**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.