In a rhombus ABCD, prove that AC² + BD² = 4AB².

In rhombus, AB = BC = CD = AD (All sides of a rhombus are equal)

Diagonal AC and BD intersect at point O.

Diagonals bisect each other at right angles.

So, AO = OC = $\dfrac{1}{2}$ × AC and BO = OD = $\dfrac{1}{2}$ × BD

In right angled △AOB,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AO² + BO²

$\Rightarrow \text{AB}^2 = \Big(\dfrac{\text{AC}}{2}\Big)^2 + \Big(\dfrac{\text{BD}}{2}\Big)^2 \\[1em] \Rightarrow \text{AB}^2 = \dfrac{\text{AC}^2}{4} + \dfrac{\text{BD}^2}{4} \\[1em] \Rightarrow \text{AB}^2 = \dfrac{\text{AC}^2 + \text{BD}^2}{4}$

∴ 4AB² = AC² + BD²

Hence, proved that AC² + BD² = 4AB².