Law of Sines

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Theorem

For any triangle $\triangle ABC$, we have:

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

Here $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.

Construct the altitude from $B$.

It can be seen from the definition of sine that $\sin A = \dfrac h c$ and $\sin C = \dfrac h a$.

Thus $h = c\sin A$ and $h = a\sin C$.

This gives us $c \sin A = a \sin C$.

So $\dfrac a {\sin A} = \dfrac c {\sin C}$.

Similarly, constructing the altitude from $A$ gives us $\dfrac b {\sin B} = \dfrac c {\sin C}$.

$\blacksquare$

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 $O$.

Now we have $\angle ACB = \dfrac{\angle AOB} 2$ by the Inscribed Angle Theorem.

Furthermore, $AO = BO$ from the definition of the circumcenter.

Also, $\angle AEO = \angle BEO$ from the definition of altitude and the fact that all right angles are congruent

Therefore $AE = BE$ from Pythagoras's Theorem, and then $\angle AOE = \angle BOE$ from Triangle Side-Side-Side Equality.

This gives us $\angle AOE = \dfrac {\angle AOB} 2$, and thus $\angle ACB = \angle AOE$.

Then $\sin C = \sin (\angle AOE) = \dfrac {c / 2} R$ by the definition of sine, and so:

$\dfrac c {\sin C}=2R$

Because the same statement 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:

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

$\blacksquare$