Question

If $p_1,p_2,p_3$ are respectively the lengths of perpendiculars from the vertices of a triangle $ABC$ to the opposite sides, prove that:
$\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{1}{r}$

Answer 1

Clearly, $S=\Delta ABC=\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3$
$\therefore p_1=\frac{2S}{a},p_2=\frac{2S}{b},p_3=\frac{2S}{c}$
$\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{a+b+c}{2S}=\frac{2s}{2S}=\frac{s}{S}=\frac{1}{r}$