In a triangle ABC,with usual notation,if r=r_ | vedclass

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(A) Given that $r = r_1 - r_2 - r_3$. Using the standard formulas $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_

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(A) Given that $r = r_1 - r_2 - r_3$. Using the standard formulas $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$: $\frac{\Delta}{s} = \frac{\Delta}{s-a} - \frac{\Delta}{s-b} - \frac{\Delta}{s-c}$ $\frac{1}{s-b} + \frac{1}{s-c} = \frac{1}{s-a} - \frac{1}{s}$ $\frac{s-c+s-b}{(s-b)(s-c)} = \frac{s-(s-a)}{s(s-a)}$ $\frac{2s-b-c}{(s-b)(s-c)} = \frac{a}{s(s-a)}$ Since $2s = a+b+c$,we have $2s-b-c = a$: $\frac{a}{(s-b)(s-c)} = \frac{a}{s(s-a)}$ $s(s-a) = (s-b)(s-c)$ $s^2 - sa = s^2 - s(b+c) + bc$ $s(b+c-a) = bc$ Substituting $s = \frac{a+b+c}{2}$: $\frac{a+b+c}{2} 	imes (b+c-a) = bc$ $(b+c)^2 - a^2 = 2bc$ $b^2 + c^2 + 2bc - a^2 = 2bc$ $b^2 + c^2 = a^2$ This implies $	riangle ABC$ is a right-angled triangle with $\angle A = 90^{\circ}$. In a right-angled triangle,the circumradius $R = \frac{a}{2}$,so $2R = a$.

In a $\triangle ABC$,with usual notation,if $r=r_1-r_2-r_3$,then $2R=$

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