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If, 2a+2b−3c−3d2a−2b−3c+3d=a+b−4c−4da−b−4c+4d\dfrac{2a + 2b - 3c - 3d}{2a - 2b - 3c + 3d} = \dfrac{a + b - 4c - 4d}{a - b - 4c + 4d}, prove that ab=cd\dfrac{a}{b} = \dfrac{c}{d}.

Ratio Proportion

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Answer

Given,

⇒2a+2b−3c−3d2a−2b−3c+3d=a+b−4c−4da−b−4c+4d\Rightarrow \dfrac{2a + 2b - 3c - 3d}{2a - 2b - 3c + 3d} = \dfrac{a + b - 4c - 4d}{a - b - 4c + 4d}

Applying componendo and dividendo, we get :

⇒(2a+2b−3c−3d)+(2a−2b−3c+3d)(2a+2b−3c−3d)−(2a−2b−3c+3d)=(a+b−4c−4d)+(a−b−4c+4d)(a+b−4c−4d)−(a−b−4c+4d)⇒(2a+2b−3c−3d)+(2a−2b−3c+3d)(2a+2b−3c−3d)−2a+2b+3c−3d=(a+b−4c−4d)+(a−b−4c+4d)(a+b−4c−4d)−a+b+4c−4d⇒4a−6c4b−6d=2a−8c2b−8d⇒2(2a−3c)2(2b−3d)=2(a−4c)2(b−4d)⇒2a−3c2b−3d=a−4cb−4d⇒2a−3ca−4c=2b−3db−4d\Rightarrow \dfrac{(2a + 2b - 3c - 3d) + (2a - 2b - 3c + 3d)}{(2a + 2b - 3c - 3d) - (2a - 2b - 3c + 3d)} = \dfrac{(a + b - 4c - 4d) + (a - b - 4c + 4d)}{(a + b - 4c - 4d) - (a - b - 4c + 4d)} \\[1em] \Rightarrow \dfrac{(2a + 2b - 3c - 3d) + (2a - 2b - 3c + 3d)}{(2a + 2b - 3c - 3d) - 2a + 2b + 3c - 3d} = \dfrac{(a + b - 4c - 4d) + (a - b - 4c + 4d)}{(a + b - 4c - 4d) - a + b + 4c - 4d} \\[1em] \Rightarrow \dfrac{4a - 6c}{4b - 6d} = \dfrac{2a - 8c}{2b - 8d} \\[1em] \Rightarrow \dfrac{2(2a - 3c)}{2(2b - 3d)} = \dfrac{2(a - 4c)}{2(b - 4d)} \\[1em] \Rightarrow \dfrac{2a - 3c}{2b - 3d} = \dfrac{a - 4c}{b - 4d} \\[1em] \Rightarrow \dfrac{2a - 3c}{a - 4c} = \dfrac{2b - 3d}{b - 4d}

Cross - multiplying and simplifying:

⇒(2a−3c)(b−4d)=(a−4c)(2b−3d)⇒2ab−8ad−3bc+12cd=2ab−3ad−8bc+12cd⇒2ab−2ab−8ad+3ad−3bc+8bc+12cd−12cd=0⇒5bc−5ad=0⇒5bc=5ad⇒bc=ad∴ab=cd.\Rightarrow (2a - 3c)(b - 4d) = (a - 4c)(2b - 3d) \\[1em] \Rightarrow 2ab - 8ad - 3bc + 12cd = 2ab - 3ad - 8bc + 12cd \\[1em] \Rightarrow 2ab - 2ab - 8ad + 3ad - 3bc + 8bc + 12cd - 12cd = 0 \\[1em] \Rightarrow 5bc - 5ad = 0 \\[1em] \Rightarrow 5bc = 5ad \\[1em] \Rightarrow bc = ad \\[1em] \therefore \dfrac{a}{b} = \dfrac{c}{d}.

Hence, proved that ab=cd\dfrac{a}{b} = \dfrac{c}{d}.

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