Question

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

Answer 1

Given: A quadrilateral ABCD where ∠A + ∠D = 90°.

To prove: AC² + BD² = AD² + BC²

Construction: Extend AB and CD to intersect at O.

Proof:

In ΔAOD, ∠A + ∠O + ∠D = 180°

⇒ ∠O = 90° [∠A + ∠D = 90°]

Apply Pythagoras Theorem in ΔAOC and ΔBOD,

AC² = AO² + OC²

BD² = OB² + OD²

∴ AC² + BD² = (AO² + OD²) + (OC² + OB²)

⇒ AC² + BD² = AD² + BC²

This proves the given relation.