In $\mathrm{△}ABC$ show that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ where $R$is the circum radius 

Case (i) : $\angle A$ is acute 

S is the centre of the circum circle and CD is its diameter.

Then $CS = SD = R$and

$CD = 2R$. Join $BD$. 

Then $DBC=\frac{\pi }{2}\text{ and }DBC$

is a right angled triangle.

Then $\angle BAC=\angle BDC,$

$(\because$angles in the same segment $)$

$\therefore \mathrm{Sin}A=\mathrm{Sin}\angle BAC=\mathrm{Sin}\mid BDC=\frac{BC}{CD}=\frac{a}{2R}\text{. }$

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

Case (ii) : $\angle A$ is right angle (figure) 

Then $BC=a=2R=2R\left(1\right)=2R\sin {90}^{\circ }$

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

Case (iii) : $\angle A$ is obtuse (figure)

$\mathrm{△}BC$ is right angled.

$\text{ ( }\because \text{ Angle in the semi circle) }$

In the cyclic quadrilateral $BACD$

$BDC={180}^{\circ }-\mathrm{\angle }BAC={180}^{\circ }-A$

In $\mathrm{△}BDC,\sin A=\sin \left({180}^{\circ }-A\right)$ 

$=\sin \angle BDC=\frac{BC}{CD}=\frac{a}{2R}$

$\text{ Hence }\frac{a}{\sin A}=2R$

In a similar way, we can prove

$\frac{b}{\sin B}=2R,\frac{c}{\sin C}=2R$

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