Law of Sines/Proof 2
Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
- $\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$
where $R$ is the circumradius of $\triangle ABC$.
Proof
Construct the circumcircle of $\triangle ABC$, let $O$ be the circumcenter and $R$ be the circumradius.
Construct $\triangle AOB$ and let $E$ be the foot of the altitude of $\triangle AOB$ from $O$.
By the Inscribed Angle Theorem:
- $\angle ACB = \dfrac {\angle AOB} 2$
From the definition of the circumcenter:
- $AO = BO$
From the definition of altitude and the fact that all right angles are congruent:
- $\angle AEO = \angle BEO$
Therefore from Pythagoras's Theorem:
- $AE = BE$
and then from Triangle Side-Side-Side Congruence:
- $\angle AOE = \angle BOE$
Thus:
- $\angle AOE = \dfrac {\angle AOB} 2$
and so:
- $\angle ACB = \angle AOE$
Then by the definition of sine:
- $\sin C = \map \sin {\angle AOE} = \dfrac {c / 2} R$
and so:
- $\dfrac c {\sin C} = 2 R$
The same argument holds for all three angles in the triangle, and so:
- $\dfrac c {\sin C} = \dfrac b {\sin B} = \dfrac a {\sin A} = 2 R$
$\blacksquare$