[tex3]{\boxed{\boxed{S_{\triangle}=\frac{a.b.c}{4.R}}}}[/tex3]
Demonstração:
[tex3]S_{ABC}=\frac{1}{2}.a.h[/tex3]
Para determinar [tex3]h[/tex3] construímos o [tex3]{\triangle}_{ABE}[/tex3] com [tex3]AE=2.R[/tex3]
[tex3]\newenvironment{rcases}{\left.\begin{aligned}}{\end{aligned}\right\rbrace}\begin{rcases}\hat{D}=\hat{B} (reto) \\\hat{C}=\hat{E}=\frac{AB}{2}\end{rcases}\Rightarrow\triangle_{ADC}\thicksim\triangle_{ABE}[/tex3]
Assim:
[tex3]\frac{h}{c}=\frac{b}{2.R}[/tex3] [tex3]\Rightarrow[/tex3] [tex3]h=\frac{b.c}{2.R}[/tex3]
Logo:
[tex3]S_{ABC}=\frac{1}{2}.a.\frac{b.c}{2.R}[/tex3] [tex3]\Longrightarrow[/tex3] [tex3]\boxed{\boxed{S_{ABC}=\frac{a.b.c}{4.R}}}(c.q.d)[/tex3]