If $P_{1}, P_{2}, P_{3}$ are the altitudes drawn from vertices A, B,C to the opposite sides of a triangle respectively then show that
$i) \frac{1}{P_{1}} + \frac{1}{P_{2}} + \frac{1}{P_{3}} = \frac{1}{r}$
$ii) \frac{1}{P_{1}} + \frac{1}{P_{2}} - \frac{1}{P_{3}} = \frac{1}{r_{3}}$
$iii) P_{1} P_{2} P_{3} = \frac{(a b c)^{2}}{8 R^{3}}$

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Detailed Solution

$Δ = \frac{1}{2} a p_{1}, Δ = \frac{1}{2} b p_{2}, Δ = \frac{1}{2} c p_{3}$

$\Rightarrow P_{1} = \frac{2 Δ}{a}, P_{2} = \frac{2 Δ}{b}, P_{3} = \frac{2 Δ}{c}$

$i) \frac{1}{P_{1}} + \frac{1}{P_{2}} + \frac{1}{P_{3}} = \frac{a}{2 Δ} + \frac{b}{2 Δ} + \frac{c}{2 Δ}$

$= \frac{a + b + c}{2 Δ} = \frac{2 s}{2 Δ} = \frac{1}{r}$

$ii) \frac{1}{P_{1}} + \frac{1}{P_{2}} - \frac{1}{P_{3}} = \frac{1}{r_{3}}$

$LHS = \frac{a}{2 Δ} + \frac{b}{2 Δ} - \frac{c}{2 Δ} = \frac{a + b - c}{2 Δ}$

$= \frac{(a + b + c) - c - c}{2 Δ}$

$= \frac{2 s - 2 c}{2 Δ} = \frac{2 (s - c)}{2 Δ} = \frac{s - c}{Δ} = \frac{1}{r_{3}}$

$iii) P_{1} P_{2} P_{3} = \frac{2 Δ}{a} \cdot \frac{2 Δ}{b} \cdot \frac{2 Δ}{c} = \frac{8 Δ^{3}}{a b c}$

$= \frac{8}{a b c} {[\frac{a b c}{4 R}]}^{3} = \frac{(a b c)^{2}}{8 R^{3}}$

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