Mathematics - Properties of Triangle Question with Solution | TestHub

Let $\Delta$ be the area of $△ABC$. The area can be expressed in terms of altitudes as:$\Delta =\frac{1}{2}a{h}_a=\frac{1}{2}b{h}_b=\frac{1}{2}c{h}_c$. From these relations, we can write $h_a=\frac{2\Delta }{a}$, $h_b=\frac{2\Delta }{b}$, $h_c=\frac{2\Delta }{c}$. So, $\frac{1}{h_a}=\frac{a}{2\Delta }$, $\frac{1}{h_b}=\frac{b}{2\Delta }$, $\frac{1}{h_c}=\frac{c}{2\Delta }$. Adding these reciprocals: $\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{a+b+c}{2\Delta }$. We know that the semi-perimeter $s=\frac{a+b+c}{2}$, so $a+b+c=2s$. Thus, $\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{2s}{2\Delta }=\frac{s}{\Delta }$. Also, the inradius $r$ is given by $r=\frac{\Delta }{s}$. Therefore, $\frac{s}{\Delta }=\frac{1}{r}$. Hence, $\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{r}$.