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

Or, $\mathrm{\angle }CBM=\mathrm{\angle }MAF$

Here, $\mathrm{\angle }CMB+\mathrm{\angle }CBM+\mathrm{\angle }BCM=180$°

Or, $\mathrm{\angle }CMB+\mathrm{\angle }BCM=180$° $-\mathrm{\angle }CBM$

Again, $\mathrm{\angle }CME+\mathrm{\angle }CEM+\mathrm{\angle }ECM=180$°

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

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

Or, $\mathrm{\angle }CME=\mathrm{\angle }CBM$

Or, $\mathrm{\angle }AMF=\mathrm{\angle }MAF$ [As, \angle AMF = \angle CME; Vertical Angles]

Therefore, $AF=MF$

In the similar way, $\mathrm{\angle }MDF=\mathrm{\angle }DMF$ and $MF=DF$

Or, $AF=DF$ [Proved]