If ABC is an isosceles triangle with base BC, | vedclass

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(D) Given $r = \frac{\Delta}{s}$ and $r_1 = \frac{\Delta}{s-a}$.$rr_1 = \frac{\Delta^2}{s(s-a)} = \frac{s(s-a)(s-b)(s-c)}{s(s-a)} = (s-b)(s-c)$.Since

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$R^2 \sin^2 A$

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(D) Given $r = \frac{\Delta}{s}$ and $r_1 = \frac{\Delta}{s-a}$.$rr_1 = \frac{\Delta^2}{s(s-a)} = \frac{s(s-a)(s-b)(s-c)}{s(s-a)} = (s-b)(s-c)$.Since the triangle is isosceles with base $BC$,we have $b = c$.Thus,$rr_1 = (s-b)^2$.Since $2s = a+b+c = a+2b$,we have $s-b = \frac{a+2b-2b}{2} = \frac{a}{2}$.Therefore,$rr_1 = (\frac{a}{2})^2 = \frac{a^2}{4}$.Using the sine rule,$a = 2R \sin A$,so $rr_1 = \frac{(2R \sin A)^2}{4} = \frac{4R^2 \sin^2 A}{4} = R^2 \sin^2 A$.

If $ABC$ is an isosceles triangle with base $BC$,then $rr_1=$

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