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:
- Assign coordinates: W = (0,0), X = (s,0), Y = (s,s), Z = (0,s).
- 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)
- 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)
- 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.
- 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.
- 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.