Length of Chord of Circle/Proof 1

Proof

 * LengthOfChord.png

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 :


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

Rearranging, we get:


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

as desired.