Law of Sines

Theorem
$$ \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}= 2R$$

Where $$a$$,$$b$$ and $$c$$ are the side the triangle $$ABC$$ and $$O$$ the circincenter

Proof
Join $$A$$ and $$B$$ with $$O$$ and be $$OE$$ the altitud of triangle $$AOB$$ from $$O$$

the $$OE$$ is the simetral of $$AB$$ (By the definition of the circuncenter)



Now we have

$$2 \cdot \measuredangle ACB = \measuredangle AOB$$ (By the angle in and a circunference)

$$2 \cdot \measuredangle AOE = \measuredangle AOB$$ (OE is the simetral)

$$\measuredangle ACB = \measuredangle AOE \Rightarrow \angle ACB \cong \angle AOE$$

Let be ACB : $$\gamma$$

With the definition of sine on triangle AOE

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

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

A similar argument can be used to show that the statement holds for the others angles