sines law proof

Let $ABC$ a triangle. Let $T$ a point in the circumcircle such that $BT$ is a diameter.

So $\mathrm{\angle }A=\mathrm{\angle }CAB$ is equal to $\mathrm{\angle }CTB$ (they subtend the same arc). Since $\mathrm{△}CBT$ is a right triangle, from the definition of sine we get

$$\sin \mathrm{\angle }CTB=\frac{BC}{BT}=\frac{a}{2R}.$$

On the other hand $\mathrm{\angle }CAB=\mathrm{\angle }CTB$ implies their sines are the same and so

$$\sin \mathrm{\angle }CAB=\frac{a}{2R}$$

and therefore

$$\frac{a}{\sin A}=2R.$$

Drawing diameters passing by $C$ and $A$ will let us prove in a similar way the relations

$$\frac{b}{\sin B}=2R\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\frac{c}{\sin C}=2R$$

and we conclude that

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

Q.E.D.