In a quadrilateral ABCD- given that - A - D -90- - Prove that AC 2- BD 2- AD 2- BC 2-

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 ...

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Standard X

Mathematics

Pythagoras Theorem

In a quadrila...

Question

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

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Solution

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.

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

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