Thébault's Problem I

The applet suggests the following theorem (Victor Thébault, 1937):

On the sides of parallelogram ABCD erect squares -- all either on the outside or the inside of the parallelogram. Their centers then form another square.

Proof 1

Consider the case of the squares erected on the outside of the parallelogram.

Triangles AEH, BFE, CGF, and DHG are obviously equal, so that the quadrilateral EFGH is a rhombus. Further, DHG = AHE, which implies GHE = DHA = 90^o. Q.E.D.

Proof 2

Place the origin at the center of the parallelogram ABCD, and let u and v be complex numbers associated with D and A. Then DA is associated with v - u, while AB is corresponds to -(v + u). Further,

OH corresponds to u + (v - u)/2 + i·(v - u)/2 = (u + v)/2 + i·(v - u)/2.

Similarly we can compute OE, OF, OG. For example,

The difference of the two numbers corresponds to the side EH of the quadrilateral EFGH and equals u + iv. If the squares were constructed inwardly, EH would correspond to u - iv.

Similarly, v - iu corresponds to GH. Since v - iu = -i·(u + iv), one is obtained from the other with the rotation through 90^o. In addition, |v - i·u| = |i·(u + iv)| = |(u + iv)|. Therefore the quadrilateral EFGH is a square.

Proof 3