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).

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 $$AF = DF$$.

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, $$\angle CMB + \angle CBM + \angle BCM = 180$$&deg;

Or, $$\angle CMB + \angle BCM = 180$$&deg; $$- \angle CBM$$

Again, $$\angle CME + \angle CEM + \angle ECM = 180$$&deg;

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

Or, $$\angle CME + 180$$&deg; $$- \angle CBM = 180$$&deg;

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]