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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 , , and be four points on a circle such that the lines and are perpendicular. Denote the intersection of and by . Drop the perpendicular from to the line , calling the intersection . Let be the intersection of the line and the edge . Then, the theorem states that is the midpoint of .
We need to prove that . We will prove that both and are in fact equal to .
To prove that , first note that the angles and are equal, because they are inscribed angles that intercept the same arc of the circle (). Furthermore, the angles and are both complementary to angle (i.e., they add up to 90°), and are therefore equal. Finally, the angles and are the same. Hence, is an isosceles triangle, and thus the sides and are equal.
The proof that goes similarly: the angles , , and are all equal, so is an isosceles triangle, so . It follows that , as the theorem claims.