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Q.
In $\triangle ABC$, with usual notation, if $r=r_{1}-r_{2}-r_{3}$, then $2 R =$
TS EAMCET 2020
Solution:
In a $\triangle A B C$, it is given that,
$r=r_{1}-r_{2}-r_{3} $
$\Rightarrow \frac{\Delta}{s}=\frac{\Delta}{s-a}-\frac{\Delta}{s-b}-\frac{\Delta}{s-c}$
$\Rightarrow \frac{1}{s-b}+\frac{1}{s-c}=\frac{1}{s-a}-\frac{1}{s}$
$\Rightarrow \frac{2 s-b-c}{(s-b)(s-c)}=\frac{a}{s(s-a)} $
$\Rightarrow s(s-a)=s^{2}-(b+c) s+b c $
$\Rightarrow s^{2}-s a=s^{2}-(b+c) s+b c$
$\Rightarrow (b+c-a) s=b c $
$\Rightarrow (b+c)^{2}-a^{2}=2 b c$
$\Rightarrow b^{2}+c^{2}+2 b c-a^{2}=2 b c$
$ \Rightarrow b^{2}+c^{2}=a^{2}$
$\therefore \triangle A B C$ is the right angled triangle with angle $A=90^{\circ}$,
so $2 R=a$.