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