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:

$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$

$\blacksquare$