# 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.[^1]

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, $AF = DF$ ***\[Proved\]***

## Reference


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