Law of Sines/Proof 1

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 altitude from $B$.

From the definition of sine:

$\sin A = \dfrac h c$ and $\sin C = \dfrac h a$

Thus:

$h = c \sin A$

and:

$h = a \sin C$

This gives:

$c \sin A = a \sin C$

So:

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

Similarly, constructing the altitude from $A$ gives:

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

$\blacksquare$