Brahmaguptra's 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,D be four points on a circle such that the lines AC,BD are perpendicular. Denote the intersection of AC,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 of Brahmagupta's theorem

Proof of the theorem

We need to prove that AF=FD . We will prove that both AF,FD are in fact equal to FM .

To prove that AF=FM , first note that the angles

\sphericalangle FAM=\sphericalangle CBM

because they are inscribed angles that intercept the same arc of the circle. Furthermore, the angles \sphericalangle CBM,\sphericalangle CME are both complementary to angle \sphericalangle BCM (i.e., they add up to 90°). Finally, the angles

\sphericalangle CME=\sphericalangle FMA

Hence, \triangle AFM is an isosceles triangle, and thus the sides AF=FM .

The proof that FD=FM goes similarly. The angles

\sphericalangle FDM=\sphericalangle BCM=\sphericalangle BME=\sphericalangle DMF

so \triangle DFM is an isosceles triangle, so FD=FM . It follows that AF=FD , as the theorem claims.

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