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
$i) \frac{1}{P_{1}} + \frac{1}{P_{2}} + \frac{1}{P_{3}} = \frac{1}{r}$
$ii) \frac{1}{P_{1}} + \frac{1}{P_{2}} - \frac{1}{P_{3}} = \frac{1}{r_{3}}$
$iii) P_{1} P_{2} P_{3} = \frac{(a b c)^{2}}{8 R^{3}}$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

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

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

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

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

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

$= \frac{(a + b + c) - c - c}{2 Δ}$

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

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

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

JEE

NEET

Foundation JEE

Foundation NEET

CBSE