in a quadrilateral ABCD, the diagonals AC, BD , intersect at right angles. Prove that : AB^2 + CD^2 = BC^2 +DA^2
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Answered by
31
AB^2+CD^2= OA^2+OB^2+OD^2+OC^2...(1)
BC^2+AD^2= OB^2+OC^2+OA^2+OD^2...(2)
FROM 1 AND 2
AB^2+CD^2= BC^2+AD^2
hence proved
BC^2+AD^2= OB^2+OC^2+OA^2+OD^2...(2)
FROM 1 AND 2
AB^2+CD^2= BC^2+AD^2
hence proved
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Answered by
40
From the above picture, by applying Pythagoras theorem, we can say :
AB² + CD²
= OB²+OA²+OD²+OC²
= OB²+OC²+OA²+OD²
= BC²+AD² [PROVED]
(•.• OB²+OC² = BC² and OA²+OD² = AD², by pythagoras theorem)
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PYTHAGORAS THEOREM
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In any right angled triangle, the square of the longest side of it (i.e., the hypotenuse) is equal to the sum of the squares of the other two sides (i.e., its height and base).
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