In $\triangle A B C, \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=$

Question

Aicharya · Accepted Answer

$\because$$r_1=\frac{\Delta}{s-a}$ $\therefore$ $\frac{1}{r_1}=\frac{s-a}{\Delta}$ Similarly, $\frac{1}{r_2}=\frac{s-b}{\Delta} \frac{1}{r_3}=\frac{s-c}{\Delta}$ and $\frac{1}{r}=\frac{s}{\Delta}$ Now, $\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}$ $\begin{aligned} & =\frac{(s-a)^2}{\Delta^2}+\frac{(s-b)^2}{\Delta^2}+\frac{(s-c)^2}{\Delta^2}+\frac{s^2}{\Delta^2} \ & =\frac{s^2+a^2-2 a s+s^2+b^2-2 s b+s^2+c^2-2 s c+s^2}{\Delta^2} \ & =\frac{4 s^2+a^2+b^2+c^2-2 s(a+b+c)}{\Delta^2} \ & =\frac{4 s^2+a^2+b^2+c^2-4 s^2}{\Delta^2}=\frac{a^2+b^2+c^2}{\Delta^2} \end{aligned}$

In \(\triangle A B C, \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=\)

Solution