Problem : If a, b and c are the lengths of the sides of a triangle ABC and  if \( p_1 , p_2 \) and \( p_3 \)   are the lengths of the perpendiculars drawn from the circumcentre onto the sides BC, CA and AB respectively, then show that

{\displaystyle{\frac{a}{p_1}} + {\frac{b}{p_2}} + {\frac{c}{p_3}} = {\frac{abc}{4{p_1}{p_2}{p_3}}}} .

Solution : Let  O be the circum centre

{\displaystyle{\frac{a}{p_1}}} = 2 {\displaystyle \left({\frac{\frac{a}{2}}{p_1}}\right) = 2 \tan \left({\frac{1}{2}}{\angle{BOC}}\right)}

Similarly, {\displaystyle{\frac{b}{p_2}} =  2 \tan \left({\frac{1}{2}}{\angle{AOC}}\right)}

& {\displaystyle{\frac{c}{p_3}} = 2 \tan \left({\frac{1}{2}}{\angle{AOB}}\right)}
Now, {\displaystyle{\frac{1}{2}}{\angle{BOC}} + {\frac{1}{2}}{\angle{AOC}} + {\frac{1}{2}}{\angle{AOB}}} = {\displaystyle{\frac{1}{2}} ({\angle{BOC}} + {\angle{AOC}} + {\angle{AOB}})}
= {\displaystyle{\frac{1}{2}} (360^{\circ})} = {\displaystyle{180^{\circ}}}


Lemma
For {\displaystyle{\alpha},{\beta},{\gamma} > 0} & {\displaystyle{\alpha} + {\beta} + {\gamma} = {\pi}}
{\displaystyle{\tan{\alpha}} + {\tan{\beta}} + {\tan{\gamma}} = {\tan{\alpha}}{\times}{\tan{\beta}}{\times}{\tan{\gamma}}}
{\displaystyle{\alpha} + {\beta} + {\gamma} = {\pi}}
{\Rightarrow} {\displaystyle{\alpha + \beta} = {\pi - \gamma}}
{\Rightarrow} {\displaystyle{\tan (\alpha + \beta)} = {\tan (\pi - \gamma)}}
{\Rightarrow} {\displaystyle{\frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha} \tan{\beta}}} = -\tan{\gamma}}
{\Rightarrow} {\displaystyle{\tan\alpha + \tan \beta} = {\tan{\alpha}. \tan{\beta} .\tan{\gamma}} - \tan\gamma}
{\Rightarrow} {\displaystyle{\tan{\alpha}} + {\tan{\beta}} + {\tan{\gamma}} = {\tan{\alpha}}{\times}{\tan{\beta}}{\times}{\tan{\gamma}}} (lemma proved)


So, {\displaystyle{\frac{1}{2}}\left({\frac{a}{p_1}} + {\frac{b}{p_2}} + {\frac{c}{p_3}}\right) = \tan({\frac{1}{2}}\angle{AOB}) + \tan({\frac{1}{2}}\angle{BOC}) + \tan({\frac{1}{2}}\angle{COA})}
= {\displaystyle{\tan({\frac{1}{2}}\angle{AOB})\times \tan({\frac{1}{2}}\angle{BOC})\times \tan({\frac{1}{2}}\angle{COA})}} [ as {\displaystyle{\angle{AOB} + \angle{BOC} + \angle{COA} = {360}^{\circ}}} ]
= {\displaystyle{\frac{a}{2{p_1}}}\times{\frac{b}{2{p_2}}}\times{\frac{c}{2{p_3}}}}
= {\displaystyle{\frac{abc}{8{p_1}{p_2}{p_3}}}}
\Rightarrow \frac{a}{p_1} + \frac{b}{p_2} + \frac{c}{p_3} = \frac{abc}{4{p_1}{p_2}{p_3} } ( proved )