Question

in triangle ABC prove that r1+r2+r3-r=4r​

Answer 1

Answer:

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Answer 2

Given :  in triangle ABC r₁  + r₂ + r₃  - r = 4R

To Find : Prove

Solution:

Δ  is the area of triangle

s = semi perimeter

a , b , c are the side length of triangle

r₁  , r₂ and r₃ are ex radii

r = inradius

R = circumradius

r₁ = Δ /(s - a)

r₂ = Δ /(s - b)

r₃ = Δ /(s - c)

r = Δ/s

r₁  + r₂ + r₃  - r = 4R

LHS

=  Δ /(s - a) +  Δ /(s - b) +  Δ /(s - c) -  Δ /s

=  Δ [  {(s - b) + (s - a)}/{(s-a)(s-b)}  +   (s - s + c)/s(s-c)]

= Δ [  {2s - (a + b))}/{(s-a)(s-b)}  +   (c)/s(s-c)]

2s  = a + b +  c  => 2s - (a + b) = c

= Δ [  c/{(s-a)(s-b)}  +   (c)/s(s-c)]

= Δc [ { (s)(s-c) + (s -a)(s-b) } / (s(s-a)(s-b)(s-c))]

s(s-a)(s-b)(s-c) = Δ²

(s)(s-c) + (s -a)(s-b)  = s² - sc +s² - bs - as + ab

= 2s² - s(a + b + c) + ab = 2s² - s(2s) + ab = ab

=  Δc [ ab / Δ²]

=abc/Δ

= 4abc/(4Δ)

R = abc/4Δ

= 4R

= RHS

QED

Hence proved

r₁  + r₂ + r₃  - r = 4R

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