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 $BM\perp AC$ and $EF\perp BC,$then $AF=FD,$ 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]