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Please see diagram.

Let E, F, G and H be points  at  the same distance from A, B ,C and D , on the edges AB, BC, CD, and DA respectively.  That distance need not be  AB/2 only.  It can be any.

   Now, in the triangle   HAE,  let the angle EHA be y.
   As,   angle  HAE = 90,  AEH = 180 - 90 - y = 90- y.

   Similarly, in the triangle  EBF,  we have angle  BEF = y.  This is because the triangles HAE and EBF are congruent.  EB = HA.  AE = BF and  angle A = angle B.

   Hence  angle HEF = 180 - y - (90-y) = 90 deg.  All other angles  EFG,  FGH and GHE are also  90 deg.

   Now we have to prove that all sides are equal.

    HE  =  root (HA² + AE²)  = root (BE² + BF²) = EF
           =  root (FC² + CG²) = FG  =  root (GD² + DH²)  = GH

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