Question

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

• A. Right angled
• B. Isosceles
• C. Equilateral
• D. none of these

Answer 1

Answer: (A)

Right angled

$(1-\frac{r_1}{r_2})(1-\frac{r_1}{r_3})=2$
$\Rightarrow (1-\frac{\frac{△}{s-a}}{\frac{△}{s-b}})(1-\frac{\frac{△}{s-a}}{\frac{△}{s-c}})=2$
$\Rightarrow (1-\frac{s-b}{s-a})(1-\frac{s-c}{s-a})=2$
$\Rightarrow (s-a-s+b)(s-a-s+c)=2(s-a)(s-a)$
$\Rightarrow (b-a)(c-a)=2{(s-a)}^2$
$\Rightarrow 2(bc-ab-ac+a^2)={(2s-2a)}^2$
$\Rightarrow 2bc-2ab-2ac+2a^2={(a+b+c-2a)}^2$
$\Rightarrow 2bc-2ab-2ac+2a^2={(b+c-a)}^2$
$\Rightarrow 2bc-2ab-2ac+2a^2=b^2+c^2+a^2+2bc-2ca-2ab$
On simplification, we get
$\Rightarrow 2a^2-a^2=b^2+c^2$
$\Rightarrow b^2+c^2=a^2$
$\Rightarrow b^2+c^2-a^2=0$
$\Rightarrow 2bc\cos A=0$
$\therefore \angle A={90}^0$