In triangle ABC,if (r_2-r_1)(r_3-r_1)=2 r_2 r | vedclass

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(D) Given that $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$. Dividing by $r_2 r_3$,we get $(1-\frac{r_1}{r_2})(1-\frac{r_1}{r_3})=2$. Using $r_1 = \frac{\Delta}{s-a

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$2 \sqrt{2} R \cos \left(\frac{B-C}{2}\right)$

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(D) Given that $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$. Dividing by $r_2 r_3$,we get $(1-\frac{r_1}{r_2})(1-\frac{r_1}{r_3})=2$. Using $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$,we have $(1-\frac{s-b}{s-a})(1-\frac{s-c}{s-a})=2$. This simplifies to $(\frac{s-a-s+b}{s-a})(\frac{s-a-s+c}{s-a})=2$,which is $(b-a)(c-a)=2(s-a)^2$. Since $2(s-a) = b+c-a$,we have $(b-a)(c-a) = \frac{1}{2}(b+c-a)^2$. Expanding this leads to $b^2+c^2-a^2=0$,so $a^2=b^2+c^2$,meaning $\angle A=90^{\circ}$. We know $2(r+R) = 2r+2R = (b+c-a) + a = b+c$. Since $b=2R \sin B$ and $c=2R \sin C$,$b+c = 2R(\sin B + \sin C) = 2R(2 \sin \frac{B+C}{2} \cos \frac{B-C}{2})$. Since $A=90^{\circ}$,$B+C=90^{\circ}$,so $\sin \frac{B+C}{2} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$. Thus,$2(r+R) = 4R(\frac{1}{\sqrt{2}}) \cos \frac{B-C}{2} = 2 \sqrt{2} R \cos \frac{B-C}{2}$.

In $\triangle ABC$,if $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$,then $2(r+R)=$

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