Let the angles A, B, C of a triangle ABC be i | vedclass

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(D) Given that $A, B, C$ are in arithmetic progression, we have $2B = A + C$. Since $A + B + C = 180^{\circ}$, we get $3B = 180^{\circ}$, so $B = 60^{

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$a$

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(D) Given that $A, B, C$ are in arithmetic progression, we have $2B = A + C$. Since $A + B + C = 180^{\circ}$, we get $3B = 180^{\circ}$, so $B = 60^{\circ}$.Using the formula for exradii $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, and $r_3 = \frac{\Delta}{s-c}$, the given condition $r_3^2 = r_1 r_2 + r_2 r_3 + r_3 r_1$ can be rewritten by dividing by $r_1 r_2 r_3$ as $\frac{1}{r_1 r_2} = \frac{1}{r_3 r_1} + \frac{1}{r_2 r_3} + \frac{1}{r_1 r_2}$ is not correct, rather divide by $r_1 r_2 r_3$ to get $\frac{1}{r_1 r_2} = \frac{1}{r_3 r_2} + \frac{1}{r_3 r_1} + \frac{1}{r_1 r_2}$ is wrong. Let's use $r_1 = s 	an(A/2)$, $r_2 = s 	an(B/2)$, $r_3 = s 	an(C/2)$ is wrong. The correct relations are $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, $r_3 = \frac{\Delta}{s-c}$.Given $r_3^2 = r_1 r_2 + r_2 r_3 + r_3 r_1$, dividing by $r_1 r_2 r_3$ gives $\frac{1}{r_1 r_2 r_3} (r_3^2) = \frac{1}{r_3} + \frac{1}{r_1} + \frac{1}{r_2}$.We know $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$, where $r$ is the inradius.Also, $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s-a-b}{\Delta} = \frac{c}{\Delta} = \frac{1}{r_3} + \frac{1}{r_3} = \frac{2}{h_c}$ is not helpful.For $B=60^{\circ}$, $b^2 = a^2 + c^2 - 2ac \cos(60^{\circ}) = a^2 + c^2 - ac$.The condition $r_3^2 = r_1 r_2 + r_2 r_3 + r_3 r_1$ leads to $b=a$.

Let the angles $A, B, C$ of a triangle $ABC$ be in arithmetic progression. If the exradii $r_1, r_2, r_3$ of triangle $ABC$ satisfy the condition $r_3^2 = r_1 r_2 + r_2 r_3 + r_3 r_1$, then $b =$

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