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Mathematics

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

show that : 2ad = 3bc.

Ratio Proportion

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Answer

Given,

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

Applying componendo and dividendo:

⇒a−2b−3c+4d+a+2b−3c−4da−2b−3c+4d−(a+2b−3c−4d)=a−2b+3c−4d+a+2b+3c+4da−2b+3c−4d−(a+2b+3c+4d)⇒2(a−3c)2(4d−2b)=2(a+3c)2(−4d−2b)⇒(a−3c)(4d−2b)=(a+3c)(−4d−2b)\Rightarrow \dfrac{a - 2b - 3c + 4d + a + 2b - 3c - 4d}{a - 2b - 3c + 4d - (a + 2b - 3c - 4d)} = \dfrac{a - 2b + 3c - 4d + a + 2b + 3c + 4d}{a - 2b + 3c - 4d - (a + 2b + 3c + 4d)} \\[1em] \Rightarrow \dfrac{2(a - 3c)}{2(4d - 2b)} = \dfrac{2(a + 3c)}{2(-4d - 2b)} \\[1em] \Rightarrow \dfrac{(a - 3c)}{(4d - 2b)} = \dfrac{(a + 3c)}{(-4d - 2b)}

Applying alternendo:

⇒(a−3c)(a+3c)=(4d−2b)(−4d−2b)⇒a−3c+a+3ca−3c−(a+3c)=4d−2b+(−4d−2b)4d−2b−(−4d−2b)⇒2a−6c=−4b8d⇒a−3c=−b2d⇒−2ad=−3bc⇒2ad=3bc.\Rightarrow \dfrac{(a - 3c)}{(a + 3c)} = \dfrac{(4d - 2b)}{(-4d - 2b)} \\[1em] \Rightarrow \dfrac{a - 3c + a + 3c}{a - 3c - (a + 3c)} = \dfrac{4d - 2b + (-4d - 2b)}{4d - 2b - (-4d - 2b)} \\[1em] \Rightarrow \dfrac{2a}{-6c} = \dfrac{-4b}{8d} \\[1em] \Rightarrow \dfrac{a}{-3c} = \dfrac{-b}{2d} \\[1em] \Rightarrow -2ad = -3bc \\[1em] \Rightarrow 2ad = 3bc.

Hence, proved that 2ad = 3bc.

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