Length of Chord of Circle/Proof 2
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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 | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 r \sin \dfrac \theta 2\) |
$\blacksquare$