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Question
If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.
Solution
`(a + 3b + 2c + 6d)/(a – 3b – 2c + 6d) = (a + 3b – 2c – 6d)/(a – 3b + 2c – 6d)`
⇒ `(a + 3b + 2c + 6d)/(a + 3b – 2c – 6d) = (a – 3b + 2c – 6d)/(a – 3b – 2c + 6d)` ...(by altenendo)
Applying componendo and dividendo
`(a + 3b + 2c + 6d + a + 3b - 2c - 6d)/(a + 3b + 2c + 6d - a - 3b + 2c + 6d)`
= `(a - 3b + 2c - 6d + a - 3b - 2c + 6d)/(a – 3b + 2c - 6d - a + 3b + 2c - 6d)`
⇒ `(2(a + 3b))/(2(2c + 6d)) = (2(a - 3b))/(2(2c - 6d)`
⇒ `(a + 3b)/(2c + 6d) = (a - 3b)/(2c - 6d)` ...(Dividing by 2)
⇒ `(a + 3b)/(a - 3b) = (2c + 6d)/(2c - 6d)` ...(By alternendo)
Again applying componendo and dividendo
`(a + 3b + a - 3b)/(a + 3b - a + 3b) = (2c + 6d + 2c 6d)/(2c + 6d - 2c + 6d)`
⇒ `(2a)/(6b) = (4c)/(12d) = (2c)/(6d)`
⇒ `a/b = c/d. ...["Dividing by" 2/6]`
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