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Let W, X, Y, Z be the vertices of the original square, and let s be its side length. Place the square on a coordinate plane for a simple algebraic proof:

  1. Assign coordinates: W = (0,0), X = (s,0), Y = (s,s), Z = (0,s).
  2. Find the midpoints of each side and name them A, B, C, D in order:
    • A, midpoint of WX = ((0+s)/2, (0+0)/2) = (s/2, 0)
    • B, midpoint of XY = ((s+s)/2, (0+s)/2) = (s, s/2)
    • C, midpoint of YZ = ((s+0)/2, (s+s)/2) = (s/2, s)
    • D, midpoint of ZW = ((0+0)/2, (s+0)/2) = (0, s/2)
  3. Compute the side vectors of quadrilateral ABCD (consecutive differences):
    • AB = B - A = (s - s/2, s/2 - 0) = (s/2, s/2)
    • BC = C - B = (s/2 - s, s - s/2) = (-s/2, s/2)
    • CD = D - C = (0 - s/2, s/2 - s) = (-s/2, -s/2)
    • DA = A - D = (s/2 - 0, 0 - s/2) = (s/2, -s/2)
  4. Show all four sides have equal length:

    |AB| = sqrt((s/2)^2 + (s/2)^2) = s/√2. The same calculation applies to BC, CD, DA, so all sides are equal.

  5. Show consecutive sides are perpendicular (so angles are right angles):

    Compute the dot product AB · BC = (s/2)(-s/2) + (s/2)(s/2) = -s^2/4 + s^2/4 = 0.

    A zero dot product means AB is perpendicular to BC. By the symmetry of the vectors above, each consecutive pair is perpendicular.

  6. Conclusion: ABCD has four equal sides and four right angles, so it is a square. In fact, ABCD is the original square rotated by 45 degrees about the same center and scaled by a factor of 1/√2.

Alternate quick reasoning: joining midpoints of any quadrilateral produces a parallelogram. For a square, symmetry makes that midpoint-parallelogram have equal adjacent sides and right angles, so it must be a square.


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