In triangle ABC,if angle C=90^{circ},then lef | vedclass

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(D) Given $\angle C=90^{\circ}$,we have $c^2=a^2+b^2$. Using the formulas $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delt

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$2$

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(D) Given $\angle C=90^{\circ}$,we have $c^2=a^2+b^2$. Using the formulas $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$: $\left(\frac{r_1-r_3}{r_1}\right)\left(\frac{r_2-r_3}{r_2}\right) = \left(1 - \frac{r_3}{r_1}\right)\left(1 - \frac{r_3}{r_2}\right) = \left(1 - \frac{s-a}{s-c}\right)\left(1 - \frac{s-b}{s-c}\right)$ $= \left(\frac{s-c-s+a}{s-c}\right)\left(\frac{s-c-s+b}{s-c}\right) = \left(\frac{a-c}{s-c}\right)\left(\frac{b-c}{s-c}\right)$ $= \frac{ab - ac - bc + c^2}{(s-c)^2} = \frac{ab - ac - bc + a^2 + b^2}{(\frac{a+b-c}{2})^2}$ $= \frac{4(a^2+b^2+ab-ac-bc)}{(a+b-c)^2} = \frac{4(a^2+b^2+ab-ac-bc)}{a^2+b^2+c^2+2ab-2bc-2ac}$ Since $c^2 = a^2+b^2$,the denominator becomes $2(a^2+b^2) + 2ab - 2bc - 2ac = 2(a^2+b^2+ab-bc-ac)$. Thus,the expression simplifies to $\frac{4(a^2+b^2+ab-ac-bc)}{2(a^2+b^2+ab-ac-bc)} = 2$.

In $\triangle ABC$,if $\angle C=90^{\circ}$,then $\left(\frac{r_1-r_3}{r_1}\right)\left(\frac{r_2-r_3}{r_2}\right)=$

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