Q2. In a quadrilateral ABCD, the diagonals AC and BD
intersect at right angles. Prove that AB^2 + CD^2 = BC^2+ DA^2
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Answer:
In △AOB,
AB 2 =OA 2 +OB 2 ......(1)
In △COD,
CD 2 =OC 2 +OD 2 .....(2)
In △OAD,
DA 2 =OA 2 +OD 2 ......(3)
In △BOC,
BC 2 =OC 2 +OB 2 ......(4)
Adding 1 & 2, we get,
AB 2 +CD 2 =OA 2 +OB 2 +OC 2 +OD 2 ....(5)
Adding 3 & 4 , we get,
DA 2 +BC 2 =OA 2 +OB 2 +OC 2 +OD 2 ......(6)
Form 5 & 6, we get,
AB 2 +CD 2 =BC 2 +DA 2
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