If $△$ ABC is right angled at A then $r_{2}$ + $r_{3}$ =

r_{1} - r

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r_{1} + r

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r - r_{1}

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Solution

The correct option is B $r_{1} - r$
$r_{2} + r_{3} = \frac{△}{S - b} + \frac{△}{S - c}$ .
$= △ (\frac{2 S - b - c}{(S - b) (S - c)} \times \frac{S (S - a)}{S (S - a)})$
$= △ (\frac{a s (S - a)}{△^{2}})$
$= \frac{a S (S - a)}{△} = \frac{a}{t a n A / 2} = a (A = \frac{π}{2})$
$r_{1} = \frac{△}{S - a} \times \frac{S}{S} = S$
$r = \frac{△}{S} \times \frac{S - a}{S - a} = S - a$
$\Rightarrow r_{1} - r = a = r_{2} + r_{3} \Rightarrow (A)$

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