# Timeless Theorems of Mathematics/Brahmagupta Theorem

The 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. If B M ⊥ A C {\displaystyle BM\perp AC} and {\displaystyle } EF\perp BC,[/itex] then A F = F D {\displaystyle AF=FD} according to the Brahmagupta's theorem

The theorem is named after the Indian mathematician Brahmagupta (598-668).

## Proof

### Statement

If any cyclic quadrilateral has perpendicular diagonals, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.

### Proof

Proposition : Let $ABCD$  is a quadrilateral inscribed in a circle with perpendicular diagonals $AC$  and $BD$  intersecting at point $M$ . $ME$  is a perpendicular on the side $BC$  from the point $M$  and extended $EM$  intersects the opposite side $AD$  at point $F$ . It is to be proved that $AF=FD$ .

## Reference

1. Michael John Bradley (2006). The Birth of Mathematics: Ancient Times to 1300. Publisher Infobase Publishing. ISBN 0816054231. Page 70, 85.