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Mathematics

If (a + b + c + d) : (a + b − c − d) = (a − b + c − d) : (a − b − c + d), prove that a : b = c : d.

Ratio Proportion

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Answer

Given,

⇒a+b+c+da+b−c−d=a−b+c−da−b−c+d\Rightarrow \dfrac{a + b + c + d}{a + b - c - d} = \dfrac{a - b + c - d}{a - b - c + d}

Applying componendo and dividendo, we get :

⇒(a+b+c+d)+(a+b−c−d)(a+b+c+d)−(a+b−c−d)=(a−b+c−d)+(a−b−c+d)(a−b+c−d)−(a−b−c+d)⇒(a+b+c+d)+(a+b−c−d)(a+b+c+d)−a−b+c+d=(a−b+c−d)+(a−b−c+d)(a−b+c−d)−a+b+c−d⇒2(a+b)2(c+d)=2(a−b)2(c−d)⇒a+bc+d=a−bc−d⇒a+ba−b=c+dc−d\Rightarrow \dfrac{(a + b + c + d) + (a + b - c - d)}{(a + b + c + d) - (a + b - c - d)} = \dfrac{(a - b + c - d) + (a - b - c + d)}{(a - b + c - d) - (a - b - c + d)} \\[1em] \Rightarrow \dfrac{(a + b + c + d) + (a + b - c - d)}{(a + b + c + d) - a - b + c + d} = \dfrac{(a - b + c - d) + (a - b - c + d)}{(a - b + c - d) - a + b + c - d} \\[1em] \Rightarrow \dfrac{2(a + b)}{2(c + d)} = \dfrac{2(a - b)}{2(c - d)} \\[1em] \Rightarrow \dfrac{a + b}{c + d} = \dfrac{a - b}{c - d} \\[1em] \Rightarrow \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}

Applying componendo and dividendo again:

⇒a+b+(a−b)a+b−(a−b)=c+d+(c−d)c+d−(c−d)⇒a+b+(a−b)a+b−a+b=c+d+(c−d)c+d−c+d⇒2a2b=2c2d⇒ab=cd.\Rightarrow \dfrac{a + b + (a - b)}{a + b - (a - b)} = \dfrac{c + d + (c - d)}{c + d - (c - d)} \\[1em] \Rightarrow \dfrac{a + b + (a - b)}{a + b - a + b} = \dfrac{c + d + (c - d)}{c + d - c + d} \\[1em] \Rightarrow \dfrac{2a}{2b} = \dfrac{2c}{2d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}.

Hence, proved that a : b = c : d.

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