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 $BM\perp AC$ and ${\displaystyle }$EF\perp BC,</math> then $AF=FD$ 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 $AF=FD$.