If p_1, p_2, p_3 are the altitudes of a trian | vedclass

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(D) We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$ and $\frac{1}{r} = \frac

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

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(D) We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$ and $\frac{1}{r} = \frac{s}{\Delta}$.Now,$\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{1}{\Delta^2} [(s-a)^2 + (s-b)^2 + (s-c)^2 + s^2]$.$= \frac{1}{\Delta^2} [4s^2 - 2s(a+b+c) + a^2 + b^2 + c^2]$.Since $a+b+c = 2s$,we have $4s^2 - 2s(2s) + a^2 + b^2 + c^2 = a^2 + b^2 + c^2$.So,the expression equals $\frac{a^2 + b^2 + c^2}{\Delta^2}$.Also,$\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$,which implies $a = \frac{2\Delta}{p_1}, b = \frac{2\Delta}{p_2}, c = \frac{2\Delta}{p_3}$.Substituting these,we get $\frac{1}{\Delta^2} [(\frac{2\Delta}{p_1})^2 + (\frac{2\Delta}{p_2})^2 + (\frac{2\Delta}{p_3})^2] = 4(\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2})$.

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

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