If a, b, c, d are in proportion, then prove that

i.

ii.

iii.

(i)

Given: a, b, c, d are in proportion.

a, b, c, d are in proportion a : b : : c : d ad = bc

i.e. Product of extremes = product of means.

∴ ad = bc

If then:

(11a² + 9ac)(b² + 3bd) = (a² + 3ac)(11b² + 9bd)

⇒ 11a²b² + 33a²bd + 9ab²c + 27abcd = 11a²b² +9a²bd + 33ab²c + 27abcd

⇒ 33a²bd + 9ab²c = 9a²bd + 33ab²c

⇒ 24a²bd = 24ab²c

⇒ a²bd = ab²c

⇒ ad = bc, which holds true as the numbers are in continued proportion.

∴

(ii)

Given: a, b, c, d are in proportion.

a, b, c, d are in proportion a : b : : c : d ad = bc

i.e. Product of extremes = product of means.

∴ ad = bc

If

then:

(a² + 5c²)(b²) = (b² + 5d²)(a²)

⇒ a²b² + 5c²b² = b²a² + 5a²d²

⇒ 5b²c² = 5a²d²

⇒ b²c² = a²d²

⇒ bc = ad

⇒ ad = bc, which holds true as the numbers are in continued proportion.

∴

(iii)

Given: a, b, c, d are in proportion.

a, b, c, d are in proportion a : b : : c : d ad = bc

i.e. Product of extremes = product of means.

∴ ad = bc

If then:

(a² + ab + b²)(c² – cd + d²) = (a² - ab + b²)(c² + cd + d²)

⇒ a²c² - a²cd + a²d² + abc² – abcd + abd² + b²c² – b²cd + b²d² = a²c² + a²cd + a²d² - abc² – abcd - abd² + b²c² + b²cd + b²d²

⇒ -2a²cd + 2abc² + 2abd² -2b²cd = 0

⇒ 2abc² - 2b²cd = 2a²cd – 2abd²

⇒ 2bc[ac – bd] = 2ad[ac - bd]

⇒ 2bc = 2ad

⇒ ad = bc, which holds true as the numbers are in continued proportion.

∴