In triangle ABC,bc - r_2 r_3 = | vedclass

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(A) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Consider the expression $

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$rr_1$

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(A) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Consider the expression $rr_1 + r_2 r_3 = \frac{\Delta^2}{s(s-a)} + \frac{\Delta^2}{(s-b)(s-c)}$.Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $\frac{\Delta^2}{s(s-a)} = (s-b)(s-c)$ and $\frac{\Delta^2}{(s-b)(s-c)} = s(s-a)$.Thus,$rr_1 + r_2 r_3 = (s-b)(s-c) + s(s-a)$.Expanding this,we get $s^2 - s(b+c) + bc + s^2 - sa = 2s^2 - s(a+b+c) + bc$.Since $2s = a+b+c$,we have $2s^2 - s(2s) + bc = 2s^2 - 2s^2 + bc = bc$.Therefore,$rr_1 + r_2 r_3 = bc$,which implies $bc - r_2 r_3 = rr_1$.

In $\triangle ABC$,$bc - r_2 r_3 =$

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