Question

If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.

Answer 1

`(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]`