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In a $\triangle A B C$, with usual notation, if $r=r_1-r_2-r_3$, then $2 R=$
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$\mathrm{a}$
In a $\triangle A B C$, it is given that,
$r=r_1-r_2-r_3$
$\Rightarrow \quad \frac{\Delta}{s}=\frac{\Delta}{s-a}-\frac{\Delta}{s-b}-\frac{\Delta}{s-c}$
$\Rightarrow \quad \frac{1}{s-b}+\frac{1}{s-c}=\frac{1}{s-a}-\frac{1}{s}$
$\Rightarrow \quad \frac{2 s-b-c}{(s-b)(s-c)}=\frac{a}{s(s-a)}$
$\Rightarrow \quad s(s-a)=s^2-(b+c) s+b c$
$\Rightarrow \quad s^2-s a=s^2-(b+c) s+b c$
$\Rightarrow \quad(b+c-a) s=b c \quad \Rightarrow(b+c)^2-a^2=2 b c$
$\Rightarrow b^2+c^2+2 b c-a^2=2 b c \Rightarrow b^2+c^2=a^2$
$\therefore \triangle A B C$ is the right angled triangle with angle $A=90^{\circ}$, so $2 R=a$.
$r=r_1-r_2-r_3$
$\Rightarrow \quad \frac{\Delta}{s}=\frac{\Delta}{s-a}-\frac{\Delta}{s-b}-\frac{\Delta}{s-c}$
$\Rightarrow \quad \frac{1}{s-b}+\frac{1}{s-c}=\frac{1}{s-a}-\frac{1}{s}$
$\Rightarrow \quad \frac{2 s-b-c}{(s-b)(s-c)}=\frac{a}{s(s-a)}$
$\Rightarrow \quad s(s-a)=s^2-(b+c) s+b c$
$\Rightarrow \quad s^2-s a=s^2-(b+c) s+b c$
$\Rightarrow \quad(b+c-a) s=b c \quad \Rightarrow(b+c)^2-a^2=2 b c$
$\Rightarrow b^2+c^2+2 b c-a^2=2 b c \Rightarrow b^2+c^2=a^2$
$\therefore \triangle A B C$ is the right angled triangle with angle $A=90^{\circ}$, so $2 R=a$.
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