If in a triangle left(1-frac{r_1}{r_2}right)l | vedclass

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(A) Given: $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$ We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_

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right angled triangle

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(A) Given: $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$ We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$. Substituting these values: $\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-b}}\right)\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-c}}\right) = 2$ $\left(1-\frac{s-b}{s-a}\right)\left(1-\frac{s-c}{s-a}\right) = 2$ $\left(\frac{s-a-s+b}{s-a}\right)\left(\frac{s-a-s+c}{s-a}\right) = 2$ $\left(\frac{b-a}{s-a}\right)\left(\frac{c-a}{s-a}\right) = 2$ $(b-a)(c-a) = 2(s-a)^2$ $bc - ab - ac + a^2 = 2\left(\frac{b+c-a}{2}\right)^2$ $bc - ab - ac + a^2 = 2\frac{(b+c-a)^2}{4}$ $2(bc - ab - ac + a^2) = b^2 + c^2 + a^2 + 2bc - 2ab - 2ac$ $2bc - 2ab - 2ac + 2a^2 = b^2 + c^2 + a^2 + 2bc - 2ab - 2ac$ $2a^2 = b^2 + c^2 + a^2$ $a^2 = b^2 + c^2$ Since $a^2 = b^2 + c^2$,the triangle is a right-angled triangle.

If in a triangle $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$,then the triangle is

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