Question

If $p_1,p_2,p_3$ are respectively the perpendicular from the vertices of a triangle to the opposite sides, then find the value of $p_1{p}_2{p}_3$

Answer 1

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Let $p_1$ be the altitude of AD $p_2$ of BE and $p_3$ of FC.

Using area of triangle ABC = $\Delta =\frac{1}{2}\times$ side length $\times$ height

Area of $\Delta ABC=\Delta =\frac{1}{2}a{p}_1$

$=\frac{1}{2}b{p}_2$

$=\frac{1}{2}c{p}_3$

$\Rightarrow p_1=\frac{2\Delta }{a},p_2=\frac{2\Delta }{b},p_3=\frac{2\Delta }{c}$

$\therefore p_1{p}_2{p}_3=\frac{2\Delta }{a}.\frac{2\Delta }{b}.\frac{2\Delta }{c}$

$=\frac{8{\Delta }^3}{abc}$

We have $\Delta =abc/4R\Rightarrow abc=4R\Delta$

$p_1{p}_2{p}_3=\frac{8{\Delta }^3}{abc}=\frac{8{\Delta }^3}{4R\Delta }=\frac{2{\Delta }^2}{R}$