In a quadrilateral ABCD, ∠A + ∠D = 90°, prove that : AC² + BD² = AD² + BC².

Produce AB and DC such that they meet at point E.

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In △ AED,

⇒ ∠A + ∠D + ∠E = 180°

⇒ 90° + ∠E = 180°

⇒ ∠E = 180° - 90° = 90°.

By pythagoras theorem,

⇒ AD² = AE² + DE² ………(1)

In △ BEC,

By pythagoras theorem,

⇒ BC² = BE² + CE² ………(2)

In △ AEC,

By pythagoras theorem,

⇒ AC² = AE² + CE² ………(3)

In △ BED,

By pythagoras theorem,

⇒ BD² = BE² + DE² ………(4)

Adding equations (1) and (2), we get :

⇒ AD² + BC² = AE² + DE² + BE² + CE²

⇒ AD² + BC² = (AE² + CE²) + (BE² + DE²)

⇒ AD² + BC² = AC² + BD².

Hence, proved that AC² + BD² = AD² + BC².