In any triangle $A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} =$

{(\frac{Δ}{r})}^{2}

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\frac{Δ}{s}

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\frac{Δ}{r}

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\frac{2 Δ}{r}

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Solution

The correct option is A ${(\frac{Δ}{r})}^{2}$
We know that
$r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{Δ^{2}}{(s - a) (s - b)} + \frac{Δ^{2}}{(s - b) (s - c)} + \frac{Δ^{2}}{(s - c) (s - a)}$
$= \frac{Δ^{2} (s - c + s - a + s - b)}{(s - a) (s - b) (s - c)}$
$= \frac{Δ^{2} (3 s - (a + b + c))}{(s - a) (s - b) (s - c)}$
$= \frac{Δ^{2} (3 s - 2 s)}{(s - a) (s - b) (s - c)}$
$= \frac{s (s - a) (s - b) (s - c) (s)}{(s - a) (s - b) (s - c)}$
$= s^{2}$
$= {(\frac{Δ}{r})}^{2}$

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Similar questions

In a $△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} =$

a (r r_{1} + r_{2} r_{3}) = b (r r_{2} + r_{3} r_{1}) = c (r r_{3} + r_{1} r_{2}) = a b c

\frac{r_{1} (r_{2} + r_{3})}{\sqrt{r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}}} =

r_{1}^{2} = r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1},

\frac{r_{1} (r_{2} + r_{3})}{\sqrt{r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}}} = a

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