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If the lengths of the perpendiculars from the vertices of a triangle ABC on the opposite sides are `p_(1), p_(2), p_(3)` then prove that `(1)/(p_(1)) + (1)/(p_(2)) + (1)/(p_(3)) = (1)/(r) = (1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3))`.

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`(1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3)) = (s -a)/(Delta) + (s -b)/(Delta) + (s-c)/(Delta) = (s)/(Delta) = (1)/(r)`
Also, `(1)/(p_(1)) + (1)/(p_(2)) + (1)/(p_(3)) = (a)/(2Delta) + (b)/(2Delta) + (c)/(2Delta) = (s)/(Delta) = (1)/(r) { :' Delta = (1)/(2) a.p_(1) = (1)/(2) b.p_(2) = (1)/(2) b.p_(3)}`
Hence, `(1)/(p_(1)) + (1)/(p_(2)) + (1)/(p_(3)) = (1)/(r) + (1)/(r_(2)) + (1)/(r_(3))`
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