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In $\triangle A B C$, if $r_1=2 r_2=3 r_3$, then
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The correct answer is:
a + c $=2 b$
Given, In a $\triangle A B C$, if $r_1=2 r_2=3 r_3$
Then, we have $\frac{\Delta}{s-a}=2 \times \frac{\Delta}{s-b}=3 \times \frac{\Delta}{s-c}$
$\Rightarrow \frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=k$ (let)
$\Rightarrow \frac{1}{s-a}=k \Rightarrow s-a=\frac{1}{k} \Rightarrow a=s-\frac{1}{k}$ ...(i)
and $\frac{2}{s-b}=k \Rightarrow \frac{s-b}{2}=\frac{1}{k} \Rightarrow b=s-\frac{2}{k}$ ...(ii)
and $\frac{3}{s-c}=k \Rightarrow \frac{s-c}{3}=\frac{1}{k} \Rightarrow c=s-\frac{3}{k}$ ...(iii)
From eqn. (i), (ii) and (iii)
$a+c=2 b$.
Then, we have $\frac{\Delta}{s-a}=2 \times \frac{\Delta}{s-b}=3 \times \frac{\Delta}{s-c}$
$\Rightarrow \frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=k$ (let)
$\Rightarrow \frac{1}{s-a}=k \Rightarrow s-a=\frac{1}{k} \Rightarrow a=s-\frac{1}{k}$ ...(i)
and $\frac{2}{s-b}=k \Rightarrow \frac{s-b}{2}=\frac{1}{k} \Rightarrow b=s-\frac{2}{k}$ ...(ii)
and $\frac{3}{s-c}=k \Rightarrow \frac{s-c}{3}=\frac{1}{k} \Rightarrow c=s-\frac{3}{k}$ ...(iii)
From eqn. (i), (ii) and (iii)
$a+c=2 b$.
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