Question

$if$ $r_1=r_2+r_3+r,$ then the $\Delta$ is

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

Answer 1

Answer: (C)

Right angled

$r_1=r_2+r_3+r$

$\Rightarrow r_1-r=r_2+r_3$

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

$\frac{r-r+a}{r(r-a)}=\frac{r-b+r-c}{(r-b)(r-c)}$

$\frac{a}{r(r-a)}=\frac{a+b+c-b-c}{(r-b)(r-c)}$

$\frac{a}{r(r-a)}=\frac{a}{(r-b)(r-c)}$

$\frac{1}{r(r-a)}=\frac{1}{(r-b)(r-c)}$

$(r-b)(r-c)=r(r-a)$

$r^2-br-cr+bc=r^2-ar$

$(a-b-c)r+bc=0$

$(b+c-a)r=bc$

$(b+c-a)\frac{(a+b+c)}{2}=bc$

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

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

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

$\therefore$ The triangle is a right triangle.