# Length of Chord of Circle

## Contents

## Theorem

Let $C$ be a circle of radius $r$.

Let $AB$ be a chord which joins the endpoints of the arc $ADB$.

Then:

- $AB = 2 r \sin \dfrac \theta 2$

where $\theta$ is the angle subtended by $AB$ at the center of $C$.

## Proof

Let $O$ be the center of $C$.

Let $AB$ be bisected at $E$ by $OD$.

By Conditions for Diameter to be Perpendicular Bisector, $\angle OEA$ is a right angle.

By definition of sine:

- $\sin \dfrac \theta 2 = AE / AO$

But $AB = 2 AE$ and $AO = r$.

Hence:

- $AB = 2 r \sin \dfrac \theta 2$

$\blacksquare$

## Historical Note

This result was the basis of the calculations that Hipparchus of Nicaea used when creating his trigonometrical tables.

## Sources

- 2008: Ian Stewart:
*Taming the Infinite*... (previous) ... (next): Chapter $5$: Eternal Triangles