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

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(A) We know that $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, and $r_3 = \frac{\Delta}{s-c}$.Then, $r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{\De

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

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(A) We know that $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, and $r_3 = \frac{\Delta}{s-c}$.Then, $r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}$.Taking $\frac{\Delta^2}{(s-a)(s-b)(s-c)}$ as a common factor, we get:$\frac{\Delta^2}{(s-a)(s-b)(s-c)} [(s-c) + (s-a) + (s-b)]$.Since $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$, the expression becomes:$\frac{\Delta^2}{\Delta^2/s} [3s - (a+b+c)]$.Using $a+b+c = 2s$, we have:$s [3s - 2s] = s^2$.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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