Brahmagupta theorem

$\begin{matrix}\overline{BD}\perp\overline{AC},\overline{EF}\perp\overline{BC}\\\Rarr|\overline{AF}|=|\overline{FD}|\end{matrix}$

Brahmagupta's theorem is a result in geometry. It states that if a cyclic quadrilateral has perpendicular diagonals, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. It is named after the Indian mathematician Brahmagupta.

More specifically, let $A, B, C, D$ be four points on a circle such that the lines $AC,BD$ are perpendicular. Denote the intersection of $AC,BD$ by $M$ . Drop the perpendicular from $M$ to the line $BC$ , calling the intersection $E$ . Let $F$ be the intersection of the line $EM$ and the edge $AD$ . Then, the theorem states that $F$ is in the middle of $AD$ .

Proof

Proof of the theorem

We need to prove that $AF=FD$ . We will prove that both $AF,FD$ are in fact equal to $FM$ .

To prove that $AF=FM$ , first note that the angles

$\sphericalangle FAM=\sphericalangle CBM$

because they are inscribed angles that intercept the same arc of the circle. Furthermore, the angles $\sphericalangle CBM,\sphericalangle CME$ are both complementary to angle $\sphericalangle BCM$ (i.e., they add up to 90°). Finally, the angles

$\sphericalangle CME=\sphericalangle FMA$

Hence, $\triangle AFM$ is an isosceles triangle, and thus the sides $AF=FM$ .

The proof that $FD=FM$ goes similarly. The angles

$\sphericalangle FDM=\sphericalangle BCM=\sphericalangle BME=\sphericalangle DMF$

so $\triangle DFM$ is an isosceles triangle, so $FD=FM$ . It follows that $AF=FD$ , as the theorem claims.

External links

• Weisstein, Eric W., "Brahmagupta's theorem" from MathWorld.
• Brahmagupta's Theorem at cut-the-knot

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