# Brahmagupta theorem

Brahmagupta theorem

Brahmagupta's theorem is a result in geometry. It states that if a cyclic quadrilateral has perpendicular 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 mathematician Brahmagupta.

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 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°). 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.

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

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