Law of Sines/Proof 2
Theorem
For any triangle $\triangle ABC$:
- $\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C}$
where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.
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 Equality:
- $\angle AOE = \angle BOE$
Thus:
- $\angle AOE = \dfrac {\angle AOB} 2$
and so:
- $\angle ACB = \angle AOE$
Then by the definition of sine:
- $\sin C = \sin \left({\angle AOE}\right) = \dfrac {c / 2} R$
and so:
- $\dfrac c {\sin C} = 2 R$
Because the same argument holds for all three angles in the triangle:
- $\dfrac c {\sin C} = 2 R = \dfrac b {\sin B} = 2 R = \dfrac a {\sin A}$
Note that this proof also yields a useful extension of the law of sines:
- $\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$
$\blacksquare$