SOLUTION: prove the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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Question 420398: prove the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Answer by richard1234(7193) About Me  (Show Source):
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Suppose we have a rhombus, with diagonals AC and BD intersecting at E.



Let AE = EC = x, BE = ED = y, and AB = z. The diagonals of a rhombus are perpendicular, so x%5E2+%2B+y%5E2+=+z%5E2. The sum of the squares of the diagonals is given by %282x%29%5E2+%2B+%282y%29%5E2. This is equal to 4x%5E2+%2B+4y%5E2+=+4%28x%5E2+%2B+y%5E2%29+=+4z%5E2, which is also equal to the sum of the squares of the side lengths (since each side has length z).