Length of Chord of Circle

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

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


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

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

Proof 1


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

Let $AB$ be bisected by $OD$.

Consider the pair of triangles $\triangle AOE$ and $\triangle BOE$.

We see that:

$AE = ED$ since $AB$ is bisected by $OD$
$AO = BO$ since they are radii
$OE = OE$ since they are common sides.

By Triangle Side-Side-Side Equality, $\triangle AOE = \triangle BOE$.

Then we have:

$\angle AOE = \angle BOE = \dfrac \theta 2$
$\angle OEA = \angle OEB = \dfrac {180 \degrees} 2 = 90 \degrees$

By Definition of Sine Function:

$\sin \dfrac \theta 2 = \dfrac {AE} {AO} = \dfrac {\frac 1 2 AB} r$

Rearranging, we get:

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

as desired.


Proof 2

We have $AO = BO$ since they are radii.

Therefore $\triangle AOB$ is isosceles.

By Isosceles Triangle has Two Equal Angles:

$\angle OAB = \angle OBA$

By Sum of Angles of Triangle equals Two Right Angles:

$\angle OAB + \angle OBA + \theta = 180 \degrees$

Therefore $\angle OAB = \dfrac {180 \degrees - \theta} 2 = 90 \degrees - \dfrac \theta 2$.


\(\ds \dfrac {AB} {\sin \theta}\) \(=\) \(\ds \dfrac {BO} {\sin \angle OAB}\) Law of Sines
\(\ds {AB}\) \(=\) \(\ds \dfrac {BO \sin \theta} {\sin \angle OAB}\)
\(\ds \) \(=\) \(\ds \dfrac {2 r \sin \frac \theta 2 \cos \frac \theta 2} {\map \sin {90 \degrees - \frac \theta 2} }\) Double Angle Formula for Sine
\(\ds \) \(=\) \(\ds \dfrac {2 r \sin \frac \theta 2 \cos \frac \theta 2} {\cos \frac \theta 2}\) Sine of Supplementary Angle
\(\ds \) \(=\) \(\ds 2 r \sin \dfrac \theta 2\)


Historical Note

The result Length of Chord of Circle was the basis of the calculations that Hipparchus of Nicaea used when creating his trigonometrical tables.