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

1. right angled triangle
2. equilateral triangle
3. $\angle B=60^{\circ}$
4. $\angle C=45^{\circ}$

Solution

$\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$
$ \begin{aligned} & \text { Now, } r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c} \\ & \therefore\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-b}}\right)\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-c}}\right) \\ & =\left(1-\frac{s-b}{s-a}\right)\left(1-\frac{s-c}{s-a}\right)=\frac{(b-a)(c-a)}{(s-a)^2}=2 \\ & \therefore b c+a^2-a b-a c=\frac{1}{2}(2 s-2 a)^2 \\ & \Rightarrow 2 b c+2 a^2-2 a b-2 a c=(b+c-a)^2 \\ & =b^2+c^2+a^2-2 a b-2 a c+2 b c \\ & \Rightarrow \quad a^2=b^2+c^2 \end{aligned} $ $\therefore A B C$ is right angled triangle

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

Practice more Properties of Triangles questions on Aicharya