If $d_{1}, d_{2}, d_{3}$ are the diameters of the three inscribed circles of a triangle ABC,
then $d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}$ is equal to?

a b + b c + c a

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\frac{a}{b} + \frac{b}{c} + \frac{c}{a}

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(a + b + c)^{2}

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a^{2} + b^{2} + c^{2}

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Solution

The correct option is C $(a + b + c)^{2}$
Consider, $t = d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1} = 4 (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1})$
$\Rightarrow t = \frac{4 △^{2}}{(s - a) (s - b)} + \frac{4 △^{2}}{(s - b) (s - c)} + \frac{4 △^{2}}{(s - c) (s - a)}$
$\Rightarrow t = 4 △^{2} (\frac{s - c + s - a + s - b}{(s - a) (s - b) (s - c)}) = 4 △^{2} [\frac{3 s - (a + b + c)}{(s - a) (s - b) (s - c)}]$
$\Rightarrow t = 4 \frac{s (s - a) (s - b) (s - c) (3 s - 2 s)}{(s - a) (s - b) (s - c)} = {(2 s)}^{2}$
Therefore, $t = (a + b + c)^{2}$
Ans: C

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