In geometry, **Brahmagupta's theorem** states that if a cyclic quadrilateral is orthodiagonal (that is, has perpendicular diagonals), 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 *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°), 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