If a, b, c, d are in continued proportion, then prove that ` (b -c)^(2) + (c -a)^(2) + (b-d)^(2) = (a -d)^(2)`

Explanation

Since a,b,c,d are in continued proportion, ` a/b = b/c = c/d = k " (let)" [k ne 0]`
` :. A = bk, " "b = ck, " "c = k`
` = ck.k" =dk.k`
` = ck^(2) " "= dk^(2)`
` = dk.k^(2)`
` = dk^(3)`
Now, LHS ` = (b-c)^(2) + (c - a)^(2) + (b -d)^(2)`
` = (dk^(2) - dk)^(2) + (dk - dk^(3))^(2) + (dk^(2)-d)^(2)`
` = d^(2) k^(4) - 2d^(2) k^(3) + d^(2)k^(2) + d^(2)k^(2) - 2d^(2)k^(4) + d^(2)k^(6) + d^(2)k^(4)-2d^(2)k^(2)+d^(2)`
` = d^(2)k^(6) - 2d^(2)k^(3)+d^(2)`
RHS ` = (a-d)^(2)`
` = (dk^(3)-d)^(2)`
` = d^(2)k^(6) - 2d^(2)k^(3) +d^(2)`
` :. ` LHS = RHS .