Question

$\mapsto$ If a:b = c:d then prove that, (a²+c²)(b²+d²) = (ab+cd)²
$\bigstar$ Wrong Answer will be reported.✓​

Answer 1

Answer:-

Given:

a : b = c : d

⟶ a/b = c/d

On cross multiplication we get,

⟶ ad = bc

⟶ ad - bc = 0 -- equation (1)

We have to prove:

(a² + c²) * (b² + d²) = (ab + cd)²

using (a + b)² = a² + b² + 2ab in RHS we get,

⟶ a² (b² + d²) + c² (b² + d²) = (ab)² + (cd)² + 2abcd

⟶ a²b² + a²d² + b²c² + c²d² = (ab)² + (cd)² + 2abcd

Using aⁿ * bⁿ = (ab)ⁿ we get,

⟶ (ab)² + (ad)² + (bc)² + (cd)² - (ab)² - (cd)² - 2abcd = 0

⟶ (ad)² + (bc)² - 2abcd = 0

using (a - b)² = a² + b² - 2ab we get,

⟶ (ad - bc)² = 0

⟶ ad - bc = 0

Putting the value of ad - bc from equation (1) we get,

⟶ 0 = 0.

Hence, Proved.

Answer 2

☘ Given Question :-

If If a:b = c:d then prove that, (a²+c²)(b²+d²) = (ab+cd)².

☘ Solution :-

a : b = c : d.

So, We can say,

a/b = c/d.

By Cross Multiplying, We get,

➙ ad - bc = 0.

LHS ⇒

(a²+c²)(b²+d²)

On Further Multiplying, We get,

➙ a²b² + a²d² + b²c² + c²d²

➙ (ab)² + (ad)² + (bc)² + (cd)²

RHS ⇒

(ab + cd)²

We can use identity :- (a+b)² = a² + b² + 2ab.

On Further Calculation, We get,

➙ (ab)² + (cd)² + 2abcd.

➙ a²b² + c²d² + 2abcd.

➙ (ab)² + (cd)² + 2abcd.

Putting them equal,

(ab)² + (ad)² + (bc)² + (cd)² = (ab)² + (cd)² + 2abcd.

➙ (ad)² + (bc)² = 2abcd.

➙ (ad)² + (bc)² – 2abcd = 0.

➙ (ad - bc)² = 0.

➙ 0² = 0.

➙ 0 = 0.

LHS = RHS.

Hence, Proved.