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Question
If (pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd) prove that a : b : : c : d
Sum
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Solution
(pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd)
⇒ `"pa + qb"/"pc + qd" = "pq - qb"/"pc - qd"`
⇒ `"pa + qb"/"pc - qd" = "pq + qb"/"pc - qd"`
Applying componendo and dividendo
⇒ `"pa + qb + pa - qb"/"pa + qb - pa + qb" = "pc + qs + pc - qd"/"pc - qd - pc + qd"`
⇒ `(2pa)/(2qb) = (2pc)/(2qd)`
⇒ `a/b = c/d ...("Dividing by" (2p)/(2q))`
Hence a : b :: c = d.
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