Length of Chord of Circle/Proof 1

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 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 Congruence:

$\triangle AOE = \triangle BOE$

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.

$\blacksquare$