If d_1, d_2, d_3 are the diameters of three e | vedclass

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(A) The diameters of the ex-circles are given by $d_1 = \frac{2\Delta}{s-a}$,$d_2 = \frac{2\Delta}{s-b}$,and $d_3 = \frac{2\Delta}{s-c}$.We need to ca

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$(a+b+c)^2$

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(A) The diameters of the ex-circles are given by $d_1 = \frac{2\Delta}{s-a}$,$d_2 = \frac{2\Delta}{s-b}$,and $d_3 = \frac{2\Delta}{s-c}$.We need to calculate $d_1 d_2 + d_2 d_3 + d_3 d_1 = \frac{4\Delta^2}{(s-a)(s-b)} + \frac{4\Delta^2}{(s-b)(s-c)} + \frac{4\Delta^2}{(s-c)(s-a)}$.Taking $4\Delta^2$ as a common factor,we get $4\Delta^2 \left[ \frac{(s-c) + (s-a) + (s-b)}{(s-a)(s-b)(s-c)} \right]$.Since $a+b+c = 2s$,the numerator becomes $3s - (a+b+c) = 3s - 2s = s$.Thus,the expression becomes $\frac{4\Delta^2 \cdot s}{(s-a)(s-b)(s-c)}$.Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.Substituting this,we get $\frac{4\Delta^2 \cdot s}{\Delta^2 / s} = 4s^2 = (2s)^2 = (a+b+c)^2$.

If $d_1, d_2, d_3$ are the diameters of three ex-circles of a $\triangle ABC$,then $d_1 d_2 + d_2 d_3 + d_3 d_1 =$

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