem

* [*http://www.cut-the-knot.org/Curriculum/Geometry/Brahmagupta.shtml Brahmagupta's Theorem*] atcut-the-knot

*Wikimedia Foundation.
2010.*