Question

 If a, b, c, d are in proportion, prove that abcd(a⁻² + b⁻² + c⁻² + d⁻²) = a² + b² + c² + d².

Answer 1

`a/b = c/d` = k

a = bk and c = dk

L.H.S.

= abcd(a⁻² + b⁻² + c⁻² + d⁻²)

= (bk)(b)(dk)(d)((bk)⁻² + b⁻² + (dk)⁻² + d⁻²)

= `b^2d^2k^2(1/(b^2k^2) + 1/b^2 + 1/(d^2k^2) + 1/d^2)`

= `(b^2d^2k^2)/(b^2k^2) + (b^2d^2k^2)/b^2 + (b^2d^2k^2)/(d^2k^2) + (b^2d^2k^2)/d^2`

= d² + d²k² + b² + b²k²

= (b² + d²) (1 + k²)

R.H.S.

= a² + b² + c² + d²

= (bk)² + b² + (dk)² + d²

= b²k² + b² + d²k² + d²

= b²(k² + 1) + d²(k² + 1)

= (b² + d²) (1 + k²)

L.H.S. = R.H.S.