Question

Prove that in triangle $\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=\frac{1}{2Rr}$. Where the symbols have their usual meaning.

Answer 1

$\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$

$=\frac{a^{2}bc+b^{2}ac+c^{2}ac}{abc}$

$=\frac{a+b+c}{abc}$

$=\frac{2s}{abc}$ $\{\because$ $(a+b+c)=2s\}$

Also, we knwo that,

$R=\frac{abc}{4△}$ $\{$ where$(R=circumradius)$$\}$

and , $r=\frac{△}{s}$ $\{$ where$(r=inradius)$$\}$

$⟹△=rs$

and $abc=4△R=4Rrs$

$\therefore \dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}=\dfrac{2s}{abc}$

$=\frac{2s}{4Rrs}$

$=\frac{1}{2Rr}$

hence proved