Question

The base of a triangle is divided into three equal parts. If t₁ , t₂ , t₃ be the tangents of the angles subtended by these parts at the opposite vertex, prove that

4(1 + 1/t₂²) = (1/t₁ + 1/t₂)(1/t₂ + 1/t₃)

Answer 1

Let point D and E divides the base BC into three equal parts i.e. BD = DE = DC = d (Let) and let α, β and γ be the angles subtended by BD, DE and EC respectively at their opposite vertex. 

⇒ t₁ = tanα, t₂ = tanβ and t₃ = tanγ

Now in ∆ABC 

∵ BE : EC = 2d : d = 2 : 1 

∴ from m-n rule, we get (2 + 1) cotθ = 2 cot (α + β) – cotγ 

⇒ 3cotθ = 2 cot (α + β) – cotγ .........(i)

again 

∵ in ∆ADC 

∵ DE : EC = x : x = 1 : 1 

∴ if we apply m-n rule in ∆ADC, we get 

(1 + 1) cotθ = 1. cotβ – 1 cotγ 2cotθ = cotβ – cotγ .........(ii) 

from (i) and (ii), we get