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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+d)(a−b−c+d)=(a+b−c−d)(a−b+c−d)⇒a+b+c+da+b−c−d=a−b+c−da−b−c+d\Rightarrow (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} = \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+a+b+b+c−c+d−da−a+b−b+c+c+d+d=a+a−b−b+c−c−d+da−a−b+b+c+c−d−d⇒2(a+b)2(c+d)=2(a−b)2(c−d)⇒(a+b)(c+d)=(a−b)(c−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 + a + b + b + c - c + d - d}{a - a + b - b + c + c + d + d} = \dfrac{a + a - b - b + c - c - d + d}{a - a - b + b + c + c - d - 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}

Again applying componendo and dividendo, we get :

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

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

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