In any triangle ABC,r_1 r_2 + r_2 r_3 + r_3 r | vedclass

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(A) We know that the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Substituting these into

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$\frac{\Delta^2}{r^2}$

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(A) We know that the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Substituting these into the expression $r_1 r_2 + r_2 r_3 + r_3 r_1$:$= \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}$$= \frac{\Delta^2(s-c + s-a + s-b)}{(s-a)(s-b)(s-c)}$Since $2s = a+b+c$,we have $s-a+s-b+s-c = 3s - (a+b+c) = 3s - 2s = s$.Also,we know that $\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.Substituting these values:$= \frac{\Delta^2 \cdot s}{\frac{\Delta^2}{s}} = s^2$Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r}$,so $s^2 = \frac{\Delta^2}{r^2}$.

In any $\triangle ABC$,$r_1 r_2 + r_2 r_3 + r_3 r_1$ is equal to

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