In a triangle ABC,if frac{2 r_2 r_3}{r_2-r_1} | vedclass

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(D) Given,$\frac{2 r_2 r_3}{r_2-r_1} = r_3-r_1$. Using $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$,we get: $2 \cdot

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$2R$

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(D) Given,$\frac{2 r_2 r_3}{r_2-r_1} = r_3-r_1$. Using $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$,we get: $2 \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} = \left(\frac{\Delta}{s-b} - \frac{\Delta}{s-a}\right) \left(\frac{\Delta}{s-c} - \frac{\Delta}{s-a}\right)$ $\frac{2}{(s-b)(s-c)} = \frac{(s-a)-(s-b)}{(s-b)(s-a)} \cdot \frac{(s-a)-(s-c)}{(s-c)(s-a)}$ $2(s-a)^2 = (b-a)(c-a)$ $2(\frac{b+c-a}{2})^2 = (b-a)(c-a)$ $\frac{(b+c-a)^2}{2} = (b-a)(c-a)$ $b^2+c^2+a^2+2bc-2ab-2ac = 2(bc-ab-ac+a^2)$ $b^2+c^2+a^2+2bc-2ab-2ac = 2bc-2ab-2ac+2a^2$ $b^2+c^2 = a^2$. Now,$\sqrt{r_1 r_2+r_2 r_3+r_3 r_1} = \sqrt{\frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}} = \sqrt{\frac{\Delta^2(s-c+s-a+s-b)}{(s-a)(s-b)(s-c)}} = \sqrt{\frac{\Delta^2(3s-2s)}{\frac{\Delta^2}{s}}} = s$. Thus,$\frac{r_1(r_2+r_3)}{s} = \frac{\frac{\Delta}{s-a}(\frac{\Delta}{s-b} + \frac{\Delta}{s-c})}{s} = \frac{\Delta^2(2s-b-c)}{s(s-a)(s-b)(s-c)} = \frac{\Delta^2(a)}{\Delta^2} = a = 2R$.

In a $\triangle ABC$,if $\frac{2 r_2 r_3}{r_2-r_1}=r_3-r_1$,then $\frac{r_1(r_2+r_3)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} = $

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