Mathematics - Properties of Triangle Question with Solution | TestHub

We know the formulas for exradii $r_1=\frac{\Delta }{s-a}$, $r_2=\frac{\Delta }{s-b}$, $r_3=\frac{\Delta }{s-c}$, and for inradius $r=\frac{\Delta }{s}$, where $\Delta$ is the area and $s$ is the semi-perimeter. Therefore, $\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{s-a}{\Delta }+\frac{s-b}{\Delta }+\frac{s-c}{\Delta }=\frac{s-a+s-b+s-c}{\Delta }=\frac{3s-(a+b+c)}{\Delta }$. Since $a+b+c=2s$, the expression becomes $\frac{3s-2s}{\Delta }=\frac{s}{\Delta }$. As $r=\frac{\Delta }{s}$, it follows that $\frac{s}{\Delta }=\frac{1}{r}$. Thus, $\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}$.