# 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.

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

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 $AF=DF$ .

Proof: $\angle CBD=\angle CAD$ [As both are inscribed angles that intercept the same arc $CD$ of a circle]

Or, $\angle CBM=\angle MAF$ Here, $\angle CMB+\angle CBM+\angle BCM=180$ °

Or, $\angle CMB+\angle BCM=180$ ° $-\angle CBM$ Again, $\angle CME+\angle CEM+\angle ECM=180$ °

Or, $\angle CME+\angle CMB+\angle BCM=180$ ° [As $\angle CMB=\angle CEM=90$ ° and $\angle BCM=\angle ECM$ ]

Or, $\angle CME+180$ ° $-\angle CBM=180$ °

Or, $\angle CME=\angle CBM$ Or, $\angle AMF=\angle MAF$ [As, \angle AMF = \angle CME; Vertical Angles]

Therefore, $AF=MF$ In the similar way, $\angle MDF=\angle DMF$ and $MF=DF$ Or, $AF=DF$ [Proved]

## Reference

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