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Question
If a, b, c and d are in proportion, prove that: `(a^2 + ab)/(c^2 + cd) = (b^2 - 2ab)/(d^2 - 2cd)`
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Solution
∵ a, b, c, d are in proportion
`a/b = c/d` = k(say)
a = bk, c = dk.
L.H.S. = `(a^2 + ab)/(c^2 + cd)`
= `(k^2b^2 + bk.b)/(k^2d^2 + dk.d)`
= `(kb^2(k + 1))/(d^2k(k + 1)`
= `b^2/d^2`
R.H.S. = `(b^2 - 2ab)/(d^2 - 2cd)`
= `(b^2 - 2.bkb)/(d^2 - 2dkd)`
= `(b^2(1 - 2k))/(d^2(1 - 2k)`
= `b^2/d^2`
∴ L.H.S. = R.H.S.
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