Proof the ratio of a side and the sines of its opposite angle of a triangle is equal to the diameter of the circumscribed circle of the triangle

The circumcenter is inside the triangle

In this case, angle C and the central angle ∠AOB intersect the arc, so

∠C = \dfrac{1}{2} ∠AOB

As △AOB is an isosceles triangle, ∠AOD is the bisector of ∠AOB ,

therefore,

∠AOD = ∠C , \sin C = \sin∠AOD

\sin∠AOD= \dfrac{\dfrac{c}{2} }{AO} = \dfrac{c}{2R}

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

The circumcenter is outside the triangle

In this case, ∠C is an obtuse angle, so the central angle intersecting the arc ∠AOB is an angle more than 180 \degree.

According to the Inscribed Angle Theorem,

∠C = \dfrac{1}{2} ∠AOB =∠AOG

\because ∠AOF = 180 \degree - ∠AOG

\therefore \sin ∠AOF = \sin ∠AOG= sin ∠C

\therefore sin ∠C = = \dfrac{\dfrac{c}{2} }{AO} = \dfrac{c}{2R}

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

The circumcenter is on one side of the triangle

In this case,

∠C = 90\degree

and

c=2R

\dfrac{c}{\sin C} =\dfrac{2R}{\sin 90\degree } = 2R