Question

ABCD is a parallelogram, M is the mid-point of BC and AM ⊥ BC. Prove that AD² = 4(CD² – AM²).

Answer 1

Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given,

Applying Pythagoras theorem as AM ⊥ BC

BM² = AB² − AM²

BM = `(BC)/2` or `(AD)/2`

BC = AD and AB = CD, the opposite sides of a parallelogram are equal.

⇒ BM² = CD² − AM²

⇒ `((AD)/2)^2 = CD^2 - AM^2`

⇒ `(AD^2)/4 = CD^2 - AM^2`

⇒ AD² = 4(CD² − AM²)

Hence, proved.