Length of Chord of Circle/Proof 2

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

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

Thus:

\(\ds \dfrac {AB} {\sin \theta}\)

\(=\)

\(\ds \dfrac {BO} {\sin \angle OAB}\)

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

\(\ds \)

\(=\)

\(\ds 2 r \sin \dfrac \theta 2\)

$\blacksquare$