relation between sides of a triangle and circumradious prove that $a.b.c=4R. S_{\bigtriangleup ABC}$, ($S$ stands for surface) and ($R$ stands for circumradious)

any response would be appreciated
 A: 
Hint: Use similarity on triangles $AFC$ and $BAD$.
A: A good starting point is to use the formula for area $S$ of $\bigtriangleup ABC$: 
$S = \dfrac{1}{2} \cdot ab \cdot sinC$. We now prove this formula:Let $H$ be the point on side $BC$ such that $AH$ is perpendicular to $BC$, and let $h = AH$. So $S = \dfrac{1}{2}\cdot AH \cdot BC = \dfrac{1}{2} \cdot h \cdot BC = \dfrac{1}{2} \cdot (b\cdot sinC)\cdot a = \dfrac{1}{2} \cdot ab \cdot sinC$. Thus: $4S = 2\cdot ab \cdot sinC$ , and use the law 
of sine for the same triangle: $sinC = \dfrac{c}{2R}$ and get: $4S = 2 \cdot ab \cdot \dfrac{c}{2R} = \dfrac{abc}{R}$. So $4R\cdot S_{\bigtriangleup ABC} = abc$.
Note: As for a quick proof of the law of sines in triangle $\bigtriangleup ABC$:Let $O$ be the center of the circumscribing circle of the $\bigtriangleup ABC$, and let $K$ be midpoint of side $AB$, then $\bigtriangleup AOB$ is isosceles since $OA = OB = R$. So $OK$ is the altitude of $\bigtriangleup ABC$. So $\angle C = \dfrac{\angle AOB}{2} =  \angle AOK$, and then $sinC = sin\angle AOK = \dfrac{\frac{c}{2}}{R} = \dfrac{c}{2R}$.
