If $p_1,p_2,p_3$ are the altitudes of a triangle $ABC$ from the vertices $A\mathit{,}B\mathit{,}C$ respectively, then with the usual notation, $\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=$

$\frac{1}{r_1}=\frac{s-a}{∆}, \frac{1}{r_2}=\frac{s-b}{∆}$

$\frac{1}{r_3}=\frac{s-c}{∆} \& \frac{1}{r}=\frac{s}{∆}$

Now, $\frac{1}{{r_1}^2}+\frac{1}{{r_2}^2}+\frac{1}{{r_3}^2}+\frac{1}{r^2}$

$=\frac{1}{{∆}^2}\left[\right(s-a{)}^2+(s-b{)}^2+(s-c{)}^2+s^2]$

$=\frac{1}{{∆}^2}[4s^2-2s(a+b+c)+a^2+b^2+c^2]$

Here, $a+b+c=2s$

$=\frac{1}{{∆}^2}[4s^2-4s^2+(a^2+b^2+c^2\left)\right]$

$=\frac{a^2+b^2+c^2}{{∆}^2}$

Also, $∆=\frac{1}{2}\left(a{p}_1\right)$

              $=\frac{1}{2}\left(b{p}_2\right)$

               $=\frac{1}{2}\left(c{p}_3\right)$

So, $\frac{1}{{r_1}^2}+\frac{{\displaystyle 1}}{{\displaystyle {r_2}^2}}+\frac{{\displaystyle 1}}{{\displaystyle {r_3}^2}}+\frac{{\displaystyle 1}}{{\displaystyle r^2}}$

$=\frac{1}{{∆}^2}\left[\right(\frac{2∆}{p_1}{)}^2+(\frac{2∆}{p_2}{)}^2+(\frac{2∆}{p_3}{)}^2]$

$=4\left(\frac{1}{{p_1}^2}+\frac{1}{{p_2}^2}+\frac{1}{{p_3}^2}\right)$