Question

ABCD is a rhombus. Prove that AC² + BD² = 4AB².

Answer 1

Given, ABCD is a rhombus.

We know that diagonals of a rhombus bisect each other at 90°.

⇒ ∠AOB = 90°

⇒ `AO = 1/2 AC`

⇒ `BO = 1/2 BD`

Now, In ΔAOB

AO² + BO² = AB²   ...(Pythagoras Theorem)

⇒ `1/2 AC^2 + 1/2 BD^2 = AB^2`

⇒ `(AC^2)/4 + (BD^2)/4 = AB^2`

⇒ AC² + BD² = 4AB²

Hence, proved.