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Prove with complex number: the sum of squares of parallelogram diagonal is equal to the sum of squares of four sides
Set four vertices a = 0, B = x, C = x, C = m + min (M-X) + N, then the edge is B-A = B-A = C-D, C-B = (M-X) + n = D-A diagonal, C-A = m + min-b-b = (m-2x (m-2x) (m-2x) and four vertices a = 0, B = x, and then the edge is B-A = B-A = 0, and the edge is B-A = B-A = 0-c-d-d-d, and c-d-c-c-c-c-c-c-c-c-c-a-a-a-a-a-a-a \124;\124;||||124;|||||||||||124;2n & # 178; - 4m
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