Brahmagupta theorem
In geometry, Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has diagonals that are perpendicular), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.[1] It is named after the Indian mathematician Brahmagupta (598-668).[2]
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 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 (CD). Furthermore, the angles ∠CBM and ∠CME are both 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
- ↑ Michael John Bradley (2006). The Birth of Mathematics: Ancient Times to 1300. Publisher Infobase Publishing. ISBN 0816054231. Page 70, 85.
- ↑ Coxeter, H. S. M.; Greitzer, S. L.: Geometry Revisited. Washington, DC: Math. Assoc. Amer., p. 59, 1967