# If r_1=r_2+r_3+r prove that the triangle is right angled.?

Jan 16, 2018

We know

${r}_{1} = \frac{\Delta}{s - a}$

${r}_{2} = \frac{\Delta}{s - b}$

${r}_{3} = \frac{\Delta}{s - c}$

$r = \frac{\Delta}{s}$

Inserting these values in the given relation.

${r}_{1} = {r}_{2} + {r}_{3} + r$

$\implies {r}_{1} - r = {r}_{2} + {r}_{3}$

$\implies \frac{\Delta}{s - a} - \frac{\Delta}{s} = \frac{\Delta}{s - b} + \frac{\Delta}{s - c}$

$\implies \frac{1}{s - a} - \frac{1}{s} = \frac{1}{s - b} + \frac{1}{s - c}$

$\implies \frac{2}{2 s - 2 a} - \frac{2}{2 s} = \frac{2}{2 s - 2 b} + \frac{2}{2 s - 2 c}$

$\implies \frac{1}{b + c - a} - \frac{1}{b + c + a} = \frac{1}{c + a - b} + \frac{1}{a + b - c}$

$\implies \frac{\left(b + c + a\right) - \left(b + c - a\right)}{{\left(b + c\right)}^{2} - {a}^{2}} = \frac{c + a - b + a + b - c}{{a}^{2} - {\left(b - c\right)}^{2}}$

$\implies \frac{2 a}{{\left(b + c\right)}^{2} - {a}^{2}} = \frac{2 a}{{a}^{2} - {\left(b - c\right)}^{2}}$

$\implies {\left(b + c\right)}^{2} - {a}^{2} = {a}^{2} - {\left(b - c\right)}^{2}$

$\implies {\left(b + c\right)}^{2} + {\left(b - c\right)}^{2} = {a}^{2} + {a}^{2}$

$\implies 2 {b}^{2} + 2 {c}^{2} = 2 {a}^{2}$

$\implies {b}^{2} + {c}^{2} = {a}^{2}$

This means the triangle is right angled.