Question

In a $△ABC$, shows that : $a+b+c=\frac{abc}{4Rr}$

Answer 1

To prove: $a+b+c=\frac{abc}{4Rr}$

In a $\Delta ABC$: $R=\frac{abc}{4K}$ and $r=\frac{K}{s}$

where $s=\frac{a+b+c}{2}$, $K=\sqrt{s(s-a)(s-b(s-c)}$

$\frac{abc}{4Rr}=\frac{abc}{2\times \frac{abc}{4K}\times \frac{K}{s}}$

$\frac{abc}{4Rr}=2s$

$\Rightarrow \frac{abc}{4Rr}=a+b+c$

Hence proved.