Question

If in a $△ABC,(r_2-r_1)(r_3-r_1)=2r_2{r}_3$, then the $△ABC$ is

• A. Acute angles
• B. Obtuse angled
• C. Equilateral triangle
• D. Right angled triangle

Answer 1

The correct option is D Right angled triangle

Given $(r_2-r_1)(r_3-r_1)=2r_2{r}_3$

$\Rightarrow [\frac{\Delta }{s-b}-\frac{\Delta }{s-a}][\frac{\Delta }{s-c}-\frac{\Delta }{s-a}]=2\frac{\Delta }{s-b}.\frac{\Delta }{s-c}$

$\Rightarrow \left[\frac{(s-a)-(s-b)}{(s-a)(s-b)}\right]\left[\frac{(s-a)-(s-c)}{(s-c)(s-a)}\right]=\frac{2}{(s-b)(s-c)}$

$\Rightarrow \left[\frac{(b-a)(c-a)}{{(s-a)}^2}\right]=2$

$\Rightarrow (b-a)(c-a)=2{(s-a)}^2\Rightarrow 2(b-a)(c-a)={(2s-2a)}^2$

$\Rightarrow 2(b-a)(c-a)={(b+c-a)}^2$

$\Rightarrow 2(bc-ab-ac+a^2)=b^2+c^2+a^2+2bc-2ca-2ab$

$\Rightarrow a^2=b^2+c^2$

$\therefore △ABC$ is a right angled triangle.