With usual notations in a triangle ABC,if r_1 | vedclass

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(C) Given $r_1 = 2r_2 = 2r_3$. Using the formulas for exradii,$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{2\Delta}{s-c}$. From $2r_2 = 2r_3$,we

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$4b = 3a$

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(C) Given $r_1 = 2r_2 = 2r_3$. Using the formulas for exradii,$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{2\Delta}{s-c}$. From $2r_2 = 2r_3$,we get $s-b = s-c$,which implies $b = c$. From $r_1 = 2r_2$,we have $\frac{\Delta}{s-a} = \frac{2\Delta}{s-b}$,so $s-b = 2(s-a)$. Substituting $s = \frac{a+b+c}{2}$ and $b=c$: $\frac{a+c-b}{2} = 2(\frac{b+c-a}{2})$. Since $b=c$,this simplifies to $\frac{a}{2} = 2(\frac{2b-a}{2}) = 2b-a$. $a = 4b - 2a$,which gives $3a = 4b$.

With usual notations in a triangle $ABC$,if $r_1 = 2r_2 = 2r_3$,then:

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